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HIGHER ALGEBRA Ernest E. Shult 1 Contents 1 Basics 1.1 Presumed results and conventions . . . . . . . . 1.1.1 Presumed Jargon . . . . . . . . . . . . . 1.1.2 Sets and maps . . . . . . . . . . . . . . . 1.1.3 Partitions . . . . . . . . . . . . . . . . . 1.1.4 Zorn’s Lemma and the Axiom of Choice. 1.1.5 Cardinal Numbers . . . . . . . . . . . . 1.1.6 Special results on cardinalities . . . . . . 1.2 Binary operations and Monoids. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Partially Ordered Sets and Dependance Theories 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Basic definitions. . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Definition of a partially ordered set. . . . . . . . . . . . 2.2.2 Induced subposets. . . . . . . . . . . . . . . . . . . . . 2.2.3 Dual posets and dual concepts. . . . . . . . . . . . . . 2.2.4 Maximal and minimal elements of induced posets . . . 2.2.5 Global maxima and minima: zero’s and one’s in posets. 2.2.6 Total orderings and chains . . . . . . . . . . . . . . . . 2.2.7 Antichains. . . . . . . . . . . . . . . . . . . . . . . . . 2.2.8 Zornification . . . . . . . . . . . . . . . . . . . . . . . . 2.2.9 Order ideals and filters. . . . . . . . . . . . . . . . . . . 2.2.10 Products of posets. . . . . . . . . . . . . . . . . . . . . 2.2.11 Examples and exercises. . . . . . . . . . . . . . . . . . 2.3 Mappings among posets: closure operators and Galois connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Subposets. . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Order-preserving mappings . . . . . . . . . . . . . . . . 2 13 13 13 13 18 19 20 21 23 24 24 26 26 26 27 27 28 28 29 30 30 32 33 41 41 41 2.4 2.5 2.6 2.7 2.8 2.9 2.3.3 2.3.4 Chain 2.4.1 2.4.2 2.4.3 Closure operators. . . . . . . . . . . . . . . . . . . . . Closure operators and Galois connections. . . . . . . . conditions . . . . . . . . . . . . . . . . . . . . . . . . . Saturated chains . . . . . . . . . . . . . . . . . . . . . Algebraic intervals and the height function. . . . . . . The ascending and descending chain conditions in arbitrary posets. . . . . . . . . . . . . . . . . . . . . . . . Posets with meets and/or joins. . . . . . . . . . . . . . . . . . 2.5.1 Meets and joins . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Lower semilattices with the descending chain condition. 2.5.3 Lower semilattices with both chain conditions . . . . . 2.5.4 Examples and exercises . . . . . . . . . . . . . . . . . . The Jordan-Hölder Theory . . . . . . . . . . . . . . . . . . . . 2.6.1 Lower semi-lattices and semimodularity . . . . . . . . . 2.6.2 Interval Measures on Posets . . . . . . . . . . . . . . . 2.6.3 The Jordan-Hölder Theorem . . . . . . . . . . . . . . . Möbius inversion . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 More locally finite posets. . . . . . . . . . . . . . . . . 2.7.2 The incidence algebra . . . . . . . . . . . . . . . . . . . 2.7.3 The Möbius function of a locally finite poset . . . . . . 2.7.4 Special cases . . . . . . . . . . . . . . . . . . . . . . . . 2.7.5 Remarks on other applications of the Möbius function . Point measures in lattices; the modular law. . . . . . . . . . . 2.8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2.8.2 Dual lattices and sublattices. . . . . . . . . . . . . . . 2.8.3 Kolmogoroff functions on lattices . . . . . . . . . . . . 2.8.4 The “inclusion-exclusion” principle . . . . . . . . . . . 2.8.5 Point-measure of a lattice . . . . . . . . . . . . . . . . 2.8.6 Modular lattices . . . . . . . . . . . . . . . . . . . . . . 2.8.7 Exercises concerning lattices. . . . . . . . . . . . . . . Dependence Theories . . . . . . . . . . . . . . . . . . . . . . . 2.9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2.9.2 Dependence . . . . . . . . . . . . . . . . . . . . . . . . 2.9.3 The extended definition . . . . . . . . . . . . . . . . . 2.9.4 Independence and spanning. . . . . . . . . . . . . . . . 2.9.5 Dimension. . . . . . . . . . . . . . . . . . . . . . . . . 2.9.6 Matroids . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.7 The rank function . . . . . . . . . . . . . . . . . . . . . 3 43 43 45 45 46 47 50 50 52 54 55 57 58 58 60 62 62 64 65 67 69 69 69 70 71 73 74 77 81 83 83 83 84 85 86 87 89 2.9.8 Dependence theories and matroids are the same thing . 91 2.9.9 Other formulations of dependence . . . . . . . . . . . . 94 2.9.10 Exercises on Matroids . . . . . . . . . . . . . . . . . . 96 3 Review of Elementary Group Properties 97 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 3.2 Definition and examples. . . . . . . . . . . . . . . . . . . . . . 98 3.2.1 Orders of elements and groups . . . . . . . . . . . . . . 102 3.2.2 Subgroups. . . . . . . . . . . . . . . . . . . . . . . . . 103 3.3 Homomorphisms of Groups . . . . . . . . . . . . . . . . . . . . 106 3.3.1 Definitions and basic properties. . . . . . . . . . . . . . 106 3.3.2 Automorphisms as right operators. . . . . . . . . . . . 108 3.3.3 Examples of homomorphisms. . . . . . . . . . . . . . . 109 3.4 Factor Groups and the Fundamental Theorem of Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 3.4.1 Miscellaneous Exercises. . . . . . . . . . . . . . . . . . 121 3.5 Direct products. . . . . . . . . . . . . . . . . . . . . . . . . . . 123 3.6 Other Variations: Semidirect Products and Subdirect Products.125 3.6.1 Semidirect products. . . . . . . . . . . . . . . . . . . . 125 3.6.2 Subdirect products. . . . . . . . . . . . . . . . . . . . . 128 3.7 Minimal normal subgroups of finite groups. . . . . . . . . . . . 128 4 Permutation Groups and Group Actions 4.1 Group actions and orbits . . . . . . . . . . . . . . . . . . . . 4.2 Permutations. . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Cycle Notation. . . . . . . . . . . . . . . . . . . . . 4.2.2 Even and odd permutations. . . . . . . . . . . . . . 4.2.3 Transpositions and Cycles: The Theorem of Feit, Lyndon and Scott . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Action on right cosets. . . . . . . . . . . . . . . . . . 4.2.5 Equivalent actions. . . . . . . . . . . . . . . . . . . . 4.2.6 The Fundamental Theorem of Transitive Actions. . . 4.2.7 Normal subgroups of transitive groups. . . . . . . . . 4.2.8 Double cosets. . . . . . . . . . . . . . . . . . . . . . 4.3 Applications of transitive group actions. . . . . . . . . . . . 4.3.1 Finite p-groups. . . . . . . . . . . . . . . . . . . . . . 4.3.2 The Sylow Theorems . . . . . . . . . . . . . . . . . . 4.3.3 Calculations of Group Orders. . . . . . . . . . . . . 4 131 . 131 . 134 . 134 . 136 . . . . . . . . . . 138 140 141 141 142 143 144 144 145 149 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 152 154 157 161 162 168 172 5 Normal Structure of Groups 5.1 The Jordan-Holder Theorem for Artinian Groups 5.1.1 Subnormality. . . . . . . . . . . . . . . . . 5.2 Commutators. . . . . . . . . . . . . . . . . . . . . 5.3 The Derived Series and Solvability. . . . . . . . . 5.4 Central Series and Nilpotent Groups. . . . . . . . 5.4.1 The upper and lower central series. . . . . 5.4.2 Finite nilpotent groups. . . . . . . . . . . 5.5 Coprime Action . . . . . . . . . . . . . . . . . . . 5.6 Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 177 177 180 183 188 188 192 196 199 4.5 Primitive and Multiply Transitive Actions. 4.4.1 Primitivity . . . . . . . . . . . . . . 4.4.2 The rank of a group action. . . . . 4.4.3 Multiply transitive group actions. . 4.4.4 Permutation characters . . . . . . . Exercises for Chapter 3. . . . . . . . . . . 4.5.1 First proof: . . . . . . . . . . . . . 4.5.2 Burnside’s original proof. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Generation in Groups 207 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 6.2 The Cayley graph . . . . . . . . . . . . . . . . . . . . . . . . . 207 6.2.1 Definition. . . . . . . . . . . . . . . . . . . . . . . . . . 207 6.2.2 Morphisms of various sorts of graphs. . . . . . . . . . . 208 6.2.3 Morphisms of groups induce morphisms of Cayley graphs.209 6.3 Free groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 6.3.1 The Universal Property . . . . . . . . . . . . . . . . . 216 6.4 The Brauer-Ree Theorem . . . . . . . . . . . . . . . . . . . . 224 6.5 Appendix to Chapter 6: A Detective Story: An application of (k, l, m)-groups . . . . . . . . . . . . . . . . . . . . . . . . . . 226 7 Elementary Properties of Rings 7.1 Elementary Facts about Rings . 7.1.1 Definitions . . . . . . . . 7.1.2 Units in rings . . . . . . 7.1.3 Homomorphisms . . . . 7.1.4 Ideals and factor rings. . 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 . 235 . 236 . 238 . 239 . 240 7.1.5 7.1.6 7.1.7 The fundamental theorems of ring homomorphisms . . 241 The poset of ideals. . . . . . . . . . . . . . . . . . . . . 243 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 246 8 Elementary properties of Modules 8.1 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Modules over rings. . . . . . . . . . . . . . . . . . . . 8.1.2 Submodules . . . . . . . . . . . . . . . . . . . . . . . 8.1.3 Factor modules. . . . . . . . . . . . . . . . . . . . . . 8.1.4 The fundamental theorems of homomorphisms for Rmodules. . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.5 The Jordan-Hölder Theorem for R-modules. . . . . . 8.1.6 Direct sums and products of modules. . . . . . . . . . 8.1.7 Free modules . . . . . . . . . . . . . . . . . . . . . . 8.1.8 Vector spaces. . . . . . . . . . . . . . . . . . . . . . . 8.1.9 Examples and exercises . . . . . . . . . . . . . . . . . 8.2 Consequences of chain conditions on modules . . . . . . . . 8.2.1 Definitions: Noetherian and Artinian modules . . . . 8.2.2 Effects of the chain conditions on endomorphisms. . . 8.2.3 Noetherian Modules and finite generation. . . . . . . 8.2.4 Right and left Noetherian rings. . . . . . . . . . . . . 8.2.5 Algebraic integers form a ring. . . . . . . . . . . . . . 8.3 The Hilbert Basis Theorem . . . . . . . . . . . . . . . . . . . . . . 254 254 254 257 259 . . . . . . . . . . . . . 259 261 263 264 266 268 271 271 273 278 279 282 285 9 The Arithmetic of Integral Domains. 288 9.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 9.2 Localization in Commutative Rings. . . . . . . . . . . . . . . . 294 9.2.1 Localization by a multiplicative set. . . . . . . . . . . . 294 9.2.2 Special features of localization in domains. . . . . . . . 295 9.2.3 Localization at a prime ideal. . . . . . . . . . . . . . . 296 9.3 Unique Factorization. . . . . . . . . . . . . . . . . . . . . . . . 297 9.3.1 A Special Case. . . . . . . . . . . . . . . . . . . . . . . 301 9.4 Euclidean Domains . . . . . . . . . . . . . . . . . . . . . . . . 302 9.4.1 Examples of Euclidean Domains . . . . . . . . . . . . . 304 9.5 Principal Ideal Domains are Unique Factorization Domains . . 306 9.6 If D is a UFD, then so is D[x]. . . . . . . . . . . . . . . . . . 306 9.7 Appendix to Chapter 8: The Arithmetic of Quadratic Domains.313 9.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 313 6 9.7.2 9.7.3 Which quadratic domains are Euclidean, principal ideal or factorial domains? . . . . . . . . . . . . . . . . . . . 315 Subdomains of Euclidean domains which are not UFD’s 317 10 Principal Ideal Domains and Their Modules 321 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 10.2 Quick Review of PID’s . . . . . . . . . . . . . . . . . . . . . . 321 10.3 Free modules over a PID. . . . . . . . . . . . . . . . . . . . . . 323 10.4 A technical result . . . . . . . . . . . . . . . . . . . . . . . . . 324 10.5 Finitely generated modules over a PID . . . . . . . . . . . . . 327 10.6 The uniqueness of the Invariant Factors. . . . . . . . . . . . . 331 10.7 Applications of the Theorem on PID Modules. . . . . . . . . . 333 10.7.1 Classification of Finite Abelian Groups. . . . . . . . . . 333 10.7.2 The Rational canonical form of a linear transformation 334 10.7.3 The Jordan Form . . . . . . . . . . . . . . . . . . . . . 338 10.7.4 Information Carried by the Invariant Factors . . . . . . 340 11 Theory of Fields. 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Elementary Properties of Field Extensions. . . . . . . . . . . 11.2.1 Algebraic and Transcendental Extensions. . . . . . . 11.2.2 Indices Multiply: Ruler and Compass Problems. . . . 11.2.3 The Number of Zeros of a Polynomial . . . . . . . . . 11.3 Splitting Fields and Their Automorphisms . . . . . . . . . . 11.3.1 Extending Isomorphisms Between Fields . . . . . . . 11.3.2 Splitting Fields. . . . . . . . . . . . . . . . . . . . . . 11.3.3 Normal Extensions. . . . . . . . . . . . . . . . . . . . 11.4 Some Applications to Finite Fields . . . . . . . . . . . . . . 11.4.1 Prime Subfields . . . . . . . . . . . . . . . . . . . . . 11.4.2 Finite fields . . . . . . . . . . . . . . . . . . . . . . . 11.4.3 Automorphisms of finite fields . . . . . . . . . . . . . 11.4.4 Polynomial equations over finite fields: the ChevalleyWarning Theorem . . . . . . . . . . . . . . . . . . . . 11.5 Separable Elements and Field extensions. . . . . . . . . . . . 11.5.1 Separability. . . . . . . . . . . . . . . . . . . . . . . . 11.5.2 p-th Power Mappings in Characteristic p . . . . . . . 11.5.3 Simple Roots . . . . . . . . . . . . . . . . . . . . . . 11.5.4 Separable Elements and Separable Field Extensions . 7 343 . 343 . 344 . 345 . 347 . 349 . 351 . 351 . 353 . 357 . 359 . 359 . 360 . 361 . . . . . . 362 364 364 366 367 370 11.6 Galois Theory . . . . . . . . . . . . . . . . . . . . . . 11.6.1 Galois field extensions . . . . . . . . . . . . . 11.6.2 Independence of linear characters . . . . . . . 11.6.3 Group orders and degrees in Galois extensions 11.7 Solvability of equations by radicals . . . . . . . . . . 11.7.1 Introduction . . . . . . . . . . . . . . . . . . . 11.7.2 Roots of unity . . . . . . . . . . . . . . . . . . 11.7.3 Radical extensions . . . . . . . . . . . . . . . 11.7.4 Galois’ criterion for solvability by radicals . . 11.8 Transcendental extensions . . . . . . . . . . . . . . . 11.8.1 Simple transcendental extensions . . . . . . . 11.8.2 Transcendence Degree . . . . . . . . . . . . . 11.9 Appendix to Chapter 10: Fields with a Valuation . . 11.9.1 Introduction . . . . . . . . . . . . . . . . . . . 11.9.2 The Language of Convergence . . . . . . . . . 11.9.3 Completions exist . . . . . . . . . . . . . . . . 11.9.4 The non-archimedean p-adic valuations . . . . 11.10Appendix to Chapter 10: Finite Division Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 375 376 378 383 383 386 387 390 394 394 399 402 402 405 406 413 418 12 Semiprime Rings 424 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 12.2 Complete Reducibility . . . . . . . . . . . . . . . . . . . . . . 424 12.3 Homogeneous Components. . . . . . . . . . . . . . . . . . . . 429 12.3.1 The action of the R-endomorphism ring. . . . . . . . . 429 12.3.2 The Socle of a Module is a Direct Sum of the Homogeneous Components. . . . . . . . . . . . . . . . . . . . 429 12.4 Semiprime rings. . . . . . . . . . . . . . . . . . . . . . . . . . 431 12.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 431 12.4.2 The semiprime condition . . . . . . . . . . . . . . . . . 432 12.4.3 Completely Reducible and Semiprimitive Rings are Species of Semiprime Rings . . . . . . . . . . . . . . . . . . . . 433 12.4.4 Idempotents and Minimal Right Ideals of Semiprime Rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 12.4.5 Idempotents and Left-Right Symmetries . . . . . . . . 435 12.5 Completely Reducible Rings . . . . . . . . . . . . . . . . . . . 439 12.6 Exercises for Chapter 12 . . . . . . . . . . . . . . . . . . . . . 443 12.6.1 Exercises relating to Sections 12.1 and 12.2 . . . . . . . 443 12.6.2 Warm up Exercises . . . . . . . . . . . . . . . . . . . . 446 8 12.6.3 12.6.4 12.6.5 12.6.6 Exercises concerning the Jacobson radical . . . . . . The Jacobson radical of Artinian rings . . . . . . . . Quasiregularity and the radical . . . . . . . . . . . . Exercises involving nil one-sided ideals in Noetherian rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Categorical Notions 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.1 Universal Mapping Properties. . . . . . . . . . . . . . 13.1.2 Duality. . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.3 Functors. . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Special Features of the Category of R-modules. . . . . . . . 13.2.1 Exact Sequences . . . . . . . . . . . . . . . . . . . . 13.2.2 The Direct Sum as a Universal Mapping Property . . 13.2.3 The Direct Product as a Universal Mapping Property 13.3 Projective Modules . . . . . . . . . . . . . . . . . . . . . . . 13.4 Injective Modules . . . . . . . . . . . . . . . . . . . . . . . . 13.5 hom(A, B) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5.1 hom(A, B) is an R-module. . . . . . . . . . . . . . . 13.5.2 Functorial properties of Hom . . . . . . . . . . . . . 13.5.3 Dual modules . . . . . . . . . . . . . . . . . . . . . . 13.6 Tensor Products. . . . . . . . . . . . . . . . . . . . . . . . . 13.6.1 Elementary Consequences of the Definition. . . . . . 13.7 Important applications of the tensor product . . . . . . . . . 13.7.1 Exchange of Rings . . . . . . . . . . . . . . . . . . . 13.7.2 Tensor products of R-algebras. . . . . . . . . . . . . . 13.7.3 The tensor product of modules as a module for a tensor product of rings. . . . . . . . . . . . . . . . . . . . . 13.8 Graded Algebras . . . . . . . . . . . . . . . . . . . . . . . . 13.8.1 Morphisms of graded modules . . . . . . . . . . . . . 13.8.2 Subalgebras of graded R-algebras . . . . . . . . . . . 13.8.3 Some examples of graded R-algebras . . . . . . . . . 13.8.4 Ideals and homomorphisms in associative graded Ralgebras . . . . . . . . . . . . . . . . . . . . . . . . . 13.9 Exercises for Chapter 12 . . . . . . . . . . . . . . . . . . . . 13.10Appendix to Chapter 12: Important R-algebras . . . . . . . 13.10.1 Other applications . . . . . . . . . . . . . . . . . . . 9 . 447 . 448 . 449 . 451 454 . 454 . 458 . 460 . 461 . 464 . 464 . 465 . 467 . 468 . 471 . 472 . 472 . 474 . 475 . 476 . 476 . 480 . 480 . 481 . . . . . 486 488 488 489 491 . . . . 492 494 495 500 14 Categorical Features of Vector Spaces 14.1 Left and Right Vector Space Dimensions . . . . . . . . . . . 14.2 hom(V, W ) as a Left Vector Space; the Dual Space . . . . . 14.3 Tensor Products of Vector Spaces Over Fields . . . . . . . . 14.4 Tensor Products of (F, F )-bimodules . . . . . . . . . . . . . 14.5 Tensor Products of Linear Transformations. . . . . . . . . . 14.6 The Universal Mapping Property of Multiple Tensor Products: Multilinear Mappings . . . . . . . . . . . . . . . . . . . . . . 15 Tensor Algebras and Their Offspring 15.1 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1.1 Nonassociative Algebras . . . . . . . . . . . . . . . . 15.1.2 Associative Algebras . . . . . . . . . . . . . . . . . . 15.2 The Tensor Algebra . . . . . . . . . . . . . . . . . . . . . . . 15.3 The Symmetric Algebra . . . . . . . . . . . . . . . . . . . . 15.4 The Exterior Algebra . . . . . . . . . . . . . . . . . . . . . . 15.5 Transformations Induced on the k-fold Wedge Products: The Determinant as Group Homomorphism. . . . . . . . . . . . . 15.6 Symmetric and Alternating Multilinear Mappings. . . . . . . 16 Sesquilinear Forms. 16.1 σ-sesquilinear Forms. . . . . . . . . . . . . . . . . . . . . . . 16.1.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . 16.1.2 Forms and functionals . . . . . . . . . . . . . . . . . 16.1.3 Descriptions of forms by Grammian matrices . . . . . 16.2 Reflexive forms. . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.1 Perps and radicals of reflexive forms. . . . . . . . . . 16.2.2 (σ, )-Hermitian forms. . . . . . . . . . . . . . . . . . 16.2.3 Totally isotropic subspaces; symplectic forms. . . . . 16.3 Finite-dimensional subspaces of spaces with a non-degenerate reflexive form. . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 The Witt index of Polar Systems. . . . . . . . . . . . . . . . 16.4.1 Polar systems . . . . . . . . . . . . . . . . . . . . . . 16.4.2 The radical of a polar system . . . . . . . . . . . . . 16.4.3 Finite rank, and the Witt index of a polar system . . 16.4.4 A brief interlude with graph-theoretic concepts . . . . 16.4.5 The distance metric on M(T ) . . . . . . . . . . . . . 16.5 Decompositions of Forms of Finite Rank . . . . . . . . . . . 10 506 . 506 . 508 . 510 . 513 . 514 . 514 . . . . . . 516 516 516 518 521 522 525 . 528 . 530 . . . . . . . . 536 536 536 537 538 538 538 539 541 . . . . . . . . 542 546 546 547 548 550 552 556 16.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 556 16.5.2 Orthogonal Decompositions. . . . . . . . . . . . . . . . 557 16.5.3 Hyperbolic Decompositions of Finite Rank Sesquilinear Forms Over Fields. . . . . . . . . . . . . . . . . . . 558 16.5.4 Symmetric forms in characteristic 2 . . . . . . . . . . . 561 16.6 Special cases of sesquilinear forms . . . . . . . . . . . . . . . . 565 16.6.1 Symplectic forms . . . . . . . . . . . . . . . . . . . . . 565 16.6.2 Hermitian forms . . . . . . . . . . . . . . . . . . . . . . 565 16.6.3 Special cases for orthogonal forms. . . . . . . . . . . . 568 16.7 Sesquilinear forms obtained from others by reduction to a subfield. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 16.7.1 Forms derived from a Hermitian form by field reduction 572 16.8 Exercises for Chapter 15 . . . . . . . . . . . . . . . . . . . . . 573 16.8.1 Miscellaneous warm-up problems . . . . . . . . . . . . 573 16.8.2 Forms and algebras. . . . . . . . . . . . . . . . . . . . 575 16.8.3 Forms induced on wedge products . . . . . . . . . . . . 578 16.8.4 Special constructions involving symplectic forms . . . . 580 17 Quadratic Forms 585 17.1 Quadratic Forms. . . . . . . . . . . . . . . . . . . . . . . . . . 585 17.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 585 17.1.2 Forms of finite rank . . . . . . . . . . . . . . . . . . . . 586 17.1.3 More definitions relative to a quadratic form . . . . . . 588 17.1.4 The poset of totally singular subspaces . . . . . . . . . 588 17.1.5 Radicals of quadratic forms in characteristic 2 . . . . . 590 17.2 Hyperpolic decompositions with respect to a quadratic form. . 591 17.3 The oriflame phenomenon. . . . . . . . . . . . . . . . . . . . . 592 17.4 Isometries and the Classical Groups . . . . . . . . . . . . . . . 593 17.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 593 17.4.2 Witt’s Theorem . . . . . . . . . . . . . . . . . . . . . . 595 17.5 Quadratic forms in special cases . . . . . . . . . . . . . . . . . 605 17.5.1 The real and complex cases . . . . . . . . . . . . . . . 605 17.5.2 The case of finite fields . . . . . . . . . . . . . . . . . . 606 17.6 Quadratic forms over GF(2) . . . . . . . . . . . . . . . . . . . 608 17.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 608 17.6.2 Beginnings . . . . . . . . . . . . . . . . . . . . . . . . . 608 17.6.3 The definition of an orthogonal geometry over GF(2). . 609 11 17.6.4 Non-degenerate orthogonal geometries over GF(2) in odd dimension. . . . . . . . . . . . . . . . . . . . . . 17.6.5 Non-degenerate orthogonal geometries over GF(2) in even dimension. . . . . . . . . . . . . . . . . . . . . . 17.6.6 Some low dimensional cases. . . . . . . . . . . . . . . 17.6.7 Counting the singular and non-singular vectors. . . . 17.7 Cocliques in Orthogonal Geometries over GF(2). . . . . . . 17.8 Homogeneous cocliques of maximal size. . . . . . . . . . . . 17.9 Observations obtained from homogeneous cocliques in small dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.9.1 The case n = 3. . . . . . . . . . . . . . . . . . . . . . 17.9.2 The case n = 4. . . . . . . . . . . . . . . . . . . . . . 18 The 18.1 18.2 18.3 18.4 18.5 18.6 . 610 . . . . . 611 612 613 615 616 . 619 . 619 . 620 Klein Correspondence 626 The Klein Quadric . . . . . . . . . . . . . . . . . . . . . . . . 626 The space of Symplectic Forms . . . . . . . . . . . . . . . . . 628 Pure Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 628 The Correspondence . . . . . . . . . . . . . . . . . . . . . . . 631 The correspondence and the oriflame strucure . . . . . . . . . 632 Consequences of the Klein Correspondence . . . . . . . . . . . 634 18.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 634 18.6.2 Classical isomorphisms . . . . . . . . . . . . . . . . . . 636 18.6.3 The generalized quadrangles of types Sp(4, K) and O(5, K)637 √ 18.6.4 The generalized quadrangles of types U (4, K) and O− (6, K)638 12 Chapter 1 Basics 1.1 1.1.1 Presumed results and conventions Presumed Jargon The reader is assumed to have already acquired some familiarity with groups, cosets, group homomorphisms, ring homomorphisms and vector spaces from an undergraduate abstract algebra course or linear algebra course. Since most abstract algebraic strucures in these notes are treated from first principles, we rely on these topics mostly as a source of familiar examples which can aid the intuition as well as act as points of reference that will indicate the direction various generalizations are taking. 1.1.2 Sets and maps In the area of sets and maps we should be precise about the conventions. First of all the reader should have a comfortable rapport with the following: 1. The set-theoretic notions of membership, containment and the operations of intersection and union over arbitrary collections of subsets of a set. 2. The Cartesian product construction, A × B. 3. Relations, on a set S (as a subset of S × S). Especially, equivilance relations and equivalence classes. 13 4. The fact that given a set X, there is a set 2X of all subsets of X. 5. Mappings: The word “mapping” is intended to be indistinguishable from the word “function” as it is used in most literature. A mapping f : X −→ Y is any device that associates with any element x of set X, a unique element, denoted f (x), of the set Y . We say “device that associates” here, rather than “rule that associates” to indicate the generality of the notion.1 Such a device always exists. We may define a mapping f : X → Y as a subset R ⊆ X × Y with the property that for every x ∈ X, there exists exactly one y ∈ Y such that the pair (x, y) is in R. Then the notation y = f (x) simply means (x, y) ∈ R and we metaphorically express this fact by saying “the function f sends element x to y – as if the function was actively doing something. The suggestive metaphors continue when we also render this same fact – that y = f (x) – by saying that y is the image of element x or equivalently x is a preimage of y or that x lies in the fibre f −1 (y). Two mappings are considered equal if they “do the same things”. Thus if f and g are both mappings (or functions) from X to Y we say that mapping f is equal to mapping g if and only if f (x) = g(x) for all x in X. 6. Identity mappings. A very special example is the following: The mapping 1X : X → X which takes each element x of X to itself – i.e. f (x) = x – is called the identity mapping on set X. This mapping is very special and is uniquely defined just by specifying the set X. 7. Domains, codomains, restrictions and extensions mappings. In defining a mapping f : X → Y , the sets X and Y are a vital part of the definition of a mapping or function. The set X is called the domain of the function; the set Y is called the codomain of the mapping or function. A simple manipulation of both sets allows us to define new functions from old ones. For example, if A is a subset of the domain set X, and 1 The word “rule” may elicit its ordinary-language meaning of an English sentence which instructs you how to compute an image – say, a power series formula for a function. But English sentences are finite and there are only countably many of them – while the number of functions between infinite sets is usually uncountable. You would be amazed how many dreadful Calculus texts define a function as a “rule” which associates a unique y for each x of the domain set. 14 f : X → Y is a mapping, then we obtain a mapping f |A : A → Y which sends every element a ∈ A to f (a) (which is defined a fortiori). This new function is called the restriction of the function f to the subset A. If g = f |A , we say that f extends function g. Similarly, if the codomain Y of the function f : X → Y is a subset of a set B (that is, Y ⊆ B), then we automatically inherit a function f |B : X → B just from the definition of “function”. When f : X → Y = X is the identity mapping on X, the replacement of the codomain from X by a larger set B yields a mapping 1X |B called the containment mapping2 . 8. The composition of mappings. If f : X → Y and G : Y → Z, there is a well-defined mapping g ◦ f : X → Z which takes any x ∈ X to g(f (x)) in Z. The student should understand the “associativity” of the composition operation when all compositions are defined. Thus if h : Z → W and f and g are as indicated, then h ◦ (g ◦ f ) = (h ◦ g) ◦ f . The reader should know that if f : X → Y is any mapping, then 1Y ◦ f = f, and f ◦ 1X = f. 9. Images and range. If f : X → Y is a mapping, the collection of all “images” f (x) of elements of X is clearly a subset of Y which we call the image or range of the function f and is denoted f (X). 10. One-to-one mappings (injections) and onto mappings (surjections). A mapping f : X → Y is called onto or surjective if and only if Y = f (X) as sets. That means that every element of set Y is the image of some element of set X, or, put another way, the fibre f −1 (y) of each element y of Y is nonempty. A mapping f : X → Y is said to be one-to-one or injective if any two distinct elements of the domain are not permitted to yield the same image element. This is another way of saying, that the fibre f −1 (y) of each element y of Y is permitted to have at most one element. 2 Unlike the notion of “restriction”, this construction does not seem to enjoy a uniform name. 15 11. The reader should be familar with the fact that the composition of two injective mappings is injective and the composition of two surjective mappings (performed in that chronological order) is surjective. She or he should be able to prove that the restriction f |A : A → Y of any injective mapping f : X → Y (here A is a subset of X) is an injective mapping. 12. Bijections. A mapping f : X → Y is a bijection if and only if it is both injective and surjective – that is, both one-to-one and onto (in the preBourbaki vernacular). When this occurs, every element y ∈ Y has its fibre f −1 (y) containing a unique element which can unambiguously be denoted f −1 (y). This notation allows us to define the unique function f −1 : Y → X which we call the inverse of the bijection f . Note that the inverse mapping possesses these properties, f −1 ◦ f = 1X , and f ◦ f −1 = 1Y , where 1X and 1Y are the identity mappings on sets X and Y , respectively. 13. Indexing families of subsets of a set X with the notation {Xα }I – or {Xα }α∈I . This should be understood in its guise as a mapping I −→ 2X . 3 14. The construction hom(A, B) as the set of all mappings A −→ B. (This is denoted B A in some parts of mathematics.) The reader should see that if A is a finite set, say with n elements, then hom(A, B) is just the n-fold Cartesian product of B with itself B × B × · · · × B (with exactly n factors) Also if N denotes the collection of all natural numbers, then hom(N, B) is the set of all sequences of elements of B. Notation for the image of an element in the domain of a mapping. There is a perennial akwardness cast over common mathematical notation for the composition of two maps. Mappings are sometimes written as left 3 Note that in the notation, the “α” is ranging completely over I and so does not itself affect the collection being described; it is what Logicians call a “bound” variable. 16 operators and sometimes as right operators, and the akwardness is not the same for both choices due to the assymetric fact that English is read from left to right. Because of this, right operators work much better for representing the action of sets with a binary operation as mappings with compositions. Then why are left operators used at all? There are two answers: Suppose there is a division of the operators on X into two sets – say A and B. Suppose also that if an operator a ∈ A is applied first in chronological order and the operator b ∈ B is applied afterwards; that the result is always the same had we applied b first and then applied a later. Then we say that the two operations “commute” (at least in the time scale of their application, if not the temporal order in which the operators are read from left to right). This property can often be more conveniently rendered by having one set, say A, be the left operators, and the other set B be the “right” operators, and then expressing the “commutativity“ as something that looks like an “associative law”. So one needs both kinds of operators to do that. The second reason for using left operators is not so mathematical: the answer comes from the sociological accident of English usage. In the English phrase “function of x”, the function comes first, and then its argument. This is reflected in the left-to-right notation “f (x)” so familiar from calculus. Then if the composition α◦β were to mean “α is applied first and then β”, we must have (α ◦ β)(x) = β(α(x)). That is, we must reverse their “reading” order. On the other hand, if we say α ◦ β means β is applied first and α is applied second – so (α ◦ β)(x) = α(β)(x) – then things are nice as far as the treatment of parentheses are concerned, but we seem to be reading things in the reverse chronological order (unless we compensate by reading from right to left). Either way there is an inconvenience. A (not all that overwhelming) majority in the literature have chosen the latter as the least of the two evils. Here, at least, we adopt the latter conventions insofar as left operators are concerned. So we have this (Rule for Composition of Left Operators) If α : X → Y and β : Y → Z are left operators on their respective domains X and Y , then, for each x∈X (β ◦ α)(x) := β(α(x)); — so that β ◦ α : X → Z denotes the composition of the mappings, viewed as a left operator on X. 17 The first argument certainly applies to discussions of representations of groups and rings, where on the one hand, it is really handier to think of the right action of groups or rings as induced mappings which are right operators. At the same time, one desires “morphisms” which commute with all these operators , and so it is easier to think of these morphisms as left operators: so both types of operators must be used. The convention of these notes is to write mappings as left operators, and to indicate right operators by exponential notation or with explicit apologies (as in the case of right R-modules), as right multiplications. Thus in general we adopt these rules: Rule #1 α ◦ β denotes the composition resulting from first applying β and then α in chronological order. Thus (α ◦ β)(x) = β(α(x)). Rule #2 Exponential notation indicates right operators. Thus compositions have these images: xα◦β = (xα )β . Exception to Rule #2 For right R-modules we indicate the right operators by right “multiplication”, that is right juxtaposion. Ring multiplication still gets represented the right way since we have (mr)s = m(rs) for module element m and ring elements r and s (it looks like an associative law). (The reason for eschewing the exponential notation in this case is that the law mr+s = mr + ms for right R-modules would then not look like a right distributive law.) 1.1.3 Partitions A partition of a set X is a collection {Xα }I of non-empty subsets Xα of X satisfying 1. ∪α∈I Xα = X 2. The sets Xα are pairwise disjoint – that is, Xα ∩ Xβ = ∅ whenever α 6= β. 18 The sets Xα of the union are called the components of the partition.4 A partition may be described in another way: as a surjection π : X −→ I. Then the collection of fibres – that is, the sets π −1 (α) := {x ∈ X|π(x) = α} as α ranges over I – form the components of a partition. Conversely, if {Xα }I is a partition of X, then there is a well-defined surjection π : X −→ I which takes each element of X to the index of the unique component of the partition which contains it. ˙ In these lecture notes we shall write X = A + B (rather than X = A∪B or X = A ] B) to express the fact that {A, B} is a partition of X with just two components A and B. Similarly we write X = X1 + X2 + ... + Xn when X possesses a partition with n components Xi , i = 1, 2, . . . , n. This notation goes back to Galois’ rendering of a partition of a group by cosets of a subgroup. The notation is very convenient since one doesn’t have to “doctor up” a “cup” (or “union”) symbol. Unfortunately similar notation is also used in more algebraic contexts with a different meaning. We resolve the possible ambiguity in this way: When a partition (rather than, say, a sum of ideals) is intended, the partition will simply be introduced by the words “partition” or “decomposition”. 1.1.4 Zorn’s Lemma and the Axiom of Choice. The student ought to be acquainted with and to be able to use The Axiom of Choice Suppose {Xα }I is a family of pairwise disjoint non-empty sets. Then there exists a subset R of the union of the Xα , which meets each component Xα in exactly a one-element set. This assertion is about the existence of systems of representatives. If only finitely many of the Xα are infinite sets this can be proved from ordinary set theory. But the reader should be aware that in its full generality it is independant of set theory, and yet, is consistent with it. The reader is thus 4 It does not seem to be useful to allow partitions to have empty components. 19 encouraged to think of it as an adjunct axiom to set theory, to make a note of each time it is used, and to quietly employ the appropriate guilt feelings when using it. Obviously it has many uses. For example it guarantees us a system of coset representatives for any subgroup of any group. In fact we shall see an application of it in the very next subsection on cardinal numbers. In the presence of set theory (which is actually ever-present for the purposes of these notes) the Axiom of Choice is equivilant to another assertion called Zorn’s Lemma, which should also be familiar to the reader. Since it appears in the setting of partially ordered sets, its discussion is deferred to a later section of this chapter. The reader need not be required to know a proof of the equivalence of the Axiom of Choice and Zorn’s Lemma, or a proof of their consistency with set theory. At this stage just the psychological assurances are enough to read these notes. For a good developement of the many surprising equivalent versions of the Axiom of Choice the truly curious student is encouraged to peruse the excellent and readable monograph of Prof. Mary Ellen Rudin as well as read another study in clarity: the book by Ian Stewart entitled The Joy of Sets. 1.1.5 Cardinal Numbers Not all collections of things are actually amenable to the axioms of set theory, as Russell’s paradox illustrates. Nonetheless certain operations and constructions on such collections can still exist. It is still possible that they may possess equivalence relations and that is true of the collection of all sets. We have mentioned that a mapping f : A −→ B which is both an injection and a surjection is called a bijection or a one-to-one correspondence in a slightly older language. In that case the partition of A defined by the surjection f (see above) is the trivial partiton of A into its one-element subsets. This means the fibering f −1 defines an inverse mapping f −1 : B −→ A which is also a bijection satisfying f ◦ f −1 = 1A and f −1 ◦ f = 1B as every student knows as described above. It is clear, using obvious facts about compositions of bijections and identity mappings, that the relation of two sets having a bijection connecting them is an equivalence relation. The resulting equivalence classes are called cardinal numbers and the equivalence class containing set X is denoted |X| and is called the cardinality of X. One recalls that one writes |X| ≤ |Y | if and only there exists an injection 20 f : X −→ Y . This is a well-defined relation between cardinal numbers of sets for if |X| = |X 0 | then there is a bijection g : X 0 −→ X and the composition g ◦ f defines an injection from X 0 into Y , so |X 0 | ≤ |Y | by definition. These lecture notes presume and use the following three results concerning cardinalities of sets: Theorem 1 (The Bernstein-Schröder Theorem) If for two sets X and Y , |X| ≤ |Y | and |Y | ≤ |X| then |X| = |Y | (A nice proof with pictures appears in the famous book of Birkhoff and McLane.) Theorem 2 If there exists a surjection f : X −→ Y then |Y | ≤ |X|. proof: : With the axiom of choice one gets a subset X 0 containing exactly one element xy from each fiber f −1 (y) as y ranges over Y . One then has an injection Y −→ X taking y to xy . Theorem 3 If I is an infinite set, then there exists a partition I = A + B with both A and B in bijective correspondence with I – that is, |A| = |I| = |B|. 1.1.6 Special results on cardinalities As usual, let N denote the set of natural numbers, that is, the set {0, 1, 2, . . .} of all integers which are positive or zero (non-negative). There is a useful special result whose proof may not have appeared in the student’s undergraduate experience. Theorem 4 If I is an infinite set, then |I × N| = |I|. proof: By Theorem 3 just above, there is a partition I = A + B and bijections α : I −→ A and β : I −→ B. Set B1 := B = β(I), Bk := β k (I) and Ak := β k−1 ◦ α(I). We then have diagram β β β β I → B1 → B2 → B3 → · · · α ↓ β β β β A → A1 → A2 → A3 → 21 where each Ak is contained in the Bk directly above it in the diagram. For any n ∈ N we have a partition I = A + A1 + A2 + · · · + An + Bn−1 . Setting Ã = A + A1 + A2 + · · · (an infinite union of pairwise disjoint sets) we see that the formula ψ(x, n) := β n−1 ◦ α(x) defines a mapping ψ : I × N −→ Ã whose restriction to I × {n} induces an injection I × {n} → An−1 (with the convention that A0 := A). Thus we see that ψ is an injection. Thus by definition |I × N| ≤ |I|. But as there is a bijection I → I × 1 ⊆ I × N we also have |I| ≤ |I × N|. The result now follows from the Bernstein-Schröder Theorem. Corollary 5 . Let {Xα }I be a family of finite non-empty subsets of a set S indexed by an infinite set I. Then |∪α∈I Xα | ≤ |I| proof: We may formally construct the disjoint union of Xα by replacing each Xα by Xα × {α} and taking the union F = ∪α (Xα × {α}) in the ambient set S × I. Then the projection on the first coordinate defines a surjection F → U := ∪α Xα . At each α there exists some surjection φα := N → Xα since Xα is a finite set. This sequence {φα } in turn defines a surjection ψ :I ×N→F upon setting ψ(x, m) := (φα (x), α) ∈ F . Since the composition of the two surjections presented above is a surjection I × N → U we have |I × N| ≥ |U | by Theorem 2 above. But by Theorem 4 just preceeding, |I × N| = |I|, and so |I| ≥ |U | follows. 22 1.2 Binary operations and Monoids. Suppose X is a set and let X (n) be the n-fold Cartesian product of X with itself. For n > 0, a mapping X (n) → X is called an n-ary operation on X. If n = 1 is is just a mapping of x into itself. There are certain concepts that are brought to bear at the level of 2-ary (or binary) that fade away for larger n. We say that set X admits a binary operation if there exists a 2-ary function f : X × X → X. In this case, it is possible to indicate the operation by a constant symbol (say “*”) inserted between the elements of an ordered pair — thus one might write “x*y” to indicate f ((x, y)) (which we shall write as f (x, y) to rid ourselves of one set of superfluous pqrentheses). Indeed one might even use the empty symbol in this role – that is f (x, y) is represented by “juxtaposition” of symbols: we write “xy” for f (x, y). This seems the simplest way to describe the relevant properties. The binary operation on X is said to be commutative if and only if xy = yx, for all x and y in X. The operation is associative if and only if x(yz) = (xy)z for all x, y, z ∈ X. A set admitting an associative binary operation is called a semigroup. A semigroup is said to possess an identity element or two-sided identity if and only there exists an element e ∈ X such that ex = x = xe for all x ∈ X. A semigroup with a two-sided identity is called a monoid. A semigroup (or monoid) with respect to a commutative binary operation is simply called a commutative semigroup (or commutative monoid). For example the non-negative integers, N0 , form a commutative monoid under ordinary addition of integers. It is not our intention to venture into various algebbraic structures at such an early stage. But we are forced to single out monoids, since they are always lurking around so many of the most basic definitions (for example, interval measures on posets, just to name one). 23 Chapter 2 Partially Ordered Sets and Dependance Theories 2.1 Introduction The reader is certainly familiar with examples of sets X possessing a transitive relation “≤” which is antisymmetric. Such a pair (X, ≤) is called a partially ordered set. For any subset Y of X there is a partial ordering (Y, ≤) induced on Y by the partially ordered set (X, ≤) which surrounds it: one merely restricts the relation “≤” to the pairs of X × X. We then call (Y, ≤) an induced poset of (X, ≤). Certainly one example familiar to most readers is the poset P(X) := (2X , ⊆) called the power poset of all subsets of X under the containment relation. Throughout this algebra course one will encounter sets X which are closed under various various n-ary operations subject to certain axioms – that is, some species of “algebraic object”. Each such object X naturally produces a partially ordered set whose members are the subsets of X closed under these operations. — that is, the poset of algebraic subobjects of the same species. Of course, if there are no particular operations or axioms to be preserved, we have merely redescribed the power poset P(X) := (2X , ⊆) mentioned in the preceeding paragraph. If X is some algebraic species closed under operations subject to axioms, then its poset of algebraic subobjects of the same species is certainly an induced subposet of the power poset P(X). But as a convention to distinguish the two, instead of writing “A ⊆ B” (the simple containment relation) we write “A ≤ B” to indicate that A is an 24 algebraic subobject of B of the same algebraic species. For example, If X is a group, then the poset of algebraic subobjects is the poset of all subgroups of X. If R is a ring and M is a right R-module, then we obtain the poset of submodules of M . Special cases are the posets of vector subspaces of a right vector space V and the poset of right ideals of a ring R. In turn these posets have special induced posets: Thus the poset L(V ) of all finite-dimensional vector subspaces of a vector space V is transparently a subposet of the poset of all vector subspaces of V . For example from a group G one obtains the poset of normal subgroups of G, or more generally, the poset of subgroups invariant under any fixed subgroup A of the automorphism group of G. Similarly there are posets of invariant subrings of polynomial rings, and invariant submodules for an R-module admitting operators. Finally, there are induced posets of algebraic objects which are closed with respect to a closure operator on a poset (perhaps defined by a Galois connection). All of these examples will be made precise later. The important thing to note at this stage is that 1. Partially ordered sets underly all of the algebraic structures discussed in this book. 2. Many of the crucial conditions which make arguments work are basically properties of the underlying posets alone and do not depend on the particular species of algebraic object with which one is working: Here are the main examples: (a) Zorn’s Lemma, (b) the ascending and descending chain conditions, (c) Galois connections and closure operators, (d) interval measures on semimodular semilattices (The General Jordan Hölder Theorem), (e) Möbius inversion, and (f) dependence theory (where “dimension” becomes matroid rank). The purpose of this chapter is to introduce those basic arguments that arise strictly from the framework of partially ordered sets, ready to be used for the rest of this book. 25 2.2 2.2.1 Basic definitions. Definition of a partially ordered set. A partially ordered set, (P, ≤), hereafter called a poset, is a set P with a transitive antireflexive binary relation ≤. This means that for all elements x, y and z of P , 1. x ≤ y and y ≤ z together imply x ≤ z, and 2. x ≤ y and y ≤ x together imply x = y. We write y ≥ x if x ≤ y and write x < y if x ≤ y and x 6= y. In the latter case we say that x is properly below y, or, equivalently, that y is properly above x. 2.2.2 Induced subposets. Suppose (P, ≤) is a poset, and X is a subset of P . Then we can agree to induce the relation ≤ on the subset X — that is, for any two elements x and y of the subset X, we agree to say that x ≤X y if and only if x ≤ y in the poset (P, ≤). Then any child can verify that (x ≤ y) if and only if (x ≤X y) – it is just a consequence of the definitions. The empty set is considered to be an induced subposet of any poset. Let X and Y be two subsets of P where (P, ≤) is a poset. We make these observations: 1. If (X, ≤X ) is a partial ordering on X and if (Y, ≤Y ) is a partial ordering on Y such that (X, ≤X ) is an induced subposet of (Y, ≤Y ) and (Y, ≤Y ) is an induced subposet of (P, ≤), then (X, ≤X ) is also an induced subposet of (P, ≤). This fact means that if X is any subset of P , and (P, ≤) is a poset, we don’t need those little subscripts attached to the relation “≤”: we can unambiguously write (X, ≤) to indicate the subposet induced by (P, ≤) on X. In fact, when it is clear that we are speaking of induced subposets, we may write X for (X, ≤) and speak of “the induced poset X”. 2. If (X, ≤) and (Y, ≤) are induced subposets of (P, ≤) then we can form the intersection of induced posets (X ∩Y, ≤). In fact, since this notion depends only on the underlying sets, one may form the intersection 26 of induced posets (∩σ∈I Xσ , ≤), from any family {(Xσ , ≤)|σ ∈ I} of induced posets of a poset (P, ≤). Perhaps the most important example of an induced poset is the interval. Suppose (P, ≤) is a poset and that x ≤ y for elements x, y ∈ P . Consider the induced poset [x, y]P := {z ∈ P |x ≤ z ≤ y}. This is called the interval in (P, ≤) from x to y. If (P, ≤) is understood, we would write [x, y] in place of [x, y]P . But we must be very careful: there are occasions in which one wishes to discuss intervals within an induced poset (X, ≤), in which case one would write [x, y]X for the elements z of X between x and y. Clearly one could mix the notations and write [x, y]X := [x, y] ∩ X, the intersection of two induced posets. 2.2.3 Dual posets and dual concepts. Suppose now that (P, ≤) is given. One may obtain a new poset, (P ∗ , ≤∗ ) whose elements are exactly those of P , but in which the partial ordering has been reversed! Thus a ≤ b in (P, ≤) if and only if b ≤∗ a in (P ∗ , ≤∗ ). In this case, (P ∗ , ≤∗ ) is called the dual poset of (P, ≤). We might as well get used to the idea that for every definition regarding (P, ≤), there is a “dual notion” — that is, the generically-defined property or set in (P, ≤) resulting from defining the same property or set in (P ∗ , ≤∗ ). Examples will follow. 2.2.4 Maximal and minimal elements of induced posets An element x is said to be maximal in (X, ≤) if and only if there is no element in X, which is strictly larger than x – that is x ≤ y for y ∈ X implies x = y. Note that this is quite different than the more specialized notion of a global maximum (over X) which would be an element x in X for which x0 ≤ x for all elements x0 of X. Of course defining something does not posit its existence; (X, ≤) may not even possess maximal elements, or if it does, it may not contain a global maximum. We let max X denote the full (but possibly empty) collection of all maximal elements of (X, ≤). 27 By replacing the symbol “≤” by “≥” in the preceeding definitions, we obtain the dual notions of a minimal element and a global minimum of an induced poset (X, ≤). Of course some posets may contain no minimal elements whatsoever. 2.2.5 Global maxima and minima: zero’s and one’s in posets. In some posets (P, ≤) there exists an element “0P ” (just to give it a name) which it less-than-or-equal to all other elements of P . Clearly such an element is unique. We call it the zero-element of the poset. From its definition, a zero element would always be a the global minimum. Dually, there may exist a “one-element”, denoted 1P , or something similar, which is an element in (P, ≤) with the property that all other elements are less-than-or equal to it. Obviously this the zero-element of (P ∗ , ≤∗ ). Some posets have a “zero”, some have a “one”, some have both, and some have neither. But whether or not a “zero” is present in P , one can always adjoin a new element declared to be properly below all elements of P to obtain a new poset 0(P ). For example, if P contains just one element p, then 0(P ) consists of just two elements {01 , p} with 01 < p — called a chain of length one . Iterating this construction with the same meaning of P , we see that 02 (P ) introduces a “new zero element” , 02 to produce a 3-element poset with 02 < 01 < p – that is, a chain of length two. Clearly 0k (P ) would be a poset with elements (up to a renaming of the elements) arranged as 0k < 0k−1 < · · · < 01 < p, which we call a chain of length k. Of course this construction can also be performed on any poset P so that 0k (P ) in general becomes the poset P with a tail of length k − 1 adjoined to it from below – a sort of attached “kite-tail”. Also, by dually defining 1k (P ) one can attach a “stalk” of length k − 1 above an arbitrary poset P . 2.2.6 Total orderings and chains A poset (P, ≤) is said to be totally ordered if {c, d} ⊆ C implies c ≤ d or d ≤ c. (Sometimes this notion is referred to as “simply ordered”. Obviously 28 any induced poset of a totally ordered set is also totally ordered. Any maximal (or minimal element) of a totally ordered set is in fact a global maximum (or global minimum). Familiar examples of totally ordered sets are obtained as induced posets of the real numbers under its usual ordering, for example (1) the rational numbers, (2) the integers, (3) the positive integers (4) the open, half-closed or closed intervals (a, b), [a, b), (a, b], or[a.b] of the real number system, or (5) the intersections of sets in (4) with those in (1)-(3). A chain of (P, ≤) is an induced poset (C, ≤) which happens to be totally ordered. The familiar real number system under the natural ordering is totally ordered, and so are any of its induced subposets such as the integers, the positive integers, any real open, half-open, or closed interval, or any intersection of these intervals with the integers or rational numbers etc. 2.2.7 Antichains. The reader is certainly familiar with many posets which are not totally ordered, such as the set P(X) of all subsets of a set X of cardinality at least two. Here the relation “≤” is containment of sets. Again there are many structures that can be viewed as a collection of sets, and thus become a poset under this same containment relation: for example, subspaces of a vector space, subspaces of a point-line geometry, subgroups of a group, ideals in a ring and R-submodules of a given R-module and in general nearly any admissable subobject of an object admitting some specified set of algebraic properties. Two elements x and y are said to be incomparable if both of the statements x ≤ y and y ≤ x are false. A set of pairwise incomparable elements in a poset is called an antichain.1 The set max(X, ≤) where (X, ≤) is an induced subposet of (P, ≤), is always an antichain. 1 In a great deal of the literature, sets of pairwise incomparable elements are called independant. Despite this convention, the term “independant” has such a wide usage in mathematics that little is served by employing it at this point to indicate what we have called an antichain. However, some coherent sense of the term “independance” – of the sort useful for Algebra – is exposed in a section on Matroids later in this chapter. 29 2.2.8 Zornification Although an induced poset (X, ≤) of a poset (P, ≤) may not have a maximum, it might have an upper bound – that is, an element m in P which is larger than anything in the subset X ( precisely rendered by “m ≥ x for all elements x of X”). (Of course if it happened that such an element m were already in X then it would be a global maximum of X). The dual notion of “lower bound” should be easy to formulate. Of course it would be useful to know that induced posets (X, ≤) actually have maximal elements. We mention here a criterion for asserting that such maximal elements exist. Zorn’s Lemma Suppose (P, ≤) is a poset for which any chain has an upper bound. Then (P, ≤) possesses a maximal element. It has had many equivalent formulations, and has gone by many names, such as “the Hausdorff maximum principle”, “Zorn’s Lemma” and others2 . However it is not a Lemma or even a Theorem. It does not follow from the axioms of set theory, nor does it contradict them. That is why we called it “a criterion for an assertion”. Using it can never produce a contradiction with formal set theory. But since its denial also cannot produce such a contradiction, one can apparently have it either way. The experienced mathematician, though not always eschewing its use, at least prudently keeps a running tab on how many appeals the maximum principle he or she may have invoked.3 In these notes the maximum principle will be used only very sparingly. 2.2.9 Order ideals and filters. For this subsection fix a poset (P, ≤). An order ideal of P is an induced subposet (J, ≤), with this property: If y ∈ J and x is an element of P with x ≤ y, then x ∈ J. 2 For a nice account of the history of this widely used principle, see Mary Ellen Rudin’s book. 3 Certainly using the Maximum Principle should be accompanied by the appropriate feeling of guilt. Who knows? Each indulgence in Zornification might revisit some of you in another life. 30 In other words, an order ideal is a subset J of P with the property that once some element belongs to J, then all elements of P below that element are also in J. Note that the empty subposet is an order ideal. The reader may check that the the intersection of any family of order ideals is an order ideal. (Since order ideals are a species of induced posets, we are using “intersection” here in the sense of the previous subsection 2.1.2 on induced subsposets.) Then there is the dual notion. Suppose (F, ≤∗ ) were an order ideal of the dual poset (P, ≤∗ ). Then what sort of subset of P is F ? It is characterized by being subset of P with this property: 1. If x ∈ F and y is any element of P with x ≤ y, then y ∈ F . Any subset of P with this property is called a filter. There is an easy way to construct order ideals. Take any subset X of P . Then define PX := {y ∈ P |y ≤ x for some element x ∈ X}. Note that always X ⊆ PX . In fact the order ideal PX is “generated” by X in the sense that it is the intersection of all order ideals that contain X. Also, we understand that P( ∅) = ∅. In the particular case that X = {x} we write Px for P{x} , and call Px the principle order ideal generated by x (or just a principle order ideal if x is left unspecified. Of course any order ideal has the form PX (all we have to do is set X = PX ) but in general, we do not need all of the elements of X. For example if x1 < x2 for xi ∈ X, and if we set X 0 := X − {x1 }, then PX = PX 0 . Thus if we throw out elements of X each of which is below an element left behind, the new set defines the same order ideal that X did. At first the student might get the idea that we could keep throwing out elements until we are left with an antichain. That is indeed true when X is a finite set, or more generally if every element of X is below some member of max(X, ≤). But otherwise it is generally false. The dual notion of the filter generated by X should be transparent. It is the set P X of elements of P which bound from above at least one element of X. It could be described as P X := {y ∈ P |Py ∩ X 6= ∅}. 31 By duality, the intersection of any collection of filters is a filter. An induced subposet (X, ≤) of (P, ≤) is said to be convex if, whenever x1 and x2 are elements of X with x1 ≤ x2 in (P, ≤), then in fact the entire interval [x, y]P is contained in X. 2.2.10 Products of posets. Suppose (P1 , ≤) and (P2 , ≤) are two posets4 . The product poset (P1 × P2 , ≤) is the poset whose elements are the elements of the Cartesian product P1 × P2 , where element (a1 , a2 ) is declared to be less-than-or-equal to (b1 , b2 ) if and only if a1 ≤ b1 and also a2 ≤ b2 . It should be clear that this notion can be extended to any collection of posets {(Pσ , ≤)|σ ∈ I}. Its elements are the elements of the Cartesian product Πσ∈I Pσ – that is, the functions f : I → U , where U is the disjoint union of the sets Pσ with the property that at any σ in I, f always assumes a value in Pσ . Then function f ≤ g if and only if f (σ) ≤ g(σ) for each σ ∈ I. Suppose now, each poset Pσ contains its own “zero element” element 0σ less than or equal to all other elements of Pσ . We can define a direct sum of posets {(Pσ , ≤)} as the induced subposet of the direct product consisting only of the functions f ∈ Πσ∈I Pσ for which f (σ) differs from 0σ for only P finitely many σ. This poset is denoted σ∈I Pσ . 4 Usually authors feel that the two poset relations should always have distinguished notation – that is, one should write (P1 , ≤1 ) and (P2 , ≤2 ) instead of what we wrote. At times this can produce intimidating notation that would certainly finish off any sleepy students with detectable breath. Of course that precaution certainly seems to be necessary if the two sets P1 and P2 are identical. But sometimes this is a little over-done. Since we already have posets denoted by pairs consisting of the set Pi and a symbol “≤00 , the relation “≤00 is assumed to be the one operating on set Pi and we have no ambiguity except possibly when the ground sets (Grundmengen) Pi are equal. (I just made up that word! I doubt if decent Germans use it.) Of course in the case the two “ground-sets” are equal we do not hesitate for a moment to adorn the symbol “≤00 with further distinguishing emblems. This is exactly what we did in defining the dual poset. But even in the case that P1 = P2 one could say that in the notation, the relation “≤” is determined by the name Pi of the set, rather then the actual set, so even then the “ordered pair” notation makes everything clear. I hope this makes everything clear. 32 2.2.11 Examples and exercises. Since we will need to discern whether two examples of posets are the same poset or not, we must at least define an isomorphism of posets. An isomorphism of poset (P1 , ≤1 ) to (P2 , ≤2 ) is a bijection f : P1 → P2 with the property that for any two elements x and y of P1 , x ≤1 y if and only if f (x) ≤2 f (y). As explained in the introduction of Chapter 1, the isomorphism simply amounts to a change of names – each “x’ in P1 is renamed “f (x)” and the relation “≤1 ” is renamed “≤2 ”. We also say that (P1 , ≤1 ) is isomorphic to (P2 , ≤2 ) when there is an isomorphism from the former to the latter. Among the vast collection of all posets, the relation of being isomorphic is an equivalence relation. Also, as noted in Chapter 1, algebraists tend to abuse the language by saying that two posets “are the same poset” if and only if they belong to the same equivalence class under isomorphisms. This abuse can be rendered tautologous by thinking of an (abstract) poset as the the isomorphism class of posets itself, and the examples of posets as presentations or models of the general (abstract) poset. In general, in algebra, we want two objects to be isomorphic if and only one can get a complete description of the second object simply by changing the names of the operating parts of the first object and the names of the relations among them that must hold – and vice versa. This is the “alias” point of view; the two structures are really the same thing with some names changes imposed.5 The other way is to form a bijection of the relevant parts of the first object with the second such that relations hold among parts in the domain if and only if the corresponding relations hold among their images.6 There is no logical distinction between the two approaches – only a psychological one. My wish is to get across to the student all that is expected when we speak of isomorphisms – we are merely dealing with the same situation, but under a new management which has renamed everything. Unfortunately “renaming” is a subjective human conceptualization that is akward to define precisely – that is why, at that beginning, we like to make the defi5 I must confess that this is the view I normally prefer. I have been deliberately vague in talking about parts rather than “elements” for the sake of generality. 6 33 nitions in terms of bijections rather than “renamings” even though the latter is how many of us secretly think of it. We shall meet this over and over again; it is the “abstractness” of Abstract Algebra.7 Example 1. (Examples of totally ordered sets) Isomorphism classes of these sets are the ordinal numbers. The most familiar examples are these: 1. The natural numbers This is the set of positive integers, N := {1, 2, 3, 4, 5, 6, . . .}, with the usual ordering 1 < 2 < 3 < 4 < 5···. Rather obviously, as an ordered set, it is isomorphic to the chain 0 < 1 < 2 < ··· or even k < k + 1 < k + 2 < k + 3 < ···, k any integer, under the shift mapping n → n + k − 1.8 2. The system of integers Z := {. . . < −2 < −1 < 0 < 1 < 2 < . . .}. 3. The system of the rational numbers with respect to the usual ordering — that is a/b > c/d if and only if ad > bd, an inequality of integers. 7 There is a common misunderstanding of this word “abstract” that mathematicians seem condemned to suffer. To the man on the street, “abstract” seems to mean “having no relation to the world, no applications”. Unfortunately this is the overwhelming view of State Legislatures, pundits of Education, and University Administrators everywhere (one hears words like “Ivory Tower”, “Intellecuals on welfare”, etc.). On the contrary, a concept is “abstract” precisely because is has more than one application – not that that it hasn’t any application, as one would expect from Ivory-Tower concepts remote from the world. They have got it just backwards! 8 This isomorphism explains why it is commonplace to do an induction proof with respect to the second of these examples rather than the first. In enumerative combinatorics, for example, the “natural numbers” are define to be all non-negative integers, not just the positive integers. 34 4. The real number system R. 5. Any induced subposet of a totally ordered set. We have already mentioned intervals of the real line. (Remark: the word “interval” here is used for the moment as it is used in Freshman College Algebra, open, closed, and half-open intervals such as (a, b] or [a, ∞). In this context, the intervals of posets that we have define earlier, becom the closed intervals, [a, b], of the real line, with a consistency of notation. Here is an example: Consider the induced poset of the rational numbers (Q, ≤) consisting of those postive fractions less than or equal to 1/2 which (in lowest terms) have a denominator not exceeding the positive integer d in absolute value. For d = 7 this is the chain 1 1 1 1 2 1 2 3 1 < < < < < < < < . 7 6 5 4 7 3 5 7 2 This is called a Farey series. A curiosity is that if ab and dc are adjacent members from left to right in this series, then bc − ad = 1! Example 2. (Examples of the classical locally finite (or finite) posets which are not chains) A poset (P, ≤) is said to be a finite poset if and only if it contains only finitely many elements – that is, |P | < ∞. It is said to be locally finite if and only if every one of its intervals [x, y] is a finite poset. 1. The Boolean poset B(X) of all finite subsets of a set X, with the containment relation (⊆) between subsets as the partial-ordering. 2. The divisor poset D = (N, |) of all positive integers N under the divisor relation: – that is, we say a|b if and only if integer a divides integer b evenly – i.e. b/a is an integer. 9 9 There are variations on this theme: In an integral domain a non-unit a is said to be irreducible if and only if a = bc implies one of b or c is a unit. Let D be an integral domain in which each non-unit is a product of finitely many irreducible elements, and let U be its group of units. Let D∗ /U be the collection of all non-zero multiplicative cosets U x. Then for any two such cosets, U x and U y, either every element of U x divides every element of U y or else no element of U x divides any element of U y. In the former case write U x ≤ U y. Then (D∗ /U, ≤) is a poset. If D is a unique factorization domain, then, as above, (D∗ /U, ≤) is locally finite for it is again a product of chains (one factor in the product for each association class U p of irreducible elements). One might ask what this poset looks like when D is not a unique factorization domain. Must it be locally finite? It’s something to think about. 35 3. Posets of vector subspaces. The partially ordered set L<∞ (V ; q) of all finite-dimensional vector subspaces of a vector space V over a finite field of q elements is a finite poset. (There is a generalization of this: the poset L<∞ (M ) of all finitely generated submodules of a right Rmodule M and in particular the poset L<∞ (V ) of all finite-dimensional subspaces of a right-vector space V over some division ring.) 10 4. The partition set: Πn . Suppose X is a set of just n elements. Recall that a partition of X is a collection π := {Y1 , . . . , Yk } of non-empty subsets Yj whose join is X but which pairwise intersect at the empty set. The subsets Yj are called the components of the partition π. Suppose π1 := {Yi |i ∈ I} is a partition of X and π 0 = {Zk |k ∈ K} is a second partition. We say that partition π 0 refines partition π if and only if there exists a partition I = J1 + · · · Jk of the index set, such that Yi := ∪{Z` |` ∈ Ji }. [We can state this another way: A partition π can be associated with a function fπ : X → I where the preimages of the points are the fibres partitioning X: the same being true of π 0 and a function fπ0 : X → I. We say that partition π 0 refines a partition π if and only if fπ (x) = fπ0 (x).] For example {6}{4, 9}{2, 3, 8}{1, 5, 10}{9, 10}, with four components refines {1, 5, 6, 9, 10}, {2, 3, 4, 8, 9} with just two components. Then Πn is the partially ordered set of all partitions of the n-set X under the refinement relation.11 5. The Poset of Finite Multisets. Suppose X is any non-empty set. A multiset is essentially a sort of inventory whose elements are drawn from X. For example: if X = {oranges, apples, and bananas} then m = { three oranges, two apples} is an inventory whose elements are multiple instances of elements from X. Letting O = oranges, A = 10 In Aigner’s book [?, page 203], L<∞ (V, q) is denoted L(∞, q) in the case that V has countable dimension. This makes sense when one’s plan is to relate the structure to certain types of generating functions (the q-series). But of course, it is a well-defined locally finite poset whatever the dimension of V . 11 Its cardinality |Πn | is called the nth Bell number and will reappear later in the context of permutation characters. 36 apples, and B = bananas , one may represent the multiset m by the symbol 3 · O + 2 · A + 0 · B or even the sequence (3, 2, 0) (where the order of the coordinates corresponds to a total ordering of X). But both of these notations can become inadequate when X is an infinite set. The best way is to think of a multiset as a mapping. Precisely, a multiset is a mapping f : X → N0 := N ∪ {0}, from X into the the set N0 of non-negative integers. A multiset f is dominated by a multiset g (written f ≤ g) if and only if f (x) ≤ g(x) for all x ∈ X (where the “≤” in the presented assertion relects the standard total ordering of the integers). The collection M(X) of all multisets of a set X forms a partially ordered set under the dominance relation. Since the multisets are actually mappings from X to N0 the dominace relation is exactly that used in comparing mappings in products. We are saying that the definition of the poset of multisets shows us that (M(X), ≤) = Y x∈X N0 , (2.1) – that is a product in which each “factor” Pσ in the definition of product of posets is the constant poset (N0 , ≤) of non-negative integers. The multiset f is said to be a finite multiset of magnitude |f | if and only if f (x) > 0 for only finitely many values of x, and X |f | = f (x), x∈X,f (x)>0 (2.2) (2.3) where the sum in the second equation is understood to be the integer 0 when the range of summation is empty (i.e. f (x) = 0 for all x ∈ X). Thus in the example concerning apples and oranges above, the multiset m is finite of magnitude 3 + 2 = 5. Thus the collection M≤∞ (X) of all finite multisets forms an induced poset of (M(X), ≤). Next one observes that a mapping f : X → N0 37 is a finite multiset if and only if f (x) = 0 for all but a finite number instances of x ∈ X. This means equation (2.1) above has a companion with the product replaced by a sum: M≤∞ (X) = X x∈X N0 . (2.4) Example 3. Simplicial complexes. We begin with a definition that is a little more general than is common. A poset (P, ≤) is called a simplicial complex if and only if 1. The poset (P, ≤) contains a zero element, 0. 2. Given any element x ∈ P , the interval [0, x] is isomorphic to the Boolean poset B(Sx ) where Sx is any set of finite cardinality r(x). The non-negative integer r(x) is called the (simplicial) rank of the element x, and, as may be expected, the function r : P → N0 taking x to r(x) is called the simplicial rank function. Clearly the zero element 0 is the unique element of rank zero. For each non-negative integer k, let Pk denote the elements of the simplicial complex (P, ≤) which have rank exactly k. Then P = ∪k Pk and P0 = {0}. The elements of P1 are called vertices. We can then form a graph ∆ = (P1 , P2 ) where vertex x ∈ P1 is incident with “edge” y ∈ P2 if and only if x ≤ y in (P, ≤). One sees that for any element x in P , that the order ideal Px meets P1 in exactly r(x) elements — i.e. each poset element x dominates exactly r(x) vertices. Indeed, we see that we can take Sx := Px ∩ P1 in the isomorphism [0, x] ' B(Sx ). [Note that S0 is the empty set.] With that convention we have: For all x, y ∈ P , x ≤ y if and only if Sx ⊆ Sy . Now let S(P ) be the full collection of all subsets of P1 of the form Sx = Px ∩ P1 for some element x of P . All of the sets in S(P ) are finite and so define an induced poset (S(P ), ≤) of the Boolean poset B(P1 ). Lemma 6 The following holds for any simplicial complex (P, ≤): 38 1. The poset (S(P ), ≤) is an order ideal of B(P1 ). 2. The mapping P → S(P ) defined by x → Sx is a poset isomorphism (P, ≤) ' (S(P ), ≤). 3. A poset (P, ≤) is a simplicial complex if and only if it is an order ideal of some Boolean poset B(X) for some set X. ¯ is a reflexive and transitive relation on a set X. Exercise 1. Suppose < (Such a relation is called a pseudo-order.) For any two elements x and y ¯ and also y <x. ¯ of X, let us write s ∼ y if and only if x<y 1. Show that “∼” is an equivalence relation. 2. For each element x of X, let [x] be the ∼-equivalence class containing ¯ then a<b ¯ for every element a ∈ [x] and element x. Show that if x<y element b ∈ [y]. (In this case we write “[x] ≤ [y]”.) 3. Let X/ ∼ denote the collection of all ∼-classes of X. Show that (X/ ∼ , ≤) is a poset. Exercise 2. Make a full list of the posets defined in Examples 1 and 2 above which 1. Have a zero element. 2. Have a one-element. 3. Are locally finite. Exercise 3. Let P be a fixed poset. If there is a bijection f : X → Y prove that there exists an isomorphism of posets: Y P → x∈X Y X P → X x∈X y∈Y y∈Y P, P, if P has a zero element. (This means that the index set X has only a weak effect on the definition of a product.) Exercise 4. Prove the following 39 Theorem 7 The divisor poset D is isomorphic to the poset M<∞ (N) of all finite multisets over the set N of positive integers. That is, we have an isomorphism X : (D, ≤) → N N0 . (2.5) [Hint: We sketch the argument: Notice that the right side of (2.5) consists of all sequences of non-negative integers of which only finitely many are positive (equivalently, all finite sequences of positive integers). The mapping is easy to define. First place the positive prime numbers in ascending (natural) order 2 = p1 < p2 < · · ·. For example p6 = 17. Now any integer d ∈ D greater than one posseses a unique factorization into positive prime powers, with the primes placed from left to right in ascending order – d = Πσ∈Jd paσσ , for some (possibly empty) ascending sequence of positive integers J. If one declares (d) to be the function f : N → N0 such that f (σ) = ( aσ 0 if σ ∈ Jd , if σ ∈ 6 Jd , and declare (1) := 0, the constant function with all values 0, then one sees that f (c) ≤ f (d) in the sum on the right side of equation (2.5) if and only if c divides d evenly – i.e. c ≤ d in (D, ≤). Clearly is a bijection.] Exercise 5. Recall that if (P, ≤) and X ⊆ P , then the symbol PX denoted the order ideal PX := {z ∈ P |z ≤ x for some x ∈ X}. Show that PX = ∩{Px |x ∈ X}. Exercise 6. Show that PX ∩ PY = PX∪Y . Exercise 7. Give an example showing that if X and Y are subsets of P , then PX ∪ PY and PX∩Y may not be equal. Exercise 8. Give an example of a poset (P, ≤) and an order ideal J of P such that J does not have the form P (X) for any antichain X. Exercise 9. Give formal proofs of all three conclusions in Lemma 6. Exercise 10. In any undirected graph G = (V, E) a clique is any collection of pairwise adjacent vertices. Suppose (P, ≤) is a simplicial complex and let ∆ = (P1 , P2 ) be its graph on vertices. Show that (P, ≤) is isomorphic to the poset of all finite cliques of the graph ∆ under set-inclusion. 40 Exercise 11. Let V be a vector space of infinite dimension. Let I be the collection of subsets of V consisting of the empty set together with every finite linearly independant set of vectors. Show that with respect to the containment relation, (I, ≤) is a simplicial complex. 2.3 2.3.1 Mappings among posets: closure operators and Galois connections Subposets. Now let (P, ≤) be a poset, and suppose X is a subset of P with a relation ≤X for which (X, ≤X ) is a partially ordered set. In general, there may be no relation between (X, ≤X ) and the poset (X, ≤) that is induced on X. But if it is true that for any x and y in X, x ≤X y implies x ≤ y, then we say that (X, ≤X ) is a subposet of (P, ≤). How does this compare with the notion of induced subposet that we have been using so far? If (X, ≤X ) is an induced subposet of (P, ≤), then for any x and y in X, x ≤X y if and only if x ≤ y. Thus in a general subposet it might happen that two elements x1 and x2 of (X, ≤X ) are incomparable with respect to the ordering ≤X even though one is bounded by the other (say, x1 ≤ x2 ) in the ambient poset (P, ≤). 2.3.2 Order-preserving mappings Let (P, ≤) and (Q, ≤) be posets. A mapping f : P → Q is said to be order preserving if and only x ≤ y implies f (x) ≤ f (y), for all elements x and y of P .12 Now there are two ways to view the image set f (P ) as a subset of (Q, ≤): (1) First one may view it as the induced poset (f (P ), ≤). (2) Secondly one may think of it as an image poset (f (P ), ≤f ) where, for a, b ∈ f (P ), one 12 The student familiar with such things from a topology class may recognize that the order preserving mappings among simplicial complexes are what are usually called simplicial mappings in that context. 41 writes a ≤f b if and only if there are elements x and y in the respective fibers f −1 (a) and f −1 (b) such that x ≤ y. In this case we still have that a ≤f b implies a ≤ b with respect to poset (Q, ≤). So we have the following: If f : P → Q is an order preserving mapping between posets (P, ≤) and (Q, ≤), then the image poset (f (P ), ≤f ) is a subposet (but not necessarily an induced suposet) of (Q, ≤). If the order preserving mapping f : P → Q is injective then the posets (P, ≤) and (f (P ), ≤f ) are isomorphic, and we say that f is an embedding of poset (P, ≤) into (Q, ≤). The following result, although not used anywhere in this book, at least displays the fact that one is interested in cases in which the image poset of an embedding is not an induced subposet of the co-domain. Any poset (P, ≤) can be embedded in a totally ordered set.13 We need a few other definitions related to order preserving. A mapping f : P → Q is said to be order-reversing if and only x ≤ y implies f (x) ≥ f (y). Example 4. Suppose X = ∪σ∈I Aσ , a union of non-empty sets indexed by I. Now there is a natural order reversing mapping among power posets: α : P(X) → P(I), which takes each subset Y of X, to α(Y ) := {σ ∈ I|Y ⊆ Aσ }. For example, if Y is contained in no Aσ , then α(Y ) = ∅. Finally an order preserving mapping f : P → P is said to be monotone non-decreasing if p ≤ f (p) for all elements p of P . 13 Many books present an equivalent assertion “any poset has a linear extension”. The proof is an elementary induction for finite posets. For infinite posets it requires some grappling with Zorn’s Lemma and ordinal numbers. 42 2.3.3 Closure operators. A closure operator is a monotone non-decreasing mapping τ : P → P which is idempotent, that is, τ (τ (x)) = τ (x) for all x ∈ P . We call the images of τ the closed elements of P . There are many contexts in which closure operators arise, and we list a few. 1. The ordinary topological closure in the poset of subsets of a topological space. 2. The mapping which takes a subset of a group (ring or R-module) to the subgroup (subring or submodule, resp.) generated by that set in the poset of all subsets of a group (ring or R-module). 3. The mapping which takes a set of points to the subspace which they generate in a point-line geometry (P, L). 2.3.4 Closure operators and Galois connections. One interesting context in which closure operators arise are Galois connections. Let (P, ≤) and (Q, ≤) be posets. A Galois connection (P, Q, α, β) is a pair of order-reversing mappings α : P → Q and β : Q → P , such that the two compositions β ◦ α : P → P and α ◦ β : Q → Q are both monotone non-decreasing. Galois connections arise in several contexts, especially when some algebraic object acts on something: For example • Groups acting on a set X. • Groups acting on a field as in the classical Galois Theory. • Rings acting as a ring of endomorphisms of an abelian group. Another example arises in algebraic geometry. We say that a polynomial p(x1 , . . . , xn ) vanishes at a vector v = (a1 , . . . , an ) in the vector space F (n) of n-tuples over the field F if and only if p(v) := p(a1 , . . . , an ) = 0 ∈ F . Let (P, ≤) be the poset of ideals in the polynomial ring F [x1 , . . . , xn ] with respect to the containment relation and for each ideal I let α(I) be the set of vectors in F (n) at which each polynomial of I vanishes. Let (Q, ≤) be the poset of all subsets of F (n) with respect to containment. For any subset X 43 of F (n) , let β(X) be the set of all polynomials which vanish simultaneously at every vector in X. Then (P, Q, α, β) is a Galois connection. The Corollary following the Lemma below shows how closure operators can arise from Galois connections. Lemma 8 Suppose (P, Q, α, β) is a Galois connection. Then for any elements p and q of P and Q respectively α(β(α(p))) = α(p) and β(α(β(q))) = β(q) proof: For p ∈ P , p ≤ β(α(p)) since β ◦ α is monotone non-decreasing. Thus α(β(α(p))) ≤ α(p) since α is order reversing. On the other hand α(p) ≤ α(β(α(p))) = (α ◦ β)(α(p)) since (α ◦ β) is monotone non-decreasing. By antisymmetry, we have the first equation of the statement of the Lemma. The second statement then follows from the symmetry of the definition of Galois connection – that is (P, Q, α, β) is a Galois connection if and only if (Q, P, β, α) is. Corollary 9 If (P, Q, α, β) is a Galois connection, then τ := β ◦ α is a closure operator on (P, ≤). proof: Immediate upon taking the β-image of both sides of the first equation of the preceeding Lemma. Example 5. Once again consider the order reversing mapping α : P(X) → P(I) of Example 4, where X was a union of non-empty sets Aσ indexed by a parameter σ ranging over a set I. For every subset Y of X, α(Y ) was the set of those σ for which Y ⊆ Aσ . Now there is another mapping β : P(I) → P(X) defined as follows: If J ⊆ I set β(J) := ∩σ∈J Aσ , with the understanding that if J = ∅, then β(J) = X. Then β is easily seen to be an order reversing mapping between the indicated power posets. Now the mapping τ = β ◦ α : P(X) → P(X) takes each subset Y of X to the intersection of all of the Aσ which contain it (with the convention that empty intersections denote the set X itself). 44 2.4 2.4.1 Chain conditions Saturated chains Recall that a chain C of poset (P, ≤) is simply a (necessarily induced) subposet (C, ≤P ) which is totally ordered. In case C is a finite set, |C| is called the length of the chain. We say that a chain C2 refines a chain C1 if and only if C1 ⊆ C2 . The chains of a poset (P, ≤) themselves form a partially ordered set under inclusion relation (the dual of the refinement relation) which we denote (ch(P, ≤), ⊆). Suppose now that C0 ≤ C1 ≤ · · · is an ascending chain in the poset (ch(P, ≤), ⊆). Then the set-theoretic union ∪Ci is easily seen to be totally ordered and so is an upperbound in (ch(P, ≤), ⊆). An easy application of Zorn’s Lemma then shows that every element of the poset (ch(P, ≤), ⊆) lies below a maximal member. These maximal chains are called unrefineable (or saturated) chains. Thus Theorem 10 If C is a chain in any poset (P, ≤), then C is contained in a saturated chain of (P, ≤). Of course we can restrict this in special ways to an interval. In a poset (P, ≤), chain from x to y is a chain (C, ≤) with x as a “zero element” and y as a “one element”. Thus x ≤ y and {x, y} ⊆ C ⊆ [x, y]. The collection of all chains of (P, ≤) from x to y itself becomes a poset (ch[x, y], ⊆) under the containment ( or “corefinement”) relation. (Note that (ch[x, y], ⊆) not quite the same as ch([x, y], ≤) since the latter may contain chains which do not contain x or y. We can still apply Theorem 10 for P = [x, y] and the chains C in it that do contain {x, y} to obtain: Corollary 11 If C is chain from x to y in a poset (P, ≤), then there exists a saturated chain C 0 from x to y which refines C. In particular, given an interval [x, y] of a poset (P, ≤), there exists an unrefinable chain in P from x to y. 45 2.4.2 Algebraic intervals and the height function. We have just observed a rather strong condition (FC) that implies that every saturated chain is finite. In general, nothing of the sort occurs. Still, enough occurs to produce strong theories: I mention two examples: 1. The general Jordan-Hölder Theorem whose context is algebraic interval measures in semimodular lower semilattices. 2. Locally finite posets, the natural domain of the theory of the Mobius function. Let (P, ≤) be any poset. An interval [x, y] of (P, ≤) is said to be algebraic (or of finite height) if and only if there exists a saturated chain from x to y of finite length.14 Note that to assert that (a, b) is an algebraic interval, does not preclude the simultaneous existence of infinite chains from a to b. The height of an algebraic interval [x, y] is then defined to be the minimal length of a saturated chain from x to y.15 If (a, b) is an algebraic interval its height, the smallest length of a finite unrefinable chain from a to b, is denoted h(a, b). Thus for an algebraic interval [a, b], h(a, b) is always a non-negative integer. We denote the collection of all algebraic intervals of (P, ≤) by the symbol AP . Proposition 12 The following hold: 1. If (a, b) and (b, c) are algebraic intervals of poset (P, ≤), , then (a, c) is also an algebraic interval. 2. The height function h : AP → N0 from the algebraic intervals of (P, ≤) to the non-negative integers, satisfies this property: If (a, b) and (b, c) are algebraic intervals, then h(a, c) ≤ h(a, b) + h(b.c) 14 This adjective “algebraic” does not enjoy uniform usage. In Universal Algebras , elements which are the join of finitely many atoms are called algebraic elements (I suppose by analogy with the theory of field extensions). Here we are applying the adjective to an interval, not an element of a poset. 15 The term “height” is used here instead of “length” which is appropriate when all unrefineable chains have the same length, as in semimodular lower semilattices that appear in the Jordan-Hölder Theorem or in the case of simple chains, where we have already used the term. 46 Example 6. Recall that a poset (P, ≤) is said to be a simplicial complex if 1. It contains an element 0 such that 0 ≤ p, ∀p ∈ P . 2. For each interval (0, p), the subposet [0, p] := ({r ∈ P |0 ≤ r ≤ p}, ≤) is the poset of all subsets of some finite set Fp . Clearly every interval (a, b) of a simplicial complex is algebraic with height function h(a, b) = |Fb | − |Fa |. 2.4.3 The ascending and descending chain conditions in arbitrary posets. Let (P, ≤) be any partially ordered set. We say that P satisfies the ascending chain condition or ACC if and only if every ascending chain of elements of P stabilizes after a finite number of steps – that is, for any chain p1 ≤ p2 ≤ p3 ≤ · · · , there exists a positive integer N , such that pN = pN +1 = · · · = pk , for all integers k greater than N. Put another way, (P, ≤) has the ACC if and only if every properly ascending chain p1 < p 2 < · · · terminates after some finite number of steps. Finally, it could even be put a third way: (P, ≤) satisfies the ascending chain condition if and only if there is no countable sequence {pn }∞ n=1 with pn < pn+1 for all natural numbers n. Lemma 13 The poset P = (P, ≤) satisfies the ascending chain condition (ACC) if and only if it satisfies 1. (The Maximum Condition) Every non-empty subset X of P (regarded as an induced poset) contains a maximal member. 2. (The Second form of the Maximum Condition) In particular, for every induced poset X of (P, ≤), and every element x ∈ X, there exists an element y ∈ X such that x is bounded by y and y is maximal in X. (To just make sure that we understand this: there exists an element y ∈ X such that 47 (a) y ≤ x. (b) If u ∈ X and u ≤ y implies that u = y. remark: The student is reminded: to say that “x is a maximal member of a subset X of a poset P ” simply means x is an element of X which is not properly less than any other member of X. proof of the lemma: 1. (The Maximum Condition implies the ACC). Suppose the ACC fails. Then there must exist a properly ascending countably infinite chain p1 < p2 < · · ·.. . Then X := {pi |i = 1, 2, } is a set with no maximal element. So the Maximum condition fails. ( The ACC implies the Maximum Condition). Let X be any nonempty subset of P . By way of contradiction, assume that X contains no maximal member. Choose x1 ∈ X. Since x1 is not maximal in X, there exists an element x2 ∈ X with x1 < x2 . By the same token x2 < x3 for some x3 ∈ X. Continuing in this way we obtain an infinite properly ascending chain x1 < x 1 < x 3 < . . . , contrary to the assumption of ACC.16 16 In Devlin’s wonderful book The Joy of Sets one has the exact argument of this last line with the phrase “Proceeding inductively” replacing the initial phrase “Continuing in this way”. The graduate student has probably encountered arguments like this many times, where a sequence with certain properties is said to exist because after the first n members of the sequence are constructed, it is always possible to choose a suitable n + 1-st member. This has always struckthe author as a leap of faith, for the sequence alleged to exist must exemplify infinitely many of these choices. At the very least we are involved in the Axiom of Choice in choosing the xi — but in a sense it appears worse. The sets are not just sitting there as if we had prescribed non-empty sets of socks in closets lined up in an infinite hallway (the traditional folk-way of explaining the Axiom of Choice). Here it as if each new closet was being defined by our choice of sock in in a previous closet. All I can feebly tell you is that it is basically equivalent to the Axiom of Choice – it does not matter whether the hallway of closets is replaced by a “tree” of closets, each full of socks, each sock bearing a prescription of a “successor closet” (indeed, almost a model of Wheeler’s “many-worlds” view of the Universe) – it is a theorem about the existence of infinite paths in suitable infinite graphs. Accepting this, the Axiom of Choice serves up an entire properly-ascending sequence that never ends. I just wanted you to know that the Axiom of Choice may be involved whenever you (as a graduate student) purport to produce a chain on the basis that every member of a finite truncation of a proposed chain can be extended. 48 It remains to show that the two versions of the Maximum condition are equivalent. Clearly the second (apparently stronger) version implies the first; one need only pick x ∈ X (X is non-empty), and the second version produces the desired y maximal in X. Now assume only the first version of the maximum condition. Take a subset X and an element x ∈ X. Then set X 0 = X ∩ P x where P x := {z ∈ P |x ≤ z} is the principal filter generated by x. Then X 0 is non-empty (it contains x) and so we obtain an element y maximal in X 0 from the first version of the maximum condition. Now if y were not a maximal member of X there would exist an element y 0 ∈ X with y < y 0 . But in that case x < y 0 so y 0 ∈ X ∩ P X = X 0 . But that would contradict y being maximal in X 0 . Thus y is in in fact maximal in X and dominates x as desired. Of course, merely by replacing the poset (P, ≤) by its dual P ∗ := (P, ≥), and applying the above we have the dual development: We say a poset (P, ≤) possesses the descending chain condition or DCC if and only if every descending chain p1 ≥ p2 ≥ · · · , pj ∈ P, stabilizes after a finite number of steps. That is, there exists a positive integer N such that pN = pN +1 = · · · pN +k for all positive integers k. We say that x is a minimal member of the subset X of P if and only if x ≥ y for y ∈ X implies x = y. A poset (P, ≤) is said to possess the minimum condition if and only if (Minimum Condition) Every nonempty subset of elements of the poset P = (P, ≤) contains a minimal member. Then we have: Lemma 14 A poset P = (P, ≤) has the descending condition (DCC) 1. if and only if it satisfies the minimum condition or 2. if and only if it satisfies this version of the minimimum condition: for any induced poset X any x ∈ X, there exists an element y minimal in X which is bounded above by x. Proof: Just the dual statement of Lemma 13. 49 Corollary 15 (The chain conditions are hereditary) If X is any induced poset of (P, ≤), and P satisfies the ACC (or DCC), then X also satisfies the ACC (or DCC, respectively). proof: This is not really a Corollary at all. It follows from the definition of “induced poset” and the chain-definitions directly. Any chain in the induced poset is a fortiori a chain of its ambient parent. We mention it only to have a signpost for future reference, allowing us to alert the reader to when this principle is invoked. Later, Lemma 14 will have a surprising consequence for lower semilattices with the DCC. Any poset satisfying both the ACC and the DCC also satisfies (FC) Any saturated chain of P has finite length. This is because patently one of the two chain conditions is violated by an infinite saturated chain. Conversely, if a poset P satisfies condition (FC) then there can be no properly ascending or descending chain of infinite length since by Theorem 10 it would then lie in a saturated chain which was also infinite against (FC). Thus Lemma 16 The condition (FC) is equivalent to the assumption of both DCC and ACC. 2.5 2.5.1 Posets with meets and/or joins. Meets and joins Let W be any subset of a poset (P, ≤). The join of the elements of W is an element v in (P, ≤) with these properties: 1. w ≤ v for all w ∈ W . 2. If v 0 is an element of (P, ≤) such that w ≤ v 0 for all w ∈ W , then v ≤ v0. Similarly there is the dual notion: the meet of the elements of W in P would be an element m in P such that 50 1. m ≤ w for all w ∈ W . 2. If m0 is an element of (P, ≤) such that m0 ≤ w for all w ∈ W , then m0 ≤ m. Of course, P may or may not possess a meet or a join of the elements of W . But one thing is certain: if the meet exists it is unique; if the join exists, it is unique. Because of this uniqueness we can give these elements names. V V We write P (W ) (or just (W ) if the ambient poset P is understood) for W the meet of the elements of W in P (if it exists). Similarly, we write P (W ) W (or (W )) for the join in P of all of the elements of W (if that exists). In the case that W is the set {a, b}, we render the meet and join of a and b by a ∧ b and a ∨ b, respectively. The reader may verify a∧b = b∧a a ∧ (b ∧ c) = (a ∧ b) ∧ c (a ∧ b) ∨ (a ∧ c) ≤ a ∧ (b ∨ c) (2.6) (2.7) (2.8) and the three dual statements when the indicated meets and joins exist. We say that (P, ≤) is a lower semilattice if it is “meet-closed” – that is, the meet of any two of its elements exist. In that case (Exercise ???) it is easy to show that meets exist for finite subsets of P . Dually we can define an upper semilattice (it is join closed). A lattice is a poset (P, ≤) that is both a lower semilattice and an upper semilattice – that is the meet and join of any two of its elements both exist in P . Thus, a lattice is a self-dual concept: If P = (P, ≤) is a lattice, then so is its dual poset P ∗ = (P, ≤∗ ). If every non-empty subset U of P has a meet (join) we say that arbitrary meets (joins) exists. If both arbitrary meets and joins exist we say that P is a complete lattice. Example 7. 1. Any totally ordered poset is a lattice. The meet of a finite set is its minimal member; its join is its maximal member. Considering the open real interval (0, 1) with its induced total ordering from the real number system, it is easy to see (a) that there are lattices with no “zero” or “one”, 51 (b) that there can be (infinite) subsets of a lattice with no lower bound or no upper bound. 2. The poset B(X) of all subsets of a set X is a lattice with intersection of two sets being their meet, and the union of two sets being their join. This lattice is self dual and is a complete lattice, meaning that any subset of elements of B(X), whether infinite or not, has a least upper bound and gretest lower bound – i.e. a meet and a join. Of course the lattice has X as its “one“ and the empty set as its “zero”. 2.5.2 Lower semilattices with the descending chain condition. The proof of the following theorem is really due to Stanley, who presented it only for the case that P is finite[?, Proposition 3.3.1, page 103]. Theorem 17 Suppose P is a meet-closed poset (that is a lower semilattice) with the (DCC) condition. Then the following assertions hold: 1. If X is a meet-closed induced poset of P , then X contains a unique minimal member. 2. Suppose, for some subset X of P , the filter P X := {y ∈ P |x ≤ y for all x ∈ X} is non-empty. Then the join W (X) exists in P . In particular if (P, ≤) possesses a one-element 1̂ then for any arbitrary W subset X of P , there exists a join (X). Put more succinctly, if 1̂ exists, unrestricted joins exist. V 3. For any non-empty subset X of P , the universal meet (X) exists and is expressible as a meet of a finite number of elements of X. 4. The poset P contains a 0̂. 5. If P contains a 1̂, then P is a complete lattice. 52 proof: 1. The set of minimal elements of X is non-empty (Lemma 14 of the section on chain conditions.) Suppose there were at least two distinct minimal members of X, say y1 and y2 . Then y1 ∧ y2 is also a member of X by the meet-closed hypothesis. But by minimality of each yi , one has y1 = y1 ∧ y2 = y2 . Since any two minimal members of the set X are now equal (and the set of them is non-empty) there exists a unique minimal member. The proof of Part 1 is complete. 2. Let X be any subset of P for which the filter P X is non-empty. One observes that P X is a meet-closed induced poset and so by part 1, contains W a unique minimal member j(X). Then, by definition, j(X) is the join (X). If 1̂ ∈ P then of couse P X is non-empty for all subsets X and the result follows. 3. Let W (X) be the collection of all elements of P which can be expressed as a meet of finitely many elements of X, viewed as an induced poset. Then W (X) is meet-closed, and so possesses a unique minimal member x0 by Part 1. In particular x0 ≤ x for all x ∈ X. Now suppose z ≤ x for all x ∈ X. Then z is less than or equal to any finite meet of elements of X, and so is V less than or equal to x0 . Thus x0 = (X), by definition of a global meet. V Suppose (X) < x0 . Then there exists some element x ∈ X such that V x ∧ x0 < x0 (otherwise x0 ≤ x for all x ∈ X forcing x0 = (X)). Then patently x ∧ x0 is expressible as a meet of finitely many members of X and so lies in W (X). This contradicts x0 being the minimum element of W (X). V Thus we must conclude x0 = (X), and we are done. 4. By part 1, P contains minimal elements (often called “atoms”). Suppose m is such an atom. If m ≤ x for all x ∈ P then x would be the element 0̂, against our definition of “minimal element”. Thus there exists an element y ∈ P such that y is not greater than or equal to x. Then y ∧ m (which exists by meet-closure) is strictly less than m. Since m is an atom, we have y ∧ m = 0̂ ∈ P and the proof is complete. 5. If 1̂ exists, P enjoys both unrestricted joins by part 2. By part 4, 0̂ exists. Thus for any subset X of P , the order ideal PX that it generates is W non-empty. So we can form the join (PX ). This element – by definition – is the unique minimum of the filter P PX generated by PX . Since X ⊆ P PX , the W aforesaid minimum belongs to the order ideal PX . Thus we see that (PX ) is itself a member of PX and so is the unique maximal element of this order 53 W V ideal. This means (PX ) is (X) and exists. Thus unrestricted meets exist and so now P is a complete lattice. The proof is complete. Example 8. The following posets are all lower semilattices with the DCC: 1. N+ = {0 < 1 < 2 < . . .} of all non-negative integers with the usual ordering. This is a chain. 2. D, the positive integers under the partial ordering of “dividing” – that is, a ≤ b if and only if a divides the integer b. 3. B(X), the poset of all finite subsets of a set X. 4. L<∞ (V ), the poset of finite-dimensional subspaces of a vector space V . 5. M<∞ (I), the finite multisets over a set I. It follows that arbitrary meets exist and a 0̂ exists. remarkOf course, also, we could adjoin a global maximum 1̂ to each of these examples and obtain complete lattices in each case. 2.5.3 Lower semilattices with both chain conditions Recall from Section 1, that a poset has both the ACC and the DCC if and only if it possesses condition (FC) Every chain in P has finite length. An observation is that if P satisfies (FC), then so does its dual P ∗ . Similarly, if P has finite height, then so does its dual. This is trivial. It depends only on the fact that rewriting a finite saturated chain in descending order produces a saturated chain of P ∗ . We obtain at once Lemma 18 Suppose P is a lower semilattice satisfying (FC) . 1. P contains both a 0̂. If P contains a 1̂ then P is a complete lattice. 2. Every element of P − {1̂} is bounded by a maximal member of P − {1̂} – that is, an element of max(P ). 54 3. Every element of P − {0̂} is above a minimal element of P − 0̂ – that is, an element of min(P ). V 4. The meet of all maximal elements of P – that is (max(P )) (called the Frattini element or radical of (P. ≤)) exists and is the meet of just finitely many elements of max(P ). 5. If 1̂ ∈ P , the join of all atoms, (min(P )) ( called the socle of P and denoted soc(P )) exists and is the join of just finitely many atoms. W proof: Parts 1. and 4.follow from the previous subsection. Parts 2. and 3. are consequences of the chain-conditions (see the First Section) and Part 5. follows from Parts 1. and 3. Example 9. Hopefully the reader will notice that the hypothesis that P contained the element 1̂ played a role – even took a bow – in Parts 1. and 5. of the above Lemma. Is this necessary? After all, we have both the DCC and the ACC. Well, there is an asymmetry in the hypotheses. P is a lower semilattice, but not an upper semilattice, though this symmetry is completely restored once 1̂ exists in P , because then we have arbitrary joins. Consider any one of the posets D, B(X), L<∞ (V ). These are ranked posets, with the rank of an element being (1) the number of prime factors, (2) the cardinality of a set, or (3) the dimension of a subspace – respectively. Select a positive integer r and consider the induced poset trr (P ) of all elements of rank at most r in P where P = D, B(X), L<∞ (V ). Then trr (P ) is still a lower semilattice with both the DCC and the ACC. But it has no 1̂. Indeed the examples conceivably proliferate for unranked lower semilattices P when we replace trr (P ) by requiring this to be the order ideal generated by an antichain of elements x ∈ P of finite ultra length ul[0̂, x]. You see how easy it is? 2.5.4 Examples and exercises Exercise 12. We let [n] be the chain {1 < 2 < · · · < n} is the usual total ordering of the natural numbers. Show that the infinite sum [1] ∪ [2] ∪ [3] ∪ · · · satisfies (FC) but does not possess finite height, and possesses neither a 1̂ nor a 0̂. 55 Exercise 13. If we adjoin 0̂ to the poset of Exercise 11, show that the Frattini element exists, and is 0̂. Exercise 14. Show that [1] × [2] × [3] × · · · does not satisfy (FC). Exercise 15. Let P be a lower semilattice with 0̂ and 1̂ satisfying (FC). Set B := max(P ) and let P̄ be the induced subposet generated by B – that is, the set of elements of P expressible as a finite meet of elements of B. We understand the empty meet to be the element 1̂, so the latter is an element of P̄ . 1. Show that P̄ has the Frattini element φ(P ) as its zero element, 0̂P̄ . 2. For each x ∈ P − {1̂}, show that the induced poset P x ∩ P̄ has a unique minimal member (which we shall call σ(x)). 3. Defining σ(1̂) = 1̂ show the following: (a) The mapping σ : P → P̄ is a surjective morphism of posets. (b) For each x ∈ P , one has σ(σ(x)) = σ(x). (Any morphism onto an induced subposet which satisfies these last two conditions is called a closure operator.) Exercise 16. Must the induced subposet of all elements of finite ultralength in a lower semilattice with DCC necessarily have the ACC? Exercise 17. Suppose L is a lattice. Give an induction proof showing that for any finite collection {a1 , . . . , an } of elements of L, the elements a1 ∨ · · · ∨ an and a1 ∧ · · · ∧ an , exist and are respectively the greatest lower bound and greatest upper bound in L of the set {a1 , . . . , an }. Exercise 18. A lattice L := (L, ≤) is said to be distributive if and only if, for any elements a and b in L, one always has: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c). (2.9) An example of a distributive lattice is the power poset P(X) of all subsets of a set X. 56 1. Prove that if L is a distributive lattice and (M, ≤) is an induced subposet closed under taking pair-wise meets and joins, then M is also a distributive lattice. 2. Use the result of item 1. to prove the following: (a) The Boolean poset B(X) - of all finite subsets of X is a distributive lattice. (b) Let J (P ) be the poset of all order ideals of a poset (P, ≤) under the containment relation. Use item 1. to prove that J (P ) is a distributive lattice. [One must define “meet” and “‘join” of order ideals and set L = P(P ) in item 1.] 2.6 The Jordan-Hölder Theory In the previous two sections we defined the notions of “algebraic interval”, “height of an algebraic interval” and the meet-closed posets which we called “lower semilattices”. They shall be used with their previous meanings without further comment. This section concerns a basic theorem that emerges from an property of lower semilattices, that of semimodularity. Many important algebraic objects give rise to lower semilattices which are semimodular (for example the posets of subnormal subgroups of a finite group, or the submodule poset of an Rmodule) and each enjoyed its own “Jordan-Holder Theorem” – it always being understood there was a general form of this theorem. It is in fact a very simple theorem about extending “semimodular functions” on the set of covers of a semimodular lower semilattice to an interval measure on that semilattice.17 Sounds like a mouthful, but it is really quite simple. Fix a poset (P, ≤). An unrefinable chain of length one is called a cover and is denoted by (a, b) (which is almost the name of its interval [a, b] – altered to indicate that we are talking about a cover). Thus (a, b) is a cover if and only if a < b and there is no element c in P with a < c < b — i.e. a is a maximal element in the induced poset of all elements prperly below 17 The prefix “semi-” is justified for several reasons. The term “modular function” has quite another meaning as a certain type of meromorphic function of a complex variable. Secondly the function in question is defined in the context of a semimodular lower semilattice. So why not put in the “semi”?. I do not guarentee that every term coined in this book has been used before. 57 b (the principal order ideal Pb minus {b}). The collection of all covers of (P, ≤) is denoted CovP . Note that CovP is a subset of AP , the collection of all algebraic intervals of (P, ≤). 2.6.1 Lower semi-lattices and semimodularity A lower semilattice (P, ≤) is said to be semimodular if, whenever (x1 , b) and (x2 , b) are both covers with x2 6= x2 , then both (x1 ∧ x2 , x1 ) and (x1 ∧ x2 , x2 ) are also covers. Lemma 19 Suppose (P, ≤) is a semimodular lower semilattice. Suppose (x, a) is algebraic and (b, a) is a cover, where x ≤ b. Then (x, b) is algebraic. Proof: Since (x, a) is algebraic, there exists a finite unrefinable chain from x to a, say A = (x = a0 , a1 , . . . , an = a). Clearly each interval (x, aj ) is algebraic. By hypothesis, x ≤ b and so there is a largest subscript i such that ai ≤ b. Clearly i < n. If i = n − 1 then an−1 = b so (x, b) is algebraic by the previous paragraph. Thus b 6= an−1 . Since both (an−1 , a) and (b, a) are covers, then by semimodularity, both (an−1 ∧ b, an−1 ) and (an−1 ∧ b, b) are covers. Continuing in this way we obtain that (ak ∧ b, ak ) and (ak ∧ b, ak−1 ∧ b) are covers for all k larger than i. Finally, as ai ≤ b, this previous statement yields ai+1 ∧ b = ai and (by semimodularity) (ai , ai+2 ∧ b) must be a cover (see Figure 1). Now (x = a0 , . . . , ai , ai+2 ∧ b, ai+3 ∧, . . . , an−1 ∧ b, b) is an finite unrefinable chain since its sucessive pairs are covers. This makes (x, b) algebraic, completing the proof. 2.6.2 Interval Measures on Posets Let M be a commutative monoid. This means M possesses a binary operation, say “◦” which is commutative and associative, as well as an identity element, say “e” such that m = e ◦ m = m ◦ e for all m ∈ M . There is one monoid that plays an important role in the applications of our main theorem and that is the monoid M<∞ (X) of all finite multisets over a set X. We have met this object before in the guise of a locally finite poset (see the last item under Example 2 of this Chapter). However it is also a commutative semigroup. We have seen that any finite multiset over X can be represented as a function f : X → N0 from X to the non-negative integers 58 Figure 2.1: The poset showing (x, b) is algebraic. The symbol “cov” on a depicted interval indicates that it is a cover. N0 whose “support” is finite – i.e. the function achieves a non-zero value in only finitely instances as x wanders over X. Now, if f and g are two such functions, we may let “f + g” denote the function that takes x ∈ X to the non-negative integer f (x) + g(x), the sum of two integers. Clearly f + g has finite support, and so (M (X), +) (under this definition of “plus”) becomes a commutative semigroup. But as the constant function 0X : X → {0} is an identity element with respect to this operation, M (M(X), +) is actually a monoid (a commutative semigroup with an identity element). An interval measure µ of a poset (P, ≤) is a mapping µ : AP → M from the set of algebraic intervals of P into a commutative monoid M with identity element e such that µ(a, a) = e for all a ∈ P. (2.10) µ(a, b) ◦ µ(b, c) = µ(a, c) whenever (a, b) and (b, c) are in AP (2.11) [Notice that we have found it convenient to write µ(a, b) for µ([a, b]).] Here are some examples of interval measures on posets: Example 10. Let M be the multiplicative monoid of positive integers. Let (P, ≤) be the set of positive integers and write x ≤ y if x divides y evenly. 59 Then every interval of (P, ≤) is algebraic. Define µ by setting µ(a, b) := b/a for every interval (a, b). Example 11. Let (P, ≤) be as in Example 10, but now let M be the additive monoid of all non-negative integers. Now if we set µ(a, b); = the total number of prime divisors of b/a, then µ is a measure. Example 12. Let (P, ≤) be the poset of all finite-dimensional subspaces of some (possibly infinite dimensional) vector space V , where “≤” is the relation of “contained in”. (i) Let M be the additive monoid of all non-negative integers. Then if we define µ(A, B) := dim(B/A), then µ is a measure on (P, ≤) (ii) If M is the multiplicative group {1, −1}, then setting µ(A, B) := (−1)dim(A/B) for every algebraic interval (A, B) also defines a measure. 2.6.3 The Jordan-Hölder Theorem Now we can prove Theorem 20 (The Jordan-Hölder Theorem) Let (P, ≤) be a semimodular lower semilattice. Suppose µ : CovP −→ (M, +) is a mapping from the set of covers of P to a commutative monoid (M, +), and suppose this mapping is “semimodular” in the sense that µ(b, y) = µ(a ∧ b, a) whenever (a, y) and (b, y) are distinct covers. (2.12) Then for any two finite unrefinable chains U = (u = u0 , . . . , un = v) and V = (u = v0 , . . . , vm = v) from u to v, we have n−1 X µ(ui , ui+1 ) = i=0 m−1 X µ(vi , vi+1 ) (2.13) i=0 and n = m (the finite summation is taking place in the additive monoid M ). In particular, µ extends to a well-defined measure µ̂ : AP −→ M . Proof: If n = 0 or 1, then U = V , and the conclusion holds. We therefore proceed by induction on the minimal length h(x, y) of an unrefinable chain from x to y for any algebraic interval (x, y) – that is, the height of (x, y). 60 Figure 2.2: The main figure for the Jordan Hölder Theorem If um−1 = vn−1 , h(u, vn−1 ) = n − 1, so by induction, n−2 i=0 µ(ui , ui+1 ) = µ(v , v ) and m − 1 = n − 1 and the conclusion follows. i i+1 i=0 So assume um−1 6= vn−1 . Set z = um−1 ∧ vn−1 . Since (um−1 , v) and (vn−1 , v) are both covers, by semi-modularity, so also are (z, um−1 ) and (z, vn−1 ) (see Figure 2). Since u ≤ z, (z, vn−1 ) is a cover, and (u, vn−1 ) is algebraic, we see that Lemma 1 implies that (u, z) is also algebraic. So a finite unrefinable chain P Pm−2 Z = (u = z0 , z1 , . . . , zr = z) from u to z exists. Since h(u, vn−1 ) = n − 1 < n, by induction r−1 X ( µ(zi .zi+1 )) + µ(z, vn−1 ) = i=0 m−1 X µ(ui , ui+1 ) (2.14) i=0 and r + 1 = n − 1. But h(um−1 ) ≤ r + 1 so induction can also be applied to the algebraic interval (u, um−1 ) so r−1 X ( µ(zi , zi+1 )) + µ(z, um−1 ) = m−1 X µ(ui , ui+1 ). (2.15) i=0 i=0 and r + 1 = n − 1. Thus n = m. But by (3), µ(z, um−1 ) = µ(vn−1 , v) and µ(z, vn−1 ) = µ(um−1 , v). The result (4) now follows from (5), (6) and the commutativity of M . The proof is complete. 61 There are many applications of this theorem. In the case of finite groups (R-modules) we let (M, +) be the commutative monoid of multisets of all finite simple groups (or all irreducible R-modules, resp.). These are the classical citations of the Jordan-Hölder Theorem for Groups and R-modules.18 Exercise 19. (The Jordan-Hölder Theorem implies the Fundamental Theorem of Arithmetic) Let (P, ≤) be the poset of positive integers where a ≤ b if and only if integer a divides integer b evenly (Example 1, of this Section). 1. Show that (P, ≤) is a lower semimodular semilattice with all intervals algebraic. Let µ : CovP → Z+ be the function which records the prime number b/a at every cover (a, b) of (P, ≤). Indicate why µ is a semimodular function in the sense given in the Jordan-Hölder Theorem. 2. Suppose integer a properly divides integer b. For every factorization b/a = p1 p2 · · · pr of b/a into primes show that there exists a finite unrefineable chain (a = a0 , a1 , a2 , . . . ar = b) such that ai /ai−1 = pi for i = 1, . . . r. 3. Conclude from the Jordan-Hölder Theorem that every positive integer possesses a factorization into a unique collection of positive prime numbers up to the order of the factors. 2.7 Möbius inversion We pass now from interval measures on posets to more general functions on the intervals of a poset. If our functions assume values in a commutative ring A, and the poset is locally finite, an interesting theory emrges which encodes many subtleties of the poset. 2.7.1 More locally finite posets. Recall that a poset (P, ≤) is said to be locally finite if and only if, given any interval [a, b] (where a ≤ b in P ) there are only finitely many distinct poset 18 I beg the reader to notice that in the case of groups there is no need for Zassenhaus’ famous “butterfly lemma”, or for the need to prove that subnormal subgroups of a finite group form a lattice – a paper by Wielandt in an early issue of the Journal of Algebra, whose introduction thought this was a necessity. Even if it were needed, that result is immediate from the one- or two-line proof in Theorem 17. 62 elements z between a and b. In Example 2 of this chapter we gave a number of examples of locally finite posets. The following examples are a little more “generic” – they tell you how to produce new examples from known ones. Example 13. If X is any subset of a locally finite poset, then the poset induced by X is always locally finite. In particular, the induced poset [a, b] of an interval is locally finite. The interval posets of the first example are all just finite chains; moreover any finite chain (like any finite poset) is locally finite, and any such chain can be realized as an interval poset of the natural numbers. An easy example could be formed by taking any subspace-closed collection S of finite-dimensional subspaces of a vector space V , that is, an order ideal in L<∞ (V ). For example S could be: 1. all subspaces of V which meet a fixed subspace W trivially. (This is sometimes called “the geometry of attenuated spaces”.) 2. all finite-dimensional subspaces which are totally isotropic with respect to a non-degenerate reflexive σ-sesquilinear form or toally singular with respect to a quadratic form. (These are the “polar space geometries which we shall meet these in Chapters ** and **.) If V is over a finite field, then these two examples for S yield locally finite posets. Example 14. The dual P ∗ of any locally finite poset (P, ≤) is also locally finite. Example 15. Let (Pi , ≤), i = 1, 2, . . . , n, be a family of locally finite posets. The elements of the product of these posets is the Cartesian product P1 × · · · × Pn of sequences of length n , whose i-th member is an element of Pi . A sequence a = (a0 , a1 , . . . , an ) is less than a sequence b = (b0 , b1 , . . . , bn ) if and only ai ≤ bi for all i. It follows that the interval poset defined by (a, b) is the poset induced on Y [a, b] = i [ai , bi ], a finite set. Thus the product poset is also locally finite. 63 2.7.2 The incidence algebra Let (P, ≤) be a locally finite poset and let R be any ring. From these two objects we are going to construct a new ring A = A(P, R), called the incidence algebra of (P, ≤) over ring R. Let IP be the collection of all intervals of P . (By local finiteness, they are all algebraic intervals.) The elements of A are the functions f : IP → R, from the set of intervals into the ring R. Addition in A is point-wise – that is, if f and g are functions from intervals to R, then f + g is the function whose value at interval (a, b) is f (a, b) + g(a, b). Clear with respect to addition, (A, +) is a commutative group. (It is in fact a direct product of copies of (R, +) indexed by IP .) Multiplication is not defined pointwise. Instead it is a sort of “convolution”, which we denote as a binary operation “◦”. If f and g are functions from intervals to R, that is, elements of A, then f ◦ g is defined to be the function whose value at interval (a, b) is X a≤x≤b f (a, x)g(x, b). (2.16) Notice that the definition of f ◦ g makes sense, since the sum in (2.16) is finite, and the multiplication within the summands is occurring in the ring R. The operation “◦” is distributive with respect to addition, since f ◦ (g + h)(a, b) = X a≤x≤b = X f (a, x)g(x, g) + a≤x≤b f (a, x)((g(x, b) + h(x, b)) X = (f ◦ g)(a, b) + (f ◦ h)(a, b) = (f ◦ g + f ◦ h)(a, b) a≤y≤b f (a, y)h(y, b) (2.17) for all intervals (a, b). Thus f ◦ (g + h) = f ◦ g + f ◦ h. The proof of the left distributive law is very similar. Also the operation “◦” is associative, for both f ◦ (g ◦ h) and (f ◦ g) ◦ h have value X f (a, x)g(x, y)h(y, b) a≤x≤y≤b at interval (a, b). 64 The function : IP → R defined by (a, b) := ( 1 0 if a = b if a < b at every interval (a, b) is easily seen to be the identity element of A. Thus A = A(P, R) is a ring. 2.7.3 The Möbius function of a locally finite poset There are two more very special elements in A. The zeta function, ζ : IP → R is the function which has constant value 1, the identity element of the ring R, at every interval. Theorem 21 The zeta function has a 2-sided inverse in A. Proof: We are going to inductively define a function µ` : IP → R which is a left inverse of ζ in A. Fix an element a in P . It suffices to give an appropriate definition of µ` (a, b) for all elements b such that b ≥ a (so that (a, b) is an interval). First set µ` (a, a) = 1. If (a, x) ∈ CovP , set µ` (a, x) = −1. Suppose, at some stage that we have defined µ` (a, x) for all x ∈ [a, b] such that x < b. Then we define X µ` (a, b) := − µ (a, x). a≤x<b ` 19 (2.18) Notice that (2.18) provides an algorithm for calculating µ` (a, b) for any b ≥ a. [If we change the value of the lower end of the interval from a to some other element a0 , a whole new series of calculations must ensue. These new calulations are in no way influenced by our previous calulations of the µ` (a, z)’s.] Now from (2.18), we have µ` ◦ ζ = . Similarly, there exists in A a right inverse µr for zeta. It too is inductively defined, this time by the equations: µr (b, b) = 1, and X µr (a, b) = − a<y≤b µr (y, b). 19 (2.19) The algorithm is “greedy” in the sense that it ignores the global envoirnment of (a, b) in the poset P . Instead it only looks at the first values at hand, the µ` (a, x) for x in the interval poset [a, b] properly below b. 65 Thus ζ ◦ µr = is easily verified. Now why is µr = µ` ? In any ring, if both left and right inverses of an element exist, then they are equal. In this particular case, the argument is this: µ` = µ` ◦ = µ` ◦ (ζ ◦ µr ) = (µ` ◦ ζ) ◦ µr = ◦ µr = µr . From this point onward we write µ or µP (if we wish to emphasize the poset P ) for µ` = µr . The element µP of A is called the Möbius function of the poset (P, ≤). Corollary 22 The Möbius function of the poset P over the ring R takes values in the subring of R generated by the multiplicative identity. These values are in the center of the ring R. We now observe Theorem 23 (Möbius Inversion) Let (P, ≤) be a locally finite poset, and let f : IP → R be any R-valued function on the set of intervals. Suppose the functions F1 and F2 from IP to R are define by setting F1 (a, b) : = X a≤x≤b F2 (a, b) : = X a≤x≤b f (a, x), (2.20) f (x, b), (2.21) for any interval (a.b) in (P, ≤). Then for any interval (c, d), we have f (c, d) = X F (c, y)µ(y, d) = c≤y≤d 1 X c≤y≤d µ(c, y)F2 (y, d). Proof: The definitions of F1 and F2 assert that F1 = f ◦ ζ, and, F2 = ζ ◦ f. Then F1 ◦ µ = (f ◦ ζ) ◦ µ = f ◦ (ζ ◦ µ) = f ◦ = f. Similarly µ ◦ F2 = (µ ◦ ζ) ◦ f = ◦ f = f. 66 (2.22) 2.7.4 Special cases Let C = (x0 , x1 , . . . , xn ) be a chain – that is, a finite totally ordered poset. This is certainly one of the simplest posets that one could imagine. What is it’s Möbius function? By (2.19), µ(x0 , x0 ) = 1 and µ(x0 , x1 ) = −1. But then (2.18) implies µ(x0 , xi ) = 0 if i > 1. Since any induced poset [xi , xj ], with i ≤ j, is also a chain, we see that 1 µ(xi , xj ) = −1 0 if i = j if j = i + 1 if j > i + 1 Recall that if {(Pσ , ≤σ )|σ ∈ J} is a family of posets, each with a “zero” element 0σ ,then the direct sum of these posets is the poset (P, ≤) whose elements are the maps f : J → ∪σ∈J Pσ from the index set into the dijoint union of the Pσ such that f (σ) ∈ Pσ for all σ ∈ J and such that f (σ) 6= 0σ for only finitely many indices σ. The partial ordering is such that f ≤ g if and only if f (σ)≤σ g(σ) for every σ ∈ J. Then it is easy to see that a direct sum of locally finite posets is also locally finite. We now show Theorem 24 Suppose (P, ≤) is a direct sum of locally finite posets {(Pσ , ≤σ )|σ ∈ J}. Suppose (P, ≤) has Möbius function µ and each poset (Pσ , ≤σ ) has Möbius function µσ . For each interval (f, g) in P , one has µ(f, g) = Y µ (f (σ), g(σ)). σ∈J,f (σ)6=g(σ) σ (2.23) Proof: Fix interval (f, g) in (P, ≤), and let K := {1, . . . , n} be the set of indices σ ∈ J for which g(σ) > 0σ ∈ Pσ . We understand this to mean K = ∅ if n = 0. But the equation (2.23) is true if f = g, for then 1 = µ(f, g) and Q f (σ) = g(σ) so the right side is µσ (f (σ), g(σ)) = 1n . So we may assume that f < g and n = |K| is at least one. As an induction hypothesis, we may assume that for each x ∈ [f, g], with x < g, we have Yn µ(f, x) = σ=1 µσ (f (σ), x(σ)). (2.24) 67 Now by definition of the Möbius function, 0 = µ(f, g) + X f ≤x<g = µ(f, g) + X f ≤x<g µ(f, x) Yn ( µ (f (σ), x(σ))), σ=1 σ (2.25) where the second equality applies the induction hypothesis (2.24). On the other hand 0 = 0n = Yn X ( σ=1 µ (f (σ), z(σ))), f (σ)≤z(σ)≤g(σ) σ which upon “multiplying out” (that is, using the distributive laws of R) is the sum of all possible n-fold monomials, Yn µ (f (σ), z(σ)), σ=1 σ with the the z(σ) ranging independantly over their respective interval posets [f (σ), g(σ)]. But the summation in the right side of (2.25) is also the sum of all such monomials except for one: the one monomial summand Yn µ (f (σ), g(σ)) σ=1 σ is missing. It follows from (2.25) that this missing monomial is µ(f, g), and the theorem is proved. We look at two classic applications. We saw in Theorem 7, that the poset D for the divisor relation among the positive integers was a direct sum of copies of the poset N+ (of the natural order relation among the positive integers) indexed by the natural numbers. All these posets are locally finite (indeed, this was our second item in Example 2, page 35). Since each interval (a, b) in D (representing number a dividing number b) is a product of finite chains (representing the divisors of those maximal prime-power divisors of b/a, we see that µ(a, b) = ( 0 (−1)n if b/a is divisible by the square of a prime, if b/a is the product of n distinct primes. The function µ(n) := µ(1, n) is the traditional Möbius function from number theory. 68 Now let (P (n), ≤) be the poset of all subsets of an n-set X = {1, 2, . . . , n}, say, partially ordered by containment. This poset has a “zero” (the empty set), a “one” (the whole setX) and 2n elements. Then (P (n), ≤) can be viewed as a direct sum of n copies of the 2-element poset {0 < 1}. This is because any subset Y of X can be regarded as the characteristic function y : X → Z which has value 1 at elements of X and value 0 otherwise. Then if A ⊆ B ⊆ X, so that (A, B) is an interval in (P (n), ≤), our theorem shows that µ(A, B) = (−1)|B−A| . 2.7.5 Remarks on other applications of the Möbius function The relations are boundless. Here is a sampling: 1. If (P, Q, α, β) is a Galois connection, then under mild conditions, the value of µ(0, 1) (the Euler Characteristic minus 1) can be determined by just partial knowledge of the Möbius function of the other side. 2. When f : P → Q is monotone, and the fibre over every interval (0, q) contains a global maximum q 0 , then X {y|f (y)=s} µ(0, y) = 0, provided s > 0. 3. Similarly, if F is a family of subsets of a finite set containing the empty set, closed under intersection, if X is their union, and if µ is the Möbius function of the poset (F ∪ {X}, ⊆), then µ(0, 1) = µ(∅, X) = q2 − q3 + q4 · · · where qk is the number of k-subsets of F which together intersect at the empty set. 2.8 2.8.1 Point measures in lattices; the modular law. Introduction In a recent article in the American Math Monthly [1] one finds this 69 Proposition 25 Let V1 , . . . , Vk be a collection of subspaces of an n-dimensional space V . If dim V1 + dim V2 + · · · + dim Vk > (k − 1)n, then dim(V1 ∩ V2 ∩ · · · ∩ Vk ) is positive. Proof: Let U = V1 ⊕ · · · ⊕ Vk be the formal direct sum of the Vi and let W be the direct sum of k − 1 copies of V . Define the transformation T : U → W by the rule that if vi ∈ Vi , i = 1, . . . k, then T ((v1 , . . . , vk )) = (v1 − v2 , v2 − v3 , . . . , vk−1 − vk ). Then the kernel of T is just the space of all vectors (v, v, . . . , v) where v ∈ V1 ∩ · · · ∩ Vk . Thus as (k − 1)n < Xk i=1 dim Vi = dim U = dim T (U ) + dim(ker T ) ≤ dim(W ) + dim(ker T ) = (k − 1)n + dim(ker T ), the result follows. Remark: Of course one can always take V = V1 + V2 + · · · + Vk in the Proposition. As one can see, this is a theorem about dimensions. Dimensions measure something about elements of the poset of subspaces of a vector space. In fact, the dimension obeys a rule called “Kolmogoroff’s Axiom” which applies to measures in many other contexts, such as probability, area, cardinality etc. The object of this section is to display this fundamental concept, and in particular, to generalize the above lemma in this wider context. In this spirit, the vector-subspace poset is replaced by a “lattice”. 2.8.2 Dual lattices and sublattices. Recall that a lattice is a poset (L, ≤) for which any two elements a and b of the lattice possess a greatest lower bound or meet a ∧ b and also possess a least upper bound or join a ∨ b. One has the following two “associative laws” hold: a ∨ (b ∨ c) = (a ∨ b) ∨ c, and a ∧ (b ∧ c) = (a ∧ b) ∧ c, 70 (2.26) (2.27) for any elements a, b and c of L. Thus one may omit parentheses in compositions that use only one of the two binary operations “∨” or “∧”. One sees that for any finite collection {a1 , . . . an } of elements of L, the elements a1 ∨ · · · ∨ an and a1 ∧ · · · ∧ an , exist and are respectively the greatest lower bound and greatest upper bound in L of the set {a1 , . . . , an }. It follows from the definitions that the dual of a lattice (L, ≤) is also a lattice L∗ . A lattice is said to possess a zero element if and only if it has a global minimum – usually denoted by “0”. Similarly, it possesses a “one” if and only it contains a global maximum – denoted by the symbol “1” as well as the name “one”. A lattice is said to be self-dual if and only if it is isomorphic to its dual. Suppose L = (L, ≤) is a lattice. A sublattice of L is a subposet (P, ≤P ) of L that is closed under the pairwise meets and joins that are taking place in L.. In Exercise ?? the student is asked to prove the following Lemma 26 1. Any sublattice of a lattice is an induced poset. 2. Suppose P = (P, ≤P ) is a subposet of a lattice (L, ≤). Suppose further that (P, ≤P ) is a lattice in its own right. Then (a) It is possible for P not to be an induced subposet of L. (b) It is possible P to be an induced subposet of L and still not be a sublattice of L. 2.8.3 Kolmogoroff functions on lattices Let (L, ≤) be any lattice. In turn let (A, +) be any commutative monoid with with respect to the operation “+” having identity element 0. Moreover A is called a cancellation monoid if ac ≤ bc always implies a ≤ b, for all elements a, b and c in A. A Komolgoroff function is a mapping f : L → A, such that the following axiom holds: 71 (Komolgoroff’s Axiom) For any two elements a and b of L f (a ∨ b) + f (a ∧ b) = f (a) + f (b) (2.28) Note that the “+” in these equations is occuring in the additive monoid A. The reader should recall from a previous linear algebra course, that for the poset of all subspaces of a finite dimensional vector space V , the dimension function W → dim W is a Komolgoroff function into the additive monoid of the non-negative integers. A basic observation is this: Lemma 27 (Duality for Kolmogoroff functions.) Suppose f : L → A is a Komolgoroff function from lattice L into a commutative monoid A. Let L∗ be the dual lattice of (L, ≤). Both of the lattices possess the same set of elements; and so f may be regarded as a mapping from the set of elements of the dual lattice into A. In this sense f is a Komolgoroff measure of the dual poset L∗ . It is almost an insult to ask anyone for a proof! The Komolgoroff axiom is clearly self-dual. Consider first the following Lemma 28 Suppose f is a Komolgoroff measure on the lattice L into the commutative monoid A. For elements x1 , . . . , xk of lattice L, we have the following two identities: 1. (The basic identity) Xk i=1 f (xi ) = f (x2 ∨ x1 ) + f (x3 ∨ (x1 ∧ x2 )) + +f (x4 ∨ (x1 ∧ x2 ∧ x3 )) + . . . + f (x1 ∧ x2 ∧ · · · ∧ xk ) Xk = [ ^j−1 f (xj ∨ ( j=2 x ))] r=1 r + f (x1 ∧ · · · ∧ xk ). (2.29) 2. (Vriar’s Identity) Xk i=1 f (xi ) = f (x2 ∧ x1 ) + f (x3 ∧ (x1 ∨ x2 )) + +f (x4 ∧ (x1 ∨ x2 ∨ x3 )) + . . . + f (x1 ∨ (x2 ∧ · · · ∨ xk )) Xk = [ _j−1 f (xj ∧ ( j=2 x ))] r=1 r 72 + f (x1 ∨ · · · ∨ xk ). (2.30) Proof: The second identity is the complete dual of the first in the sense of Lemma 27. Thus it suffices only to prove the first one. For k = 2 this is equation (2.28). Assume the result for k = n − 1 and add f (xn ) to both sides. The left side is then the sum of the measures of the xi from i = 1 to n. The last two terms on the right side are f (x1 ∧ · · · ∧ xn−1 ) + f (xn ). (2.31) By equation (1) with a = x1 ∧ · · · ∧ xn−1 and b = xn , (4) becomes f (xn ∨ (x1 ∧ · · · ∧ xn−1 )) + f (x1 ∧ · · · ∧ xn ) and the sum on the right now has the desired form. 2.8.4 The “inclusion-exclusion” principle A lattice L = (L, ≤) is said to be distributive if and only if, for any three elements a, b and c of L, one always has a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c). (2.32) In this subsection we consider a Kolmogoroff function f : L → A from a distributive lattice L with values in a commutative monoid A. An “inclusionexclusion” principle seeks to express the value of the Kolmogoroff function at the join of several elements in terms of its values at meets of various subsets of these elements. In order to express this we need to introduce some notation. Fix a finite non-empty subset P = {`1 , . . . , `n } of elements of a lattice L. This set is indexed by [n] := {1, . . . , n}. For every non-empty subset J = {a1 , . . . , ak } of [n], let ^ J (P ) := `a1 ∧ `a2 ∧ . . . ∧ `ak . Next, for any integer k ∈ [n], set Σk f (P ) := X ^ f( |J|=k J (P )), where the sum on the right hand-side ranges over all subsets J of [n] of cardinality k – that is the sum (in monoid A) of the f -values at all possible meets of k-subsets of [n]. Thus Σ1 f (P ) = f (`1 ) + f (`2 ) + · · · + f (`n ) X f (`i ∧ `j ) Σ2 f (P ) = 1≤i<j≤n Σn f (P ) = f (`1 ∧ `2 ∧ · · · ∧ `n ). 73 (2.33) Now we may state Theorem 29 (Inclusion-Exclusion) Let f : L → A be a Kolmogoroff measure from the distributive lattice L to the commutative monoid (A, +). Let P = {`1 , . . . , `n } be any set of n elements chosen from L. Then, in the notation just proceeding, f (`1 ∨ · · · ∨ `n ) + X (k even) Σk f (P ) = X (k odd ) Σk f (P ). (2.34) If morover, A is a cancellation monoid, one may write this: f (`1 ∨ · · · ∨ `n ) = Xn k=1 (−1)(k+1) Σk f (P ). (2.35) The proof of (2.34) is left as an exercise, an easy induction proof utilizing the distributive law ( 2.32) and the Kolmogoroff axiom. Then (2.35) follows from the fact that in a cancellation monoid x +y = z, x and y being elements of A, imply that z − y is an element of A. 2.8.5 Point-measure of a lattice In order to even state an analogue of Proposition 25 we need to be able to compare images of the Komolgoroff measure f : L → A from a lattice L to a commutative monoid. That means that the monoid A itself must have a partial ordering. A commutative monoid M is said to be a partially ordered monoid if and only if there is a partial ordering “≤” such that (M, ≤) is a poset and multiplication (or addition) by any element is order-preserving. Thus a ≤ b implies ac ≤ bc, for any elements a, b and c in M . It is said to be a partially ordered monoid with cancellation if, in addition ac ≤ bc always implies a ≤ b, for all elements a, b and c in M . A good example of a partially ordered monoid with cancellation is the additive monoid R+ of non-negative real numbers. We shall see from the 74 exercises that the additive monoid of multisets is also a partially ordered cancellation monoid. Let L be a lattice with a “0”. (As usual, we let a ∧ b and a ∨ b denote the meet and join of the elements a and b in L.) A point-measure on the lattice L, is a function µ : L → R from the set of lattice elements to a partially-ordered cancellation monoid A such that 1. µ(0) = 0, the identity of A, 2. µ is monotone non-decreasing in that a ≤ b implies µ(a) ≤ µ(b), and 3. µ is a Kolmolgoroff measure, that is, µ(a ∨ b) + µ(a ∧ b) = µ(a) + µ(b), (2.36) for any lattice elements, a and b. We can now present the generalization of the Theorem at the beginning of this section: Theorem 30 Suppose µ : L → A is an A-valued point measure on the lattice L. If for k lattice elements a1 , . . . , ak , one has µ(a1 ) + µ(a2 ) + · · · + µ(ak ) > (k − 1)µ(a1 ∨ · · · ∨ ak ), (2.37) then µ(a1 ∧ · · · ∧ ak ) > 0. Proof of Theorem 30: Let x1 , . . . xk satisfy the hypothesis of the Theorem and set X := x1 ∨ · · · ∨ xk . By the hypothesis and Lemma 28, we have (k − 1)µ(X) < X i µ(xi ) = µ(xk ) + Xk−1 i=1 µ(xi ) Xk−1 = µ(xk ) + [ ^j−1 µ(xj ∨ ( j=2 x ))] r=1 r + µ(x1 ∧ · · · ∧ xk−1 ) ≤ µ(xk ) + (k − 2)µ(X) + µ(x1 ∧ · · · ∧ xk−1 ) (2.38) (Equation (2.38) follows from the monotonicity of µ.) If (k − 2)µ(X) is cancelled from both sides, one obtains µ(X) < µ(xk ) + µ(x1 ∧ · · · ∧ xk−1 ) = µ(xk ∨ (x1 ∧ · · · ∧ xk−1 )) + µ(x1 ∧ · · · ∧ xk ). 75 (2.39) But µ(X) ≥ µ(xk ∨ (x1 ∧ · · · ∧ xk−1 )) by monotonicity, so µ(x1 ∧ · · · ∧ xk ) > 0, completing the proof. When strict inequality is replaced by equality, it is possible for the meet to have trivial measure. For example, if n = k and the Vi of Proposition 25 are hyperplanes of V , it is possible for the Vi to intersect at the zero subspace. Nonetheless, with equality, the situation is still very tight. Essentially the same proof as that of Theorem 30 shows Theorem 31 Suppose the inequality of Theorem 2 is replaced by equality. Then either ^ µ( j xj ) > 0 (2.40) or, for each index i, _ µ( x) j j = µ(xi ∨ (x1 ∧ · · · ∧ xi−1 ∧ xi+1 · · · ∧ xk )). (2.41) If moreover µ is order-injective – that is, a ≤ b and µ(a) = µ(b) always imply a = b – then the µ’s may be removed in equations (2.40) and (2.41). In a lattice of finite dimensional vector spaces, “dimension” is always an order-injective measure. Because L and µ are general, there are many applications of the inequality of Theorem 2 beyond the case that L is the lattice of all subspaces of a vector space and the measure is dimension as in Proposition 25. 1. There are all sorts of height functions on matroids – for example, subsets of a finite set with cardinality as the measure. Here is another example of this type: Let N denote the non-negative integers, and let M = N(n) be the collection of all n-tuples (m1 . . . , mn ) the mi ∈ N. Under coordinate-wise addition, M is the monoid of multisets over a given n-set. M is a lattice under the order relation (a1 , . . . , an ) ≤ (b1 , . . . , bn ) iff ai ≤ bi for each i. The theorem applies with the measure defined by µ(a1 , . . . , an ) = 76 X ai . 2. Let L be any Boolean algebra of Lesbsque-measurable subsets in Euclidean space, and let µ be their measure (area, volume etc.) 3. Let L be the lattice of finite dimensional submodules of an R-module, where R is a K-algebra, K any field. Let the measure µ(N ) be the number of composition factors of the submodule N which belong to any specified class of irreducible R-modules. As a result one has, for example: Corollary 32 1. Let A1 , . . . , Ak be finite subsets of a set X. If |A1 | + · · · + |Ak | > (k − 1)|A1 ∪ · · · ∪ Ak |, then there exists an element x in every Ai . 2. Let A1 , . . . , Ak be Lesbegue-measurable subsets of Euclidean space. If the sum of their volumes exceeds k − 1 times the volume of their union, then their intersection has a positive volume. 2.8.6 Modular lattices Consider for the moment the following example: Example 16. The poset (P, ≤) contains as elements P = {a, b, c}∪Z, where Z is the set of integers, with ordering defined by these rules: 1. a ≤ p for all p ∈ P –i.e. a is the poset “zero”, 2. c ≥ p for all p ∈ P , so c is the poset “one”, 3. b is not comparable to any integer in Z, and 4. the integers Z are totally ordered in the natural way. Notice that the meet or join of b and any integer are a and c respectively. Thus (P, ≤) is a lattice with its “zero” and “one” connected by an unrefinable proper chain of length 2 and at the same time by an infinite unrefinable chain. The lattice is even lower semimodular since the only covers are (a, b), (b, c) and (n, n + 1), for all integers n. 77 For the purposes of the Jordan-Hölder theory, such examples did not bother us, for the J-H theory was phrased as an assertion about measures which take values on algebraic intervals – that is to say, non-algebraic intervals could be ignored. However, many of the most important posets in Algebra are actually lattices with a property, the modular law, which prevents elements x and y from being connected by both a finite unrefinable chain chain and an infinite chain. This modular law always occurs in posets of congruence subalgebras which are subject to the so-called “Second Fundamental Theorem of Homomorphisms”. A lattice L is said to be a modular lattice or to be modular if and only if (M) for all elements a, b, c of L with a ≥ b, a ∧ (b ∨ c) = b ∨ (a ∧ c). The dual condition is (M∗ ) for all elements a, b, c of L with a ≤ b, a ∨ (b ∧ c) = b ∧ (a ∨ c). But by permuting the roles of a and b it is easy to see that the two conditions are equivalent. Thus Lemma 33 A lattice satisfies (M) if and only if it satisfies (M∗ ). Put another way: a lattice L is modular if and only if its dual lattice L∗ is. The most important property of a modular lattice is given in the following Lemma 34 Suppose L is a modular lattice. Then for any two elements a and b of L, there is a poset isomorphism µ : [a, a ∨ b] → [a ∧ b, b], taking each element x of the domain to x ∧ b. 78 Proof: As µ is defined, µ(a) = a ∧ b, µ(a ∨ b) = (a ∨ b) ∧ b = b, and if a ≤ x ≤ y ≤ b, then µ(x) = x ∧ b ≤ y ∧ b = µ(y). Thus µ is monotone and takes values in [a ∧ b, b]. Suppose now that x, y ∈ [a, b] and µ(x) = µ(y). Then x ∧ b = y ∧ b and so a ∨ (x ∧ b) = a ∨ (y ∧ b). Applying the law (M) to each side of the last equation, one obtains x = x ∧ (a ∨ b) = y ∧ (a ∨ b) = y, since a ∨ b dominates both x and y. Thus µ is injective. Now suppose a ∧ b ≤ x ≤ b. Then by the first inequality, x = x ∨ (b ∧ a). Since x ≤ b, applying (M∗ ) (with x and a in the role of a and c in (M∗ )), x = b ∧ (x ∨ a) = µ(x ∨ a). Thus µ is onto. Thus µ is a poset isomorphism. Remark: Just on the general principle that modularity is a relation between formulas in three variables with a partial order relation, it follows that any sublattice of a modular lattice is modular.20 Corollary 35 Any modular lattice is a semi-modular lower semilattice and so is subject to the Jordan-Hölder theory. In particular, any two finite unrefinable chains that may happen to connect two elements x and y of the lattice, must have the same length, one depending only on the pair (x, y). Proof: Apply the previous Lemma for the case [a, a ∨ b] and [b, a ∨ b] are both covers. The preceeding Corollary only compared two finite unrefinable chains from x to y. Could one still have both a finite unrefinable chain from x to y as well as an infinite one as in Example 16 at the beginning of this subsection? The next result shows that such examples are banned from the realm of modular lattices. Theorem 36 Suppose L is a modular lattice and suppose a = a0 < a1 < · · · < an = b is an unrefinable proper chain of length n preceeding from a to b. Then every other proper chain proceeding from a to b has length at most n. 20 Recall from the definition of “sublattice” on page 71 that the latter is closed under the join and meet operations of the parent lattice. 79 Proof: Of course, by Zorn’s lemma, any proper chain is refined by a proper unrefinable chain. So, if we can prove that all proper chains connecting a and b are finite, such a chain possesses a well-defined measure by the JordanHölder Theory. In particular all such chains possess the same fixed length n. So it suffices to show that any proper chain connecting a and b must be finite. On the other hand, we propose to accomplish this by induction on n. But this means that we shall have to keep track of the bound on length asserted by the induction hypothesis, in order to obtain new intervals [a0 , b0 ] to which induction may be applied. So we begin by considering a (possibly infinite) totally ordered subposet X of (P, ≤) having a as its minimal member and b as its maximal member. We must show that X is a finite proper chain. If n = 1,{a, b} = X is a cover, and we are done. If each x ∈ X satisfies x ≤ an−1 , then X − {b} is finite by induction on n, and so X is finite. So we may suppose that for some xβ ∈ X, we have xβ ∨ an−1 = b. By Lemma 34 (an−1 ∧ xβ , xβ ) is a cover. Now the distinct members of the set Y = {z ∧ an−1 |z ∈ X} form a totally ordered set whose maximal member is an−1 = b ∧ an−1 , and whose minimal member is a = a ∧ an−1 . By induction on n, Y is a finite set of at most n − 1 elements. On the other hand, it is the concatenation of two proper chains Y + : = {y ∈ Y |y ≥ an−1 ∧ xβ } Y − : = {y ∈ Y |y ≤ an−1 ∧ xβ }, whose lengths are non-zero and sum to at most n − 1. Now the poset isomorphism µ : [xβ , b] → [an−1 ∧ xβ , an−1 ] takes the members of X dominating xβ to Y + . Thus X ∩ [xβ , b] is a chain of the same length as Y + . It remains only to show that the remaining part of X, the chain X ∩ [a, xβ ] is finite. 80 Let index i be maximal with respect to the condition ai ≤ xβ . Then for each index j exceeding i but bounded by n − 1, aj = aj−1 ∨ (aj ∧ xβ ) and (aj−1 ∧ xβ , xj ∧ xβ ) is a cover or is length zero. It follows that a is connected to xβ by an unrefinable chain of length at most n − 1. By induction X ∩ [a, xβ ] is finite. The proof is complete. Corollary 37 Let L be a modular lattice with minimum element 0 and maximum element 1. The following conditions are equivalent: 1. There exists a finite unrefinable chain proceeding from 0 to 1. 2. Every proper chain has length below a certain finite universal bound. 3. L has both the ascending and descending chain conditions. Proof: The Jordan-Hölder theory shows all chains connecting 0 to 1 are bounded by the length of any unrefinable one. How this affects all proper chains is left as an exercise. Any modular lattice which satisfies the conditions of Corollary 37 is said to possess a composition series. Example 17. Let L be the lattice D of all positive integers where a ≤ b if and only if a divides b. Then D is a modular lattice which possesses the DCC but not the ACC. There is a “zero” (the integer 1), but no lattice “one”. However, every interval [a, b] possesses a composition series. 2.8.7 Exercises concerning lattices. Exercise 20. Among the posets of Examples 2 through 5, which are lattices? Exercise 21. 1. Prove statement 1. of Lemma 26 – i.e any sublattice of a lattice is an induced poset. 2. Suppose P = (P, ≤P ) is a subposet of a lattice L = (L, ≤) such that P is itself a lattice. 81 (a) Give an example of such P and L such that P is not an induced poset. (b) Give an example of such P and L such that P is an induced poset but is still not a sublattice. [Hint: The examples requested need not have more than five elements. In the second example, the joins x ∨P y and meets x ∧P y in P somehow don’t coincide with those in L.] Exercise 22. Show that any multiplicatively closed subset M of the positive real numbers R+ is a totally ordered cancellation monoid with respect to multiplication and the order induced by the natural ordering of the reals. In particular, any multiplicatively closed subset of the positive integers is a totally ordered monoid with cancellation. Show that the set S of integers which are the sum of two integer squares (at least one of which is not zero) is such a set. [Later on, the reader might be interested to know that the set Sp of numbers that can be the number of p-Sylow subgroups of some finite group is also multiplicatively closed. This is not always all positive integers congruent to one modulo the prime p (also multiplicatively closed) since it will emerge that no finite group has 22 7-Sylow subgroups.] Exercise 23. Let (M1 , ≤) and (M2 , ≤) be two partially ordered semigroups. Define a binary operation on the Cartesian product M1 × M2 be coordinatewise multiplication: that is, (a1 , a2 ) ◦ (b1 , b2 ) := (a1 b1 , a2 b2 ). Define a partial ordering on M1 × M2 by the rule: (a1 , a2 ) ≤ (b1 , b2 ) if and only if both a1 ≤ b1 and a2 ≤ b2 . Show that with respect to these definitions of product and order, M1 × M2 is also a partially ordered monoid. Moreover, show that if both M1 and M2 are partially order cancellation monoids, then so is M1 × M2 . Exercise 24. Regarding Z+ as a totally ordered monoid, show that Z+ × Z+ × · · · × Z+ (k factors), is the commutative monoid of multisets over k objects. One distribution of P P objects ai Oi is less than or equal to another bi Oi if and only if integer 82 ai is less than or equal to integer bi for each i. That is, the inventories of the second distribution are at least as large as those of the first in every category. Show how this result extends to infinite products. Conclude that the monoid of multisets, with this partial ordering is a partially-ordered cancellation monoid. Exercise 23. Show that the product L1 × L2 of two modular lattices is modular. If the lattices Lσ are modular and possess a “zero”, does the modularity extend to their direct sum? 2.9 2.9.1 Dependence Theories Introduction If we say A “depends” on B, what does this mean? In ordinary language one might intend several things: “A depends entirely on B” or that “A depends just a little on B” – a statement so mild that it might suggest only that B has a “slight influence” on A. But some syntax seems to be applied all the same: thus if we say that A depends on B and that B in turn depends on C, then A (we suppose to some small degree) depends on C. In mathematics we also usually use the word “depends” in both senses. When one asserts that f (x, y) depends (to some degree) upon the variable x, one means only that f might be influenced by x. After all, it could be that f is a “constant” function as far as x is concerned. But on the other hand the mathematician also intends that f is entirely determined by the pair (x, y). Thus we may consider the phrase “f (x, y) depends on x” as one borrowed from ordinary everday speech. In its various guises, the stronger idea of total and entire dependence appears thoughout mathematics with such a common strain of syntactical features as to deserve codification. But as you will see, the theory here is highly special. 2.9.2 Dependence Fix a set S. Let F(S) be the set of all finite subsets of the set S . A dependence relation is a relation “depends” on the Cartesian product S × F(S) such that 83 1. (The Reflexive Condition) Any element s of S depends upon any finite set which contains it as a member. 2. (The Transitivity Condition) If element x depends on the finite set A and every member of A depends on the finite set B, then x depends on B. 3. (The Exchange Condition) If z, a1 , . . . , an are all elements of S such that z depends on the set {a1 , . . . , an } but does not depend on the set {a1 , . . . .an−1 }, then an depends on the set {a1 , . . . , an−1 , z}. . Before concluding anything from this, let us consider some examples. Example 18. (Linear dependence). Suppose V is a a vector space over some field or division ring D. Let us say that a vector v depends on a finite set of vectors {v1 , . . . , vn } if and only if v can be expressed as a linear combination P of those vectors, – that is, ni=1 δi vi = v for some choice of δi in D. One checks that this defines a dependence relation on V × F(V ). Example 19. (Algebraic dependence) Let K be a field containing k as a subfield. (One might let K and k respectively be the complex and rational number fields). We say that an element b of K depends on a finite subset X = {a1 , . . . .an−1 } of K if and only if b is the root of a polymonial equation whose coefficients are expressible as a k-polynomial expressions in X. This means that there exists a polynomial p(x, x1 , . . . , xn ) in the polynomial ring k[x, x1 , . . . , xn ] when evaluated at x = b, xi = ai , 1 ≤ i ≤ n yields 0. We shall see in Chapter 10 that all three axioms of a dependence theory hold for this definition of dependence among elements of the field K. 2.9.3 The extended definition Our first task will be to extend the definition of “depends” from being a relation on S × F(S) to a relation on S × 2S – so that it will be possible speak of element x depending on a (possibly infinite) subset A of S. For any subset A of S, we say that element x depends on A if and only there exists a finite subset A1 of A such that x depends on A1 (in the old sense). We leave it as an exercise to the reader to prove the following : Lemma 38 In the following the sets A and B may be infinite sets: 84 1. Element x of S depends on any set A which contains it. 2. If A and B are subsets of S, if x depends on A and every element of A depends on set B, then x depends on B. 3. If a ∈ A, a subset of S, and x is an element of S which depends on A but does not depend on A − {a}, then a depends on (A − {a}) ∪ {x} 2.9.4 Independence and spanning. Now let A be any subset of S. The flat generated by A is the set hAi := {y|y depends on A}. A subset X of S is called a spanning set if and only if hXi = S. Next we say that a subset Y of S is independent if and only if for each element x of X, x does not depend on X − {x}. Now the collections of all spanning sets and all independent sets, each form a partially ordered set under the containment relation. Thus it makes perfect sense to speak of a minimal spanning set and a maximal independant set (should such sets exist). We wish to show that these two concepts coincide. Theorem 39 1. Every minimal spanning set is a maximal independent set. 2. Every maximal independent set is a minimal spanning set. proof: We prove 1. Suppose U is a minimal spanning set. If U is not independent then there exists an element u in U such that u depends on a finite subset U1 of U − {u}. Thus every element of S depends on U and by Lemma 10, part 1, and what we know of u, every element of U depends on U − {u}. Thus by Lemma 10, part 2, every element of S depends on U − {u}. Thus U − {u} is a spanning set, against the minimality of U . Thus U must be independent. If it were not a maximal independent set there would exist an element z in S − U which did not depend on U . But that is impossible as U spans S. Next we show 2. If X is a maximal independant set, then by maximality, X spans S. If a proper subset X1 of X also spanned S then any element x of X − X1 would depend on X1 and so would depend on X − {x} by the 85 definition of dependence. But this contradicts X independent. Thus X is actually minimal as a spanning set. The proof is complete. Theorem 40 Maximal independant subsets of S exist. proof: This is a straightforward application of Zorn’s Lemma. As remarked above, the collection J of all independent subsets of S form a partially ordered set (J , ≤) under the inclusion relation21 . Now consider any chain C = {Jα } of independent sets and form their union C̄ := ∪α Jα . Then if C̄ were not an independent set, there would exist an element x in C̄ and a finite subset F of C̄ − {x} such that x depends on F . But since the chain C is totally ordered and F ∪ x is finite, there exists an index σ such that F ∪ x ⊆ Jσ . But this contradicts Jσ independent. Thus we see that C̄ must be an independent set. Thus every chain C in J possesses an upper bound C̄ in (J , ≤). By the Zorn Principle, J contains maximal elements. The proof is complete. 2.9.5 Dimension. The purpose of this section is to establish Theorem 41 If X and Y are two maximal independent sets, then |X| = |Y |. proof: By the Bernstein-Schroeder theorem, it suffices to show only that |X| ≥ |Y | for any two maximal independent sets X and Y . The proof has a natural division into two cases: the case X is a finite set and the case that X is infinite. First assume X is finite. We proceed by induction on the parameter k := |X| − |X ∩ Y |. If k = 0, then X = X ∩ Y ⊆ Y . But as Y is independent and each of its elements depends on its subset X, we must have X = Y , whence |X| = |Y | and we are done. Thus we assume k > 0. Assume X = {x1 , . . . , xn } with xi pairwise distinct elements of S and the indexing chosen so {x1 , . . . , xr } = X ∩ Y , thereby making k = n − r. From maximality of the independent set X, every element of Y − X depends on X. Similarly, every element of X depends on Y . If every element of Y depended on X0 := {x1 , . . . , xn−1 } then xn , which depends on Y , would depend on X0 21 Actually, it forms a simplicial complex since it is an easy exercise to show that a subset of an independent set must be independent. 86 by Lemma 38 (1) above, against the independence of X. Thus there is some element y ∈ Y − X which depends on X but does not depend on X0 . By the exchange condition, xn depends on X1 := {x1 , . . . , xn−1 , y}. Moreover, if X1 were not independent, then either y depends on X1 − {y} = {x1 , . . . , xn−1 } or some xi depends on (X0 − {xi }) ∪ {y}. The former alternative is ruled out by our choice of y. If the latter dependence held, then as the independence of X prevents xi from depending on X0 − {xi }, the exchange condition would force y to depend on X0 , again contrary to the choice of y. Thus X1 is an independent set of n distinct elements. But as xn depends on X1 , each element of X depends on X1 . Thus as all elements of S depend on X, they all depend on X1 as well. Thus X1 is a maximal independent set. But since |X1 ∩ Y | = 1 + |X ∩ Y |, induction on k yields |X1 | ≥ |Y |. The result now follows from |X| = |X1 |. Now assume X is an infinite set. Since Y is a maximal independent set, it is a spanning set. Thus for each element x in X there is at least one finite non-empty subset Yx of Y on which it depends. Choose, if possible y ∈ (Y − ∪X Yx ). Since y depends on X and by construction every element of X depends on ∪X Yx , we see that y depends on ∪X Yx ⊆ Y , contradicting the independence of Y . Thus we see that no such y can exist and so Y = ∪X Yx . Now by Corollary 5 |∪X Yx | ≤ |X| so |Y | ≤ |X| as required. The proof is complete. This common cardinality of maximal independent sets is called the dimension of the dependence system S. 2.9.6 Matroids We present here another view of dependence theory. At the same time the developement in this section illustrates many of the principles and defined structures encountered earlier in this chapter. Fix a set X and let I be a family of subsets of X. The pair (X, I) is called a matroid if and only if the following axioms hold: (M1) The family I is closed under taking subsets. (M2) (Exchange axiom) If A, B ∈ I with |A| > |B|, then there exists an element a ∈ A − B, such that {a} ∪ B ∈ I. 87 (M3) If every finite subset of a set A belongs to I, then A belongs to I. Note that if A and B are maximal elements of (I, ⊆), then |A| = |B| is an immediate consequence of axiom (M2). However, from axioms (M1) and (M2) alone, it does not follow that maximal elements even exist. Consider the following Example 20.Let X be be the set of natural numbers and let I be all finite subsets of X. Then certainly I is closed under taking subsets, so (M1) holds. Also if A and B are finite sets with |A| > |B| there is certainly an element a ∈ A − B such that {a} ∪ B is finite, Thus axiom (M2) holds. But now it should be clear that I contains no maximal elements at all. However, axiom (M3) avoids this problem. Lemma 42 If (X, I) is a matroid, then every element of I lies in a maximal element of I. Proof: This proof utilizes Zorn’s lemma. Let A0 be an arbitrary member of I. We examine the poset of all subsets in I which contain A0 . Let A1 ⊆ A2 ⊆ · · · be a chain in this poset and let B be the union of all the sets Ai . Consider any finite subset F of B. Then each element f ∈ F lies in some member Ai(f ) of the chain. Setting m to be the maximal index in the finite set {i(f )|f ∈ F }, we see that F ⊆ Am . Since Am ∈ I, (M1) implies F ∈ I. But since F was an arbitrary finite subset of B, (M3) shows that B ∈ I. Thus every finite chain in the poset of elements of I which contain A0 has an upper bound in that poset, so that poset contains a maximal element M . Clearly M is a maximal member of I, since any member of I which contains M also contains A0 , and so lies in the poset of which M is a maximal member. Thus A0 lies in a maximal member of I, as required. Example 21. Let V be a vector space and let I be all linearly independant subsets of V . Then axioms (M1) through (M3) hold so (V, I) is a matroid. Example 22. Suppose (V, E) is a simple graph with vertex set V and edge set E. (The adjective “simple” just means that edges are just certain unordered pairs of distinct vertices.) A cycle is a sequence of edges (e0 , e1 , . . . , en ) such that ei shares just one vertex with ei+1 , indices i taken modulo n. Thus en−1 shares a vertex with en = e0 , and the “length” of the cycle, n, cannot be one or two. A graph with no cycles is called a forest – 88 its connected components are called trees. Now let I be the collection of all subsets A of E such that the graph (V, A) is a forest. Then (V, I) is a matroid. Example 23. Let F be a fixed family of subsets of a set X. A finite subset {x1 , x2 , · · · , xn } of X is said to be a partial transversal of F if and only if there are pairwise distinct subsets A1 , . . . An of F such that xi ∈ Ai , i = 1, . . . , n — that is, the set {x1 , . . . , xn } is a “system of distinct representatives” of the sets {Ai }. Now let I be the collection of all subsets of X all of whose finite subsets are partial transversals. Then (X, I) is a matroid. Example 24. Suppose (X, I) is any given matroid, and Y is any subset of X. Let I(Y ) be all members of I which happen to be contained in the subset Y . Then it is straightforward that the axioms (M1), M(2) and (M3) all hold for (Y, I(Y )). The matroid (Y, I(Y )) is called the induced matroid on Y . 2.9.7 The rank function For any matroid M = (X, I), a maximal element of I is called a basis of M . We have seen that maximal elements of I exist and that any two of them have the same cardinality, which we denote by ρ(M ). But also any subset Y of X can be regarded as an induced matroid Y = (Y, I(Y )) as in Example 24, and it too has a rank ρ(Y ) – the cardinality of a maximal member of I(Y ) (basis of Y ). Thus rank is a function ρ : subsets of X → cardinal numbers . This is a useful function, and before going onward, we collect a few of its properties. Lemma 43 The rank function is monotone, that is, if A ⊆ B ⊆ X then ρ(A) ≤ ρ(B) (as inequality is understood among cardinal numbers). This is immediate from the fact that a maximal element of I(A) lies in some maximal element of I(B). Lemma 44 Suppose x and y are elements of a matroid M = (X, I) and B is a finite subset of X. Then ρ({x} ∪ B) = ρ({y} ∪ B) = ρ(B) 89 implies ρ({x, y} ∪ B) = ρ(B). Proof: Let F be a maximal element of I({x, y} ∪ B), and suppose that |F | = ρ({x, y} ∪ B) > ρ(B). Then by (M2), for any basis B0 for B there exists an element z ∈ F − B0 , such that {z} ∪ B0 ∈ I. Now if z were in B, the maximality of B0 in I(B) would be in peril. So z is x or y. Suppose z = x ∈ / B. Then as B0 is finite and {x} ∪ B0 ∈ I, we would have ρ({x} ∪ B0 ) ≤ ρ({x} ∪ B) = ρ(B), which says |{x} ∪ B0 | = |B0 |, which absurd for these finite sets. By symmetry, the alternative that z = y leads to a similar absurdity. Thus the assumption that |F | exceeds the rank of B is untenable. The monotonicity of the rank function shows that ρ(B) ≤ |F |, so we must have equality. Corollary 45 Suppose A and B are finite subsets of X, and for each element a in A, ρ({a} ∪ B) = ρ(B). Then ρ(A ∪ B) = ρ(B). Proof: This just follows by iterating the lemma through the finitely many elements of A. Theorem 46 Let ρ be the rank function of the matroid M = (x, I). Then it satisfiies these three rules: (R1) For each subset Y of X, 0 ≤ ρ(Y ) ≤ |Y |. (R2) ( Monotonicity) If Y1 ⊆ Y2 ⊆ X, then ρ(Y1 ) ≤ ρ(Y2 ). (R3) (The submodular inequality) If A and B are subsets of X, then ρ(A ∪ B) + ρ(A ∩ B) ≤ ρ(A) + ρ(B). 90 proof: The first two statements follow immediately from the definition of the rank function. So we need only prove the submodular inequality. Let A and B be two subsets of X. Let F be a basis for A ∩ B, and let G be a basis of A ∪ B containing F . Then G ∩ A and G ∩ B belong to I(A) and I(B), respectively, so |G ∩ A| ≤ ρ(A) and |G ∩ B| ≤ ρ(B). Then ρ(A) + ρ(B) ≥ = = = = |G ∩ A| + |G ∩ B| |(G ∩ A) ∪ (G ∩ B)| + |(G ∩ A) ∩ (G ∩ B)| |G ∩ (A ∪ B)| + |G ∩ A ∩ B| |G| + |F | ρ(A ∪ B) + ρ(A ∩ B). The proof is complete. The properties (R1) – (R3) actually characterize matroids. Theorem 47 Suppose r is a map from the set of all subsets of a set X, into the cardinal numbers, satisfying the three rules (R1), (R2), (R3). Let I the the collection of a sets X with the property (*) For every finite subset FX of X we have |FX | = r(FX ). Then (X, I) is a matroid. 2.9.8 Dependence theories and matroids are the same thing For any matroid (X, I), the subsets in I are usually called independent sets – a name we have avoided up to now, since independent sets have already been defined in the context of a dependence theory. This suggestive name is in fact justified by the following theorem: Theorem 48 Let X be a set and let F(X) be all finite subsets of X. Let D ⊆ X × F(X) be a dependence theory in the sense of the previous subsections. (We say “x depends on A” if xDA.) Let I be the collection of all subsets Y of X independant with respect to D. Then (X, I) is a matroid. 91 Proof: Recall that in our dependence theory any (possibly infinite) subset Y of X is said to be an “independent set” if and only if: For each element y ∈ Y , y does not depend on any finite subset of Y − {y}. It is immediate that a subset of an independent set is independant. Equally immediate from this definition is the fact that a subset is independant if and only if every finite subset of it is independent. So (M1) and (M3) hold. Now suppose A and B are independent subsets of X with |A| > |B|. Then in particular A − B is non-empty. Now assume by way of contradiction that the conclusion (M2) were false. Then for every element a ∈ A − B, {a} ∪ B is not independent. Since B is independent by hypothesis, the Exchange Axiom forces a to depend on B. Thus every element of A − B depends on B. But also, every element of A ∩ B depends on B, simply by the Reflexive Property. Thus A ⊆ hBi, the flat spanned by B. Since B is an independent spanning set for hBi, B is a basis for hBi. But as A is also an independent subset of hBi, it lies in a maximal independent subset A1 of hBi. By Theorem 41, |A| ≤ |A1 | = |B|. This contradicts |A| > |B|. The proof is complete. Now suppose we are given a matroid (X, I). Can we recover a dependence theory from this matroid? Clearly we need to have a definition of “dependence”. Consider this definition: Definition Let x be an element of X and let A be a finite subset of X. We say that x depends on A if and only if there exists a subset A0 of A which lies in I, such that either (i) x ∈ A0 , or (ii) {x} ∪ A0 is not in I. One consequence of this definition is Lemma 49 If B is a finite set, x depends on B if and only if ρ({x} ∪ B) = ρ(B). (2.42) Proof: Suppose x depends on B. Then there exists a finite set B0 ∈ (B), such that {x}∪B0 ∈ / I. Suppose by way of contradiction that ρ({x}∪B)ρ(B). Then every basis of {x} ∪ B contains x. Since B0 is in I(B) it extends to a basis B 0 of {x}∪B. But, as remarked, x ∈ B 0 , where (M1) forces {x}∪B0 ∈ I, a contradiciton. Thus ρ({x} ∪ B) = ρ(B). Now we must prove the converse statement. Suppose B is a finite set such the ρ({x} ∪ B) = ρ(B). We must show that x depends on B. 92 Suppose first that x ∈ B. If {x} ∈ / I, set Z = ∅ ∈ I, and if {x} ∈ I, set Z = {x}. Then in either case, Z ∈ I(B), while x ∈ Z or {x} ∪ Z ∈ / I. Thus x depends on B, by our definition. Thus we may assume that x is not in B. Let B0 be a maximal member of I(B), that is a basis for B. Now as x is not in B,and B is finite, {x} ∪ B0 ∈ I would imply that ρ({x} ∪ B) > |B0 | = ρ(B), contrary to assumption. Thus {x} ∪ B0 ∈ / I, while B0 ∈ I, so x depends on B, according to our definition. Theorem 50 Given a matroid (X, I), let the relation of “dependence” between elements of X and finite subsets of X be given as in the preceeding definition. Then this notion satisfies the axioms of a dependence theory. Proof: First suppose x ∈ A. Then ρ({x} ∪ A) = ρ(A) and so, as A is finite, Lemma 49 shows that x depends on A. Thus the Reflexive Property holds. The Transitivity Property: Suppose x depends on set finite set A, and that every every element of A depends on set B. We must establish the conclusion that x depends on B. If x ∈ A, the conclusion is already part of the hypothesis. If x ∈ B, the conclusion follows from the reflexive property already established. So without loss of generality, assume x ∈ / A ∪ B. Now for each element a ∈ A, there exists a finite subset Ba ∈ I(B) , such that {a} ∪ Ba ∈ / I. Let B0 be the finite union of these finite sets, that is B0 := ∪a∈A Ba . Then by Lemma 49 it is correct to say that ρ({a} ∪ B0 ) = ρ(B0 ) for each a ∈ A, as well as ρ({x} ∪ A) = ρ(A). Since both A and B0 are finite, Corollary 45 shows ρ(A ∪ B0 ) = ρ(B0 ). Now, as x depends on A, x depends on the finite set A ∪ B0 . So ρ({x} ∪ A ∪ B) = ρ(A ∪ B). 93 It follows that ρ({x} ∪ B0 ) ⊆ ρ({x} ∪ A ∪ B) = ρ(A ∪ B0 ) = ρ(B0 ) Thus by Lemma 49, x depends on B0 and so x depends on B. The Exchange Condition: Suppose B = {b1 , . . . , bn } and B 0 := B − {xn }, and the element x depends on B but not B 0 . Then ρ({x} ∪ B) = ρ(B) while ρ({x} ∪ B 0 ) = 1 + ρ(B 0 ). Now suppose ρ(B 0 ) = ρ(B). Then ρ({x} ∪ B 0 ) ≤ ρ({x} ∪ B) = ρ(B) = ρ(B 0 ), from which Lemma 49 would imply that x depends on B 0 , against the hypothesis. Thus we have 1 + ρ(B 0 ) = ρ(B) = ρ({x} ∪ B 0 ). (2.43) Now everything can be assembled. ρ({bn } ∪ ({x} ∪ B 0 )) = ρ({x} ∪ B) = ρ(B) = ρ({x} ∪ B 0 . From the equality of the first and last terms and Lemma 49 applied to the finite set {x} ∪ B 0 , we see that bn depends on {x} ∪ B 0 . That, of course, is the desired conclusion of the exchange condition. 2.9.9 Other formulations of dependence There are purely combinatorial ways to express the gist of a dependence theory, and these are the various formulations of the notion of a matroid in terms of flats, in terms of axioms on circles, or in terms of closure operators. 94 Lets consider flats, for a moment. We have already defined them from the point of view a dependence theory. From a matroid point of view, the flat hAi spanned by subset A in matroid (X, I), is the set of all elements x for which {x} ∪ Ax ∈ / I for some finite subset Ax of A. It is an easy exercize to show that hhAii = hAi, for all subsets A of X. In fact Theorem 51 The mapping τ which sends each subset A of Xto the flat hAi spanned by A, is a closure operator on the lattice of all subsets of X. The image sets – or “closed” sets – form a lattice: 1. The intersection of two closed sets is closed, and is the meet in this lattice. 2. The closure of the set-theoretic union of two sets is a join, that is, a global minimum in the poset of all flats above the two sets. The reader should be able to prove this. Proposition 52 Suppose P = P (X) is the poset of all subsets of a set X and τ : P → P satisfies (Increasing) For each subset A, A ⊆ τ (A). (Closure) The mapping τ is a closure operator – that is, it is an idempotent monotone mapping of a poset into itself. (The Steinitz-MacLane Exchange Property) If X ⊆ P and y, z ∈ P − τ (X) then y ∈ τ (X ∪ {z}) implies z ∈ τ (X ∪ {y}). Then, setting I = {A ⊆ X|x ∈ / τ (A − {x}) for all x ∈ A}, we have that M := (X, I) is a matroid. Conversely, if (X, I) is a matroid, then the mapping τ which takes each subset of A to the flat hAi which it spans, is a monotone increasing closure operator P → P possessing the Steinitz-MacLane Exchange Property. 95 2.9.10 Exercises on Matroids Exercise 24. Prove the three parts of Lemma 38 for the extended definition of dependence allowing infinite subsets of S. Exercise 25. Verify the axioms of a matroid for Example 22, on the edge-set of a graph. Exercise 26. Verify the axioms of a matroid for Example 23, on finite partial transversals. Exercise 27. Let Γ = (V, E) be a simple graph. Let M = (E, I) be the matroid of Example 22. For any subset F of the edge set E, let (V, F ) = ∪σ∈K (Vσ , Fσ ) be a decomposition of Γ into connected components. For each connected component (Vσ , Fσ ), let Eσ be the collection of edges connecting two vertices in Vσ . Thus (Vσ , Eσ ) is the subgraph induced on the vertex set Vσ . Prove that in the matroid, the flat spanned by F is the union ∪σ∈K Eσ . Exercise 28. Let M = (X, I) be a matroid. A set which is minimal with respect to not lying in I is called a circuit. Show that any circuit is finite. [Remark: There is also a characterization of matroids by circuits. See Oxley, Prop. 1.3.10.] 96 Chapter 3 Review of Elementary Group Properties 3.1 Introduction In mathematics we often know various ways to construct something by inputing some data D to some procedure P . If we wish to classify these same somethings up to isomorphism, we are faced with two problems: Does the construction procedure P give all such objects? Can I tell when two distinct data, D1 and D2 yield isomorpic objects when inputted to P ? If the answer to the first question is “No” then P is not an interesting procedure, and one might think of discarding it. But if the answer to the first question is “yes ” – or even almost “yes” – the second question becomes interesting. But this question comes down to all the ways some one object might be “realized” as the consequence of datum D and procedure P . Often it happens that each way to do this is obtained by first appying an automorphism to the something and then performing a “canonical” realization. In that case, to classify the somethings, we need to know about their automorphisms. This is just one of the reasons that group theory is needed, whatever field of mathematics you might choose to enter. There are of course many other reasons having to do with special uses of groups, such as “Polya counting”1 , quantum physics, invariant theory etc.). But overall, the need to classify things seems the broadest reason that group theory is pertinent to all of mathematics. 1 which has a long pre-Polya history 97 Groups are defined by a hallowed set of axioms which are useful only for the purpose of banishing all ambiguities from the subject. Somehow, staring at the axioms of a group mesmerizes one away from their basic and natural habitat. Every group that ever exists in the world is in fact the full group of symmetries of something. That is what the theory of groups is: the study of symmetries. But it is useful to know this only on a philosophical plane. Knowing that there is a set of objects such that every group – or even every finite group –is the full group of automorphisms of at least one of these objects, is not very helpful for classifying groups. Moreover it is evidence that the converse problem of classifying the objects which exhibit these groups is not only not brought nearer to a classification; it points in the opposite direction: these objects probably cant be classified.2 3.2 Definition and examples. This is probably a good place to explain a convention. A group is a set G with a binary operation with certain properties – of which there are a good many examples. But familiar examples breed familiar customs. In some of these examples the binary operation is written as “+”. in others as “◦” or “×”. How then does one speak generally about all possible groups? The standard solution is this: The group operation of an arbitrary group, will be indicated by simple juxtaposition, the act of writing one symbol directly after the other – thus, the group operation, applied to the ordered pair (x, y) 2 I hope I will not belabor the point in recalling the day that a colleague of mine announced to me that he had just proved that for any finite group G, there existed a projective plane P whose automorphism group was G. I know that I dissapointed my colleague by my lack of enthusiasm for the theorem – as much as I wished to encourage him. The problem is that this just shows how hopeless the subject of projective planes is. Every single projecive plane he could exhibit whose automorphism group was a finite group G was in fact an infinite projective plane. Far from being the tight object which everyone had imagined it was, projective planes (especially the infinite ones) are such a loose subject that every finite group can be presented as their automorphism group. This is a theorem stating how bad the subject of projective planes really is. (I am sure his proof was not easy, but it is doomed to be a non-useful theorem. My remarks apply in fact to any Theorem of the form “For any finite group G, there exists an object X(G) in class C ( a class which has not been classified (such as finite groups or it subclass p-groups)), such that Aut(X(G) ' G. In short, it does not advance the classification problem for groups to construct a Zoo of objects in some equally unclassified category, exhibiting each of these groups as their automorphism group. Enough of sermons! 98 will be denoted xy. A group is a set G equipped with a binary operation G×G → G (denoted here by juxtaposition) such that the following statements hold: 1. (The Associative Law) The binary operation is associative – that is (ab)c = a(bc) for all elements a, b, c of G. (The parentheses indicate the temporal order in which operations were performed. This axiom more-or-less says that the temporal order of applying the group operations doesn’t matter, while, of course, the order of the elements being operated on does matter.) 2. (Identity Element) There exists an element, say e, such that eg = g = ge for all elements g in G. 3. (The Existence of Inverses) For each element x in G, there exists an element x0 such that x0 x = xx0 = e where e is the element referred to in Axiom 2.3 One immediately deduces the following Lemma 53 For any group G one has: 1. The identity element e is the unique element of G possessing the property of Axiom 2. 2. Given element g in G, there is at most one element g 0 such that gg 0 = g 0 g = e. 3. In light of the preceeding item 2, we may denote the unique inverse of g by the symbol g −1 . Then note that (i) (ab)−1 = (b−1 )(a−1 ) for all elements a, b ∈ G. (ii) (a−1 )−1 = a, for each element a ∈ G. (iii) For any element x and natural number k, the product xx · · · x (with k factors) is a unique element which we denote as xk . We have (xk )−1 = (x−1 )k . ( As a notational convention, this element is also written as x−k .) 3 People have figured out how to “improve” these axioms, by hypothesizing only right identities and left inverses and so on. There is no attempt here to make these axioms independent or logically neat. – indeed they over-state their case. Yet, a little redundancy won’t hurt at this stage. 99 . Example 1. Familiar Examples: (a) The group (Z, +) of integers under the operation of addition. (b) The group of non-zero rational numbers under multiplication. (c) The group of non-zero complex numbers under multiplication. (d) The cyclic group of order n. This can be thought of as the group of rotations of a regular n-gon. This group consists of the elements {1, x, x2 . . . , xn−1 } where x is a clockwise rotation of 2π/n radians and “1” is the identity transformation. All multiplications of the powers of x can be deduced solely from the algebraic identity xn = 1. (e) The dihedral group of order 2n. This is the group of all rigid symmetries of the regular n-gon. In addition to the group of n rotations just considered, it also contains n reflections which are symmetries of the n-gon. If n is an odd number, these reflections are about an axis through a corner or vertex of the polygon and bisecting an opposite side. If n is even, the axes of the reflections are of two sorts: those which go from a vertex through its opposite vertex, and those which bisect a pair of opposite sides. There are then n/2 of each type. The elements of the group all have the form ti xj , where t is any reflection, x is the clockwise rotation by 2π/n radians and 0 ≤ t ≤ 1 and 0 ≤ j ≤ n − 1. The results of all multiplications of such elements can be deduced entirely from the relations xn = 1, t2 = 1, tx = x−1 t (and its consequence, tx−1 = xt). (f) The symmetric groups. Let X be a set. A permutation of the elements of X is a bijective mapping X → X. This class of mappings is closed under composition of mappings, inverses exists, and it is easy to verify that they form a group with the identity mapping 1X which takes each element to itself, as the group identity element. This group is called the symmetric group on X and is denoted Sym(X). If |X| = n it is well known from elementary counting principles that there are exactly n! permutations of X. In this case one writes Sym(n) for Sym(X), since the names or identities of the elements of X do not really affect the nature of the group of all permutations. 100 (g) The group of rigid motions of a (regular) cube. Just imagine a wooden cube on the desk before you. We consider the ways that cube can be rotated so that it achieves a position congruent to the original one. The result of doing one rotation after another is still some sort of rotation. It should be clear that these rotations can have axes which are situated in three different ways with respect to the cube. The axis of rotation may pass through the centers of opposite faces, it may pass though the midpoints of oppositie edges, or it could be passing through opposite vertices. (h) Let G = (V, E) be a simple graph. This means the edges E are pairwise distinct 2-subsets of the vertex set V . Two (distinct) vertices are said to be adjacent if and only if they are the elements of an edge. Now the group of automorphisms of the graph G is the set of permutations of the set of vertices V which preserve the adjacency relation. The group operation is composition of the permutations. (i) Let R be any ring with identity element e.4 An element u of R is said to be a unit if and only there exists a two-sided multiplicative inverse in R – that is, an element u0 such that u0 u = e = uu0 . (Observe in this case, that u0 itself must be a unit.) Then the set U (R) of all units of ring R forms a group under ring multiplication. Some specific cases: (a) The group U (Z) of units of the ring of integers Z, is the set of numbers {+1, −1} under multiplication. Clearly this group behaves just like the cyclic group of order 2, one of those introduced in Example (d) above. (b) Let D = Z ⊕ Zi, where i2 = −1, the ring of Gaussian integers {a + bi|a, b ∈ Z} as a multiplicatively and additively closed subset (that is a sub-ring) of the ring of all complex numbers. Then the reader may check that U (D) is the set {±1, ±i} under multiplication. This is virtually identical with the cyclic group of order four as defined in Example (d). (c) Let V be a left vector space over a division ring D. Let hom(V, V ) be the collection of all linear transformations f : V → V (viewed as right operators on the set of vectors V ). This is an additive 4 It is presumed that the reader has met rings before. Not much beyond the usual definitions are presumed here. 101 group under the operation “+” where, for all f, g ∈ hom(V, V ), (f + g) is the linear transformation which takes each vector v to vf + vg. With composition of such transformations as “multiplication” the set hom(V, V ) becomes a ring 5 Now the group of units of hom(V, V ) would be the set GL(V ) of all linear transformations t : V → V where t is bijective on the set of vectors (i.e. a permutation of the set of vectors. (V is not assumed to be finite or even finite-dimensional in this example.) The group GL(V ) is called the general linear group of V . (j) The group GL(n, F ). Let F be a field, and let G = GL(n, F ) be the set of all invertible n × n matrices with entries in F . 6 This set is closed under multiplication and forms a group. In fact it is the group of units of the ring of all n ×n matrices with respect to ordinary matrix multiplication and matrix addition. 3.2.1 Orders of elements and groups Suppose x is an element of a group G. We inductively define xn+1 := x(xn ). Thus x1 = x, x2 = xx, x3 = xx2 = xxx = x2 x (because of the associative law), and so on. If there exists a natural number n such that xn = e, then x is said to have finite order. Of course if there is not such a natural number, then it is an easy exercise to see that the the members of G represented by the infinite set {1, x, x2 , x3 , . . .} are pairwise distinct. In this case we say that x has infinite order. But clearly the set of natural numbers Nx := {n ∈ N|xn = e} is an ideal in the ring Z. Since Z is a priciple ideal domain with only {±1} as its group of units, it follows that there is a unique least natural number n in Nx , which we call the order of the element x. We denote this number by the sympbol o(x). Note that because Nx is a principal ideal generated by o(x), we see that xm = e implies that o(x) divides m. If x has finite order, the number o(x) is called the order of the element x. 5 Actually it is more, for hom(V, V ) can be made to have the structure of a vector space, and hence an algebra if D possesses an anti-automorphism – e.g. when D is a field. But we can get into this later. 6 Recall from your favorite undergraduate linear algebra or matrix theory course that if n is a natural number, then a matrix M has a right inverse if and only if it has a left inverse (this is a consequence of the equality of row and column ranks). 102 Elements of order 2 play a special role in finite group theory, and so are given a special name: any element of order two is called an involution. The order of a group G is the number |G| of elements within it. 3.2.2 Subgroups. Let G be a group. For any two subsets X and Y of G, the symbol XY denotes the set of all group products xy where x ranges over X and y ranges independantly over Y . (It is not immediately clear just how many group elements are produced in this way, since a single element g might be expressible as such a product in more than one way. At least we have |XY | ≤ |X| · |Y |.) A second convention is to write X −1 for the set of all inverses of elements of the subset X. Thus X −1 := {x−1 |x ∈ X}. This time |X −1 | = |X|, since the correspondence x → x−1 defines a bijection X → X −1 (using Lemma 53 3(ii) here.) A subset H of G is said to be a subgroup of G if and only if 1. HH ⊆ H, so that by restriction H admits the group operation of G, and 2. with respect to this operation, H is itself a group. One easily obtains the useful Lemma 54 (Subgroup Criterion.) For any subset X of G, the following are equivalent: 1. X is a subgroup of G. 2. X = X −1 and XX ⊂ X. 3. XX −1 ⊆ X. . A remark on notation. Whenever subset X is a subgroup of G, we write X ≤ G instead of X ⊆ G. This is in keeping with the general practice in abstract algebra of writing A ≤ B whenever A is a subobject of the same algebraic species as B. For example we write this if A is an R-submodule of the R-module B, and its special case: when A is a vector subspace of a vector space B. Usually the context should make it clear what species of algebraic object we are talking about. Here it is groups. 103 Corollary 55 1. The set-intersection over any family of subgroups of G is a subgroup of G. 2. If A and B are subgroups of G, then AB is a subgroup of G if and only if AB = BA (an equality of sets). 3. For any subset X of G, the set hXiG of all finite products y1 y2 · · · yn , n ∈ N where the yi range independantly over X ∪ X −1 , is a subgroup of G, which is contained in any subgroup of G which contains subset X. Thus we can also write hXiG = ∩X⊆H≤G H where the intersection on the right is taken over all subgroups of G which contain X. The subgroup hXiG (which is often written hXi when the “parent” group G is understood) is called the subgroup generated by X. As is evident from the Corollary, it is the smallest subgroup (in the poset of all subgroups of G) which contains set X. Example 2. Examples of Subgroups. (a) Let Γ(V, E) be a graph with vertex set V and edge set E, and set H := Aut(Γ), the group of automorphisms of the graph Γ, as in Example (h) of groups in the previous subsection. Then H is a subgroup of Sym(V ), the symmetric group on the vertex set. (b) Cyclic Subgroups Let x be an element of the group G. The set hxi := {xn |n ∈ Z} (where we adopt the convention of Lemma 53 3(iii) that x−n = (x−1 )n ) is clearly a subgroup of G (Corollary 55). Now one sees that the order of the element x is the (group) order of the subgroup hxi. (c) Let Y be a subset of X. The set of all permutations of X which map the subset Y onto istelf forms a subgroup of Sym(X) called the stabilizer of Y . If G is any subgoup of Sym(X), then the instersection of the stabilizer of Y with G is called the stabilizer of Y in G and is denoted StabG (Y ). This is clearly a subgroup of G. Thus it makes sense to speak of the stabilizer of a specified vertex in the automorphism group Aut(Γ) of a graph Γ. (By convention one writes Stab(y) instead of Stab({y}). Note also that all these definitions apply when G = Sym(X), as well.) 104 (d) Let us return once more to the example of the group of rotations of a regular cube (example (g) of the previous subsection). We asserted that each non-identity rigid motion was a rotation in ordinary Euclidean 3space about an axis symmetrically placed on the cube. One can see that the group of rotations about an axis through the center of a square face and through the center of its opposite face, is a cyclic group generated by a rotaion y of order 4. There are 6/2 = 3 such axes: in fact, they can be taken to be the three pair-wise orthogonal coordinate axes of the surrounding Euclidean space. This contributes 6 elements y of order 4 and 3 elements of order 2 (such as y 2 ). Another type of axis extends from the midpoint of one edge to the midpoint of an opposite edge. Rotations about such an axis form a subgroup of order two. The generating rotation t of order two does not stabilize any face of the cube, and so is not any of the involutions (elements of order two) stabilizing any of the previous “face-to-face” axes. Since there are 12 edges in six opposite pairs, these “edge-to-edge” axes contribute 6 new involutions to the group. Finally there are 8/2 = 4 “vertex-to-vertex axes, the rotations about which form cyclic subgoups generated by a rotation of order 3. Thus each of these 4 groups contribute two elements of order three. Thus the group of rotations of the cube contains 1 identity element, 6 elements of order four, 3 elements of order two stabilizing a face, 6 involutions not stabilizing a face, and 8 elements of order three – a total of 24 elements. Cosets. Suppose H is a subgroup of a group G. If x is any element of G, we write Hx for the product set H{x} introduced in the last subsection. Such a set Hx is a right coset of H in G. If x and y are elements of G and H ≤ G, then y is an element of Hx if and only if Hy = Hx. This is any easy exercise. It follows that all the elements of G are partitioned into right cosets as G = ∪x∈T Hx a disjoint union of right cosets, for appropriate T. Here T is merely a set consisting of one element from each right coset. Such a set is called a system of right coset representives of H in G, or sometimes a (right) transversal of H in G. 105 The components of this partition – that is the sets {Hx|x ∈ T } for any transversal T – is denoted G/H. It is just the collection of all right cosets themselves. The cardinality of this set is called the index of H in G and is denoted [G : H]. One notes that right multiplication by element x induces a bijection H → Hx. Thus all right cosets have the same cardinality. Since they partition all the elements of G we have the following: Lemma 56 (Lagrange’s Theorem) 1. If H ≤ G, then |G| = [G : H] · |H|. 2. The order of any subgroup divides the order of the group. 3. The order of any element of G divides the order of G. We conclude with a useful result Lemma 57 Suppose A and B are subgroups of the finite group G. Then |AB| · |A ∩ B| = |A| · |B|. proof: Consider the mapping f : A × B → AB, which maps every element (a, b) of the Cartesian product to the group product ab. This map is surjective, and the fibre f −1 (ab) contains all pairs {(ax, x−1 b|x ∈ A ∩ B}. Thus |A × B| ≥ |A ∩ B| · |AB|. On the other hand, if ab = a0 b0 for (a, b) and (a0 , b0 ) in A × B, then a−1 a0 = b(b0 )−1 = x ∈ A ∩ B. But then ax = a0 , x−1 b = b0 . So the fibres are no larger than |A ∩ B|. This gives the inequality in the other direction. 3.3 Homomorphisms of Groups 3.3.1 Definitions and basic properties. Let G and H be groups. A mapping f : G → H is called a homomorphism of groups if and only if f (xy) = f (x)f (y) for all elements x, y ∈ G 106 (3.1) Here as was our convention, we have represented the group operation of both abstract groups G and H by justaposition. Of course in actual practice, the operations may already possess some other notation customary for familiar examples. For any subset X, we set f (X) := {f (x)|x ∈ X}. In particular, the set f (G) is called the homomorphic image of G. We have the usual glossary for special properties of homomorphisms. Suppose f : G → H is a homomorphism of groups. Then 1. f is an epimorphism if f is onto – that is, f is a surjection of the underlying sets of group elements. 2. f is an embedding of groups whenever f is an injection of the underlying set of group elements. 3. f is an isomorphism of groups if and only if f is a bijection of the underlying set of elements – that is, f is both an embedding and an epimorphism. 4. f is an endomorphism of groups if it is a homorphism into itself. 5. f is an automorphism of a group G if it is an isomorphism G → G of G to itself. The following is an elementary exercise. Lemma 58 Suppose f : G → H is a homomorphism of groups. Then 1. If 1G and 1H denote the unique identity elements of G, then f (1G ) = 1H . 2. For any element x of G, f (x−1 ) = (f (x))−1 . 3. The homomorphic image f (G) is a subgroup of H. . Lemma 59 Suppose f : G → H and g : H → K are group homomorphisms. Then the composition g ◦ f : G → K is also a homomorphism of groups. Moreover: 1. If f and g are epimorphisms, then so is g ◦ f . 107 2. If f and g are both embeddings of groups, then so is g ◦ f , 3. If f and g are both isomorphisms, then so is g ◦ f , and the inverse mapping f −1 : H → G, 4. If f and g are both endomorphisms (i.e. G = H = K), then then so is g ◦ f , and, 5. if f and g are both automorphisms of G then so are g ◦ f and f −1 . Thus the set of automorphisms of a group G form a group under composition of automorphisms. (This is called the automorphism group of G and is denoted Aut(G)). Finally we introduce an invariant associated with every homomorphism of groups. The kernel of the group homomorphism f : G → H is the set ker f := {x ∈ G|f (x) = 1H }. The beginning reader should use the subgroup criterion to verify that ker f is a subgroup of G. If f (x) = f (y) for elements x and y in G, then xy −1 ∈ ker f , or equivalently, (ker f )x = (ker f )y as cosets. Thus we see Lemma 60 The group homomorphism f : G → H is an embedding if and only if ker f = 1, the identity subgroup of G. We shall have more to say about kernels later. 3.3.2 Automorphisms as right operators. As noted just above, homomorphisms of groups may be composed when the arrangement of domains and codomains allows this. In that case we wrote (g ◦ f )(x) for g(f (x)) – that is, f is applies first, then g. As remarked in Chapter 1, that notation is not very convenient if composition of mappings is to reflect a binary operation on the set of mappings itself. We have finally reached such a case. The automorphisms of a group G themselves form a group Aut(G). To represent how the group operation is realized by composition of the induced mappings, it is notational convenient to represent then in “exponential notation” and view the composition as a composition of right operators. 108 (The exponential convention) If σ is an automorphism of a group G, and g ∈ G, we rewrite σ(g) as g σ . This way, passing from the group operation (in Aut(G) ) to composition of the automorphisms does not entail a reversal in the order of the group arguments: We have, xστ = (xσ )τ . 7 3.3.3 Examples of homomorphisms. Symmetries that are induced by group elements on some object X are a great source of examples of group homomorphisms. Where possible in these examples we write these as left operators with ordinary composition “◦” — but we will begin to render these things in exponentional notation here and there, to get used to it. In the next Chapter on group actions, we will be using the exponential notation uniformly when a group acts on anything. Example 3. Examples of homomorphisms. (a) Suppose there is a bijection between sets X and Y . Then there is an isomorphism Sym(X) → Sym(Y ). This just amounts to changing the names of the objects being permuted. (b) Let R∗ be the multiplicative group of all nonzero real numbers, and let R+∗ be the multiplicative group of the positive real numbers. Then the “squaring” mapping, which sends each element to its square, defines an epimorphism which is an endomorphism R∗ → R+∗ , and an endomorphism R∗ → R∗ . The kernel is the multiplicative group consisting of the real numbers ±1 in both cases. (c) In any intermediate algebra course (cf. books of Dean or Herstein, for example), one learns that complex conjugation (which sends any complex number z = a + bi to z̄ := a − bi, a and b real) is an automorphism of the field of complex numbers. It follows that the norm mapping N : C∗ → R+∗ from the multiplicative group of non-zero complex 7 This is part of a general scheme in which elements of some ‘abstract group’ G (with its own multiplication) induce a group of symmetries Aut(X) of some object X so that group multiplication is represented by composition of the automorphisms. These are called “group actions” and are studied carefully in the next chapter. 109 numbers to the multiplicative group of positive real numbers, satisfies N (z1 z2 ) = N (z1 )N (z2 ) and hence is a homomorphism of groups. The kernel of the homomorphism is the group C1 of complex numbers of norm 1 – the so-called circle group . (When one considers that N (a + bi) = a2 + b2 , a, b ∈ R, it is not mysterious that the set of integers which are the sum of two perfect squares is closed under multiplication.) (d) (Part 1.) Now consider the group of rigid rotations of the (regular) cube. There are four diagonal axes intersecting the cube from a vertex to its opposite vertex. These four axes intersect at the center of the cube, which we take to be the origin of Euclidean 3-space. The angle α formed at the origin by the intersection of any of these two axes, satisfies cos(α) = ±1/3. Let us label these four axes 1, 2, 3, 4 in any manner. As we rotate the cube to a new congruent position, the four axes are permuted among themselves. Thus we have a mapping rotations of the cube → permutations of the labeled axes which defines a group homomorphism rigid rotations of the cube → Sym(4), the symmetric group on the four labels of the axes. The kernel would be a group of rigid motions which stabilizes each of the four axes. Of course it is conceivable that some axes are reversed (sent end-to-end) by such a motion while others are fixed pointwise by the same motion. In fact if we had used the three face-centered axes, it would be possible to reverse two of the axes while rigidly fixing the third. But with these four vertex-centered axes, that is not possible. (Can you show why? It has to do with the angle.) So the kernel here is the identity rotation of the cube. Thus we have an embedding of the rotations of the cube into Sym(4). But we have seen in the previous subsection that both of these groups have order 24. Thus by the “pidgeon-hole principle” the homomorhism we have defined is an isomorphism. (d) (Part 2.) Again G is the group of rigid rotations of the cube. There are exactly three face-centered axes which are at right angles to one another. A 90o rotation about one of these three axes fixes it, but 110 transposes the other two. Thus if we label the three face-centered axes by the letters {1, 2, 3}, and send each rotation in the group G to the permutation of the labels of the three face-centered axes which it induces, we obtain an epimorphism of groups G → Sym(3), or, in view of part 1 of (d), a homomorphism Sym(4) → Sym(3). The kernel is the group K which stabilizes each of the three face-centered axes. This group consists of the identity element, together with three involutions, each being a 180o rotation about one of the three face-centered axes. Multiplication in K is commutative. (e) Linear groups to matrix groups. Now let V be a vector space over a field F of finite dimension n. We have seen in the previous subsection that the bijective linear transformations from V into itself form a group which we called GL(V ), the general linear group on V . Now fix a basis A = {v1 , . . . , vn } of V . Any linear transformation T : V → V , viewed as a right operator of V can be represented as a matrix AT A := (pij ) where T (vi ) = pi1 v1 + pi2 v2 + · · · + pin vn (so that the rows of the matrix depict the fate of the vector vi under T )8 . Then we see that A (T ◦ S)A = A (T )A · A (S)A (where chronologicaly, the notation intends that T is applied first, then S, being right operators.) This way the symbolism does not transpose the order of the arguments, so in fact the mapping T →A TA 8 Thanks to the analysts’ notation for functions, combined with our habit of reading from left to right, many linear algebra books make linear transformations left operators of their vector spaces, so that their matrices are then the transpose of those you see here. That is, the columns of their matrices record the fates of their basis vectors. However as algebraists are aware, this is actually a very awkward procedure when one wants to regard the composition of these transfomations as a binary operations on any set of such transformations. 111 defines a group homomorphism fA : GL(V ) → GL(n, F ) of the group of linear bijections into the group of n × n invertible matrices under ordinary matrix multiplication. What is the kernel? This would be the group of linear transformations which fix each basis element, hence every linear combination of them, and hence every vector of V . Only the identity mapping can do this, so we see that fA is an isomorphism of groups. (f) The determinant homomorhism of matrix groups. The determinant associates with each n × n matrix, a scalar which is non-zero if the matrix is invertible. That determinants preserve matrix multiplication is not very obvious from the formulae expressing it as a certain sum over the elements of a symmetric group. Taking it on faith, for the moment, this would mean that the mapping det : GL(n, F ) → F ∗ , taking each invertible matrix to its determinant is a group homomorphism into the multiplicative group of non-zero elements of the ground field F . The kernel, then, is the group of all n × n matrices of determinant 1, which is called the special linear group and is denoted SL(n, F ). (g) Even and odd permutations and the sgn homomorphism. Now consider the symmetric group on n letters. In view of Example (a) above, the symmetric groups Sym(X) on finite sets X of size n are all isomorphic to one another, and so are given a neutral uniform description: Sym(n) is the group of all permutations of the set of “letters” {1, 2, . . . , n}. Subgroups of Sym(n) are called permutation groups on n letters and are so useful in providing an envoirnment for calculation, that many properties of finite groups are in fact proved by such calculations. The way to transport arguments with symmetric groups to general groups is to exploit homomorphisms G → Sym(n). These are called “group actions” and are fully exploited in the next subsection. 112 Now we can imagine that the neutral set of letters Ωn := {1, 2, . . . , n} are formally a basis A of an n-dimensional vector space over any chosen field F . Then any permutation becomes a permutation of the basis elements of V , which extends to a linear transformation T of V , and can then be rendered as a matrix A T A with respect to the basis A as in example (f). Thus, by a composition of several isomorphisms that we understand, together with their restrictions to subgroups, we have obtained an isomorphism Sym(n) → GL(n, F ) which represents each permutation by a matrix possessing exactly one 1 in each row and in each column, all other entries being zero. Such a matrix is called a a permutation matrix. Now even the usual sum formula for the determinant shows that the determinant of a permutation matrix is ±1. Now if we accept the thesis of example (f) just above, that the determinant function is in fact multiplicative, we obtain a useful group homomorphism: sgn : Sym(n) → {±1} into the multiplicative group Z2 of numbers ±1, which records the determinant of each permutaion matrix representing a permutation. The kernel of sgn is called the alternating group, denoted Alt(n), and its elements are called even permutations. All other permutations are called odd permutations. Since sgn is a group homomorphism, we see that An even permutation times an even permutation is even. An odd permutation times an odd permutation is even. An even permutation times an odd permutation (in any order) is odd. A transposition is a permutation which interchanges just two of the n letters, and leaves all remaining letters fixed. Such a permutation is represented by a permutation matrix having just two of its 1’s in symmetric positions off the main diagonal, and all remaining 1’s on the that diagonal. Clearly the determinant of such a matrix is −1. Thus any transposition is an odd permutation. Obviously the product of 113 an even number of transpositions is even, while the product of an odd number is odd. We shall learn later that any permutaion displacing a finite number of letters can be written as a product of transpositions, so this criterion is definitive9 . (h) The automorphism group of a cyclic group of order n. Finally, perhaps, we should consider an example of an automorphism group of a group. We consider here, the group Aut(Zn ), the group of automorphisms of the cyclic group of order n, where n is any natural number. Suppose, then,that C is the additive group of integers mod n – that is, the additive group of residue classes modulo n. Thus {[j] := j + nZ}, j = 1, 2, . . . n − 1. The addition rule becomes [i] + [j] = [i + j] or [i + j − n], whichever does not exceed n, where 1 ≤ i ≤ jn. Then element [1] generates this group. Indeed so does [m] if and only if gcd(m, n) = 1. Thus, if f : C → C is an automorphism of C it follows than f ([1]) = [m] where gcd(m.n) = 1. Moreover, since f is a homorphism, f [k] = [mk]. Thus the automorphism f is completely determined by the natural number m coprime to and less than n. The number of such numbers is called the Euler φfunction, and it’s value at n is denoted φ(n). Thus φ(n) = |Aut(Zn )|. It now follows that Aut(C) is isomorphic to the multiplicative group Φ(n) of all residues mod n which consist only of numbers which are relatively prime to n.10 (i) The inner automorphism group. Let G be a general abstract group and let x be a fixed element of G. We define a mapping ψx : G → G, called conjugation by x , by the rule ψx (g) := x−1 gx, for all g ∈ G. One can easily verify that ψx (gh) = ψx (g)ψx (h), and as ψx is a bijection, 9 Since the development of this example assumed the thesis of Example (f) — that the determinant of a product of matrices is the product of their determinants — which may not be known from first principles by some students, we shall give an alternative proof of the existence of the sgn homomorphism in the section on Permutation Groups and Group Actions in the next Chapter. 10 We will obtain a more exact structure of this group when we encounter the Sylow theorems. 114 it is an automorphism of G. Any automorphism of G which is of the form ψx for some x in G, is called an inner automorphism of G. Now if {x, y, g} ⊆ G, one always has y −1 (x−1 gx)y = (y −1 x−1 )g(xy) = (xy)−1 g(xy) (3.2) which means ψy ◦ ψx = ψxy (3.3) for all x, y. Thus the set of inner automorphisms is closed under composition of morphisms. Setting y = x−1 in Equation (3.3) we have ψx−1 = ψx−1 . (3.4) and so the set of inner automorphisms is also closed under taking inverses. It now follows that the set of inner automorphisms of a group G forms a subgroup, which we denote Inn(G), of the full automorphism group and call the inner automorphism group of G. Now Equation (3.3) would suggest that there is a homomorphism from G to Aut(G) except for one thing: the arguments of the ψ-morphisms come out in the wrong order in the right side of the equation. That is because the operation“◦” is denoting composition of left operators. This reveals the efficacy of using the exponential notation for denoting automorphisms as right operators. We employ the following: (Convention of writing conjugates in groups) If a and b are elements of a group G, we write a−1 ba in the exponential form ba . In this notation Equation (3.2) reads as follows: (g x )y = g xy (3.5) for all {g, x, y} ⊆ G. What could be simpler? Then we understand Equation (3.3) to read ψxy = ψx ψy , 115 (3.6) where the juxtaposition on the right hand side of the equation indicates composition of right operators – that is the chronological order in which the mappings are performed reads from left to right, ψx first and then ψy . Now Equation (3.6) provides us with a group homomorphism: ψ : G → Aut(G) taking element x to the inner automorphism ψx , the automorphisms of Aut(G) being composed as right operators (as in the exponential convention for isomorphisms on page 108). What is the kernel of the homorphism ψ? This would be the set of all elements z ∈ G such that ψz = 1G , the identity map on G. Thus this is the set Z(G) of elements z of G satisfying any one of the following equivalent conditions: (a) ψz = 1G , the identity map on G, (b) z −1 gz = g for all elements g of G, (c) zg = gz for all elements g of G. The subgroup Z(G) is called the center of G. The identity element is always in the center. If Z(G) = G, then multiplication in G is “commutative” – that is xy = yx for all (x, y) ∈ G × G. Such a group is said to be commutative, and is affixed with the adjective abelian. Thus G is abelian if and only if G = Z(G). Finally, we ask the student to prove that G/Z(G) is never a nontrivial cyclic group. [Hint: If false, cosets of the form Z(G)x±k partition G, and elements on one of these commute with those in any other such coset. So G can be shown to be abelian.] A Glossary Expected to be Understood from the Examples. 1. homomorphism of groups. 2. epimorphism of groups. 3. embedding of groups. 4. isomorphism of groups. 116 5. endomorphism of a group. 6. automorphism of a group. 7. the kernel of a homomorphism kerf . 8. the automorphism group of a group Aut(G). 9. the inner automorphism group, Inn(G), of a group, G. 10. an inner automorphism. 11. the center of a group, Z(G). 12. abelian groups. 3.4 Factor Groups and the Fundamental Theorem of Homomorphisms Whenever I am obligated to present before a class a Theorem labeled as some sort of “Fundamental Theorem”, I feel embarassed. Nearly all the time such a theorem is not fundamental at all. Who on earth decided that the Fundamental Theorem of Algebra should be the fact that the Complex Numbers form an algebraically closed field? Who on earth decided that the the Fundamental Theorem of Geometry should be the fact that an isomorphism between two Desarguesian Projective Spaces of sufficient dimension is always induced by a semilinear transformation of the underlying vector spaces ? The answer is obvious: it was so declared by persons outside the field who never had to assume the burden of classifying the objects in that field. So what is a “fundamental theorem”? At a minimum it is a proposition with these properties: 1. It should be used so constantly in the daily life of a scholar of the field, that quoting it becomes repetitive. 2. It’s logical distance from the “first principles” of the field should be short enough to bear a short explanation to a puzzled student (that is, the alleged “fundamental theorem” should be “teachable”.) 117 We are lucky today! The fundamental theorems of homomorphisms of groups actually meets both of these criteria. The set of all subgroups of a group are permuted among themselves by automorphisms of a group. Explicitly: if σ ∈ Aut(G), and K is a subgroup of G, then K σ := {k σ |k ∈ K} is again a subgroup of G.11 Moreover, if H is a subgroup of Aut(G), then any subgoup of G left invariant by the elements of H is said to be H-invariant. Naturally we have special cases for special choices of H. A subgroup K which is invariant under the full group Aut(G) is said to be a characteristic subgroup of G. We write this K char G. (The center which we previously encountered is certainly one of these.) But let us drop down to a larger class of H-invariant subgroups by letting H descend from the full automorphism group of G to the case that H is simply the group Inn(G), encountered in the previous Section, Example 3, part (i). A subgroup which is invariant under Inn(G) is said to be normal in G or a normal subgroup of G. Just putting definitions together one has Lemma 61 The following are equivalent conditions for a subgroup K of G: 1. ψx (K) = K, for all x ∈ G, 2. x−1 Kx = K for every x ∈ K, 3. xK = Kx, for each x ∈ G. (In this Lemma, ψx is the inner automorphism induced by the element x: see Example 3 part (i) preceeding.) As a notational convenience, the symbol K G will always stand for the assertion: “K is a normal subgroup of G” or, equivalently, “K is normal in G”. This is always a relation between a subgroup K and some subgroup G which contains it. It is not a transitive relation. It is quite possible for a group G to possess subgroups L and K, with L K and K G, for which L is not normal in G.12 Two immediate consequences of normality are the following: 11 Note the convention of regarding automorphisms as right operators whose action is denoted exponentially (see page 108). 12 In the group G of rigid rotations of a cube (Example 1 part (g) of Section 3.2), the 180o rotation about one of the three face-centered axes, generates a subgroup L which pointwise fixes its axis of rotation, but inverts the other two face-centered axes. This a normal subgoup of the abelian subgroup K of all rotations of the cube which leave each 118 Corollary 62 If N is a normal subgroup of the group G, and H is any subgroup of G, then 1. N ∩ H is a normal subgroup of H. (As a special case, if H contains N , then N is normal in H.) 2. N H = HN is a subgroup of G. The proof is left as an exercise for the beginning student. If the condition of Part 3. of the above Lemma 61 is true, one sees that the subset xK is in fact the set Kx, for each x ∈ G. But it also asserts that Kx · Ky = Kxy as subsets of G. Thus there is actually a multiplicative group G/K whose elements are the right cosets of K in G, mutiplication unambiguously given by the rule (Kx) · (Ky) = Kxy. We have a name for the group of cosets of a normal subgroup N . It is called the factor group G/N . Now let f : G → H be a homomorphism of groups. Two special groups associated with a group homomorphism f have been introduced earlier in this Chapter: the range or image, f (G), and the kernel, ker f . Now if y ∈ ker f , and x ∈ G, we see that f (x−1 yx) = (f (x))−1 f (y)f (x) = (f (x))−1 · 1 · f (x) = 1 ∈ H since f (y) = 1. Thus for all y ∈ ker f , and x ∈ G, x−1 yx ∈ ker f so ker f is always a normal subgroup. What about the corresponding factor group? Theorem 63 (The Fundamental Theorem of Homomorphisms) 1. If K is a normal subgroup of a group G, the mapping G → G/K which maps each element x of G to the right coset Kx which contains it, is an epimorphism of groups. Its kernel is K. of the three face-centered axes invariant. Then K is normal in G, being the kernel of the homomorphism of Example 3 Part (d) of Section 3.3. But clearly L is not normal in G, since otherwise its unique pointwise-fixed face-centered axis, would also be fixed. But it is not fixed as G/K induces all permutations of Sym(3) on these axes. 119 2. If f : G → H is a homomorphism of groups, there is an isomorphism η : f (G) → G/(ker f ). Thus every homomorphic image f (G) is isomorphic to a factor group of G. proof: 1. That the mapping ν : G → G/K defined by ν(x) := Kx, is a homomorphism follows from KxKy = Kxy, for all (x, y) ∈ G × G. By definition of G/K it is onto. Since the coset K = K · 1 is the identity element of G/K, the kernel of ν is precisely the subset {x ∈ G|Kx = K} = {x ∈ G|x ∈ K} = K . Thus ker ν = K. 2. Now set ker f := K. We propose a mapping η : f (G) → G/K which takes an image element f (x) to the coset Kx. Since this recipe is formulated in terms of a single element x, we must show that the proposed mapping is well-defined. Suppose f (x) = f (y). We wish to show Kx = Ky. But the hypothesis shows that f (x−1 y) = f (x−1 )f (y) = (f (x))−1 f (x) = 1 ∈ H, so x−1 y ∈ ker f = K, whence Kx = Ky, as desired. Now η(f (x)f (y)) = η(f (xy)) = Kxy = KxKy = η(f (x)) · η(f (y)). (3.7) (3.8) Thus η is a homomorphism. If f (x) were in the kernel of η, then η(f (x)) = Kx = K, the identity element of G/K. Thus x ∈ K, so f (x) = f (1), the identity element of f (G). Finally, η is onto, since, for any g ∈ G, the coset Kg = η(f (g)). Thus η is a bijection. The proof is complete There are two further consequences, contained in the following Corollary 64 (The So-called Second and Third Fundamental Theorems of Homorphisms) 1. Suppose K and N are both normal subgroups of G, with K contained in N . Then N/K is a normal subgroup of G/K, and G/N is isomorphic to the factor group (G/K)/(N/K). 2. If N is a normal subgroup of G, and H is any subgroup of G, HN/N is isomorphic to H/(H ∩ N ). proof: 1. By Corollary 62, part 1, K is normal in N , so N/K is the multiplicative group of cosets {Kn|n ∈ N }. For any coset Kx of G/K, and coset 120 Kn of N/K, (Kx)−1 KnKx = K(x−1 nx), which is in N/K. Thus N/K G/K. Thus there is a natural epimorphism f2 : G/K → (G/K)/(N/K) onto the factor group as described in the Fundamental Theorem (Theorem 63, part 1). By the same token, there is a canonical epimorphism f1 : G → G/N . By Lemma 59 of Section 3.2. the composition of these epimorphisms is again an epimorphism: f2 ◦ f1 : G → (G/K)/(N/K). An element x of G is first mapped to the coset Kx and then to the coset (Kx){Kn|n ∈ N } (which, viewed as a set of elements of G is just N n). Thus x maps to the identity element (N/K)/(N/K) of (G/K)/(N/K) if and only if x ∈ N . Thus the kernel of epimorphism f2 ◦f1 is N . The result now follows from the Fundamental Theorem of Homorphisms (Theorem 63 Part 2.). 2. Clearly N HN and N ∩ H H by Corollary 62, part1. We propose to define a morphism f : H → HN/N by sending element h of H to coset N h ∈ HN/N . Clearly hh0 is sent to N hh0 = (N h)(N h0 ), so f is a group homomorphism. Since every coset of N in HN = N H has the form N h for some h ∈ H, f is an epimorphism of groups. Now if h ∈ H, then N h = N , the identity of HN/N , if and only if h ∈ H ∩N . Thus ker f = H ∩N and now the conclusion follows from the Fundamental Theorem of Homomorphisms. 3.4.1 Miscellaneous Exercises. Exercise 1. Suppose N is a normal subgroup of the group G. 1. Show that there is an isomorphism between the poset of all subgroups H of G which contain N , and the poset of all subgroups of G/N (both posets partially ordered by the containment relation). [Hint: The isomorphism takes H in the first poset to H/N in the second.] 2. Show that H is normal in G if and only if H/N is normal in G/N . Thus the isomorphism of part 1 and its inverse both preserve the normality relation. (Note that they need not preserve the property of being characteristic in either direction. The group H containing N is called the inverse image of H/N .) 3. Using the fundamental theorem of homomorphisms conclude that if f :G→L 121 is an epimorphism of groups, then there is a 1-1 correspondence of the subgroups of L with the subgroups of G containing ker f preserving containment and normality. The correspondence takes a subgroup L1 of L to the subset {g ∈ G|f (g) ∈ L1 }, also called the inverse image of L1 . Exercise 2. If X is a subset of the group G (X is not necessarily a subgroup), define the centralizer in G of X to be the set of all group elements g of G such that gx = xg for all x ∈ X. This set is denoted CG (X). Similarly, the normalizer in G of X is the set of elements {g ∈ G|g −1 Xg = X}. Such elements may permute the elements of X by conjugation, but they stabilize X as a set. A subgroup H is said to normalize X if and only if H ≤ NG (X). If NG (X) = G, then X is called a normal set in G. 1. Show that CG (X) and NG (X) are subgoups of G. 2. Show that if X is a normal set in G, then CG (X) is also normal in G. 3. Show that if N is a normal subgroup of G, then the inner automorphism ψg : x → g −1 xg induces an automorphism of N . Exercise 3. 1. Conclude that a characteristic subgroup of a normal subgroup of G is normal in G, that is, K char N G implies K G. 2. Conclude that a characteristic subgroup of a characteristic subgroup of G is characteristic in G, that is, L char K char G implies L char G. 3. Show that the mapping which sends element g to ψg |N , the restriction of the inner automorphism conjugation-by-g to N , defines a group homomorphism ψ N : G → Aut(N ) whose kernel is CG (N ). Conclude that the group of automorphisms induced on N by the inner automorphisms of G is isomorphic to G/CG (N ). Exercise 4. Show that for any group G, Inn(G)Aut(G). (The factor group Out(G) := Aut(G)/Inn(G) is called the outer automorphism group of G.) Exercise 5. Prove the assertions of Corollary 62. 122 Exercise 6. A group is said to be a simple group if and only if its only proper normal subgroup is the identity subgroup. (Note that the definition forbids the identity group to be a simple group.) 1. Prove that any group of prime order is a simple group. 2. Suppose G is a group. A subgroup M is a maximal subgroup of G if and only if it is maximal in the poset of all proper subgroups of G. A subgroup M is a maximal normal subgroup of G if and only if it is maximal in the poset of all proper normal subgroups of G. (Note that it does not mean that it is a maximal subgroup which is normal. A maximal normal subgroup may very well not be normal.) Prove that a factor group G/N of G is a simple group if and only if N is a maximal normal subgroup. [Hint: Use the result of Exercise 1. just above.] 3.5 Direct products. A direct product is a formal construction for getting new groups in a rather easy way. First let G = {Gσ |σ ∈ I} be a family of groups indexed by the index set I. We can write elements of the cartesian product of the Gσ as functions [ f : I → σ∈I Gσ such that f (σ) ∈ Gσ . Of course, when I is countable, we can represent elements f in the usual way, as sequences (f (1), f (2), . . .). Define a binary operation on the cartesian product by declaring the “product” f1 f2 of f1 and f2 to be the function whose value (f1 f2 )(σ) at σ ∈ I is f1 (σ)f2 (σ) – that is, the values fi (σ) at the “coordinate” σ are multiplied (in Gσ ) to yield the σ-coordinate of the “product”. This is termed “coordinatewise multiplication”, since for sequences, the product of (a1 , a2 , . . .) and (b1 , b2 , . . .) is (a1 b1 , a2 b2 , . . .), by this definition. The cartesian product with this coordinatewise multiplication clearly forms a group, which we call the direct product over G and is denoted by Y Y G (G ) or σ G σ σ∈I or, when I = {1, 2, . . . , n}, simply by G1 × G2 × · · · × Gn . 123 It contains a subgroup consisting of all maps f in the definition of direct product, for which f (σ) fails to be the identity element of the group Gσ only finitely many times. This subgroup is called the weak direct product or direct sum over G and is denoted X (Gσ ) or σ∈I X G (Gσ ). When I is finite, there is no distinction between the direct product and direct sum. P For every index τ in I, there is clearly a homomorphism πτ : G (Gσ ) → Gτ which takes f to f (τ ), for all f in the direct product. Such an epimorphism is called a projection onto the coordinate indexed by τ . This epimorphism retains this name even when it is restricted to the direct sum. Any permutation of the index set induces an obvious isomorphism of direct products and direct sums. Thus G1 × G2 is isomorphic to G2 × G1 even though they are not formally the same group. Similarly, any (legitimate) rearrangement of parentheses involved in constructing direct products yields isomorphisms; specifically G1 × G2 × G3 ' (G1 × G2 ) × G3 ' G1 × (G2 × G3 ) Now suppose A and B are two subgroups of a group G with A ∩ B = 1, and B normalizing A (that is, B ≤ NG (A)). Then together they generate a subgroup AB with order |A| · |B| (Corollary 55, part 2, and Lemma 57). Now suppose in addition that A normalizes B. Then for any element a ∈ A and element b ∈ B, the factorizations (aba−1 )b−1 = a(ba−1 b−1 ) show that aba−1 b−1 ∈ A ∩ B = {1} and so ab = ba. Thus all elements of A commute with all elements of B. In that case the mapping A × B → AB which takes (a, b) ∈ A × B to ab, is a group homomorphism whose kernel is {(x, x−1 )|x ∈ A ∩ B}. Since A ∩ B = {1}, this map is an isomorphism. This is generalized below. Summarizing, Lemma 65 1. For any permutation π in Sym(I), There is an isomorphism G1 × G2 × · · · × Gn → Gπ(1) × Gπ(2) × · · · × Gπ(n) although neither of the groups are necessarily formally the same. 124 2. Any two well-formed groupings of the factors of a direct product into parentheses yields groups which are isomorphic to the original direct product and hence are isomorphic to each other. 3. (Internal Direct Products) Suppose A1 , A2 , . . . is a countable sequence of subgroups of a group G. (a) Suppose Aj normalizes Ai whenever i 6= j, and (b) A1 A2 · · · Ak−1 ∩ Ak = 1, for all k ≥ 2. Then A1 A2 · · · An is isomorphic to the direct product A1 × · · · × An for any finite initial segment {A1 , . . . , An } of the sequence of subgroups. 3.6 3.6.1 Other Variations: Semidirect Products and Subdirect Products. Semidirect products. Suppose N and H are two abstract groups and that we are given a homomorphism f : H → Aut(N ). Then f defines a formal construction of a group N : H , which is called semidirect product of N by H.13 The elements of N.H are the elements of the cartesian product N × H. Multiplication proceeds according to this rule: For any two elements (n1 , h1 ) and (n2 , h2 ) in N × H: −1 (n1 , h1 )(n2 , h2 ) := (n1 · (n2 )f (h1 ) , h1 h2 ). Verification of the associative law for a triple product (n1 , h1 )(n2 , h2 )(n3 , h3 ) comes down to the calculation f ((h1 h2 )−1 ) n3 −1 f (h−1 2 ) f (h1 ) = (n3 ) . Example 4. 13 In some older books there is other notation for the semidirect product – some sort of decoration added to the “” symbol. The notation N : H is the notation for a “split extension of groups” used in the Atlas for Finite Groups [?]. 125 1. If f : H → Aut(N ) is the trivial homomorphism – that is, f maps every element of H to the identity automorphism of N – then N H is just the ordinary direct product N × H of the previous section. 2. If A and B are subgroups of a group G for which A normalizes B and A ∩ B = 1, then the subgroup AB of G, is isomorphic to the semidirect product B.A with respect to the morphism f : A → Aut(B) which takes each element a of A to the automorphism of B induced by conjugating the elements of B by a – that is ψa |B . 3. Let N be the additive group of integers mod n, and let H be any subgroup of the multiplicative group of residues coprime to n (for example, the quadratic residues coprime to n). Let G be the group of all permutations of N of the form µ(m, h) : x → hx + m, for all x ∈ N. Then G is the semidirect product N : H. 4. Let F be any field, let F ∗ be the multiplicative group of all nonzero elements of F , and let H be any subgroup of F ∗ . Then the group of all transformation of F of the form x → hx + a, a ∈ F, h ∈ H, is the semidirect product (F, +) : H, where (F, +) denotes the group on F whose operation is addition in the field F . 5. (The Frobenius group of order 21.) The group is generated by elements x and y subject only to the relations x3 = 1, y 7 = 1, x−1 yx = y 2 and is a semidirect product Z7 : Z3 . (This is an example of a “presented group”, which we shall meet in Chapter 6.) 6. A group is said to be a normal extension of a group N by H (written G = H.N ) if and only if N G and G/N ' H. 126 The extension is said to be split of and only of there is a subgroup H1 of G such that G = N H1 , N ∩ H1 = 1. In this case, H1 ' H. One can see that G is a split extension of G by H if and only if G is a semidirect product of N by H. First, if the extension is split, every element of G has the form g = nh, (n, h) ∈ N × H. Because N ∩ H1 = 1, the expression g = nh is unique for a given element g. We thus have a bijection σ : G → N × H1 which takes g = nh to (n, τ (h)), where τ is the isomorphism H1 → H = G/N . Now if we multiply two elements of G, we typically obtain −1 (h1 (n1 h1 )(n2 h2 ) = n1 (h1 n2 h−1 1 )h1 h2 = n1 (n2 ) ) · h1 h2 , where, by our convention on inner automorphisms, g x := x−1 gx. The σ-value of this element is −1 ((n1 )(n2 )(h1 ) , τ (h1 )τ (h2 )). Thus σ is an isomorphism of G with the semidirect product of N by H, defined by composing τ −1 with the homomorphism H1 → Aut(N ) induced by conjugation. (The latter is just the restriction to H1 of the homomorphism ψ of Example 3, part (i).) Conversely, if G is the semidirect product of N by H defined by some homomorphism ρ : H → Aut(N ), then, setting N0 := {(n, 1)|n ∈ M } H0 := {(1, h)|h ∈ H} then N0 G = N0 H0 and N0 ∩ H0 = 1, so G is the split extension N0 by H0 and conjugation in G by element h0 of H0 induces automorphism ρ(h0 ) on N0 . (Clearly, N0 ' N and H0 ' H.) 127 3.6.2 Subdirect products. Something a little less precise and more general is the notion of a subdirect product. We say that a group H is a subdirect product of the groups Q {Gσ |σ ∈ I} if and only if (i) H is a subgroup of the direct product I Gσ and (ii) for each index τ , the restriction of the projection map πτ to H is onto – i.e. πτ (H) = Gτ . There may be many such subgroups, so the isomorphism type of H is not determined by (i) and (ii) above. Subdirect products naturally arise in this situation: Suppose M and N are normal subgroups of a group G. Then G/(M ∩ N ) is a subdirect product of G/N and G/M . This is because G/(M ∩ N ) can be embedded as a subgroup of (G/M )×(G/N ) by mapping each coset (N ∩M )x to the ordered pair (M x, N x). In fact something very general happens: If {Nσ |σ ∈ I} is a family of normal subgroups of a group G, then • G/(∩I Nσ ) is the subdirect product of the groups {G/Nσ |σ ∈ I}. • If F is “closed under taking subdirect products” – that is, any subdirect product of members of F is in F (for example the class of abelian groups is closed under subdirect products) – then there is a “smallest” normal subgroup GF yielding a factor G/GF in F. Precisely G/GF ∈ F and if N is normal with G/N ∈ F, then GF ≤ N . 3.7 Minimal normal subgroups of finite groups. In this section the reader is reminded of the following definitions and result from Exercise 6, page 123: 1. A simple group is a group G whose only proper normal subgroup is the identity subgroup. 2. A maximal normal subgroup of a group G is a maximal element in the partially ordered set of proper normal subgroups of G (ordered by inclusion, of course). 3. A factor group G/N of G is a simple group if and only if N is a maximal normal subgroup of G. We begin with a an elementary Lemma 128 Lemma 66 Suppose {Mi |i ∈ I} is a finite collection of maximal normal subgroups of a group G. Then for some subset J of I, G/(∩i∈I Mi ) = G/(∩j∈J Mj ) ' X j∈J (G/Mj ), a direct product of simple groups. proof: The result is true for |I| = 1, as G/M1 is simple. We use induction on |I|. Renumbering the subscripts if necessary, the induction hypothesis allows us to assume that for some subset J = {1, . . . , d} ⊆ I 0 = {1, . . . , k − 1}, N : = ∩i∈I 0 Mi = ∩j∈J Mj . G/N ' (G/M1 ) ⊕ · · · ⊕ (G/Md ). (3.9) (3.10) We set I := {1, . . . , k} = I 0 + {k} and assume, without loss, that Mk is not contained in N . Then since N Mk is a normal subgroup of G properly containing Mk , we must have G = N Mk . Then G/(N ∩ Mk ) ' (G/N ) ⊕ (G/Mk ), and the result for |I| = k follows upon substitution for G/N in the right-hand side. One may conclude Corollary 67 1. Suppose G is a finite group with no non-trivial proper characteristic subgroups. Then G is a direct product of pairwise isomorphic simple groups. 2. If N is a minimal normal subgroup of a finite group G, then N is a direct product of isomorphic simple groups. proof: Part 1. Let M be a maximal normal subgroup of the finite group G. Then G has a finite automorphism group, and so {M σ |σ ∈ Aut(G)} is a finite collection of maximal normal subgroups of G whose intersection N is a characteristic subgroup of G properly contained in G. By hypothesis, 129 N = 1. Then by the above Lemma 66, G = G/N is the direct product of of simple groups. If G = 1, we have an empty direct product and there is nothing to prove. Otherwise, the product of those direct factors isomorphic to the first direct factor clearly form a non-trivial characteristic subgroup, which, by hypothesis must be the whole group. Part 2. Since N is a minimal normal subgroup of the finite group G, N is a non-trivial finite group with only the identity subgroup as a proper characteristic subgroup, so the conclusion of Part 1. holds for N . 130 Chapter 4 Permutation Groups and Group Actions 4.1 Group actions and orbits As defined earlier, a permutation of a set X is a bijective mapping X → X. Because we will be dealing with groups of permutations it will be convenient to use the “exponential ” notation which casts permutations in the role of right operators. Thus if π is a permutation of X, we write xπ for the image of element x under the permutation π. Recall that composition of bijections and inverses of bijections are bijections, so that the set of all bijections of set X into itself form a group under composition which we have called the symmetric group of X and have denoted Sym(X). Recall also that any bijection ν : X → Y induces a group isomorphism ν̄ : Sym(X) → Sym(Y ) by the rule: ν(xπ ) = (ν(x))ν̄(π) . In order to emphasize the distinction between the elements of Sym(X) and the elements of the set X which are being permuted, we often refer to the elements of X by the neutral name “letters”. In view of the isomorphism just recorded, the symmetric group on any finite set of n elements can be thought of as permuting the set of symbols (or ‘letters’) Ωn := {1, 2, . . . , n}. This group is denoted Sym(n) and has order n!. Now suppose H is any subgroup of Sym(X). Say for the moment that two elements of X are H-related if and only if there exists an element of H which takes one to the other. It is easy to see that “H-relatedness” is 131 an equivalence relation on the elements of X. The equivalence classes with respect to this relation are called the H-orbits of X. Within an H-orbit, it is possible to move from any one letter to any other, by means of an element of H. Thus X is partitioned into H-orbits. We say H is transitive on X if and only if X comprises a single H-orbit. We now extend these notions in a very useful way. We say that a group G acts on set X if and only if there is a group homomorphism f : G → Sym(X). We refer to both f and the image f (G) as the action of G, and we shall borrow almost any adjective of a subgroup f (G) of Sym(X), and apply it to G. Thus we call an f (G)-orbit, a G-orbit, we say “G acts in k orbits on X” if X partitions into exactly k f (G)-orbits, and we say “G is transitive on X”, or “acts transitively on X” if and only if f (G) is transitive on X. If the action f is understood, we write xg instead of xf (g) . Also the unique G-orbit containing letter x is the set {xg |g ∈ G}, and is denoted xG . It’s cardinality is called the length of the orbit. The group action is said to be faithful if and only if ker f : G → Sym(X) = 1 ∈ G – that is, f is an embedding of G as a subgroup of the symmetric group. The power of group actions derives from the fact that the same group can have several actions, each of which can yield new information about a group. We have already met several examples of group actions, although not in this current language. We list a few Examples of Group Actions 1. Recall that each rigid rotation of a regular cube induces a permutation on the following sets: (a) the four vertex-centered axes. (b) the three face-centered axes. (c) the six edge-centered axes. (d) the six faces. (e) the eight vertices. (f) the twelve edges. 132 If Y is any one of the six sets just listed, and G is the full group of rigid rotations of the cube, then we obtain a transitive action fY : G → Sym(Y ) in each of these cases. Except for the case Y is the three perpendicular face-centered axes (Case (b)), the action is faithful. The action is an epimorphism onto the symmetric group only in Cases (a) and (b). 2. If N is a normal subgroup of G, then G acts on the elements of N by inner automorphisms of G. Thus ng := g −1 ng for each element g of G and each element n of N , to give the group action: f : G → Sym(N ). Clearly the identity element of G forms one of the G-orbits and has length one. If N = G, the G-orbits xG := {g −1 xg|g ∈ G}, x ∈ G are called conjugacy classes of G. If one is so inclined, one can then myopically concentrate on the transitive action G → Sym(xG ) of a single conjugacy class xG . 3. The image H σ of any subgroup H of G under an automorphism σ is again a subgroup of G, and this is certainly true of inner automorphisms. Thus H g := g −1 Hg is a subgroup of G called a G-conjugate of H or just conjugate of H. Thus there is an action G → Sym(S(G)) where S(G) is the collection of all subgroups of G. As in the previous case, this restricts to the transitive action G → Sym(H G ) on the set H G := {H g |g ∈ G} of all conjugates of the subgroup H. 4. Finally we can take any fixed subset X of elements of G and watch the action G → Sym(X G ) on the collection X G := {g −1 Xg|g ∈ G} of all G-conjugates of X. 5. Let Pk = Pk (V ) be the collection of all k-dimensional subspaces of the vector space V , where k is a natural number. We assume dim V ≥ k to avoid the possibility that Pk is empty. Since an invertible linear transformation must preserve the dimension of any finite dimensional subspace, we obtain an action f : GL(V ) → Sym(Pk ). It is necessarily transitive. 133 6. Similarly, if we have an action f : G → Sym(X), where |X| ≥ k, we also inherit an action fY : G → Sym(Y ) on the following sets Y : (a) The set X (k) of ordered k-tuples of elements of X. (b) the set X (k)∗ of ordered k-tuples with pairwise distinct entries. (c) the set X(k) of all (unordered) subsets of X of size k. 4.2 Permutations. A permutation π ∈ Sym(X) is said to displace letter x ∈ X if and only if xπ 6= x. A permutaion is said to be finitary if and only if it displaces a finite number of letters. The products among, and inverses and conjugates of finitary permutations is always finitary, and so the set of all finitary permutations always form a normal subgroup of the symmetric group. Any finitary permutation obviously has finite order. In this subsection we shall study certain arithmetic properties of group actions which are mostly useful for finite groups. 4.2.1 Cycle Notation. Now suppose π is a finitary permutation in Sym(X). Letting H be the cyclic subgroup of Sym(X) generated by π, we may partition X into Horbits. By our self-imposed finitary hypothesis, there are only finitely many H-orbits of length greater than one. Let O be one of these, say of length n. Then for any letter x in O, we see that O = xH must consist of the set 2 n−1 {x, xπ , xπ , . . . xπ }. We represent this permutation of O by the symbol: 2 (x xπ xπ . . . xπ n−1 ) called a cycle. Stated in this generality, the notation is not very impressive. But in specific instances it is quite useful. For example the notation (1 2 4 7 6 3) denotes a permutation which takes 1 to 2, takes 2 to 4, takes 4 to 7, takes 7 to 6, takes 6 to 3, and takes 3 to 1. Thus everyone is moved forward one position along a circular trail indicated by the cycle. In particular the cycle notation is not unique since you can begin anywhere in the cycle: thus (7 6 3 1 2 4) represents the same permutation just given. Moreover, 134 writing the cycle with the reverse orientation of the circuit yields the inverse permutation, (4 2 1 3 6 7) in this example. Now the generator π of the cyclic group H acts on each H-orbit as a cycle. This can be restated as the assertion that any finitary permutation can be represented as a product of disjoint cycles, that is, cycles which pairwise displace no common letter. Such cycles commute with one another so it doesn’t matter in which order the cycles are written. Also, if the set X is known, the permutation is determined only by its list of disjoint cycles of length greater than one – the cycles of length one (indicating letters that are fixed need not be mentioned. Thus in Sym(9), the folloing permutations are the same: (14597)(26) = (62)(14597)(8). Now to multiply two such permutations we simply compose the two permutations – and here the order of the factors does matter. (It is absolutely necessary that the student learn to compute these compositions.) Suppose π = (7 6 1)(3 4 8)(2 5 9) σ = (7, 10, 11)(9 2 4). (4.1) (4.2) Then applying π first and σ second. the composition of these right operators is πσ = (1, 10, 11, 7 6)(2 5)(3 9 4 8). We simply compute hπσi-orbits.1 The computation begins with an involved letter – say “1” – and asks what πσ did to it? 1π = 7 and 7σ = 10, so 1πσ = 10. Then it asks what πσ did to 10 in turn? One gets 10πσ = 11, 11πσ = 7,and so on, until the hπσi-orbit is completed upon returning to 1. One then looks for a new letter displaced by at least one of the two permutations, but which is not involved in the orbit just calculated – say, “2” – and repeats the process. We make this observation: Lemma 68 The order of a finitary permutation expressed as a product of disjoint cycles is the least common multiple of the lengths of those cycles. proof: This is an immediate consequence of two facts: (i) disjoint cycles commute, and (ii) and the order of a cycle is its length. (See Section 2.3 for 1 I will follow Burnside in explaining that commas are introduced into the cycle notation only for the specific purpose of distinguishing a 2-or-more-digit entry from other entries. 135 the fact that if x and y are commuting elements of finite order in a group, then o(xy) = lcm(o(x), o(y)).) 4.2.2 Even and odd permutations. We begin with a technical result: Lemma 69 If k and l are non-negative integers, then 1. (a b)(a x1 . . . xk b y1 . . . yl ) = (a y1 . . . yl )(b x1 . . . xk ), and 2. (a b)(a y1 . . . yl )(b x1 . . . xk ) = (a x1 . . . xk b y1 . . . yl ). proof: The permutation on the left side of the first equation sends a to y1 , sends yi to yi+1 for i < l, and yl to a. Similarly it sends b to x1 , sends xj to xj+1 for j < k, and sends xk to b. Thus the left side has been expressed as the product of the two disjoint cycles on the right side of the first equation. The second equation follows from the first my multiplying both sides of the first equation by the 2-cycle (a b). Let FinSym(X) denote the group of all finitary permutations of the set of ‘letters’ X. Each element π of FinSym(X) is expressible as a finite product of disjoint cycles of lengths d1 , . . . dk greater than one, together with arbitrarily many cycles of length one. Define the sign of π to be the number Yj=k j=1 (−1)dj −1 . Then the function sgn : FinSym(X) → {±1} is well-defined. A permutation of Sym(X) which displaces exactly two letters is called a transposition. Thus every transposition can be expressed in cycle notation by (a b), for the two displaced letters, a and b. Clearly, the sign of a transposition is (−1)2−1 = −1. We now can state Lemma 70 If g is a permutation in FinSym(X), and if t is a transposition, then sgn(tg) = −sgn(g) = sgn(gt). 136 proof: The permutation g is a product of disjoint cycles cj of length dj . Q Thus sgn(g) = j (−1)dj −1 . Suppose t is the transposition (a b). If the letters a and b are not involved in any of the cycles gi , then the result follows from the formula for the sign of gt = tg since we have simply tacked on the disjoint cycle (a b) in forming gt. Suppose on the other hand, that a and b appear in the same hgi-orbit, say the one represented by the cycle gj . Then gj has the form (a x1 . . . xk b y1 . . . yl ), where k and l are non-negative. By Lemma 69, tgj = (a y1 . . . yl )(b x1 . . . xk ). Thus dj = k + l + 2 so sgn(gj ) = (−1)k+l+1 while sgn(tgj ) = (−1)k (−1)l , so sgn(tgj ) = −sgn(gj ). Since tg = g1 · · · (tgj ) · · · gi · · ·, the result follows from the formula. Now suppose a occurs in one cycle, say g1 = (a x1 , . . . xk ) and b occurs in another cycle, which we can take to be g2 = (b y1 . . . yl ) . Then by Lemma 69, tg = (tg1 g2 )g3 · · · gm = (a y1 . . . yl b x1 . . . xk )g3 · · · gm . Thus Y sgn(tg) = sgn(tg1 g2 ) i>2 (−1)(di −1) Y = (−1)sgn(g1 )sgn(g2 ) i>2 (−1)(di −1) = −sgn(g). So the result follows in this case. Now any cycle (x1 . . . xm ) is a product of transpositions (x1 xk )(xk xk−1 ) · · · (x3 x2 ). Since each element of FinSym(X) is the product of disjoint cycles we see that FinSym(X) is generated by its transpositions. It now follows from Lemma 70 that no element can be both a product of an even number of transpositions as well as a product of an odd number of transpositions. Thus we have a partition FinSym(X) = A+ + A− 137 of all the elements of FinSym(X) into the set A+ of all finitary permutations which are a product of an even number of transpositions, and A− , all finitary permutations which are a product of an odd number of transpositions. Right multiplication by any finitary permutation at best permutes the two sets A+ and A− , and multiplication by a transposition clearly transposes them. Thus we have a transitive permutation action sgn : FinSym(X) → Sym({A+ , A− }). The right side is Sym(2) ' Z2 , isomorphic to the multiplicative group {±1}. The kernel of this action is the normal subgroup FinAlt(X) := A+ , the finitary alternating group. Its elements are called the even permutations. The elements in the other coset A− , are called odd permutations. These terms are used only for finititary permutations. If X is a finite set, there is no distinction gained by singleing out the finitary permutations, and the prefix “Fin” is dropped throughout. So in that case A+ is simply called the alternating group and is denoted Alt(X) or Alt(n) when |X| = n. The factorization of a cycle of length n given above shows that any cycle of even length is an odd permutation and any cycle of odd length is an even permutation. It only “sounds” confusing; by the formula, the sign of an n-cycle is (−1)n−1 . 4.2.3 Transpositions and Cycles: The Theorem of Feit, Lyndon and Scott Recall that a transposition of a symmetric group Sym(X) is an element which dispaces exactly two of the letters of the set X. These two letters must clearly exchange places, otherwise we are dealing with the identity permutation. Let T be any collection of transpositions of Sym(X). We can then construct an simple2 graph G(T ) := (X, T ) with vertex set X and edge set consisting of the transposed 2-subsets determined by each involution. 2 In the context of graphs (as opposed to groups) “simple” actually means something rather simple. A simple graph is one which is undirected without loops or multiple edges. That is edges connect only distinct vertice (no loops), there is at most one edge connecting two distinct vertices, and no orientation to the edge – i.e. edges are 2-subsets of the vertex set. 138 Lemma 71 Let T be a set of transpositions of Sym(X). Suppose G(T ) := hT iSym(x) , the subroup generated by the transpositions of T , acts on X with orbits Xσ of finite length. Then G(T ) ' X σI Sym(Xσ ) where Xσ are the G(T )-orbits on X. proof: First of all G(T ) is a finitary permutation group on X. Let Tσ be the set of transpositions which form an edge of the connected component Xσ of G(T ). It is clear that G(T ) is the direct sum of the groups Gσ := hTσ i. so that all that remains is to show is that Gσ acts on Xσ as the full symmetric group on Xσ , a question in finite groups of actions. So, without loss we assume that G(T ) is transitive on the finite set X of cardinality n. If |T | = 1, then |X| = 2 and there is nothing to prove. We may now proceed by induction on |T | and conclude that G(T ) is a tree. Now we are free to choose the transposition t ∈ T so that as an edge of the graph G(T ), t connects an “end point” a of valence one, and a vertex b. Thus G(T − {t}) is a tree on the vertices X − {a}, and so by induction the stabilizer in G(T ) of the vertex a has order (n − 1)!, and so as G(T ) is clearly transitive, it has order n!. Since it is faithful, (for its elements fix every single letter of every other orbit), we have here the full symmetric group. Theorem 72 (Feit, Lyndon and Scott) Let T be a minimal set of transpositions generating Sym(n). Then their product in any order is an n-cycle. proof: We have seen from above that the minimality of T implies that the graph G(T ) is a tree. Let a = t1 t2 · · · tn−1 be the product of these transpositions in any particular order. Consider t1 and remove the edge t1 from the graph G(T ) to obtain a new graph G 0 which is not connected, but has two connected component trees G1 and G2 , connected in G by the “bridging” edge t1 . Now by an obvious induction any order of multiplying the transpositions of Ti := {s ∈ T |s is an edge in Gi }, 139 i = 1, 2, . . ., yields an n -cycle. Thus a = t(x1 , . . . xk )(y1 , . . . yl ), where the second cycle b is the product of the transpositions of T1 as they are encountered in the factorization a = t1 · · · tn , and the third cycle c above is the product of the transpositions of T2 in the order in which they are encountered in the product given for a. (Recall that every transposition in T1 commutes with every transposition in T2 .) Now these two cycles involve two seperate orbits bridged by the transposition t1 . That the product t1 bc is an n-cycle follows directly from Lemma 69 Part 2. above. The proof is complete. This strange theorem will bear fruit in the proof of the Brauer-Ree Theorem, which takes a bow in our chapter on generation of groups (Chapter 6). 4.2.4 Action on right cosets. Let H be any subgroup of the group G, and let G/H be the collection{Hg|g ∈ G} of all right cosets of H in G. Note that we have extended out previous notation. G/H used to mean a group of right cosets of a normal subgroup H of G. It is still the same set (of right cosets of H in G) even when H is not normal in G. But it no longer has the structure of a group. Nonetheless, as we shall see, it still admits a right G-action. For any element g of G and coset Hx in G/H, Hxg is also a right coset of H in G. Moreover, Hxg = Hyg implies Hx = Hy, while Hx = (Hxg −1 )g. Thus right multiplication of all the right cosets of H in G by the element g induces a bijection πg : G/H → G/H. Moreover for elements g, h ∈ G, the identity Hx(gh) = ((Hx)g)h, forced by the associative law, implies πg πh = πgh as right operators on G/H. Thus we have a group action πH : G → Sym(G/H) which takes g to the permutation πg , induced by right multiplication of right cosets by g. This action is always transitive for coset Hx is mapped to coset Hy by πx−1 y . 140 4.2.5 Equivalent actions. Suppose a group G acts on sets X and Y – that is, there are group homomorphisms fX : G → Sym(X) and fY : G → Sym(Y ). We say that the two actions are eqivalent if and only if there is a bijection e : X → Y such that for each letter x ∈ X, and element g ∈ G, e(xfX (g) ) = (e(x))fY (g) . That is, the action is the same except that we have used the bijection e to “change the names of the elements of X to those of Y ”. 4.2.6 The Fundamental Theorem of Transitive Actions. First we make an observation: Lemma 73 Let f : G → Sym(X) be a transitive action of the group G. For each letter x ∈ X, let Gx be the subgroup of all elements of G which leave the letter x fixed – i.e. Gx := {g ∈ G|xg = x}. 1. If x, y ∈ X, then Gx and Gy are conjugate subgroups of G. Precisely, Gy = g −1 Gx g whenever xg = y. 2. The set {g ∈ G|xg = y} of all elements of G taking x to y, is a right coset of Gx . −1 −1 proof: 1. If xg = y, y g Gx g = xgg Gx g = xGx g = xg = y, so g −1 Gx g ⊆ Gy .. −1 But since x = y g , we have Gy ⊆ gGx g −1 , by the same token. Thus the first containment is reversible and g −1 Gx g = Gy . 2. If xg = y then clearly, xh = y for all h ∈ Gx g. Conversely, if xg = xh = y, then gh−1 ∈ Gx , so Gx g = Gx h. Theorem 74 (The Fundamental Theorem of Transitive Group Actions.) Suppose f : G → Sym(X) is a transitive group action. Then for any letter x in X, f is equivalalent to the action πGx : G → Sym(G/Gx ), of G on the right cosets of Gx by right multiplication. 141 proof: We must construct the bijection e : X → G/Gx compatible with both actions. For each letter y ∈ X set e(y) := {g ∈ G|xg = y}. We have seen in Lemma 73 that the latter is a right coset Gx h of Gx . Now for any element g in G, e(y g ) is the set of all elements of G which take x to y g . This clearly contains Gx hg; yet by Lemma 73, it must itself be a right coset of Gx . Thus e(y g ) is the right multiple e(y)g, establishing the compatibility of the actions. Corollary 75 1. If G acts on the set X, then every G-orbit has length dividing the order of G. 2. If G acts transitively on X, then for any subgroup H of G which also acts transitively on X, one has G = Gx H, for any x ∈ X. proof: The action of G on any G-orbit O is transitive. Applying the fundamental theorem we see that |O| = |G/Gx | for any letter x in O. 4.2.7 Normal subgroups of transitive groups. If G acts transitively on a set X, we say that G acts regularly on X if and only if for some x ∈ X, Gx = 1. Lemma 76 Suppose G acts transitively on X. 1. If N is a normal subgroup of G, then all N -orbits on X have the same length. 2. In particular, if N is a non-identity normal subgroup of G lying in a subgroup Gx , for some x in X, then N acts trivially on X, and the action is not faithful. 3. The faithful transitive action of an abelian group is always a regular action. 142 4. Suppose N is a normal sybroup of G which acts regularly on X. Then for any x ∈ X, G = Gx N , with Gx ∩ N = 1, and the action of Gx on X − {x} (by restriction) is equivalent to that action of Gx on N − {1}, the set of all non-identity elements, induced by conjugation of the elements of N by the elements of Gx . proof: 1. For any g ∈ G and x ∈ X, the mapping xN → (xN )g = xN g = xgN = (xg )N is a bijection of N -orbits. Since G is transitive, all N -orbits can be gotten this way. 2. This is immediate from 1, since N has an orbit of length 1. 3. If A is an abelian subgroup acting transitively on X, for any x ∈ X, Ax acts trivially on X by part 2. Since A acts faithfully, Ax = 1, so A has the regular action. 4. That G = Gx N and Gx ∩ N = 1 follows from Lemma 75, part 2, and the fact that N is regular. Since the normal subgroup N is regular, there is a bijection, ex : N → X = xN which sends element n ∈ N to xn . For each element h ∈ Gx , e(h−1 nh) = xh −1 nh = xn h = e(n)h . So e defines an equivalence of the action of Gx on N by conjugation and the action of Gx on X obtained by restriction. Throwing away the corresponding Gx -orbits {1} and {x} of length 1 yields the last statements of 4. 4.2.8 Double cosets. Let H and K be subgroups of a group G. Any subset of the form HgK is called an (H, K)-double coset. Such a set is, on the one hand a disjoint union of right cosets of H, and on the other, a disjoint union of left cosets of K. So it’s cardinality must be a common multiple of |H| and |K|. The precise cardinality is given in the following: Lemma 77 Let H and K be subgroups of G. 1. |HgK| = |H| · [K : K ∩ g −1 Hg] = |H||K|/|gKg −1 ∩ H| = [H : H ∩ gKg −1 ]|K|. 143 2. The (H, K)-double cosets partition the elements of G. proof: In the action πH of G on G/H by right multiplication, the K-orbits are precisely the right cosets of H within an (H, K)-double coset. Since G/K is partitioned by such orbits, and the right cosets of H partition G, part 2. follows. The length of a K-orbit on G/H, is the index of its subgroup fixing one of its “letters”, say Hg. This subgroup would be {k ∈ K|Hgk = Hg} = K ∩ gHg −1 }. Since each, right coset of H has |H| elements of G, the second and third terms of the equations in Part 1. give |HgK|. The last term appears from the symmetry of H and K. 4.3 Applications of transitive group actions. For any subset X of a group G, the normalizer of X in G is the set of all elements NG (X) := {g ∈ G|g −1 Xg = X}. Now G acts transitively by conjugation on X G = {g −1 XG g ∈ G}, the set of distinct conjugates of X, and the normalizer NG (X) is just the subgroup fixing one of the “letters” of X G . Thus Lemma 78 1. Let X be a subset of G. The number |X G | of distinct conjugates of X in G is the index [G : NG (X)] of the normalizer of X in G. This holds when X is a subgroup, as well. 2. The size of a conjugacy class xG in G is the index of the centralizer CG (x) in G. 4.3.1 Finite p-groups. A finite p-group is a group whose order is a power of a prime number p. The trivial group of order 1 is always a p-group. We have Lemma 79 144 1. If a p-group acts on a set X of cardinality not divisible by the prime p, then it fixes a letter. 2. A non-trivial finite p-group has a nontrivial center. 3. If H is a proper subgroup of a finite p-group P , then H is properly contained in its normalizer in P , that is NP (H). proof: 1. Let P be a finite p-group, say of order pn . Then any P -orbit on X has length a power of p. Thus if no orbit had length 1, p would divide |X|, since X partitions into such orbits. But as p does not divide |X|, some P -orbit must have length 1, implying the conclusion. 2. Let P be as in part 1, but of order pn > 1. Now P acts by conjugation on the set P − {1} of pn − 1 non-identity elements. Since p does not divide the cardinality of this set, part 1 implies there is a non-identity element z left fixed by this action. That is, g −1 zg = z, for all g ∈ P . Clearly z is a non-identity element of the center Z(P ), our conclusion. 3. Suppose H is a proper subgroup of a p-group P . By way of contradiction assme that H = NP (H). Then certainly, Z(P ) ≤ H. By Part 2. Z(P ) 6= 1. If H = Z(P ), the H P , a contradiction to H proper. Thus, under the homomorphism f : P → P̄ := P/Z(P ), H maps to a non-trivial proper subgroup f (H) := h̄ of the p-group P̄ . By induction on |P̄ |, NP̄ (H̄) properly contains H̄. But the preimage of the left side is {x ∈ P |x−1 Hx ⊆ some cosets of Z(H) in H = NP (H) ⊆ H. this contradicts NP (H) ≤ H. 4.3.2 The Sylow Theorems Theorem 80 (Sylow’s Theorem) Let G be a group of order n and let k be the highest power of the prime p such that pk divides the group order n. Then the following hold: 1. (Existence) G contains a subgroup P of order pk . 145 2. (Covering and Conjugacy) Every p-subgroup R of G is contained in some conjugate of P . In particular, any subgroup of order pk is conjugate in G to P . 3. (Arithmetic) The number of subgroups of order pk is congruent to 1 mod p. proof: 1. Let Σ denote the collection of!a subsets of G of cardinality pk . n The number of such subsets is |Σ| = . Since n − r and pk − r are both pk divisible by the same highest power of p, the number |Σ| is not divisible by the prime p. Now G acts on Σ by right multiplication – that is, if X ∈ Σ then Xg ∈ Σ, for any group element g. Thus Σ partitions into G-orbits under this action, and not all of these orbits can have length divisible by p, since p doesn’t divide |Σ|. Let Σ1 be such a G-orbit of length not divisible by p. By the Fundamental theorem of transitive group actions (Theorem 74), the number of sets in Σ1 is the index of the subgroup GX fixing the “letter” X in Σ1 . Since n = |G| = [G : GX ]|GX | = |Σ1 ||Gx |, we see that pk divides |GX |. But by definition, XGX = X, so X is a union of left cosets of GX , so |GX | divides |X| = pk . It follows that GX has order pk exactly. So GX is the desired subgroup P of statement 1. 2. Let R be any p-subgroup of G. Then R also acts on Σ1 by right multiplication. Since p does not divide |Σ1 |, Lemma 79, part 1., shows that R must fix a letter Y in Σ1 . Thus R ≤ GY . But as G is transitive on Σ1 , Lemma 73 shows that GY is conjugate to P = GX . Thus R lies in a conjugate of P as required. 3. Now let S denote the collection of all subgroups of G having order pk exactly. By parts 1. and 2., already proved, S is non-empty, and G acts transitively on S by conjugation. Thus by Lemma 78 part 1., |S| = [G : N ], where N := NG (P ), the normalizer in G of a subgroup P in S. Now P itself acts on S by conjugation, fixing itself, and acting on S − {P } in orbits of p-power length. Suppose {R} were such a P -orbit in S − {P } of length one. Then P normalizes R so P R is a subgroup of G. Then |P R| = |P |·[R : P ∩R], (Lemma 63, Part 2.) a product of pk and [R : P ∩ R], another power of p. Since |P R| divides n and pk is the largest p-power dividing n, one must conclude that [R : P ∩ R] = 1, which forces P = R, contrary to the choice of R in S{P }. Thus P acts on S − {P } in P -orbits of lengths divisible by p. This yields |S| ≡ 1 mod p. 146 The subgroups of G of maximal p-power order are called the p-sylow subgroups of G, and the single conjugacy class which they form is denoted Sylp (G). Corollary 81 (The Frattini Argument) Suppose N is a normal subgroup of the finite group G, and select P ∈ Sylp (N ). Then G = NG (P )N . proof: We know that conjugation by elements of G induces automorphisms of the normal subgoup N . Since Sylp (N ) is the full collection of subgroups of N of their order, G acts on Sylp (N ) by conjugation. But by Sylow’s Theorem (Theorem 80) the subgroup N is already transitive on Sylp (N ). The result now follows from Lemma 75, part 2. Corollary 82 1. Any normal p-Sylow subgroup of a finite group is in fact characteristic in it. 2. Any finite abelian group is the direct product of its p-sylow subgroups, that is, A ∼ S1 × S2 · · · × Sn , where Si is the pi -sylow subgoup of A, and {p1 , . . . , pn } lists all the distinct prime divisors of |A| exactly once each. Moreover, we have: Aut(A) ' Aut(S1 ) × Aut(S2 ) × · · · , Aut(Sn ). 3. The Euler phi-function is multiplicative, that is, φ(n) = φ(a)φ(b) whenever n = ab and gcd(a, b) = 1. proof: 1. Any normal p-sylow subgroup is the unique subgroup of its order, and hence is characteristic. 2. Since in a finite abelian group, each p-Sylow subgoup is normal, and has order prime to the direct product of the remaining r-sylow subgroups (r 6= p), A is the described internal direct product by Lemma 65 , part 3. Since by 1. each Si is characteristic in A, each automorphism σ of A 147 induces an automorphism σi of Si . But conversely, if we apply all possible automorphisms σi to the direct factor Si and the identity automorphism to all other p-sylow subgroups Sj , j 6= i, we obtain a subgroup Bi of Aut(A). Now the internal direct product characterization of Lemma 65 applies to the Bi to yield the conclusion. 3. Applying 2. when A is the cyclic group of order n, Zn , one obtains φ(n) = Yi=n when n has prime factorization n = ders. φ(pai i ) i=1 a1 a2 p1 p2 · · · pann , upon equating group or- Lemma 83 (The Tail-Wags-the-Dog Lemma.) Suppose Γ = (V, E) is a bipartite graph with vertices in two parts V1 and V2 , and each edge of E involving one vertex from V1 and one vertex from V2 . Suppose G is a group of automorphism of the graph Γ acting on the vertex set V with V1 and V2 as its two orbits. Then the following conditions are equivalent. 1. G acts transitively on the edges of Γ. 2. For each vertex v1 ∈ V1 , the subgroup G1 of G fixing v1 acts transitively on the edges on v1 3. For each vertex v2 in V2 , the subgroup G2 of G fixing v2 acts transitively on the edges of Γ on vertex v2 . proof: By the symmetry of the Vi , is suffices to prove the equivalence of 1. and 2. If G transitively permutes the edges, any edge on v1 must be moved to any other edge on v1 by some element g. But since v1 is the only vertex of the G-orbit V1 incident with these edges, such an element g fixes v1 . So 2. holds. Conversely, suppose the subgroup G1 transitively permutes the edges on v1 . Since G is transitive V1 , any edge meeting V1 at a single vertex can be taken to any other. But by hypothesis, all edges have this property. Theorem 84 (The Burnside Fusion Theorem) Let G be a finite group, and let X1 and X2 be two normal subsets of a p-sylow subgroup P of G. Then X1 and X2 are conjugate in G if and only if they are conjugate in the normalizer in G of P . In particular, any two elements of the center of P are conjugate if and only if they are conjugate in NG (P ). 148 proof: Obviously, if the Xi are conjugate in NG (P ), they are conjugate in G. So assume the Xi belong to X := X1G . We now form a graph whose vertex set is X ∪ S where S := Sylp (G). An edge will be any pair (Y, R) ∈ X × S for which the subset Y is a normal subset of the p-sylow group R. Then G acts by conjugation on the vertex set V with just two orbits, X and S (the first by construction, the second by Sylow’s Theorem). Moreover the conjugation action preserves the normal-subset relationship on X ×S, and so G acts as a group of automorphisms of our graph, which is clearly bipartite. Now by Sylow’s theorem, the subgroup G1 := NG (X1 ) acts transitively by conjugation on Sylp (G1 ). But since X1 is normal in some p-sylow subgroup (namely P ), Sylp (G1 ) ⊆ Sylp (G) Thus Sylp (G1 ) are the S-vertices of all the edges of our graph which lie on vertex X1 . We now have the hypothesis 2. of the Tail-Wags-the-Dog Lemma. So the assertion 3. of that lemma must hold. In this context, it asserts that the stabilizer G2 of a vertex of S – for example, NG (P ) – is transitive on the members of X which are normal in P . But that is exactly the desired conclusion. Theorem 85 (Thompson Transfer Theorem) Suppose N is a subgroup of index 2 in a 2-sylow subgroup S of a finite group G. Suppose t is an involution in S − N , which is not conjugate to any element of N . Then G contains a normal subgoup M of index 2. proof: Let G act on G/N , the set of right cosets of N in G by right multiplication. In this action, the involution t cannot fix a letter, for if N xt = N x, then xtx−1 ∈ N , contrary to assumption. Thus t acts as [G : S] 2-cycles on G/N . Since this makes t the product of an odd number of transpositions in Sym(G/N ), we see that the the image of G in the action f : G → Sym(G/N ) meets the alternating group Alt(G/N ) at a subgroup of index 2. Its preimage M is the desired normal subgroup of G. 4.3.3 Calculations of Group Orders. The Fundamental Theorem of Transitive Group Actions is guite useful in calculating the orders of the automorphism groups of various objects. We peruse a few examples here. 149 Figure 4.1: Two views of Petersen’s graph. (Petersen’s Graph) The reader is invited to verify that the two graphs which appear in Figure 4.1 are indeed isomorphic (vertices with corresponding numerical labels are matched under the isomorphism. Presenting them in these two ways reveals certain automorphisms. In the left hand figure, vertex 1 is adjacent to vertices 2,3, and 4. The remaining vertices form a hexagon, with three antipodal pairs, (5, 8), (6, 9) and (7, 10), which are connected to 5, 4 and 3, respectively. Any symmetry of the hexagon induces a permutation of these antipodal pairs, and hence induces a corresponding permutation of {2, 3, 4}. Thus, if we rotate the hexagon clockwise we obtain an automorphism of the graph which permutes the vertices as: y = (1)(2 3 4)(5 10 9 8 7 6). But if we reflect the graph about its vertical axis we obtain t = (1)(2 4)(3)(5 9)(6 8)(7)(10). Now any automorphism of the graph which fixes vertex 1 must induce some symmetry of the hexagon on the six vertices not adjacent to vertex 1. The automorphism inducing the identity, preserves the three antipodal pairs, and so fixes vertices 2,3 and 4, as well as vertex 1 – i.e. it is the identity element. Thus if G is the full automorphism of the Petersen graph, G1 is the dihedral group hy, ti of order 12, acting in orbits, {1}, {2, 3, 4}, {5, 6, 7, 8, 9, 10}. On the other hand, the right hand figure reveals an obvious automorphism of order five, namely u = (1 2 5 6 4)(3 8 10 7 9). That G is transitive comes from overlaying the three orbits of G1 , and the two orbits of hui of length five. Together, one can travel from any vertex to any other, by hopping a ride on each orbit, and changing orbits where they overlap. This forces [G : G1 ] = 10, the number of vertices. Since we have already determined |G1 | = 12, we have determined that the full 150 Figure 4.2: The configuration of the projective plane of order 2. automorphism group of this graph has order |G| = [G : G1 ]|G1 | = 10 · 12 = 120. (The Projective Plane of Order 2.) Now consider a system P of seven points, which we shall name by the integers {0, 1, 2, 3, 4, 5, 6}. We introduce a system L of seven triplets of points, which we call lines: L1 L2 L3 L4 L5 L6 L0 : : : : : : : = = = = = = = [1, 2, 4] [2, 3, 5] [3, 4, 6] [4, 5, 0] [5, 6, 1] [6, 0, 2] [0, 1, 3] This system is depicted by the six straight lines and the unique circle in Figure 4.3.3 above. Automorphisms of the system (P, L) are those permutaions of the seven points which take lines to lines – i.e. preserve the system L. We shall determine the order of the group G of all automorphisms. First, from the way lines were defined above, as translates mod 7 of L1 , we have an automorphism u = (0 1 2 3 4 5 6) : (L1 l2 L3 L4 L5 L6 L0 ). 151 Thus G is certainly transitive on points. Thus the subgroup G1 fixing point 1 has index 7. There are three lines on point 1, namely, L0 , L1 and L5 . There is an obvious reflection of the configuration of Figure 4.3.3 about the axis formed by line L5 , giving the automorphism s = (1)(6)(5)(0 4)(3 2) : (L0 L1 )(L3 L6 )(L2 )(L4 )(L5 ). But not quite so obvious is the fact that t = (0)(1)(3)(2 6)(4 5) induces the permutaion (L0 )(L1 L5 )(L2 L3 )((L4 )(L6 ) and so is an automorphism fixing 1 and L5 and transposing L0 and L1 . Thus G1 acts as the full symmetric group on the set {L0 , L − 1, L5 } of three lines on point 1. Consequently the kernel of this action is a normal subgroup K of G1 , at index 6. Now K stabilizes all three lines {L0 , L1 , L5 }, and contains r = (0 3)(1)(2 4)(5)(6) : (L0 )(L1 )(L2 L4 )(L3 L6 )(L5 ). So K acts as a transposition on the remaining points of L0 beyond point 1. Thus K in turn has a subgroup K1 fixing L0 pointwise, and stabilizing L1 and L5 . K1 , in turn, contains u = (0)(1)(2 4)(3)(5 6) : (L0 )(L1 )(L5 )(L2 L3 )(L4 L6 ). It in turn contains a subgroup of index 2 whose elements now fix 0, 1, 3, 5 and 6. One can see that any such element stabilizes every line (for every pair of fixed points determines a fixed line), and hence all intersections of such lines, and hence fixes all points. Thus K1 has order only 2. We now have |G| = [G : G1 ][G1 : K][K : K1 ]|K1 | = 7 · 6 · 2 · 2 = 168. 4.4 4.4.1 Primitive and Multiply Transitive Actions. Primitivity Suppose G acts on a set X. A system of imprimitivity for the action of G is a non-trivial partition of X which is stabilized by G. Thus X is a disjoint union of two or more proper subsets Xi , not all of cardinality one, such that for each g ∈ G, Xig = Xj for some j – that is, G permutes the components of the partition among themselves. Of course, if G is transitive in its action on X, the components Xi all have the same cardinality (evidently one dividing the cardinality of X), and are transitively permuted among themselves. 152 As an example, the symmetric group Sym(n) acts on all of its elements by right multiplication (the regular representation). But these elements were partitioned as Sym(n) = A+ ∪ A− , into those elements which were a product of an even number of transpositions (A+ ) and those which were a product of an odd number of transpositions (A− ). Right multiplication permutes these sets, and they form a system of imprimitivity when n > 2. We have seen that when N is a non-trivial normal intransitive subgroup of a group G acting faithfully on X, then the N -orbits form a system of imprimitivity (Lemma 76 considers the case of N -orbits when G acts transitively on X.) If a transitive group action preserves no system of primitivity, it is said to be a primitive group action. We understand this to mean that primitive groups are always transitive. The immediate characterization is this: Theorem 86 Suppose G acts transitively on X, where |X| > 1. Then G acts primitively if and only Gx is a maximal subgroup of G for some x in X. proof: Since G is transitive, all the subgroups Gx are conjugate in G (see Lemma 73, part 1). By the fundamental theorem of group actions 74, we may regard this action as that of right multiplication of the right cosets of Gx . Now if Gx < H < G, then each right coset of H is a union of right cosets of Gx . In this way the rights cosets of H partition G/Gx to form a system of imprimitivity. Thus in the imprimitive case, no such H exists, so Gx is G or is maximal in G. The former case is ruled out by |X| > 1. Conversely, if X = X1 + X2 + · · · is a system of imprimitivity, then all the Xi have the same cardinality and for x ∈ X1 , Gx stabilizes the set X1 . But by transitivity of G, if L = StabG (X1 ) is the stabilizer in G of the component X1 , then L must act transitively on X1 . Since |X1 | = 6 1, Gx is a proper subgroup of L. Since X1 6= X, L is a proper subgroup of G. Thus transitive but imprimitive action, forces Gx not to be maximal. That is, Gx maximal implies primitive action of G on G/Gx and hence X. We immediately have: Corollary 87 If G has primitive action on X, then for any normal subgroup N of G either (i) N acts trivially on X (so either N = 1 or the action is not faithful), or (ii) N is transitive on X. 153 4.4.2 The rank of a group action. Suppose G acts transitively on a set X. The rank of the group action is the number of orbits which the group G induces on the cartesian product X × X. Here an element g of G takes the ordered pair (x, y) of X × X to (xg , y g ). Since G is transitive, one such orbit is always the diagonal orbital D = {(x, x)|x ∈ X}. The total number of such “orbitals” – as the orbits on X × X are called – is the rank of the group action. Basically, the non-diagonal orbitals mark the possible binary relations on the set X which could be preserved by G. Such a relation is defined by asserting that letter x has relation Ri to letter y (which we write as XRi y) if and only if the ordered pair (x, y) is a member of the non-diagonal orbital Oi (G-orbit on X × X.) We could then represent this relation by a directed graph, Γi = (X, Ei ), where the ordered pair (x, y) is a directed edge of Ei if and only if xRi y holds. In that case G acts as a group of automorphisms of the graph Gi , – that is, G ≤ Aut(Γi ). If Oi is an orbital, there is also an orbital Oi∗ consisting of all pairs (y, x) for which (x, y) belongs to Oi . Clearly OI∗ has the same cardinality as Oi . The orbital Oi is symmetric if and only if (x, y) ∈ Oi implies (y, x) ∈ Oi , that is Oi = Oi∗ .. In this case the relation Ri is actually a symmetric relation. In that case if Oi 6= D, then the graph Γi is a simple undirected graph.3 Thus representing groups as groups of automorphisms of graphs, has a natural setting in permutation groups of rank 3 or more. Now fix a letter x in X. If (u, v) is in the orbital Oi , there exists by transitivity an element g such that ug = x. Thus the orbital Oi possesses a representative (x, y g ), with first coordinate equal to x. Moreover, if (x, v) and (x, w) are two such elements of Oi with first coordinate x, the fact that Oi is a G-orbit, shows that there is an element h in Gx taking (x, u) to (x, w). That is, u and w must belong to the same Gx orbit on X. Thus there is a one-to-one correspondence between the orbitals of the action of G on X and Gx -orbits on X. The lengths of these orbits are called the subdegrees of the permutation action of G on X. By the Fundamental Theorem of Group Actions 74, these Gx -orbits on X (or equivalently, G/Gx ) correspond to (Gx , Gx )-double cosets of G, and the length of Gx gGx is [Gx : GX ∩ g −1 Gx g]. We summarize most of this in the following; 3 For graphs, the word “simple” means is has no multiple edges (of the same orientation) or loops. 154 Theorem 88 Suppose G acts transitively on X. 1. Then the rank of this action is the number of Gx -orbits on X. 2. Their lengths are called subdegrees and are the numbers [Gx : Gx ∩ g −1 Gx g] as g ranges over a full system of (Gx , Gx )-double coset representatives. 3. The correspondence of an orbital Oi and a (Gx , Gx )-double coset is that (Gx , Gx h) ∈ Oi if and only coset Gx h is in the corresponding (Gx , Gx )double coset, namely Gx hGx . Then |Oi | = |X|[Gx : Gx ∩ g −1 Gx g]. Clearly Oi and Oi∗ correspond to Gx -orbits of the same size. In particular, if an orbital is not symmetric, then there must be two subdegrees of the same size. 4. If a double coset Gx gGx contains an involution t, then the corresponding orbital must be symmetric. remark: All but item 4 was developed in the discussion preceeding. If t is an involution in Gx gGx , then t inverts the edge (Gx , Gx t) in the graph Γi . Let’s look at a few simple examples. 1. (Grassmanian Action) Suppose V is a finite-dimensional vector space of dimension at least 2. As we have done before, let Pk (V ) be the collection of all k-dimensional subspaces, where k ≤ (dim V )/2. Then the group GL(V ) acts transitively on Pk (V ), as noted earlier. But because of the inequality on dimensions, a given k-subspace W may intersect other k-subspaces at any dimension from 0 to k. Moreover, the stabilizer H in GL(V ) of the subspace W is transitive on all further k-subspaces which meet W at a specific dimension. Thus we see that GL(V ) acts on Pk (V ) with an action which is rank k + 1. 2. (Symmetric Groups) If k ≤ n/2, the group G = Sym(n), transitively permutes A(k), the k-subsets of A = {1, 2, . . .}. Again, the stabilizer 155 H of a fixed k-subset U , is transitive on the set of all k-subsets which meet U in a subset of fixed cardinality. Thus G = Sym(n), has rank k + 1 in its action on all k-subsets. For example, G = Sym(7), acts faithfully on the 35 points of A(3), as a rank four group with subdegrees 1, 12, 18, 4. G = Sym(6), acts as a rank 4 group on the 20 letters of A(3), with subdegrees 1, 9, 9, 1. Similarly, G = Sym(5), acts as a rank 3 group on the 10 letters of A(2), with subdegrees 1, 3, 6. If Oi is the orbital of 30 ordered pairs corresponding to the subdegree 3, then, as this orbital is symmetric, it contributes 15 undirected edges on the 10 vertices, yielding a graph Γi = (A(2), Ei ) isomorphic to Petersen’s graph, whose automorphism group of order 120 we met in the last section. 3. We have mentioned that G turns out to be a subgroup of the automorphism group of the graph Γi . It need not be the full automorphism group, and whether it is or not depends on which orbital Oi one uses. For example, we have just seen above that the group G = Sym(7), has a rank 4 action on the 35 3-subsets chosen from the set A of 7 letters, with subdegrees 1, 12, 18, 4. Now it happens that the graph Γ3 corresponding to the subdegree 18 has diameter 2, and its full automorphism group is Sym(8). 4. This might be a good time to introduce an example of a rank 3 permutation group with two non-symmetric orbitals. Consider the group G = AQ of all affine transformations, x → sx + y where x ∈ Zp , s a square, where Zp denotes the additive group of residue classes of mod a prime p ≡ 3 mod 4. Here A denotes the additive group of translations {ty : x → x + y, x ∈ Zp }, and Q denotes the multiplications by quadratic residue classes modulo p. Then G acts faithfully on G/Q, with A as a normal regular abelian subgroup. Now the group G has order p(p−1)/2 which is odd. G is rank three with subdegrees 1, (p − 1)/2, (p − 1)/2, and the non-diagonal orbitals are not symmetric. By Lemma 76, the graph Γ can be described as follows: The vertices are the residue classes [m] mod p. Two of them are adjacent if and only if they differ by a 156 quadratic residue class. Since −1 is not a quadratic residue here, this adjacency relation is not symmetric. A rank 1 permutation group is so special that it is trivial, for the diagonal orbital D is the only orbital, forcing |X × X| = |X| which implies |X| = 1. A rank 2 group action has just two orbitals, the diagonal and one other. These are the doubly transitive group actions which are considered in the next subsection. In fact Corollary 89 The following are equivalent: 1. G acts as a rank 2 permutation group. 2. G transitively permutes the ordered pairs of distinct letters of X. 3. G acts transitively on X and for some letter x ∈ X, Gx acts with two orbits: {x} and X − {x}. 4.4.3 Multiply transitive group actions. Suppose G acts transitively on a set X. We have remarked in Section 3.1 that in that case, for every positive integer k ≤ |X|, that G also acts on any of the following sets: 1. X (k) , the set of ordered k-tuples of elements from X, 2. X (k)∗ , the set of ordered k-tuples of elements of X with entries pairwise distinct. 3. X(k), the set of (unordered) subsets of X of size k. For k > 1, the first set is of little interest to us; for the group can never act transitively here, and what orbits that exist tend to be be equivalent to orbits on the other two sets. A group G is said to act k-fold transitively on X if and only if it induces a transitive action on the set X (k)∗ of k-tuples of distinct letters. It is said to have a k-homogeneous action if and only if G induces a transitive action of the set of all k-sets of letters. Lemma 90 1. If G acts k-transitively on X, then it acts k-homogeneously on X. That is, k-transitivity implies k-homogeneity. 157 2. For 1 < k ≤ |X|, a group which acts k-transitively on X acts (k − 1)transitively on X. 3. For 1 < k < |X| − 1,a group G which acts transitively on X, acts k-fold transitively on X if and only if the subgroup fixing any (k − 1)tuple (x1 , . . . , xk−1 ) transitively permutes the remaining letters, X − {x1 , . . . , xk−1 }. proof: The first two parts are immediate from the definitions. The third part follows from Corollary 89 when k = 2. So we assume 2 < k < |X| − 1, and will apply induction. Assume X = {1, 2, 3, . . .} (the notation is not intended to suggest that X is countable; only that a countable few elements have been listed first), and suppose G acts transitively on X and that for any finite ordered (k − 1)-tuple (1, 2, . . . , (k−1)), the subgroup G1,...(k−1) that fixes the (k−1)-tuple pointwise, is transitive on the remaining letters, X − {1, 2, . . . , (k − 1)}. Clearly, the result will follow if we can first show that G is (k − 1)-fold transitive on X. Now set U = X − {1, . . . , (k − 2)}. The subgroup H := G1,...,k−2 , of all elements of G fixing {1, . . . , k − 2} pointwise, properly contains the group G1,...,k−1 just discussed, and has this property: • For every letter u in U , Hu fixes {u} and is transitive on U − {u}. Since k < |X| − 1 causes |U | ≥ 3, the presented property implies H acts transitively on U . Thus we have the hypotheses if part 3 for k − 1, and so we can conclude by induction, that G has (k − 1)-fold action on X. Then the hypothesis on G1,...,k−1 implies the k-fold transitive action. The converse implication in part 3, that k-fold transitive action causes G1,...,k−1 to be transitive on all remaining letters, is straightforward from the definitions. remark: The condition that k < |X| − 1 is necessary. The group Alt(5) acts transitively on five letters, and any subgroup fixing four of these letters, is certainly transitive on the remaining letters (or more accurately “letter”, in this case). Yet the group is not 4-transitive on the five letters: it is only 3-fold transitive. Lemma 91 Suppose f : G → Sym(X) is a 2-transitive group action. Then the following statements hold: 158 1. G is primitive in its action on X. 2. If x ∈ X, G = Gx + Gx tGx where t is congruent to an involution mod ker f . 3. Suppose N is a finite regular normal subgroup of G. Then N has no proper characteristic subgroups and so is a direct product of cylic groups of prime order p. (Such a group is called an elementary abelian pgroup. It follows that |N | = |X| = pa is a prime power. proof: 1. If X1 were a non-trivial component of a system of imprimitivity, then 1 6= X1 6= X| and for any x ∈ X1 , The subgroup Gx stabilizes X1 . But Gx acts transitively on X − {x}, forcing X1 = X, a contradiction. So there is no such system of imprimitivity, as was to be shown. 2. Since n(n − 1) = |X (2)∗ | divides |G/ker f |, the latter number is even. So by Sylow’s theorem applied to G/ker f , there must exist an element z of G which acts as an involution on X. Then z must induce at least one 2-cycle, say (a, b) on X. From transitivity, we can choose g ∈ G such that ag = x. Then t := g −1 zg is the involution displacing x and can be taken as the double coset representative. 3. By Lemma 76 part 4, Gx transitively permutes the non-identity elements by conjugation. Since the conjugation action, induces automorphisms of N , N does not contain any proper characteristic subgroups. Also, all non-identity elements are G − conjugate and so have the same finite order, which must be a prime p. This means Sylr (N ) = {1} if r 6= p, so N is a p-group. By Lemma 79, N has a non-trivial center, and so is abelian. Since all its elements have order p, it is elementary abelian. Since it is regular, the statements about |X| follow. Let us discuss a few Examples 1. The groups FinSym(X) are k-fold transitive for every natural number k not exceeding |X|. However, an easy induction proof shows that Alt(X) is (k − 2)-transitive for each natural number k less than |X|. (Recall that the alternating group is by definition a subgroup of the finitary symmetric group.) 159 2. The group GL(V ) has a 2-transitive action on the set P1 (V ) of all 1-subspaces of V . This set is called the set of projective points of V ; the set P2 (V ) of all 2-dimensional spaces is called the set of projective lines. Incidence between P1 (V ) and elements of P2 (V ) is the relation of containment of a 1-subspace in a 2-subspace of V . Then P G(V ) := (P1 (V ), P2 (V )) is an incidence system of points and lines with this axiom: Linear Space Axiom. Any two points are incident with a unique line. If dim V = 2, Gl(V ) is doubly transitive on the set P1 (V ), the projective line. If the ground field for the vector space V is a finite field containing q elements, then there are q + 1 projective points being permuted. The group of actions induced – that is GL(V )/K where K is the kernel of the action homomorphism – is the group P GL(2, q), and it is 3transitive on P1 (V ). If dim V = 3, it is true that any two distinct 2-subspaces of V meet in a 1-subspace. Thus we have the additional property: (Dual linear) Any two distinct lines meet at a point. Any system of points and lines, satisfying both the Linear Space Axiom and the Dual Linear Axiom is called a projective plane. Again, if the ground field of V is a finite field of q elements then we see that |P1 (V )| = 1 + q + q 2 = |P2 (V )| and each projective point lies on q +1 projective lines, and, dually, each projective line is incident with exactly q + 1 projective points. In the case q = 2 (which means that the field is Z/(2), integers mod 2), we get the system of seven points and seven lines which we have met as an example in Section 3.4.4. We saw there that the full automorphism group of the projective plane of order 2 had order 168. 160 4.4.4 Permutation characters The results in this section only make sense for actions on a finite set X. Suppose f : G → Sym(X) is a group action on a finite set X. Then f (G) is a finite group of permutations of X. Without loss we assume G itself is finite. (We do this, so that some sums taken over the elements of G are finite. But this is no drawback, for the assertions we wish to obtain can in general be applied to f (G) as a faithfully acting finite permutation group.) Associated with the finite action of G is the function π:G→Z which maps each element g of G to the number π(g) of fixed points of g – that is, the number of letters that g leaves fixed. This function is called the permutation character of the (finite) action f , and, of course, is completely determined by f . The usefulness of this function is displayed in the following: Theorem 92 (Burnside’s Lemma) 1. The number of orbits on which G acts on X is the average: X (1/|G|) g∈G π(g). In particular, G is transitive, if and only if the average value of π over the group is 1. 2. The number of orbits of G on X (k)∗ is X (1/|G|) g∈G π(g)(π(g) − 1) · · · (π(g) − k + 1). 3. In particular, if G acts 2-transitively, then X (1/|G|) π(g)2 = 2. Also, if G is 3-transitive, X (1/|G|) g∈G 161 (π(g))3 = 5. proof: 1. Count pairs (x, g) ∈ X × G with xg = x. One way is to first choose x ∈ X and then find g ∈ Gx . This count yields X |Gx | = x∈X X X orbits |Gx ||Ox | = orbits |G| = |G| · no. of orbits. On the other hand we can count these pairs by choosing the group element g first, then x ∈ X in π(g) ways. thus |G| · no. of orbits = X g∈G π(g). 2. The number of ordered k-tuples left fixed by element g is π(g) · (π(g) − 1) · · · (π(g) − k + 1) (since such a k-tuple must be selected from the fixed point set of g). We then simply apply part 1. with X replaced by X (k)∗ . 3. If G is 2-transitive then the average value of π(π − 1) is 1. But as such a group is also transitive, the average value of π is also 1. Thus the average value of π 2 is the average value of (π(π − 1) + π) = 1 + 1 = 2. Similarly, if G is 3-transitive, X (1/|G|) g∈G (π(g))3 = (1/|G|) X + X g∈G 2 (π(g)(π(g) − 1)(π(g) − 2)) X 3π − 2 g∈G g∈G π(g) = 1 + 2 · 3 − 2 · 1 = 5. 4.5 Exercises for Chapter 3. Exercise 1. Show that in any finite group G of even order, there are an odd number of involutions. [Hint: Choose t ∈ I, the set I = inv(G) of all involutions in G. Then I partitions as {t} + Γ(t) + ∆(t), where Γ(t) := CG (t) ∩ I − {t} and ∆(t) := tG − CG (t). Show why the last two sets have even cardinality.] Exercise 2. (Herstein) Show that if exactly one-half of the elements of a finite group are involutions, then G is dihedral of order 2n, n odd. [Hint: By exercise 1, the number of involutions is an odd number equal to onehalf the group order, and by Sylow’s theorem all involutions are conjugate, 162 forcing CG (t) = hti. The result follows from any Transfer Theorem such as Thompson’s.4 ] Exercise 3. This has several parts: 1. Show that if p is a prime, then any group of order p2 is abelian. [Hint: It must have a non-trivial center, and the group mod its center cannot be a non-trivial cyclic group.] 2. Let G be a finite group of order divisible by the prime p. Show that the number of subgroups of G or order p is congruent to one modulo p. [Hint: Make the same sort of decomposition of the class of groups of order p as was done for groups of order 2 in Exercise 1. You need to know something about the number of subgroups of order p in a group of order p2 .] 3. If prime number p divides |G| < ∞, show that the number of elements of order p in G is congruent to −1 modulo p. 4. (McKay’s proof of Exercises 1 and 3, part 2.) Let G be a finite group of order divisible by the prime p. Let S be the set of p-tuples Q (g1 , g2 , . . . , gp ) of elements of G such that gi = 1. (a) Show that S is invariant under the action of σ, a cyclic shift which takes (g1 , g2 , . . . , gp ) to (gp , g1 , . . . , gp−1 ). (b) As H = hσi ' Zp acts on S, suppose there are a H-orbits of length 1 and b H-orbits of length p. Show that |G|p−1 = |S| = a + bp. As p divides the order of G conclude that p divides a. (c) Show that a 6= 0. Conclude that a is the number of elements of order p in G, and is a non-zero multiple of p. (d) Let P1 , . . . , Pk be a full listing of the subgroups of order p in G. Show that | ∪ Pj | = a ≡ 0 mod p, and deduce that k ≡ 1 mod p. 5. Suppose pk divides the order of G and P is a subgroup of order pk−1 . Show that p divides the index [NG (P ) : P ]. [Hint: Set N := NG (P ) and suppose, by way of contradiction that p does not divide [N : P ]. Then 4 I am convinced that this is the elementary proof that Herstein had in mind rather than a number of diatribes on this subject which have appeared in the problems section of the AMA monthly. 163 P is the unique subgroup of N of order pk−1 and so is characteristic in N . As pk divides |G|, we see that p divides the index [G : N ]. Now P acts on the left cosets N in G by left multiplication, with N comprising a P -orbit of length one. Show that there are at least p such orbits of length one so that there is a left coset bN 6= N , with P bN = bN . Then b−1 P b ≤ N forcing b−1 P b = P from the uniquness of its order. But then b ∈ N , a contradiction.] 6. Suppose the prime power pk divides the order of the finite group G. Define a p-chain of length k to be a chain of subgroups of G, P1 < P 2 < · · · < P k , where Pj is a subgroup of G of order pj , j = 1, . . . , k. Show that the number of p-chains of G of length k is congruent to one modulo p. [Hint: Proceed by induction on k. The case k = 1 is part 4 (d) of this Exercise. One can assume the number m of p-chains of length k − 1 is congruent to 1 mod p. and these chains of length k − 1 partition the chains of of length k. If c is a chain of length k − 1 with terminal member Pk−1 , then any extension of c to a chain of length k is obtained by adjoining a group Pk of order pk in the normalizer NG (Pk−1 ). The number of such candidates is thus the number of subgroups of order p in NG (Pk−1 )/Pk−1 . Part 5 of this Exercise shows that p divides this group order, and so by part 4 (d), the number of Zp ’s in the factor N/P is congruent to one mod p. Show this forces the conclusion.] 7. From the previous result show that if pk divides |G|, for k > 0, then the number of subgroups of order pk is congruent to 1 mod p. [Remark: Note that parts 3 though 7 of this Exercdise did not require either the Sylow theorems of even Cauchy’s Theorem – indeed, they essentially reprove them. Exercise 1 and part 2 of this Exercise, on the other hand, required Cauchy’s theorem in order to have Zp ’s to discuss.] Exercise 4. Show that any group of order fifteen is cyclic. [Hint: Use Sylow’s Theorem to show that the group is the direct product of two cyclic p-sylow subgroups.] Exercise 5. Show that any group of order 30 has a characteristic cyclic subgroup of order 15. 164 Exercise 6. Show that any group N which admits a group of automorphisms which is 2-transitive on its non-identity elements is an elementary abelian 2group or a cyclic group of order 3. Extend this result by showing that if there is a group of automorphisms of N which is is 3-transitive on the the non-identity elements of N , then |N | ≤ 4. Exercise 7. Show that if G faithfully acts as a k-fold transitive group on a set X, and Gx is a simple group, then (as k is at least 1) any proper normal subgroup of G is regular on X. Using a basic lemma of this chapter, show that G is the semidirect product N Gx ( here Gx ∩ N = 1 so Gx is a complement to N in G). Moreover, Gx acts (k − 1)-fold transitively on the non-identity elements of N . Conclude that under these circumstances, G can be at most 4-fold transitive on X, and then only in the case that |X| = 5 and G = Alt(5). Exercise 8. Show that the group Alt(5) acts doubly transitively on its six 5-sylow subgroups. Exercise 9. Show that if N is a non-trivial normal subgroup of G = Alt(5), then (as G acts primitively on both 5 and 6 letters) N either (i) has order divisible by 30 or (ii) normalizes and hence centralizes (Why?) each 5-sylow subgroup. Exercise 10. Conclude that possibility (ii) of the previous exercise is impossible, by showing that a 5-sylow subgroup is its own centralizer. Exercise 11. Show in just a few steps, that Alt(5) is a simple group. Exercise 12. If G is a k-transitive group, and Gx1 ,...xk−1 is simple, then, provided k ≥ 3, and |X| > 4 , G must be a simple group. So in this case, the property of simplicity is conferred from the simplicity of a subgroup. (This often happens for primitive groups of higher permutation rank, but the situation is not easy to generalize.) Exercise 13. Organize the last few exercises into a proof that all alternating groups Alt(n) are simple for n ≥ 5. Exercise 14. (Witt’s Trick) Suppose K is a k-transitive group acting faithfully on the set X, where k ≥ 2. 165 1. Show that if H := Kα , for some letter α in X, then K = H + HtH for some involution t in K. 2. Suppose ∞ is not a letter in X, set X 0 := X ∪ {∞}, and set S := Sym(X 0 ). Suppose now that S contains an involution s such that (a) s transposes ∞ and α, (b) s normalizes H – i.e. sHs = H, and (c) o(st) = 3, that is, (st)3 = 1, while st 6= 1. Then show that the subset Y = K ∪ KsK is a subgroup of S. [Hint: One need only show that Y is closed under multiplication, and this is equivalent to the assertion that sKs ⊂ Y . Decompose K; use the fact that sH = Hs and that sts = tst.] Exercise 15. Show that in the previous Exercise Y is triply transitive on X 0. Exercise 16. Suppose G is a simple group with an involution t such that the centralizer CG (t) is a dihedral group of order 8. Show the following: 1. The involution t is a central involution – that is, t is in the center of at least one 2-Sylow subgroup of G, or, equivalently, |tG | is an odd number. Thus any 2-Sylow subgroup of G is just D8 . 2. There are exactly two conjugacy classes of fours groups – say V1G and V2G , where V1 and V2 are the unique two fours subgroups of a 2-Sylow subgroup D of G. Each of them is a TI-subgroup – that is a subgroup which intersects its distinct conjugates only at the identity subgroup. 3. There is just one conjugacy class of involutions in G. [Hints: This is basically an exercise in the use of the Burnside Fusion Theorem and Sylow’s Theorem. Part 1.: Take D = CG (t) ' D8 . Then its center is just hti, so NG (D) ≤ CG (t) = D. Part 2.: Let Vi , i = 1, 2, be the two unique fours subgroups of D ∈ Syl(G). By the Burnside Fusion theorem they could be conjugate in G only if this were true in NG (D) = D, where it is patently untrue. Clearly two conjugates of Vi could at best meet at a central involution, of which there is only one 166 class, tG . In that case they would be two fours subgroups of the same 2-Sylow subgroup, and hence could not be conjugate. Part 3.: Suppose false: Then at least one involution, say without loss v1 ∈ V1 , is not central; so its centralizer in G, C1 , contains V1 as a 2-Sylow subgroup. By the Thompson Transfer Theorem, the fact that the fours group V2 has index 2 in the 2-Sylow subgroup D and the fact that G possesses no normal subgroup of index 2, t1 is conjugate to some involution in V2 −{t}, say t2 . (At this stage t is conjugate to no further involution in D – a so-called Z ∗ -involution. 5 ) But then by the Burnside Fusion Theorem, the normal subsets Xi := {ti , ti t} are conjugate in N (D) = D, a contradiction. 6 ] Exercise 17. This exercise is really an optional student project. First show that if G if a 4-fold transitive group of permutations of the finite set X, then X (1/|G|) g∈G π(g)4 = 15. The object of the project would be to show that in general, if G is k-fold transitive on X then X (1/|G|) g∈G π(g)k = Bk , where Bk is the total number of partitions of a k-set (also known as the k-th Bell number). An analysis of the formulae involves one in expressing the number of sequences (with possible repititions) of k elements from a set X in terms of ordered sequences with pair-wise distinct entries. The latter are expressed in term of the so-called “falling factorials” x(x − 1)(x − 2) · · · (x − ` + 1) in the variable x = |X|. The coefficients involve Stirling numbers and the Bell number can be reached by known identities. But a more direct argument can be made by sorting the general k-tuples over X into those having various sorts of repitions; this is where the partitions of k come in. 5 There is a famous theorem – G. Glauberman’s Z∗-theorem –which asserts that no Z -involution can occur in a simple group of order different than 2. Of course we cannot invoke such things here. 6 Note the power here in using the Burnside Fusion theorem on normal subsets of a Sylow subgroup, not just normal subgroups. ∗ 167 Appendix to Chapter 4: Minimal normal subgroups of finite doubly transitive groups. All groups in this section are finite groups. We have seen in Section 3.7 that a minimal normal subgroup of any finite group is a direct product of isomorphic simple groups. If one of these direct factors is a cyclic group of prime order p (denoted Zp ), then the minimal normal subgroup is a direct product of finitely many Zp ’s. Such a group is called an elementary abelian p-group (cf. Lemma 22). Such a finite group A is characterized by the properties of being abelian and of exponent p – that is, xp = 1 for all x ∈ A. The main result of the section is a theorem of Burnside stating that a minimal normal subgroup of a (finite) doubly transitive group is either a regular elementary abelian p-group or a transitive finite non-abelian simple group. I have always found the proof in Burnside’s book difficult to read. Presentations of this lemma are infrequent in the textbook literature. Those rare books on permutation groups which will even refer to the theorem invariably refer to the original in Burnside’s book (Wielandt [?], or Passman [?]). We revisit this theorem for two purposes: (1) to see the theory we have presented in action, and (2) to extract the theorem into a transparent referrable form. We present below two distinct proofs, the second being a rewriting of Burnside’s original proof in modern language.7 4.5.1 First proof: The version offered here closely exploits Sylow’s theorem. 7 My own history with this theorem is nearly idiotic: Until a very few years ago, I had always beleived Burnside’s original proof was basically flawed. A few years ago, R. Abhyankar (learning group theory as fast as he could – that is, faster than most of us) asked me on the spot for a proof of Burnside’s result. It took a night but I coughed up the first proof below. It uses little more than Sylow’s theorem and the standard results about permutations groups given earlier in this chapter. Recently my graduate student (Kasikova) remarked that she thought Burnside’s proof was correct. One of the things that I have learned is to always pay attention to her slightest mathematical remarks. So I looked it over and decided that indeed, Burnside’s original proof is correct, provided there is some straightening up of the pronoun references. As it stands it is a proof which is easily baffling, and for that reason alone, needs a reintroduction into the modern literature. Similarly, this proof uses only elementary concepts about permutations groups which have been presented in Chapter 3. The proof is not different than Burnside’s; it is just translated into modern language. 168 But first, a few elementary technical facts must be set down. Lemma 93 Assume H is a proper subgroup of G which contains a p-Sylow normalizer of G. Then for some element g ∈ G, H ∩ H g does not contain a p-Sylow subgroup of G. proof: Assume, for every element g ∈ G, that H ∩ H g contains a p-Sylow −1 subgroup of G. For any g, let R ∈ Syl(H ∩ H g ) ⊆ Syl(G). Then {R, Rg } ⊆ −1 Syl(H), so by Sylow’s theorem, Rg h = R for some h ∈ H. Then g −1 h ∈ NG (R) ≤ H. So g ∈ H. But since g was arbitrarily chosen, we have H = G, a contradiction. Thus our beginning assumption must be false, and so the conclusion holds. Lemma 94 Suppose a group G is the direct product of two of its subgroups, A and B – so we write G = A × B. Suppose L is a subgroup G of the form L = A1 × B1 , where A1 ≤ A and B1 ≤ B. Then for any element g = (g1 , g2 ) ∈ A × B = G, L ∩ Lg = (A1 ∩ Ag11 ) × (B1 ∩ B1g2 ). (4.3) −1 proof: An element y of L lies in L ∩ Lg if and only if y · y g ∈ L. So, writing y = ab, where a ∈ A1 , b ∈ B1 , and g = g1 g2 where g1 ∈ A and g2 ∈ B, y ∈ L ∩ Lg is equivalent to asserting that y · yg −1 = ab(ab)g −1 −1 −1 −1 −1 = abag1 bg2 = (aag1 )(bbg2 ) is an element of L and therefore, is uniquely expressible as an element of A1 times and element of B1 as well as uniquely expressible as an element of −1 −1 A times an element of B. Thus aag1 ∈ A1 and bbg2 ∈ B1 , from which it follows that a ∈ A1 ∩ Ag11 and b ∈ B1 ∩ B1g2 . So the left side of equation ( 4.3) is contained in the right. The reverse containment is easy. If K is any finite group, and p is a prime number, the symbol |K|p will denote the highest power of p dividing the group order |K| – that is, the order of a p-Sylow subgroup of K. 169 Lemma 95 Suppose G = X1 × X2 , where the Xi are regarded as subgroups of G. Suppose H is a subgroup of G, containing neither direct summand Xi , but containing nontrivial p-Sylow subgroups of both X1 and of X2 . Finally, assume that H contains a p-Sylow normalizer of G. Then there exist elements x, y ∈ G − H such that |H ∩ H x |p 6= |H ∩ H y |p . proof: By hypothesis, H contains a full p-sylow subgroup P = P1 × P2 , where 1 6= Pi ∈ Sylp (X1 ), i = 1, 2. Set Hi := H ∩ Xi , a normal subgroup of H containing Pi as a nontrivial p-Sylow subgroup of Hi , i = 1, 2. Then P is a p-Sylow subgroup of H lying within its normal subgroup K := H1 H2 . It follows that K = H1 H2 contains every p-Sylow subgroup of H – even every p-element of H. Choose an element g = g1 g2 ∈ G − H where gi ∈ Xi , i = 1, 2. Let R be a p-Sylow subgroup of H ∩ H g . Then as R is a p-subgroup of H, R must lie in K = H1 × H2 , as well as K g . Thus R is a p-Sylow subroup of K ∩ K g = (H1 ∩ H1g1 ) × (H2 ∩ H2g2 ), by Lemma 94. Thus |H ∩ H g |p = |H1 ∩ H1g1 |p · |H2 ∩ H2g2 |p . Fix i ∈ {1, 2}. By hypothesis, Hi is a proper subgroup of Xi containing a non-trivial p-Sylow subgroup Pi of X1 . Moreover, as NXi (Pi ) ≤ NG (P1 × P2 ) ∩ X1 ≤ H ∩ X1 = H1 , we see that (Xi , Hi , Pi ) satisfy the hypotheses of (G, H, P ) in Lemma 93. Thus it is possible to choose gi so that |Hi ∩ Higi |p < |H1 |p . Now setting x = g1 and y = g1 g2 , we see that |H ∩ H y |p = |H1 ∩ H1g1 |p · |H2 ∩ H2g2 |p |H ∩ H x |p = |H1 ∩ H1g1 |P · |H2 |p 170 (4.4) (4.5) (4.6) So |H ∩ H y |p < |H ∩ H x |p < |H|p . The last inequality shows that neither x nor y lie in H. The first inequality, is the desired conclusion. This completes the proof. Now we can proceed with the main result. Theorem 96 Suppose G is a finite doubly transitive group of permutations of a set X containing at least two objects. Then any minimal normal subgroup of G is either 1. an elementary abelian p-group which acts regularly on X (so that |X| must be a prime power), or 2. a non-abelian simple group. proof: One first notes that since the class of elementary p-groups includes the identity subgroup, we may assume |X| > 1 (or equivalently, that |G| > 1). In any case, we may assume that N is a minimal normal (and hence non- identity) subgroup of G. By Corollary 20 at the end of Chapter 3, N is either an elementary abelian p-group or the direct product of isomorphic simple nonabelian groups. Since double transitivity implies primitivity, N must be a transitive group. In the former case (since G acts faithfully), N must be a regular group and the first conclusion holds. Thus we may assume N = X1 × . . . Xt , where the direct factors Xi are non-abelian groups of a single isomorphism type. It remains only to show that t = 1. Assume by way of contradiction that t > 1. Fix a “letter” α in X. Then by double transitivity, the one-point stabilizer, Gα , is transitive on X − {α}, and so its normal subgroup H := Gα ∩ N acts on X − {α} with orbits all of the same length. This means that for every element g ∈ N − H, the index [H : H ∩ H n ] does not depend on the choice of n ∈ N − H. Put another way |H ∩ H x | is constant for all x ∈ N − H. Moreover 171 (4.7) (*) H cannot contain a non-trivial normal subgroup of N , since otherwise the fact that such a normal subgroup of transitive N fixes a letter, would imply that it fixes all letters and hence is trivial. In particular H cannot contain any Xi . Now suppose P is a p-Sylow subgroup of H. If NN (P ) is not contained in H, then, by choosing x ∈ NN (P ) − H in Equation ( 4.7), we see that |P | divides |H ∩ H x | for all x ∈ N . Now if this happened for every prime p which divides |H|, we should have |H ∩ H x | = |H| for all x ∈ N – that is H = H x for all x, or H is normal in N against (*). Thus there must be some prime p dividing H for which the p-Sylow subgroup P of H satifies NN (P ) ≤ H. We fix such a p-Sylow subgroup P . Now, since a proper subroup of a finite p-group properly lies in its normalizer in that p-group, the condition NN (P ) ≤ H in fact implies that P is a full p-Sylow subgroup of N . Then P = P1 × P2 × · · · × Pt , where Pi is a full p-Sylow subgroup of Xi . Since the Xi are isomorphic, all the Pi are non-trivial groups. Now, noting that (*) prevents H from containing B = X2 × · · · × Xt , the quintette of groups (N, X1 , B, H, P ) satisfy all of the hypotheses on (G, X1 , X2 , H, P ) in Lemma 95. This means we can find x, y ∈ N − H such that |H ∩ H x |p < |H ∩ H y |p for some x, y ∈ N − H. This clearly contradicts (*). Thus t > 1 is impossible and the proof is complete. 4.5.2 Burnside’s original proof. A point-line geometry (P, L) is a set P of points together with a collection L of subsets of P called lines. Such a point-line geometry is called a linear space if and only if any two distinct points lie in exactly one line, lines contain at least two points, and there are at least two distinct lines. It follows that each point must lie on at least two distinct lines. The automorphism group of a point-line geometry (P, L), is the set Aut(P, L) of all permutations of P which induce a permutation of the lines – that is, they stabilize the collection L. We have used these as examples of groups in Chapter 3. Theorem 97 Suppose N is a point-transitive group of automorphisms of a linear space (P, L) with the property that for each point p, its stabilizer Np , 172 leaves invariant each line on point p. Then the stabilizer in N of two distinct points is the identity subgroup – i.e. Np ∩ Nq = 1, for distinct points p and q. Proof: Suppose p and q are distinct points and L is the unique line of L containing {p, q}. Then any point z not in L is the intersection of two lines (one on z and p; the other on z and q) which are fixed by Np ∩ Nq . Thus the latter fixes the non-empty set P − L pointwise. But then, any point u in L is the intersection of the line on u and z with L and both of these are also stabilized by Np ∩ Nz , which we have seen contains Np ∩ Nq . Thus Np ∩ Nq fixes every point of P, and so contains only the identity permutation. The next two lemmas involve direct products of groups and are slightly more general than needed. Lemma 98 Suppose G = A × B, where each factor is non-trivial. Suppose H is a subgroup of G with the “Frobenius property” that H ∩ H g = 1 for all elements g ∈ G − H Then H is not a proper subgroup – that is, H = 1 or H = G. Proof: We identify A with subgroup A × 1 and B with subgroup 1 × B so G = AB, and A ∩ B = 1. Assume 1 6= x ∈ H ∩ A. Then x ∈ H ∩ H b for all b ∈ B. The Frobenius property then implies B ≤ H. Since B 6= 1, the same argument with A replacing B shows A ≤ H whence H = G. So we can assume H ∩ A = H ∩ B = 1. Now if 1 6= h = ab ∈ H where a ∈ A and b ∈ B, then by the previous paragraph neither a nor b is in H. But h ∈ H ∩ H b and the Frobenius property imply b ∈ H, a contradiction. Thus H = 1, and the proof is complete. Lemma 99 Suppose G = A × B is a group of permutations of a set X, with each factor a non-identity group. 1. Suppose the subgroup A × 1 is transitive. Then the other factor 1 × B is semiregular – that is, the only element of 1 × B which fixes a letter is the identity permutation. 173 2. If G is primitive, then (a) each factor acts regularly on X, (b) each factor is a non-abelian simple group, and (c) the two simple factors A and B are isomorphic 3. Suppose φ : A → B is an isomorphism of simple groups. Let H := {(a, φ(a))|a ∈ A}. Then H is a maximal subgroup of G = A × B. Proof: Again we adopt the convention of writing A for A × 1 and B for the subgroup 1 × B. Part 1. Suppose A is transitive. Then for any x ∈ X, Bx fixes xA = X, so Bx = 1X . Part 2. If G is primitive, each of the normal subgroups A and B are transitive and so by Part 1, each is regular. Then for x ∈ X, we have G = AGx = BGx , and A ∩ Gx = B ∩ Gx = 1. Thus A ' G/B ' Gx ' G/A ' B. Now suppose N were a proper normal subgroup of A. Then N is normal in G. Since G is primitive, N must be transitive. But that is impossible since N is a proper subgroup of a regularly acting group. Thus no such N exists and so A is a simple group. So is B ' A. If A were abelian, then G would be abelian, and Gx would be central and so would fix xA = X. This would force Gx = 1 contrary to Gx ' A above. Thus A and B are both non-abelian simple groups. Part 3. Suppose H < M ≤ A × B = G. Then M ∩ A contains a non-identity element a0 . Then for any g = (a, φ(a)) ∈ H, the commutator −1 −1 a0 ga−1 = a0 aa−1 0 g 0 a , −1 belongs to M . But the subgroup of A generated the set {a0 aa−1 0 a |a ∈ A} is a normal subgroup of A (this is shown in the next Chapter). Since A is a simple group, we have A ≤ M . Similarly B ≤ M so M = G. This proves H is maximal in G. Part 3 of this Lemma is only here to show that the direct procuct of two simple groups can act as a primitive group. Now all the peices are assembled to run through Burnside’s argument. 174 Theorem 100 (Burnside) Let G be a doubly transitive group of permutations of the set X. Let N be a minimal normal subgroup of G. Then either 1. N is an elementary abelian p-group acting regularly on X, or else 2. N is a non-abelian simple group. Proof: We know that as N is minimal normal it is transitive and is either an elementary abelian p-group, or is a product of isomorphic non-abelian simple groups. In the former case N is regular. So we may assume N = S1 × · · · St , t ≥ 2, where all the Si are isomorphic to a fixed non-abelian simple group. There are two cases: Case 1: N is primitive on X. Thus setting A = S1 and B = S2 × · · · × St in Lemma 99 (Part 2) we see that A and B are isomorphic simple groups, so t = 2. Moreover each acts regularly on X. Then G contains a normal subgroup H = NG (S1 ) = NG (S2 ) = Hx S1 , for which Hx acts in at most two orbits on Xx . This means that in the semidirect product S1 Hx , Hx acts in at most two orbits on the non-identity elements of S1 . Thus, since S1 is not a p-group, we see that S1 has order pq, p and q primes. But in that case there can be only one p-Sylow subgroup for the larger prime, against S1 being simple. Case 2: N is imprimitive on X. Since N is transitive, a system of imprimitivity for N must be a partition π : X = X1 + X2 + · · · + Xm , m > 1, where each Xi has the same cardinality. Among all such non-trivial partitions stabilized by N choose π so the cardinality |X1 | is minimal (but larger than 1). Now if G stabilized π, it would not be primitive. Thus for some g ∈ G−N , π g : X = X1g + · · · + Xtg , is distinct from π, and thus their common refinement (that is, their meet π ∧ π g in the lattice of partitions of X) is the trivial partition. This means 175 for any 1 ≤ i ≤ j ≤ m and elements g1 and g2 in G, the two sets Xig1 and Xjg2 either coincide or intersect in at most one letter. Thus setting P := X and L := {Xig |1 ≤ i ≤ m.g ∈ G}, we see that (P, L) is a point-line geometry in which any two lines intersect in at most one point, and that each point lies on at least two lines (since L consists of more than the components of π). In fact, since G acts as a group of automorphisms of (P, L) which is doubly transitive on points, every pair of distinct points is on a line. Thus (P, L) is a linear space. Now the subgroup N , being normal in G and fixing π, stabilizes each “parallel class” π g , g ∈ G. It follows that Nx stabilizes each line of L on point x. Thus by Theorem 97, Nx ∩ Ny = 1 for distinct points x and y. In particular, for any element g ∈ N − Nx , Nxg = Ny where y = xg 6= x. This means that H := Nx is a subgroup of N with the Frobenius property. Now the assumption that t > 1 and Lemma 99 together imply that H = 1 or H = N . In the latter case, N fixes x so X = {x} and N = 1, against its being minimal normal. In the former case N is regular in N . But in that case Gx is a complement to N in G which acts on N − {1} in one orbit, forcing N to be an elementary p-group, contrary to the current case-division. Thus we are forced to have t = 1. This, then, is Burnside’s argument with all the smallest steps filled in, all the “i’s” dotted. Example Here is a simple example showing that a non-abelian minimal normal subgroup of G need not be simple when the hypothesis of double transitivity is weakened to being a primitive rank three permutation group. Let G be the graph of the 5-by-5 grid (sometimes called a lattice graph). Then G contains the direct product of two copies of Sym(5) as a normal subgroup M of index 2. Outside this is an involution, t, conjugation by which interchanges the two direct factors of M . (The group G = hM, ti is called the wreath product of Sym(5) by Z2 .) Now G acts as a rank 3 permutation group on the 25 vertices of the grid graph with subdegrees 1, 8 and 16, and so is primitive. But G has a unique minimal normal subgroup N ' Alt(5)×Alt(5), with each direct factor acting imprimitively. 176 Chapter 5 Normal Structure of Groups 5.1 The Jordan-Holder Theorem for Artinian Groups . 5.1.1 Subnormality. As recorded in the previous section, a group is simple if and only if it has no normal subgroups other than the identity subgroup or the entire group. That is, it has no proper normal subgroups. A subgroup H is said to be subnormal in G if and only if there exists a finite chain of subgroups H = N0 N1 · · · Nk = G. We express the assertion that H is subnormal in G by writing H G. Let SN (G) be the collection of all subnormal subgroups of G. This can be viewed as a poset in two different ways: (1) First we can say that A is “less than or equal to B” iff A is a subgroup of B, That is, SN has the inherited structure of an induced poset of S(G), the poset of all subgroups of G with respect to the containment relation. (2) We also have the option of saying that A “is less than B” if and only if A is subnormal in B. The 177 second way of defining a poset on SN would seem to be more conservative than the former. But actually the two notions are exactly the same, as can be concluded from the following Lemma. Lemma 101 (Basic Lemma on Subnormality) The following hold: 1. If A B and B G, then A G. 2. If A G and H is any subgroup of G, then A ∩ H H. 3. If A is a subnormal subgroup of G, then A is subnormal in any subgroup that contains it. 4. If A and B are subnormal subgroups of G, then so is A ∩ B. proof: 1. This is immediate from the definition of subnormality: By assumption there two finite ascending chains of subgroups, one running from A to B, and one running from B to G, each member of the chain normal in its successor in the chain. Clearly concatenating the two chains produces a chain of subgroups from A to G, each member of the chain normal in the member immediately above it. So by definition, A is subnormal in G. 2. If A G one has a finite subnormal chain A = A0 A1 · · · Aj := G. Suppose H is any subgroup of G. It is then no trouble to see that A ∩ H A1 ∩ H A2 ∩ H · · · Aj ∩ H = G ∩ H = H, is a subnormal series from A ∩ H to H. 3. This assertion follows from Part 2 where H contains A. 4. By Part 2, A ∩ B is subnormal in B. Since B is subnormal in G, the conclusion follows by an application of Part 1. At this point the student should be able to provide a proof that the two methods (1) and (2) of imposing a partial order on the subnormal subgroups lead to the same poset. We simply denote it SN (G), ≤ (or simply SN if G is understood) since it is an induced poset of the poset of all subgroups of G. From Lemma 101, Part 4, we see that SN (G) is a lower semilattice. Finally it follows from the second or third isomorphism theorem, that if 178 If A and B are distinct maximal normal subgroups of G = hA, Bi = AB = BA, then G/A ' A/(A ∩ B). Thus SN (G) is a semimodular lower semilattice and the mapping µ : Cov(SN ) → M (which records for every covering factor X/Y , X a maximal normal subgroup of Y , the simple group X/Y ) is “semimodular” in the sense of Theorem 20 Section 2.6.3. According to the general Jordan-Holder theory advanced in Section 2.6.3, µ may extended to an interval measure ASN → M. from all algebraic intervals of SN into the additive monoid M of all multisets of simple groups — that is, the free additive semigroup of all formal nonnegative integer linear combinations of simple groups: X simp aG G, of finite support . If there is a finite unrefinable ascending chain in SN from 1 to G, then such a chain of subnormal subgroups is called a composition series of G. The multiset of simple groups Ai+1 /Ai appearing in such a composition series 1 A1 · · · An = G is called the multiset of composition factors. The theorem states that these are the same multisets for any two composition series – the “measure” µ([1, G]). If you have been alert, you will notice that no assumption that the involved groups are finite has been made. The compositions factors may be infinite simple groups. But of course, in applying the Jordan-Hölder theorem, one is interested in pinning down a class of algebraic intervals. This is easy in the case of a finite group, for there, all intervals of SN (G) are algebraic. How does one approach this for infinite groups? The problem is that one can have an infinitely descending chain of subnormal subgroups, and also one can have an infinitely ascending chain of proper normal subgroups (just think of an abelian group with such an ascending chain). However one can select certain subposets of SN which will provide us with many of these algebraic intervals: Suppose from any SN (G) we extract 179 the set SN ∗ of all subnormal subgroups for which any downward chain of subgroups each normally containing its successor, must terminate in a finite number of steps. There are three other descriptions of this subcollection SN ∗ , of subnormal subgroups: (i) they are the subnormal subgroups of G which belong to a finite unrefineable subnormal chain extended downward to the identity. (ii) They are the elements M of SN for which [1, M ] is an algebraic interval of SN – see Chapter 2. This is the special class of subgroups amenable to the Jordan-Holder Theorem. (iii) They are the subnormal subgroups of G which themselves possess a finite composition series. We call the elements of SN ∗ the Artinian subgroups. They are closed under the operation of taking pairwise meets in SN and so SN ∗ is clearly a lower semilattice, all of whose intervals are algebraic. As a special case, we have: Theorem 102 (The Jordan-Holder Theorem for general groups.) Let SN ∗ be the class of all Artinian subnormal subgroups of a group G. Then for any H in SN ∗ , the list (with multiplicities) of all simple groups which might appear as a cheif factor in any finite unrefinable normal chain of H is independant of the particular chain. Corollary 103 Any two unrefinable normal series of a finite group recreate, the same list of finite simple groups with mutiplicities respected. . 5.2 Commutators. Let G be a group. A commutator in G is an element of the form x−1 y −1 xy and is denoted [x, y]. A triple commutator [x, y, z] is an element of shape [[x, y], z], which the student may happily work out to be [x, y]−1 z −1 [x, y]z = y −1 x−1 yxz −1 x−1 y −1 xyz. If we write ux := x−1 ux, the result of conjugation of u by x, then it is an exercise to verify these identities, for elements x, y, and z in any group G: [x, y] = [y, x]−1 . [xy, z] = [x, z]y [y, z] = [x, z][x, z, y][y, z]. 180 (5.1) (5.2) [x, yz] = [x, z][x, y]z = [x, z][x, y][x, y, z]. 1 = [x, y −1 , z]y [y, z −1 , x]z [z, x−1 , y]x . [x, y, z][y, z, x][z, x, y] = [y, x][z, x][z, y]x × [x, y][x, z]y [y, z]x [x, z][z, x]y . (5.3) (5.4) (5.5) A commutator need not be a very typical element in a group. For example, in an abelian group, all commutators are equal to the identity. If A, B and C are subgroups of a group G, then [A, B] is the subgroup of G generated by all commutators of the form a−1 b−1 ab = [a, b] as (a, b) ranges over A × B. Similarly we write [A, B, C] := [[A, B], C]. Theorem 104 (Basic Identities on Commutators of Subgroups.) Suppose A, B and C are subgroups of a group G: 1. A centralizes B if and only if [A, B] = 1. 2. [A, B] = [B, A]. 3. B always normalizes [A, B]. 4. A normalizes B if and only if [A, B] ≤ B. In that case, [A, B] is a normal subgroup of B. 5. If f : G → H is a group homorphism, then [f (A), f (B)] = f ([A, B]). 6. (The Three Subgroups Lemma of Phillip Hall) If at least two of the three subgroups [A, B, C], [B, C, A] and [C, A, B] are trivial, then so is the third. proof: (1): A centralizes B if and only if every commutator of shape [a, b], (a, b) ∈ A × B, is the identity. The result follows from this. (2) A subgroup generated by set X is also the subgroup generated by X −1 . Apply ( 5.1). (3) If (a, b, b1 ) ∈ A × B × B, then it follows from ( 5.3) that [a, b]b1 = [a, b1 ]−1 [a, bb1 ] ∈ [A, B]. (4) The following statements are clearly equivalent: 1. A normalizes B 181 2. [a, b] ∈ B for all (a, b) ∈ A × B 3. [A, B] ≤ B. The normality assertion follows from (3). (5) This is obvious. (6) Without loss assume that [A, B, C] and [B, C, A] are the identity subgroup. Then all commutators [a, b, c] and [b, c, a] are the identity, no matter how the triplet (a, b, c) is chosen in A × B × C. Now by Equation (5.4), any commutator [c, a, b] is the identity, for any (a, b, c) ∈ A × B × C. This proves that [C, A, B] has only the identity element for a generator. Corollary 105 Suppose again that A and B are subgroups of a group G. 1. If A normalizes B and centralizes [A, B], then A/CG (B) is abelian. In particular, if A ≤ Aut(B) and A centralizes [A, B] := ha−1 ab |(a, b) ∈ A × Bi, then A is an abelian group. 2. If A is a group of linear transformations of the vector space V , centralizing a subspace W as well as its factor space V /W , then A is an abelian group.1 proof: All these results are versions of the following: By the Three Subgroups Lemma, [A, B, A] = 1 implies [A, A, B] = 1. That is the proof. 1 This seems to be a frequently rediscovered result. As recently as 1991 a very distinguished colleague in applied linear algebra and combinatorics, relayed this discovery to me at an international meeting as a new result that he had learned from some Russians in Lie Representation Theory. With disrespect to noone, it shows how far politicians can isolate our sciences, and how far mathematicians can isolate themselves from neighboring fields. Perhaps one function of meetings is to discover that we are not as far apart as we thought we were. Philip Hall noted this theorem in his hand-written lecture notes 60 years ago. His notes were never published and not all of the world was lucky enough to have been in his class at that time. As a result they are not as widely known as they ought to be – except among group-theorists. (I got my (handwritten) copy of Phillip Hall’s notes from John Thompson, but both of my advisors (Suzuki and Walter) were well aquainted with this in 1958.) Theorems are not discovered in the order of their utility or simplicity. Thus a very useful basic theorem may very well not be a recognized principle until someone reorganizes things late in the maturity of a field. For example the Thompson transfer theorem and the Thompson Order Formula were late arrivals. That’s just the way it is. 182 Of course anytime someone has some machinery, some mathematician wants to iterate it. We have defined the commutator of two subgroups in such manner that the result is also a subgroup. Thus inductively we may define a multiple commutator of subgroups in this way: Suppose (A1 , A2 , . . .) is a (possibly infinite) sequence of subgroups of a group G. Then for any natural number k, we define [A1 , . . . Ak+1 ] := [[A1 , . . . Ak ], Ak+1 ]. 5.3 The Derived Series and Solvability. From the previous section, we see that the subgroup [G, G] is a subgroup which 1. is mapped into itself by any endomorphism f : G → G (a consequence of Theorem 104 (5)). (This condition is called fully invariant. Since inner automorphisms are automorphisms which are in turn endomorphisms, “fully invariant” implies “characteristic” implies “normal”.) 2. yields an abelian factor, G/[G, G] (let f : G → G/[G, G] and apply parts 2. and 3. of Theorem 104). Moreover, if G/N is abelian, then [G, G] ≤ N , so [G, G] is the smallest normal subgroup of G, which possesses an abelian factor (see Chapter 3). The group [G, G] is called the commutator subgroup or derived subgroup of G and is often denoted G0 . Since we have formed the group G0 := [G, G], we may iterate this definition to obtain the second derived group, 00 G := [[G, G], [G, G]]. Once on to a good thing, why not set G0 = G(1) and G00 = G(2) and define G(k+1) := [G(k) , G(k) ] for all natural numbers k? The resulting sequence G ≥ G(1) ≥ G(2) ≥, · · · . is called the derived series of G. Since each G(k) is characteristic in its successor which in turn is characteristic in its successor, etc. , we see that 183 Every G(k) is characteristic in G. Note also that G(k+l) = (G(k) )(l) . (5.6) A group G is said to be solvable if and only if the derived series G ≥ G(1) ≥ G(2) , . . . , eventually becomes the identity sugroup after a finite number of steps.2 The smallest positive integer k such that G(k) = 1 is called the derived length of G. Thus a group is abelian if and only if it has derived length 1. Theorem 106 ular 1. If H ≤ G and A ≤ G, then [A, H] ≤ [A, G]. In partic[H, H] ≤ [G, H] ≤ [G, G]. 2. If H ≤ G, then H (k) ≤ G(k) . Thus, if G is solvable, so is H. 3. If f : G → H is a group homomorphism, then f (G(k) ) = (f (G))(k) ≤ H (k) . 4. If N G, then G is solvable if and only if both N and G/N are solvable. 5. For a direct product, (A × B)(k) = A(k) × B (k) . 6. Solvable groups are closed under finite direct products and subgroups and hence is closed under taking finite subdirect products. In particular, if G is a finite group, there is a normal subgroup N of G minimal among normal subgroups with respect to the property that G/N is solvable. (Clearly this normal subgroup is characteristic; it is called the solvable residual.) 2 In this case, for once, the name “solvable” is not accidental. The reason will be explained in the Chapter 11 on Galois Theory. 184 proof: Part 1. follows from the containment {[a, h]|(a, h) ∈ A × H} ⊆ {[a, g]|(a, g) ∈ A × G}. Part 2. By Part 1, H 0 ≤ G0 . If we have H (k−1) ≤ G(k−1) , then H (k) = [H (k−1) , H (k−1) ] ≤ [G(k−1)) , G(k−1) ] = G(k) , utilizing H (k−1) and G(k−1) in the roles of H and G in the last sentence of Part 1. The result now follows by induction. Part 3. The equal sign is from Theorem 104, Part 5; the subgroup relation is from Part 1 above of this Theorem. Part 4. Suppose G is solvable. Then so are N and G/N by Parts 1 and 3. Now assume, G/N and N are solvable. Then there exists natural numbers k and l, such that (G/N )(k) = 1 and N (l) = 1. Then by Part 3, the former equation yields G(k) N/N = 1 or G(k) ≤ N. Then G(k+l) = (G(k) )(l) ≤ N (l) ≤ 1, by Part 2., so G is solvable. Part 5. If the subgroups A and B of G centralize each other, it follows from Equations 5.2 or 5.3 that [ab, a0 b0 ] = [a, a0 ][b, b0 ] for all (a, b), (a0 , b0 ) ∈ 0 0 0 A × B. This shows (A × B) ≤ A × B . The reverse containment follows from Part 1. Thus the conclusion holds for k = 1. One need only iterate this result sufficiently many times to get the stated result for all k. Part 6. This is a simple application of the other parts. A subgroup of a solvable group is solvable because of the inequality of Part 2. A finite direct product of solvable groups is solvable by Part 5. Thus, from the definition of subdirect product, a finite subdirect product of solvable groups is solvable. The last part follows from the remarks in Section 3.6.2. remark: Suppose G1 , G2 , . . . is a countably infinite sequence of finite groups for which Gk has derived length k (It is a fact that a group of derived length k exists for every positive integer k: perhaps one of those very rare applications P of wreathed products.) Then the direct sum k∈N Gk = G1 ⊕ · · · does not have a finite derived length and so is not solvable. Thus solvable groups are not closed under infinite direct products. 185 Recall that a finite subnormal series for G is a chain G = N0 N1 N − 1 · · · Nk = 1, where each Nj+1 is a normal subgroup of its predecessor, Nj , j = 0, . . . k−1, k is a natural number. (Of course what we have written is actually a descending subnormal series. We could just as well have written it backwards as an ascending series.) Recall also that an unrefineable subnormal series is called a composition series, and when such a thing exists, the factors Nj /Nj+1 are simple groups called the composition factors of G. That the list of composition factors that appears is the same for all composition series, was the substance of the Jordan-Hölder Theorem for groups. These notions make their reappearance in the following: Corollary 107 1. A group G is solvable of derived length at most k if and only if it possesses a subnormal series G = N0 N1 N1 · · · Nk = 1. such that each factor Nj /Nj+1 , j = 0, . . . k − 1, is abelian. 2. A finite group is solvable if and only if it possesses a composition series with all composition factors cyclic of prime order (of course, the prime may depend on the particular factor taken). proof: Part 1. If G is solvable, the derived series is a subnormal series with succesive factors abelian. Conversely, if the subnormal series is as given, 0 then N0 /N1 abelian implies [G, G] = G ≤ N1 . Inductively, if G(j) ≤ Nj , then Nj /Nj+1 abelian implies G(j+1) ≤ [Nj .Nj ] ≤ Nj+1 . So by induction, G(j+1) ≤ Nj+1 for all j = 0, 1, . . . k − 1. Thus G(k) ≤ Nk = 1, and G is solvable. Part 2. If G possesses a composition series with all composition factors cyclic of prime order, G is solvable by Part 1. On the other hand, if G is finite and solvable it possesses a subnormal series G = N0 N1 · · · Nk = 1, with each factor Nj /Nj+1 a finite abelian group. But in any finite abelian group (where all subgroups are normal) we can form a composition series by 186 choosing a maximal subgroup M1 of A, a maximal subgroup M2 of M1 , and so forth. Each factor will then be cyclic of prime order. Thus each interval Nj Nj+1 can be refined to a chain with successive factors of prime order. The result is a refinement of the original subnormal series to a composition series with all composition factors cyclic of prime order. There are many very interesting facts about solvable groups, especially finite ones, which unfortunately we do not have space to reveal. Two very interesting topics are these: • Phillip Hall’s theory of Sylow Systems. In a finite solvable group there is a representative Si of Sylpi (G), the pi -Sylow subgroups of G such that G = S1 S2 · · · Sm , and the Si are permutable in pairs – that is Si Sj = Sj Si where p1 , . . . pm is the list of all primes dividing the order of G. Thus in any solvable group of order 120, for example, there must exist a subgroup of order 15. (This is not true for example of Sym(5)). • The theory of formations. Basically, a formation is an isomorphismclosed class of finite solvable groups closed under finite subdirect products. Then each finite group G contains a characteristic subgroup GF which is unique in being minimal among all normal subgroups N of G which yield a factor G/N belonging to F. (GF is called the F-residual of G.) Whenever G is a finite group, let D(G) be the intersection of all the maximal subgroups of G. This is a characteristic subgroup which we shall meet shortly, called the Frattini subgroup of G. A formation is said to be saturated, if for every solvable finite group G, G/D(G) ∈ F implies G ∈ F. If F is saturated, then there exists a special class of subgroups, the F-subgroups, subgroups in F described only by their embedding in G, which form a conjugacy class. Finite abelian groups comprise a formation but are not a saturated formation. A class of groups considered in the next section are the nilpotent groups and they do form a saturated formation. As a result one gets a theorem like this: “Suppose X and Y are two nilpotent subgroups of a finite solvable group G, each of which is its own normalizer in G. Then X and Y are conjugate subgroups.”3 A bit surprising. 3 This conjugacy class of subgroups was first announced in a paper by Roger Carter; they are called the class of “Carter subgroups” of G. Usually mathematical objects bearing a mathematician’s name are not named for the person who discovered or advanced the 187 5.4 5.4.1 Central Series and Nilpotent Groups. The upper and lower central series. Let G be a fixed group. Set Γ0 (G) = G, Γ1 (G) = [G, G], and inductively define Γk+1 (G) := [Γk (G), G], for all positive integers k. Then the descending sequence of subgroups G = Γ0 (G) ≥ Γ1 (G) ≥ Γ2 (G) ≥ · · · , is called the lower central series of the group G. Note that Γk (G) = [G, G, . . . G] (with k arguments. ) Since any endomorphism f : G → G, maps the arguments of the above multiple commutator into themselves, the endomorphism f maps each member of the lower central series into itself. A similar argument for homomorphisms allows us to state the elementary Lemma 108 1. Each member Γk (G) of the lower central series is a fully invariant subgroup of G. 2. If f : G → H is a homomorphism of groups then f (Γk (G)) = Γk (f (G)). 3. If H is a subgroup of G, then Γk (H) ≤ Γk (G). idea. They got it right for once! But there is more to this business of names. An “inner automorphisms” of a finitedimensional Lie algebra is well-defined: it is one of shape i(n) := exp(ad(n)), where adn is nilpotent. In the theory of finite-dimensional simple complex Lie Algebras, there is a class of sub-algebras C which are characterized by being nilpotent subalgebras which are “self-normaling” in the sense that the only inner automorphisms which leave it invariant are those i(n) with n ∈ C. The theorem is that there is only one conjugacy class (one orbit under the inner automorpsm group) of such self-normalizing nilpotent subalgebras. They are called the “Cartan subalgebras”. Here in solvable groups we have one conjugacy class of self-normalizing nilpotent subgroups and these are called the “Carter subgroups”. 188 The proof is an easy exercise. A group G is said to be nilpotent if and only if its lower central series terminates at the identity subgroup in a finite number of steps. In this case G is said to belong to nilpotence class k if and only if k is the smallest positive integer such that Γk (G) = 1. Thus abelian groups are the groups of nilpotence class 1. The following is immediate from Lemma 108, and the definition of nilpotence. Corollary 109 1. Every homomorphic image of a nilpotent group of nilpotence class at most k is group of nilpotence class at most k. 2. Every subgroup of a nilpotent group of nilpotence class at most k is also nilpotent of nilpotence class at most k. 3. Every finite direct product of nilpotent groups of nilpotence class at most k is itself nilpotent of nilpotence class at most k. The beginning student should be able supply immaculate proofs of these assertions by this time. [ Follow the lead of the similar result for solvable groups in the previous section; do not hesitate to utilize the commutator identies in Part 3.] Clearly, the definitions show that [Γk , Γk ] ≤ [Γk (G), G] = Γk+1 (G), so the factor groups (Γk (G))/(Γk+1 (G)) are abelian. As a consequence, Corollary 110 Any nilpotent group is solvable. The converse fails drastically: Sym(3) is a solvable group of derived length 2. But is not nilpotent since its lower central series drops from Sym(3) to its subgroup N ' Z3 , and stabilizes at N for the rest of the series. Now again fix G. Set Z0 (G) := 1, the identity subgroup; set Z1 (G) = Z(G), the center of G (the characteristic subgroup of those elements of G which commute with every element of G); and inductively define Zk (G) to be that subgroup satisfying Zk (G)/Zk−1 (G) = Z(G/Zk−1 (G)). 189 That is, Zk (G) is the inverse image of the center of G/Zk−1 (G) under the natural homomorphism G → G/Zk−1 (G). For example: Z2 (G) = {z ∈ G|[z, g] ∈ Z(G), for all g ∈ G}. By definition Zk (G) ≤ Zk+1 (G), so we obtain an ascending series of characteristic subgroups 1 = Z0 (G) ≤ Z1 (G) ≤ Z2 (G) ≤ · · · , called the upper central series of G. Of course if G has a trivial center, this series does not even get off the ground. But, as we shall see below, it proceeds all the way to the top precisely when G is nilpotent. Suppose now that G has nilpotence class k. This means Γk (G) = 1 while Γk−1 (G) is non-trivial. Then as [Γk−1 (G), G] = Γk (G) = 1, G centralizes Γk−1 (G), that is Γk−1 (G) ≤ Z1 (G). Now assume for the purposes of induction, that Γk−j (G) ≤ Zj (G). Then [Γk−j−1 , G] ≤ Γk−j (G) ≤ Zj (G), so [Γk−j−1 , G] ≤ {z ∈ G|[z, G] ≤ Zj (G)} := Zj+1 (G). Thus Γk−j (G) ≤ Zj (G), for all 0 ≤ j ≤ k. (5.7) Putting j = k in ( 5.7), we see that Zk (G) = Γ0 (G) = G. That is, if the lower central series has length k, then the upper central series terminates at G in at most k steps. Now assume G 6= 1 has an upper central series terminating at G in exactly k steps – that is, Zk (G) = G while Zk−1 (G) is a proper subgroup of G (as G 6= 190 1). Then, of course G/Zk−1 is abelian, and so Γ1 (G) = [G, G] ≤ Zk−1 (G). Now for the purpose of an induction argument, assume that Γj (G) ≤ Zk−j (G). Then Zk−j (G)/Zk−j−1 (G) = Z(G/Zk−j−1 ) implies [Zk−j (G), G] ≤ Zk−j−1 (G). so Γj+1 (G) := [Γj (G), G] ≤ [Zk−j (G), G] ≤ Zk−(j+1) (G). Thus Γj (G) ≤ Zk−j (G), for all 0 ≤ j ≤ k. (5.8) Thus if j = k in (5.8), Γk (G) ≤ Z0 (G) := 1. Thus if the upper central series terminates at G in exactly k steps, then the lower central series terminates at the identity subgroup in at most k steps. We can assemble these observations as follows: Theorem 111 The following assertions are equivalent: 1. G belongs to nilpotence class k. (By definition, its lower central series terminates at the identity in exactly k steps.) 2. The upper central series of G terminates at G in exactly k steps. There is an interesting local property of nilpotent groups: Theorem 112 If G is a nilpotent group, then any proper subgroup H of G is properly contained in its normalizer NG (H). proof: If G = 1, there is nothing to prove. Suppose H is a proper subgroup of G. Then there is a minimal positive integer j such that Zj (G) is not contained in H. Then there is an element z in Zj (G) − H. Then [z, H] ≤ [z, G] ≤ Zj−1 (G) ≤ H. Thus z normalizes H, but does not lie in it. Corollary 113 Every maximal subgroup of a nilpotent group is normal. 191 remark There are many groups with no maximal subgroups at all. Some of these are not even nilpotent. Thus one should not expect a converse to the above Corollary. This raises the question, though, whether the property of Theorem 112 characterizes nilpotent groups. Again one might expect that this is false for infinite groups. However, in the next subsection we shall learn that among finite groups it is indeed a characterizing property. Here is a classical example of a nilpotent group: Suppose V is an ndimensional vector space over a field F . A chamber is an ascending sequence of subspaces S = (V1 , V2 , . . . Vn−1 ) where Vj has vector space dimension j. (This chain is unrefineable.) Now consider the group B(S) of all linear transformations T ∈ GL(V ) such that T fixes each Vj and induces the identity transformation on each 1-dimensional factor-space Vj /Vj−1 . Rendered as a group of matrices these are upper triangular matrices with “1’s” on the diagonal. The reader may verify that the k-th member of the lower central series of this group consists of upper triangular matrices still having 1’s on the diagonal, but having more and more upper subdiagonals all zero as k increases. Eventually one sees that T (n) is the identity matrix. 5.4.2 Finite nilpotent groups. Let p be a prime number. As originally defined, a p-group is a group in which each element has p-power order. Suppose P is a finite p-group. Then by the Sylow Theorem, the order of G cannot be divisible by a prime distinct from p. Thus we observe A finite group is a p-group if and only if it has p-power order. Now we have seen before, that a non-trivial finite p-group must possess a non-trivial center. Thus if P 6= 1 is a finite p-group, we see that Z(P ) 6= 1. Now if Z(P ) = P , then P is abelian. Otherwise, P/Z(P ) has a nontrivial center. Similarly, in general, if P is a finite p-group: Either Zk (P ) = P or Zk (P ) properly lies in Zk+1 (P ). As a result, a finite p-group possesses an upper central series of finite length, and hence is a nilpotent group. Obviously, a finite direct product of finite p-groups (where the prime p is allowed to depend upon the direct factor) is a finite nilpotent group. Now consider this 192 Lemma 114 If G is a finite group and S is a p-Sylow subgroup of G, then any subgroup H which contains the normalizer of S is itself self-normalizing – i.e. NG (H) = H In particular NG (S) is self-normalizing. proof: Suppose x is an element of NG (H). Since P is a p-Sylow subgroup of G, it is also a p-Sylow subgroup of NG (H) and H. Since P x ∈ Sylp (H), by Sylow’s theorem there exists an element h ∈ H such that P x = P h . Then xh−1 ∈ NG (P ) ≤ H whence x ∈ H. (Actually this could be done in one line, by exploiting the “Frattini Argument” (Chapter 3; Corollary 14). Now we have Theorem 115 Let G be a finite group. Then the following statements are equivalent. 1. is G is nilpotent. 2. G is isomorphic to the direct product of its various Sylow subgroups. 3. Every Sylow subgroup of G is a normal subgroup. proof: ( 2. implies 1.) Finite p-groups are nilpotent, as we have seen, and so too are their finite direct products. Thus any finite group which is the direct product of its Sylow subgroups is certainly nilpotent. (2. if and only 3.) Now it is easy to see that a group is isomorphic to a direct product of its various p-Sylow subgroups, (p ranging over the prime divisors of the order of G) if and only if each Sylow subgroup of G is normal. Sylow subgroups of such a direct product are certainly normal (being direct factors), and conversely, if all p-Sylow subgroups are normal, the direct product criterion of Chapter 2 (Lemma 13, Part (3)) is satisfied, upon considering the orders of the intersections S1 S2 · · · Sk ∩ Sk+1 , where {Si } is the full (one-element) collection of Sylow subgroups of G (one for each prime). (1. implies 3.) So it suffices to show that in a finite nilpotent group, each p-sylow subgroup P is a normal subgroup of G. If NG (P ) = G, there is nothing to prove. But if NG (P ) is a proper subgroup of G, then it is a proper self-normalizing subgroup by Lemma 114. This completely contradicts Theorem 112 which asserts that every proper subgroup of a nilpotent group is properly contained in its normalizer. 193 Fix a group G. A non-generator of G is an element x of G such that whenever x is a member of a generating set X (that is, when x ∈ X, and hXi = G), then X − {x} is also a generating set: hX − {x}i = G. Thus x is just one of those superfluous elements that you never need4 . Now it is not easy to see from “first principles” – that is, from the defining property alone, that in a finite group, the product of a non-generator and a non-generator is again a non-generator. But we have this characterization of a non-generator: Lemma 116 Suppose G is a finite group. Let D(G) be the intersection of all maximal subgroups of G. Then D(G) is the set of all nongenerators of G. proof: Suppose x is a non-generator, and let M be any maximal subgroup of G. Then if x is not in M we must have h{x} ∪ M i = G, while hM i = M , against the definition of “non-generator”. Thus x ∈ D(G). On the other hand assume that x is an arbitrary element of D(G), and that X is a set of generators of G which counts x among its active participants. If X − {x} does not generate G, then, as G is finite, there is a maximal subgroup M containing H := hX − {x}i. But by assumption x ∈ D(G) ≤ M , whence hXi = M , a contradiction. Thus always it is the case that if x ∈ X, where X generates G, then also X − {x} generates G. Thus x is a non-generator. The proof is complete. remark: This theorem really belongs to posets. Clearly finiteness can be replaced by the only distinctive property used in the proof: that for every subset X of this poset, there is an upper bound – that is, an element of the filter P X , which is either the 1-element of the poset of lies in a maximal member of the poset with the 1-element removed. Its natural habitat is therefore lattices with the ascending chain condition. (We need upper semi-lattices so that we can take suprema; we need something like a lower-semilattices in order to define the Frattini element of the poset as the meet of all maximal elements. Finally the ascending chain condition makes sure that the maximal elements fully cover all subsets whose supremum is not “1”. So at least lattices with the ascending chain condition (ACC) are more than sufficient for this result.) Theorem 117 (Global non-generators) If X ∪ D(G) generates G, then X generates G. 4 Strangely, some College Presidents come to mind. 194 proof: The argument is the same and has the same poset-generalization. If the supremum of X − D(G) is not G, it lies below a maximal member M , of the poset, and so it would be impossible that sup(X) = G. Thus X − D(G) has the “top” element G as it supremum. The subgroup D(G) is called the Frattini subgroup of G 5 . It is clearly a characteristic subgroup of G. In the case of finite groups, something special occurs which could be imitated only artificially in the context of posets: Theorem 118 In a finite group, the Frattini subgroup D(G) is a characteristic nilpotent subgroup of G. proof: The “characteristic” part is clear, and has been remarked on earlier. Suppose P ∈ Sylp (D(G). By the Frattini Argument, G = NG (P )D(G), It now follows from Theorem 117 that NG (P ) = G and so in particular P is normal in D(G). Thus every p-sylow subgroup of D(G) is normal in it. Since it is finite, it is a direct product of its Sylow subgroups and so is nilpotent. Corollary 119 (Wielandt) For a finite group G the following are equivalent: 1. G is nilpotent. 2. Every maximal subgroup of G is normal. 3. G0 ≤ D(G). proof: If M is a maximal subgroup of the finite group G which also happens to be normal, then G/M is a simple group with no proper subgroups, and so must be a cyclic group of order p, for some prime p. Thus if all maximal subgroups are normal, the commutator subgroup lies in each maximal subgroup and so lies in the Frattini subgroup D(G). On the other hand, if G/D(G) is abelian, every subgroup above D(G) is normal; so in particular, all maximal subgroups are normal. Thus Parts 2 and 3 are equivalent. But if M is a maximal subgroup of a nilpotent group, M properly lies in its normalizer in G, and so NG (M ) = G. Thus the assertion of Part 1. implies Parts 2. and 3.. 5 In some of the earlier literature one finds D(G) written as “Φ(G)”. I do not understand fully the reason for the change. 195 So it remains only to show that if every maximal subgroup is normal, then G is nilpotent. Since G is finite, we need only show that every Sylow subgroup is normal, under these hypotheses. Suppose P is a p-Sylow subgroup of G. Suppose NG (P ) 6= G. Then NG (P ) lies in some maximal subgroup M of G. By Lemma 114, M is self-normalizing. But that contradicts our assumption that every maximal subgroup is normal. Thus NG (P ) = G for every Sylow subgroup P . Now G is nilpotent by Theorem 115. 5.5 Coprime Action Suppose G is a finite group possessing a normal abelian subgroup A, and let F be the factor group G/A. Any transversal of A — that is, a system of coset representatives of A in G can be indexed by the elements of the group F . Any transversal T := {tf |f ∈ F } determines a function αT : F × F → A (called a factor system) by the rule that for all (f, g) ∈ F , tf · tg = tf g αT (f, g). In some sense, the factor system measures how far off the transversal is from being a group. Because of the associative law, one has αT (f g, h)αT (f, g)h = αT (f, gh)αT (g, h), (5.9) for all f, g, h ∈ F . (Notice that since A is abelian, conjugation by any element of the coset h = Ax gives the same result ax = x−1 ax which we have denoted by ah .) Now, given a transversal T = {tf |f ∈ F }, any other transversal S = {sf |f ∈ F } is uniquely determined by a function β : F → A, by the rule that sf = tf β(f ), f ∈ F. Then the new factor system αS for transversal S can be determined from the old one αT by the equations αS (f, g) = αT (f, g)[(β(f g))−1 β(f )g β(g)] for all f, g ∈ F. (5.10) The group G is said to split over A if and only if there is a subgroup H of G such that G = HA, H ∩ A = 1 — that is, if and only if some transversal 196 H forms a subgroup. Such a subgroup H is called a complement of A in G. In that case, the factor system αH for H is just the constant function at the identity element of A. Then, given G, A and transversal T = {tf |f ∈ F }, we can conclude from equation (5.10) that G splits over A if and only if there exists a function β : F → A such that αT (f, g) = β(f g)β(f )−g β(g)−1 , for all f, g ∈ F. (5.11) Now suppose A is a normal abelian p-subgroup of the finite group G and suppose P is a Sylow p-subgroup of G with the property that ““‘P splits over A”. That means that P (which necessarily contains A) contains a subgroup B such that P = AB and B ∩ A = 1. Let X be a transversal of P in G. Then T := BX := {bx|b ∈ B, x ∈ X} is a transversal of A in G whose factor system αT satisfies αT (b, t) = 1 for all (b, t) ∈ B × T. Now let m be the p-Sylow index [G : P ] = |X| in G. Since m is prime to |A|, there exists an integer k such that mk ≡ mod |A|. Since A is abelian, the factors of any finite product of elements of A can be written in any order. So there is a function γ : F → A defined by γ(f ) := Y x∈X αT (x, f ). Now, multiplying each side of equation (5.9) as f ranges over F , one obtains γ(h)γ(g)h = γ(gh)αT (gh)m . Then, setting β(g) := γ(g)−k for all g ∈ F , we obtain equation (5.11). Thus G splits over A. Noting that conversely, if G splits over A, so does P , we have in fact proved, Theorem 120 (Gaschütz) Suppose G is a finite group with a normal abelian p-subgroup A. Then G splits over A if and only if some p-Sylow subgroup of G splits over A. Next suppose A is a normal abelian subgroup of the finite group G and that H is a complement of A in G, so G = HA and H ∩ A = 1. Now 197 suppose some second subgroup K was also a complement to A. Then there is a natural isomorphism µ : H → K factoring through G/A, and a function β : H → A so that µ(h) = hβ(h) for all h ∈ H. Expressed slightly differently, K is a transversal {kh |h ∈ H} whose elements can be indexed by H, and kh = hβ(h) for all h ∈ H. Since K is a group, it is easy to see that for h, g ∈ H, kg · kh = hgh = (gh)β(gh) = gβ(g) · hβ(h) = ghβ(g)h β(h). So β(gh) = β(g)h β(h), for all g, h ∈ H. (5.12) Now assume m = [G : A] is relatively prime to |A|, and select an integer n Q such that mn ≡ mod |A|. Then as A is abelian, the product H β(g), taken as g ranges over H, is a well-defined constant b, an element of A. Forming a similar product of the terms on both sides of equation (5.12) as g ranges over H, one now has b = bh · β(h)m , so β(h) = a · a−h , where a = bn , h ∈ H. Then for every h ∈ H, kh = hβ(h) = h · a−h · a = a−1 ha. Thus K = a−1 Ha is a conjugate of H. So we have shown Lemma 121 If A is a normal abelian subgroup of a group G of finite order, and the order of A is relatively prime to its index [G : A], then any two complements of A in G are conjugate. The reader should be able to use Gaschütz’ Theorem and the preceeding Lemma to devise an induction proof of the following Theorem 122 (The lower Schur-Zassenhaus Theorem) Suppose N is a solvable normal subgroup of the finite group G whose order is relatively prime to its index [G : N ]. Then 198 1. there exists a complement H of N in G, and 2. any two complements of N in G are conjugate by an element of N . On the other hand, using little more than Sylow’s theorem, a Frattini argument, and induction on group orders one can easily prove Theorem 123 (The upper Schur-Zassenhaus Theorem) Suppose N is a normal subgroup of the finite group G, such that |N | and its index [G : N ] are coprime. If G/N is solvable, then 1. there exists a complement H of N in G, and 2. any two complements of N in G are conjugate by an element of N . Remark: Both the lower and upper Schur-Zassenhaus theorems contain a hypothesis of solvability, either on N or G/N . However this condition can be dropped altogether, because of the famous Feit-Thompson theorem which says that any finite group of odd order must be solvable. The hypothesis that |N | and [G : N ] are coprime forces one of N or G/N to have odd order and so at least one is solvable. Of course the Feit-Thompson theorem is far beyond the sophistication of this relatively elementary course. As an application of the Schur-Zassenhaus theorem, one can show Theorem 124 (Glauberman) Let A be a group of automorphisms acting on a group G leaving invariant some coset Hx of a subgroup H. Then the following hold: 1. H is itself A-invariant. 2. If A and H have coprime orders (so that at least one is solvable), then A fixes an element of the coset Hx. Proof: This is an exercise. 5.6 Exercises for Chapter 4 Exercise 1. We say that a subgroup A of G is subnormal in G if and only if there is a finite ascending chain of subgroups: A = N0 < N 1 · · · < N m = G 199 from A to G, with each member of the chain normal in its successor – i. e. Ni Ni+1 , i = 0, . . . , m−1. Using the Corollary to the Fundamental Theorem of Homomorphisms, show that SN (G) is a semi-modular lower semi-lattice. Exercise 2. Suppose N is a family of normal subgroups of G. Let hN i be defined to be the set of elements of G which are expressible as a finite product of elements each of which belongs to a subgroup in N . 1. Show that hN i is a subgroup of G. 2. We say that N is a normally closed family, if and only if for any non-empty subset M ⊆ N , hMi ∈ N . A group is said to be a torsion group if and only if every element of G has finite order. (This does not mean that the group is finite: far from it: there are many infinite torsion groups.) Show that the family T (G) of all normal torsion subgroups of G, is a normally closed family. [hint: When any element x is expressed as a finite product x = a1 · · · ak with ai ∈ Ni ∈ N , only a finite number of groups Ni are involved. So the proof comes down to the case k = 2.] Then the group Tor(G) := hT (G)i ∈ T (G) is a characteristic subgroup of G Exercise 3. Fix a prime number p. A group H is said to be a p-group if and only if evey element of H has order a power of the prime p. Show that the collection of all normal p-subgroups of a (not necessarily finite) group G is normally closed. (In this case the unique maximal member of the collection of all normally closed p-subgroups of G is denoted Op (G).) Exercise 4. Let G be any group, finite or infinite. 1. Show that any subnormal torsion subgroup of G lies in T (G). 2. Show that any subnormal p-subgroup of G lies in Op (G). Exercise 5. Let Q denote the additive group of the rational numbers. The group Zp∞ is defined to be the subgroup of Q/Z generated by the infinite set {1/p, 1/p2 , . . .}. Show that this is an infinite p-group. Also show that it is indecomposable, – that is, it is not the direct product of two of its subgroups. Exercise 6. Let F be a family of abstract groups which is closed under taking unrestricted subdirect products – that is, if H is a subgroup of the direct 200 product σ∈I Xσ , with each Xσ ∈ F, and the restriction of each canonical projection Y πτ : σ∈I Xσ → Xτ Q to H is an epimorphism, then H ∈ F. 1. Show that there exists a unique normal subgroup NF such that NF is minimal with respect to being a normal subgroup which yields a factor G/NF ∈ F 6 2. Show that there is always a subgroup G0 of G minimal with respect to these properties: (a) G0 G, (b) G/G0 is abelian. 3. Explain why there is not always a subgroup N of G minimal with respect to being normal and having G/N a torsion group (or even a pgroup). [Hint: Although torsion groups are closed under taking direct sums, they are not closed under taking direct products.] 4. If G is finite, and F is closed under finite subdirect products – that is, the set of possibilities for F is enlarged because of the weaker requirement that it is enough that any subgroup H of a finite direct product of members of F which remains surjectve under the restrictions of the canonical projections, is still in F– then there does exist a subgroup NF which is minimal with respect to the two requirements: (i) NF G, and (ii) G/NF ∈ F. (If π is any set of (non-negative) prime numbers, a group G is said to be a π-group if and only every element of G has finite order divisible only by the primes in π. In particular, a π-group is a species of torsion group, and a p-group is a species of π-group. Show that all π-groups are closed under finite subdirect products. [In this case we write 0 NF = Oπ (G), Op (G), or Op (G) 6 Note that as a class of abstract groups F can either be viewed as either (i) a collection of groups closed under isomorphisms, or (ii) a disjoint union of isomorphism classes of groups. Under the former interpetation, the “membership sign”, ∈, as in “ G/NF ∈ F” is appropriate. But in the second case, “∈” need only be replaced by “an isomorphism with a member of F”. One must get used to this. 201 where π = {p} or all primes except p, in the last two cases.] Exercise 7. Suppose A and B are normal subgroups of a group G. Prove the following: 1. If A and B normalize subgroup C, then [AB, C] = [A, C][B, C]. [Hint: Use the commutator identity (1.2) and the fact that B normalizes [A, C] to show the containment of the left side in the right.] 2. Show that the k-fold commutator [AB, . . . , AB] is a normal product Y [X1 , X2 , . . . , Xk ], where the Xi = A or B, and the product extends over all 2k possibilities for the k-tuples (X1 , . . . , Xk ). (Of course each factor in the normal product is a normal subgroup of G. But some of them occur more than once since the cases with X1 = A and X2 = B, are duplicated in the cases with X1 = B and X2 = A.) 3. Show that if A and B are nilpotent of class k and `, respectively, then AB is nilpotent of class at most k + `. [Hint: Consider Γk+`+1 (AB) and use the previous step.] 4. Prove that in a finite group, there exists a normal nilpotent subgroup which contains all other normal nilpotent suproups of G. (This characteristic subgroup is called the Fitting subgroup of G and is denoted F(G).) The next few excercizes spell out a subtle but well-known theorem known as the Baer-Suzuki theorem (Not to be confused with “Brauer-Suzuki Theorem”7 The proof considered here is actually do to Alperin and Lyons[?].) 7 Still another instance in which Mathematical History, with a Nabikovian perverseness, tends to place name-similarities in the students’s path. Theorem ??, on the existence and conjugacy of Carter subgroups of finite solvable groups has the same hypotheses for subgroups as does the theorem on the existence and conjugacy of Cartan subgroups of finite-dimensional complex Lie algebras. A folklore common in the graduate student’s “coffee-klatches” of the 50’s was that Philip Hall knew the theorem in advance, and merely 202 In the following G is a finite group. There is some new but totally natural notation due to Aschbacher. Suppose Σ is a collection of subgroups of G and H is a subgroup of G. Then the symbol Σ ∩ H denotes the collection of all subgroups in Σ which happened to be contained in the subgroup H. (Of course this collection could very well be empty.) If Γ is any collection of subgroups of G, the symbol NG (Γ) denotes the subgroup of G of all elements which under conjugation of subgroups leave the collection Γ invariant. Exercise 8. Suppose Ω is a G-invariant collection of p-subgroups of G. Let Q be any p-subgroup of G and let Γ be any subset of Ω ∩ Q for which Γ ⊆ NG (Γ). Show that either 1. Γ = Ω ∩ Q or 2. Then there is an element X ∈ Ω ∩ Q which also lies in NG (Γ). [Hint: Without loss of generality, one may assume that Γ = Ω ∩ Q ∩ NG (Γ). Then every element of NQ (NQ (Γ)) leaves Γ = Ω ∩ NQ (Γ) invariant, so NQ (NQ (Γ) = NQ (Γ). Since Q is a nilpotent group, Γ is a Q-invariant collection of subgroups of Q and the conclusion follows easily.] Exercise 9. Prove the following: Theorem 125 (The Baer-Suzuki Theorem) Suppose X is some p-subgroup of the finite group G. Either X ≤ Op (G) or hX, X g i is not a p-subgroup for some conjugate X g ∈ X G . [Hint: Set Ω = X G , choose P a p-sylow subgroup of G which contains X, and set ∆ := Ω ∩ P . Since ∆ = Ω implies hΩi is a normal p-group, forcing the conclusion X ∈ Op (G), we may assume Ω − ∆ 6= ∅. Now show 1. that for any subgroup Y ∈ Ω − ∆, hY, ∆i cannot be a p-group. [ If R is a p-Sylow subgroup containing hR, ∆i, compare |Ω ∩ R| and |Ω ∩ P |.] suggested it to a very bright student of a similar name when he or she (statiscally) should eventually happen along. I reject this hypothesis as being sociologically unlikely: (1) with at most a handful of good students alloted by fate to any Professor, it is very improbalbe that one of these has a name closely resembling a given one, (2) Carter could have proved anything he liked, and (3) such an hypothesis would interfere with the lofty thesis of this footnote that there is a natural and charming and mystical tendency for this to happen. How else should it happen that the mathematician Andrew Wiles who proved Fermat’s last theorem not have a name resembling Andre Weil? 203 Next consider the set S of all pairs (Y, Γ) where Y ∈ X G − ∆, Γ ⊆ ∆ and hY, Γi is a p-group. (We have just seen that this collection is non-empty.) Among these pairs, choose one, say (Y, Γ) ∈ S, so that |Γ| is maximal. Now if Γ = ∅, hXi is not a p-group and we are done. Thus we may assume Γ is non-empty. Now set Q := hY, Γi, a p-group. Then, noting that hY, ∆∩Qi is a p-group, the maximality of Γ forces Γ = ∆ ∩ Q. Now Γ̄ := ∪X∈Γ X ⊆ P ⊆ NG (∆) and Γ̄ ⊆ Q together imply Γ̄ ⊆ NC (Q) ∩ NG (∆) ⊆ NG (Q ∩ ∆) = NG (Γ). By the previous exercise, either Γ = Ω ∩ Q or there is a subgroup Y 0 ∈ Ω ∩ Q − Γ with Y 0 ⊆ NG (Γ). The former alternative is dead from the outset because of the existence of Y . Now Y 0 belongs to X G − ∆, and hY 0 , Γi is a p-subgroup, as it is a subgroup of Q. Thus (Y 0 , Γ) is also one of the extreme pairs in S. Similarly, since hY 0 , ∆i is not a p-group, Γ 6= ∆ and so, applying the previous exercise once more with P in place of Q, there must exist a subgroup Z ∈ ∆ − Γ, with Z ∈ NP (Γ). Thus {Y 0 , Z} ∪ Γ ⊆ Ω ∩ NG (Γ) . Now NG (Γ) is a proper subgroup of G (otherwise some element of X G lies in the normal p-subgroup hΓi). Now for the first time, we exploit induction on the group order to conclude that hY 0 , Γ ∪ {Z}i is a p-group. Clearly this contradicts the maximality of |Γ|, completing the proof.] Exercise 9. The proof just sketched is very suggestive of other generalizations, but there are limits. We say that a finite group is p-nilpotent if and only if G/Op0 (G) is a p-group. Show that the collection of finite p-nilpotent subgroups is closed under finite normal products. (In a finite group, there is then a unique maximal normal p-nilpotent subgroup, usually denoted Op0 p (G).) 204 Is the following statement true? Suppose X is a p-subgroup of the finite group G. Then either X ∈ Op0 p (G) or else there exists a conjugate X g ∈ X G such that hX, X g i is not p-nilpotent. [Hint: Think about the fact that in any group, the group generated by two involutions is a dihedral group.] Exercise 10. Suppose F is a family of groups closed under normal products. Suppose G possesses a maximal normal subgroup N belonging to F (this occurs when G is finite. Is it true that any subnormal subgroup of G lies in N ? Prove it, or give a counterexample. Finally we add an “open problem”8 . Exercise 11. Open problem: Prove the following Theorem 126 (Brodkey’s Theorem Extended) Suppose G is a finite group with Op (G) = 1 — that is, G has no nontrivial normal p-subgroups. Show that there exist two p-Sylow subgroups P and P g such that, P ∩ P g = 1. [Yes, one could reduce to the case that G is simple and use the classification of finite simple groups to solve the problem. But of course that would condemn the theorem to a rather limited usefulness. At that level, it could never be used to streamline the theory of the classification, and could be applied only in fields willing to accept the classification – say coding theory or most of the work on flag-transitive incidence geometries. Moreover, the hypotheses are so elementary that one could easily imagine that the theorem could be proved just with the theorems developed so far. Such a principle is not a good guide as the the Fermat Theorem shows. Moreover, there are other theorems in Group theory, such as the theorem concerning the Frobenius Kernel, which are equally naive in their hypotheses, yet seem to require character theory for no good reason. Nonetheless, one cannot escape the feeling, that there is some analogy between the Brodkey result and the Alperin-Lyons proof of the Baer-Suzuki theorem. An analogue in sesquilinear forms (Chapter 16) is the fact that for non-degenerate polar spaces of finite rank, there exist two maximal totally isotropic (or totally singular) subspaces which intersect at 0.] Exercise 12. There are other open generalizations of the Baer-Suzuki Theorem due to Chat-Yin Ho. Here is one of his examples: Suppose P is a 8 Probably the word “open” only means that the solutions are not apparent to the author; but this definition has such an overwhelming breadth that it dilutes all meaning. 205 p-Sylow subgroup of the finite group G. Suppose X is a p-subgroup with the property that hX, Y i is a p-subgroup for every subgroup Y in X G ∩ P (using the Aschbacher notation). Then necessarily X ⊆ P . Is this true? Exercise 13. A student of this course should be able to explain exactly why this generalization is logically as strong as the Baer-Suzuki theorem. 206 Chapter 6 Generation in Groups 6.1 Introduction One may recall that for any subset X of a given group G, the subgroup generated by X always has two descriptions: 1. It is the intersection of all subgroups of G which contain X; 2. It is the subset of all elements of G with are expressible as a finite product of elements of X or their inverses in G. The two notions are certainly equivalent, for the set of elements described in item 2, is closed under group multiplication and taking inverses and so is a subgroup of G (by the Subgroup Criterion) containing X. On the other hand it must be a subset of every subgroup containing X. So this set is the same subgroup which was denoted hXi. But there is certainly a difference in the way the two notions feel. Our point of view in this chapter will certainly be in the spirit of the second of these two notions. 6.2 6.2.1 The Cayley graph Definition. Suppose we are given a group G and a subset X. X is said to be a set of generators of G if and only if hXi = G. For any such set of generators X, we can obtain a directed graph C(G, X) without multiple edges as follows: 207 1. The vertex set of C := C(G, X) is the set of elements of G. 2. The directed edges are the ordered pairs of vertices: (g, gx), where g ∈ G and x ∈ X. (It is our option to label such an edge by the symbol “x”, since each edge could only be assigned in this way a unique label.) The directed labelled graph C(G, X) is called the Cayley graph of the generating set X of the group G. One now notices that left multiplication of all the vertices of C (that is, the elements of G) by a given element h of G induces an automorphism πh of C (in fact one preserving the labels) since (hg, hgx) is a an edge labelled by “x”. Let us look at a very simple example. The set X = {a = (12), b = (23)} is a set of generators of the group G = Sym(3), the symmetric group on the set {1, 2, 3}. The Cayley graph then has six vertices, and each edge has two outgoing edges labeled “a” and “b”, respectively, and two ingoing edges labeled “a” and “b” as well. Indeed, we could coalesce an outgoing and ingoing edge with the same label, as one undirected edge of that label. This only happens when the label indicates an involution of the group. In this case we get a hexagon with alternating sides labelled “a” and “b” alternately. 6.2.2 Morphisms of various sorts of graphs. Since we are talking about graphs, this may be a good time to discuss homomorphisms of graphs. But there are various sorts of graphs to discuss. The broad categories are 1. Simple graphs Γ = (V, E) where edges are just subsets of V , the vertex set, of size two. (These record all unreflexive symmetric relations on a set V , for example the acquaintanceship relation at a large party.) 2. Undirected graphs with multiple edges . (This occurs when the edges themselves live an important life of their own – for example when they are the bridges connecting islands, or doors connecting rooms.) 3. Simple graphs with directed edges. Here the edge set is a collection of ordered pairs of vertices. (This occurs when one is recording assymmetric relations, for example, which Football team beat which in a Football League, assuming each pair of teams plays at most once. Or the the comparison relation of elements in a poset. We even allow 208 two relations between points, for example (a, b) and (b, a) may both be directed edges. This occurs, if we have sites in the “Alt Stadt” which are connected by a network of both one-way and two-way streets.) 4. Any of the above, with labels attached to the edges. Cayley graphs, for example, can be viewed as directed labeled graphs. A graph homomorphism f : (V1 , E1 ) → (V2 , E2 ) is a mapping, f : V1 → V2 such that if (a, b) ∈ E then either f (a) = f (b) or (f (a), f (b)) ∈ E2 . If the two graphs are labeled, there are labelling mappings λi : Ei → Li assigning to each (directed) edge, a label from a set of labels Li , i = 1, 2. Then we require an auxilliary mapping f 0 : L1 → L2 , so that if (a, b) ∈ E1 carries label x and (f (a), f (b)) is an edge, then it too carries label f 0 (x). 6.2.3 Morphisms of groups induce morphisms of Cayley graphs. Now suppose f : G → H is a surjective morphism of groups. Then it is easy to see that if X is a set of generators of G, then also f (X) is a set of generators of f (G) = H. Now f is a fortiori a map from the vertex set of the Cayley graph C(G, X) to the vertex set of the Cayley graph C(H, f (X)). For (g, x) ∈ G × X, (g, gx) is a directed edge labelled x. Then clearly either f (g) = f (gx) or else, (f (g), f (gx)) is a directed edge of C(H, f (X)), labelled f (x). Thus Lemma 127 If X is a set of generators of a group G and f : G → H is an epimorphisms of groups, then f (X) is a set of generators of H and f induces a homomorphism of the Cayley graphs, f : C(G, X) → C(H, f (X)) as labeled directed graphs. 6.3 Free groups The Cayley graphs of the previous section were defined by the existence of a group G (among other things). Now we would like to reverse this; we should like to start with a set X and obtain out of it a certain group which we shall denote F (X). We do this in stages. 209 First we give ourselves a fixed abstract set X. The free monoid on X is the collection M (X), of “words” xi x2 · · · xk spelled with “letters” xi chosen from the alphabet X.1 The number k is allowed to be zero – that is, there is a unique “word” spelled with no letters at all –which we denote by the symbol φ and call the empty word.2 The number k is called the length of the word which, as we have just seen, can be zero. The concatenation of two words u = x1 · · · xk and w = y1 y2 · · · y` , xi , yj ∈ X, is the word, u ∗ w := x1 · · · xk y1 · · · y` . The word “bookkeeper” is the concatenation of “book” and “keeper”; 12345 is the cancatenation 123 ∗ 45 or 12 ∗ 345, and so forth. Clearly the concatenation operation “*” is a binary operation on the set W (X) of all words over the alphabet X, and it is associative (though very non-commutative). Moreover, for any word w ∈ W (X), φ ∗ w = w ∗ φ = w. So the empty word φ is a twosided identity with respect to the operation “∗”. Thus M (X) = (W (X), ∗) is an associative semigroup with identity, – that is a monoid. Indeed, since it is completely determined by the abstract set X alone, it is called the free monoid on (alphabet) X.3 Now let σ : X → X be an involutory fixed-point-free permutation of X. This means σ is a bijection from X to istelf such that 1. σ 2 = 1X , the identity mapping on X, and 2. σ(x) 6= x for all elements x in X. We are going to construct a directed labelled graph Γ = F (X, σ), called a frame which is completely determined by X and the involution σ. Let W ∗ (X) be the subset of W (X) of those words y1 y2 · · · yr , r any nonnegative integer, for which yi 6= σ(yi+1 ), for any i = 1, . . . , r − 1. These are 1 Of course, to be formal about it, one can regard these as finite sequences (x1 , . . . , xk ), but the analogy with linguistic devices is helpful because the binary operation we use here is not natural for sequences. 2 I suppose one could say that blank pieces of paper are literally “literally” covered with some number of these words – but a philosophical paper on the issue of how many empty words are on a blank piece of paper is a medieval project that could best be described by such words. 3 Monoids were introduced in Chapter 1. 210 the words which never have a “factor” of the form yσ(y). We call them the reduced words. We now construct a labelled directed graph Γ whose vertices are these reduced words in W ∗ (X). The set E of oriented labelled edges are of two types: 1. Ordered pairs (w, wy) ∈ W ∗ (X) × W ∗ (X) are to be directed edges labelled “y”. (Note that the nature of the domain forces the reduced word w not to end in the letter σ(y) – otherwise wy would not be reduced.) 2. Similarly those ordered pairs of the form (wy, w) (under the same hypothesis that w not end in σ(y)) are directed edges labeled by “σ(y)”. The two types of directed edges are distinguished this way: in the first kind, the “head” of the directed edges is a word of length one more than it’s “tail”, while the reverse is the case for the second kind of edge. Then the directed graph Γ = (W ∗ (X), E) which we have defined and have called a frame has these properties: 1. Let Vk be the set of words of W ∗ (X) of length k. Then V0 = {φ}, and the vertex set partitions as W ∗ (X) = V0 + V1 + V2 + . . .. 2. Each vertex has all its outgoing edges labeled by X; it also has all its ingoing edges labeled by X. Each directed edge (a, b) that is labeled by y ∈ X, has its “transpose” (b, a) labelled by σ(y). 3. For k > 0, each vertex in Vk receives just one ingoing edge and gives one outgoing edge to exactly one vertex in Vk−1 . All further edges leaving or arriving at this vertex, are to or from vertices of Vk+1 . 4. Given any ordered pair of vertices (w1 , w2 ) (recall the wi are reduced words) there is a unique directed path in Γ from vertex w1 to vertex w2 . It follows that there are no circuits other than “backtracks” – that is circular directed walks of the form (v1 , v2 , . . . vm , . . . v2m+1 ) where vj = v2m+1−j , (vj , vj+1 ) and (vj+1 , vj ) are all directed edges, j = 1, . . . m . 211 At this point we define a monoid homomorphism µ : M (X) → Aut(Γ). For each y ∈ X, and vertex w in W ∗ (X) define µ(y)(w) := ( yw if w does not begin with letter σ(y) w0 if w has the form σ(y)w0 . It is then quite clear that µ(y) is a label-preserving automorphism of Γ (this needs only be checked at vertices w of very short length where “cancellation” affects the terminal vertices. Notice that by definition µ(σ(y)) = (µ(y))−1 , the inverse of the automorphism induced by σ(y). Next, for any word m = y1 y2 · · · yk in the monoid M (X) we set µ(m)(w) := µ(yk )(µ(yk−1 )(· · · (µ(y1 )(w)) · · ·), that is, µ(m) is the composition µ(yk ) ◦ · · · ◦ µ(y1 ) of automorphisms of Γ. Clearly, µ(m1 m2 ) = µ(m1 )µ(m2 ) for all (m1 , m2 ) ∈ M (X) × M (X) and µ(φ) = 1W ∗ (X) . So µ is a monoid homomorphism µ : W (X) → Aut(Γ). Note that µ is far from being 1-to-1. For each y ∈ X, µ(yσ(y)) = 1Γ . the identity automorphism on Γ. Thus all compound expressions derived from the empty word by successively inserting factors of the form yσ(y), – such as abσ(c)deσ(e)σ(d)cf gσ(g)σ(f )σ(b)σ(a) – also induce the identity automorphism of Γ. We now have Lemma 128 Let G = Aut Γ, the group of all label preserving automorphisms of the directed graph Γ. 1. The monoid homomorphism µ : M (X) → G is onto. 212 2. G acts regularly on the vertices of Γ. 3. The group G is generated by the set µ(Y ), where Y is any system of representatives of the σ-orbits on X. 4. Identities of the form µ(m) = 1Γ occur if and only if the word m ∈ M (X) is formed by a sequence of insertions of factors yi σ(yi ) starting with the empty word. 5. Γ is the Cayley graph of G with respect to the set of generators X. proof: Suppose g is an element of G fixing a vertex w in Γ. Then, as each (in- or out-) edge on w bears a unique label, every vertex next to w is fixed. Since every vertex u is connected to the vertex φ, the empty word in W ∗ (X), by an “undirected path” of length equal to the length `(u), the undirected version of Γ is connected. This forces the elements g in the first sentence of this paragraph to fix all vertices W ∗ (X) of Γ. But now it is clear that for any word w ∈ W ∗ (X), regarded as a monoid element, µ(w) is an element of G taking vertex φ to w. Thus µ(M (X)) = G, and 1. and 2. are proved. So is the statement that µ(X) is a set of generators of G in part 3. Suppose now, that w = y1 y2 · · · yk ∈ M (X), satisfies µ(w) = 1Γ . Then sucessive premultiplications by yk , then yk−1 , and so on, take the vertex φ on a journey along a directed path with successive arrows labelled yk , yk−1 , . . .; eventually returning to φ via a directed edge (y1 , φ). At each stage of this journey, the current image of φ is either taken to a vertex that is one length further or else one length closer to its original “home”, vertex φ. So there is a position v = µ(yj yj+1 · · · yk )(φ) in which a local maximum occurs in the distance from “home”. This means that both v − = µ(yj+1 · · · yk )(φ) and v + = µ(yj−1 yj · · · yk )(φ) are one unit closer to the starting vertex φ than was v. But as pointed out in the construction of Γ, each vertex z at distance d from φ has only one out-edge and one in-edge to a vertex at distance d − 1 from φ, and the two closer vertices are the same vertex –say z1 –, so that the two edges in and out of the vertex z are two orientations of an edge on the same pair of vertices {z, z1 }. It follows therefore that v − = v + and that yj , the label of directed edge (v − , v), is the σ-image of the label yj−1 of the reverse directed edge (v, v + ). Thus 213 µ(w) = µ(y1 y2 · · · yj−2 yj−1 yj yj+1 · · · yk ) = µ(y1 · · · yj−1 yj+1 · · · yk ) = 1Γ (6.1) Now one applies induction on the length of w0 := y1 · · · yj−2 yj+1 · · · yk , to complete the proof of 3. Part 4. is obvious. Now let Y be a system of representatives of the σ-orbits on X. Then we have X = Y ∪ σ(Y ), and Y ∩ σ(Y ) = ∅. The group G above is canonically defined (one sees) by the set Y alone. To recover X one need only formally define X to be two disjoint copies of Y , and invoke any bijection Y → Y to define the involutory permutation σ : X → X. Moreover, we see from Part 3 of the Lemma 128 that the only relations that can exist among the elements of G, are consequences of the relations µ(y) ◦ µ(σ(y)) = 1, y ∈ X. Accordingly the group G is called the free group on the set of generators Y and is denoted F (Y ) 4 . remarks: 1. (Uniqueness.) As remarked F (Y ) is uniquely defined by Y , so if there is a bijection f : Y → Z, then there is an induced group isomorphism F (Y ) ' F (Z) 2. For any frame Φ = F (Z, ζ), Aut(Φ) is a free group F (Z0 ), for any system Z0 of representatives of the ζ-orbits on Z. 3. (The inverse mapping on Γ.) Among the reduced words comprising the vertices of Γ, there is a well-defined involutory “inverse mapping” superscript -1 : W ∗ (X) → W ∗ (X) 4 Although this is the customary notation, unfortunately, it is not all that distinctive in mathematics. For example F (X) can mean the field of rational functions in the indeterminants X, as well as a myriad of meanings in Analysis and Topology. 214 which fixes only the empty word φ and takes the reduce word w = y1 y2 · · · yk to the reduced word w−1 := σ(yk )σ(yk−1 ) · · · σ(y1 ). Thus the group-theoretic inverse mapping µw → (µw )−1 is extended to the vertices of Γ by the rule that (µw )−1 := µw−1 : that is, the inverse mapping of group elements induces a similar mapping on W ∗ (X) via the bijection µ from the regular action of G = Aut(Γ) on the vertices of Γ. Thus without ambiguity, we may write y −1 for σ(y) and Y −1 for σ(Y ), so Y ∩ Y −1 = ∅. (It is this mapping on X which defines the graph Γ; there is no mention of a special system of representatives of σ-orbits. Thus if Y 0 is any other such system, then F (Y ) = F (Y 0 ) with “equals”, rather than the weaker isomorphism sign.) Corollary 129 Any subgroup of a free group is also a free group. proof: Let F (Y ) be the free group on the set Y , let X = Y ∪ Y −1 , and let Γ be the Cayley graph of F (Y ) with respect to the set of generators X. We used Γ to define F (Y ), for by the preceeding lemma F (Y ) is the group of all label-and-direction-preserving automorphisms of Γ. Let H be a non-trivial subgroup of F (Y ) and let ΓH be the H-orbit on vertices of Γ which contains the vertex φ, the empty word. Now each vertex in ΓH is a reduced word w and there is a unique directed path from φ to that vertex; moreover the sequence of labels from Y encountered along this directed path, are in fact the letters which in their order “spell” the reduced word w comprising this vertex . An H-neighbor of φ, is a vertex w of ΓH such that along the unique directed path in Γ from φ to w, no intermediate vertex of ΓH is encountered. Let Y H denote the collection of all H-neighbors of φ. Note from the remarks above, and the fact that H is a subgroup of F (X) that if the reduced word w is an H-neighbor of φ, then so is w−1 . Now we construct a new labelled graph on the vertex set ΓH . There is a directed edge (x, y) ∈ ΓH × ΓH if and only if x = yw where yw is reduced and w ∈ Y H . (This means x = µ(y)(w).) In a sense this graph can be extracted from the original graph Γ in the following way: If we think of ΓH as embedded in the graph Γ, we are drawing a directed edge from x to y labelled w if an only if the reduced word formed by juxtaposing the labels 215 of the successive edges along the unique directed path in Γ from x to y is w and w is in Y H . Now the graph ΓH has as its vertices elements of W ∗ (Y H ). There is a unique directed path in ΓH connecting any two vertices. Also for every two vertices there are no directed edges or exactly two directed edges, (x, y) and (y, x) respectively labelled w and w−1 for a unique w ∈ Y H . Thus the graph ΓH has all of the properties that Γ had except that X has been replaced by Y H . Thus as H is a regularly acting orientation- and label-preserving group of automorphisms of ΓH , H ' F (Y+H ) where Y H = Y−H + Y+H is any partition of Y H such that for each w ∈ Y H , exactly one representative of {w, w−1 } lies in Y+H . (Notice that we have used the inverse map on vertices of Γ in describing ΓH .) Thus the graph ΓH is a frame F = F (Y H , E), and H, being transitive, is by Lemma 128 the free group on Y+H . Exercise 1: Show that if |X| > 1, then it is possible to have |X| finite, while |Y H | is infinite. That is, a free group on a finite number of generators can have a subgroup which is a free group on an infinite number of generators. Exercise 2: This is a project. Is it possible for a group to be a free group on more than one cardinality of generator? The question comes down to asking whether F (X) ' F (Z) implies there is a bijection X → Z. Clearly the graph-automorphism definition of a free group is the way to go. Exercise 3. Another project: What can one say about the automorphism group of the free group F (X)? Use the remarks above to show that is is larger than Sym(X). 6.3.1 The Universal Property We begin with a fundamental theorem. Theorem 130 Suppose G is any group, and suppose X is a set of generators of G. Then there is a group epimorphism F (X) → G, from the free group on the set X onto our given group G, taking X to X identically. Proof: Here X is simply a set of generators of the group G, and so a subset of G. But a free group F (X) can be constructed on any set, despite whatever 216 previous roles the set X had in the context of G. One constructs the formal set Y = X + X −1 = two disjoint copies of X, with the notational convention that the exponential operator −1 denotes both a bijection X → X −1 and its inverse X −1 → X so that (x−1 )−1 = x. Recall that any arbitrary element of the free group F (X) was a reduced word in the alphabet Y – that is, an element of W ∗ (Y ), consisting of those words which do not have the form w1 yy −1 w2 for any y ∈ Y . (Recall that these were the vertices of the Cayley graph Γ = Γ(Y ) which we constructed in the previous subsection.) Now there is a mapping f which we may think of as our “amnesiarecovering morphism”. For each word w = u1 · · · yk ∈ W ∗ (Y ), f (w) is the element of G we obtain when we “remember” that each yi is one of the generators of G or the inverse of one. Clearly f (w1 w2 ) = f (w1 )f (w2 ), since this is how elements multiply when we remember that they are group elements. Then f is onto because every element of G can be expressed as a finite product of elements of X and their inverses. There is really nothing more to prove. One should bear in mind that the epimorphism of the theorem completely depends on the pair (G, X). If we replace X by another set of generators, Y ,– possibly of smaller cardinality – one gets an entirely different epimorphism: F (Y ) → G. Generators and relations. Now consider the same situation as in the previous subsection: A group G contains a set of generators X, and there is an epimorphism f := fG,X : F (X) → G, taking X to X “identically”. But there was on the one hand, a directed labelled graph Γ (the “frame”) that came from the construction in Lemma 128. This was seen to be the Cayley graph C(F (X), X). Since X ∪ X −1 is also a set of generators of G, one may form the Cayley graph C(G, X ∪ X −1 ). Then by Lemma 127 , there is a label-preserving vertex-surjective homomorphism of directed labelled graphs f : Γ → C(G, X ∩ X −1 ), from the frame Γ = F (X ∪ X −1 , −1) to the Cayley graph.. Where Γ had no circuitous directed walks other than backtracks, the Cayley graph C(G, X ∪ X −1 ) may now have directed circuits. In fact it must have them if f is not an isomorphism. For suppose f (w1 ) = f (w2 ). If w1 6= w2 , there is in Γ a unique directed walk from w1 to w2 . Its image is then a directed circuit. 217 Since G acts regularly on its Cayley graph by left multiplication, every directed circular walk in C(G, XU X −1 ) can be translated to one that begins and ends at the identity element 1 of G. Ignoring the peripheral attached backtracks, we can take such a walk to be a circuit (1, x1 , x1 x2 , x1 x2 x3 , ..., x1 x2 · · · xk = 1), xi ∈ X ∪ X −1 , where x1 x2 · · · xk is reduced in the sense that for i = 1, . . . , k − 1, xi 6= x−1 i+1 . The equation x1 · · · xk = 1 where a reduced word in the generators is asserted to be 1, is called a relation. In a set of relations, the words that are being asserted to be the identity are called relators. Obviously, the distinction is only of grammatical significance since the relators determine the relations and vice versa. Let R be a set of relators in the set of generators X; precisely, this means that R is a set of reduced words in W ∗ (X ∪X −1 ). Then the normal closure of R in the free group F (X), is the subgroup hRF (X) iF (X) , generated by the set of all conjugates in F (X) of all relators in R. Clearly, this is the intersection of all normal subgroups of F (X) which contain R. A group G is said to be presented by the generators X and relations R, if and only if G ' F (X)/hRF (X) iF (X) . This is expressed by writing G = hX|Ri. This is the “free-est” group which is generated by a set X and satisfying the relations R in the very real sense that any other group which satisfies these relations is a homomorphic image of the presented group. Let us look at some examples: Example 1.. (The infinite dihedral group D∞ .) G = hx, y|x2 = y 2 = 1i. As you can see we allow some latitude in the notation. The set of generators here is X = {x, y}; but we ignore the extra wavy brackets in displaying the presentation. The set of relators is R = {x2 , y 2 }, but we have actually written out the explicit relations. This is a group generated by two involutions x and y. Nothing is hypothesized at all about the order of the product xy, so it generates an infinite 218 cyclic subgroup N = hxyi. Now setting u := xy, the generator of N , we see that yx = u−1 and that x(xy)x = yx so xux = u−1 . Similarly yuy = u−1 . Moreover, uy = x shows N y = N x. Thus xN = N x = N y = yN and so N is a normal cyclic subgroup of index 2 in G. One of the standard models of this group is given by its action on the integers. Example 2.. (The dihedral group of order 2n, D2n .) Here G = hx, y|x2 = y 2 = (xy)n = 1i. Everything is just as above except that the element u = xy now has order n. This means the subgroup N = hui (which is still normal) is cyclic of order n, so G = N + xN has order 2n. Note that this group is a homomorphic image of the group of Example 1. When n = 2, the group is an elementary abelian group of order 4 – the so-called Klein fours group. Example 3.. (Polyhedral or (k, l, m)-groups.) Here k, l, and m are integers greater than one. The presentation is G := hx, y|xk = y l = (xy)m = 1i. The case that k = l = 2 is the dihedral group of order 2m of the previous example. We consider a few more cases: (1) the case (k = 2, m = 2) : Here u = xy has order 2. But then x and u are involutions, and the product xu = y has order l. So G is a homomorphic image of hx, u|x2 = u2 = (xu)l i ' D2l . It is not difficult to see that this homomorphism is an isomorphism. (2) the case(k = 2, l = 3, m = 3) : There is a group satisfying the relations. The alternating group on four letters contains a permutation x = (1, 2)(3, 4) of order 2, an element y = (1, 2, 3) of order 3 for which the product u = (1, 2)(3, 4)(1, 2, 3) = (1, 3, 4) has order three, and these elements generate Alt(4). Thus there is an epimorphism of the 219 presented (2, 3, 3)-group onto Alt(4). But the reader can check that the relations imply x(yxy 2 ) = y 2 xy. The left side is the product of two involutions, while the right side is an involution. It follows that the involutions x and x1 = yxy 2 commute. But now the equations just presented show that 2 xy = x1 , and xy = xx1 . Thus the abelian fours group K := hx, x1 i is normalized by y, so the presented (2, 3, 3)-group G has coset decomposition G = K + yK + y 2 K, and so has order 12. It follows that G is isomorphic to Alt(4). (3) the case (k = 2, l = 3, m = 4) : Here x2 = y 3 = (xy)4 = 1, so xyxy = y −1 xy −1 x and yxyx = xy −1 xy−1. Now conjugation by x inverts y −1 y x since (y −1 y x ) = (y −1 )x · y = [y −1 · ((y −1 )x )−1 ]−1 = (y −1 y x )−1 . But y −1 y x = y 2 · xyx = y(yxyx) = y(xy −1 xy −1 ) = (xy −1 )−1 · y −1 · (xy −1 ) and is a conjugate of y −1 and so has order 3. But yx is conjugate to xy, so yx has order four. Thus y has order 3, yxyxh has order 2, and y · yxyx = y 2 xyx has order 3. Thus, H := hy, yxyxi is a (2, 3, 3)-group, and so by the previous case, has order at most 12. But obviously G = hx, yi = hx, Hii = H + Hx, 220 so |G| = 24. But in the case of Sym(4), setting x = (12) and y := (134), we see that there is an epimorphism G → Sym(4). (6.2) The orders force5 this to be an isomorphism, whence any (2, 3, 4) − group is Sym(4). (6.3) (4) the case (k, l, m) = (2, 3, 5) : So here goes. We certainly have G = hx, y|x2 = y3 = (xy)5 = 1i. Again set z := xy. Then y = xz has order three. Then for x and z, we have the relations x2 = z 5 = 1. (xz)3 = 1. (6.4) (6.5) xzx = z −1 xz −1 xz −1 x = zxz. (6.6) (6.7) But the latter implies Now set t = zxz −1 , an involution. Then (ty)3 = (zxz −1 · xz)3 = (z(zxz)z)3 = (z 2 xz 2 )3 = z 2 xz 4 xz 4 xz 2 = z 2 x(xzx)xz 2 = z 5 = 1. (6.8) (6.9) Thus H = ht, yi is a (2, 3, 3)-subgroup of order divisible by 6, and so is Alt(4). It contains a normal subgroup K with involutions t, r := ty 2 and ty , and setting Y = hyi, one has H = Y + Y rY . 5 Of course this could also be worked out by showing that the Cayley graph must complete itself in a finite number of vertices. But the reasons seem equally ad hoc when done this way. I suppose only a brief contest of the efficacy of presenting it both ways might be interesting from a pedagogical point of view, but given that both methods yield the same result in about the same amount of time with the same degree of sophistication, there is no advantage in this. 221 Next let s be the involution z −1 (ty )z, which, upon writing t and y in terms of x and z becomes s = z −1 xz 3 xz. Now sy = z −1 xz 3 xz · xz = (xz)−1 (z 3 xz 2 )(xz) = (xz)−1 z −1 (z −1 xz)z · xz is a conjugate of x and so has order 2. Thus s inverts y. Also sr = sty = = = = (z −1 xz 3 xz)(xz 3 xz) (xz)−1 z 3 xzxz 3 (xz) (xz)−1 z 3 (z −1 xz −1 )z 3 (xz) (xz)−1 (z 2 xz 2 )(xz) is conjugate to z 2 xz 2 which has order 3 by equation 6.8. Consider now the collection 2 M = KHK = H + HsH = H + Hs + Hst + Hsty + Hsty , a set of 60 elements closed under taking inverses. We show that M is closed under multiplication. Using the three results M = H + HsH H = Y + Y rY sY = Y s we have M M ⊆ M if and only if srs ∈ M. But this is true since (sr)3 = 1 implies srs = rsr ∈ HsH ⊆ M. Finally, to get G = M we must show that at least two of the three generators x, y or z lies in M . Now M already contains y, and s = z −1 xz 3 xz = (xz)−1 z 3 (xz) = y −1 z 3 y, 222 and so contains z 3 and z = (z 3 )2 . Now, as |G| = 60 and G contains a subgroup of index 5, there is a homomorphism ρ : G → Sym(5) whose image has order at least 30 = lcm{2, 3, 5}. Because of the known normal strucure of Sym(5) obtained in the previous chapter, the only possibility is that ρ is a monomorphism and that G ' Alt(5). Again, this could have been proved using the Cayley graph. Neither proof is particularly enlightening.6 Example 4. (Coxeter Groups.) This class covers all groups generated by involutions whose only further relations are the declared orders of the pairwise products of these involutions. First we fix abstract set X. A Coxeter Matrix M over X is a function M : X × X → N ∪ {∞} which associates with each ordered pair (x, y) ∈ X × X, a natural number m(x, y) or the symbol “∞” so that (1) m(x, x) = 1 and (2) that m(x, y) = m(y, x) – that is the matrix is “symmetric”.7 The Coxeter group W (M ) is then defined to be the presented group W (M ) := hX|Ri. where R = {(xy)m(x,y) |(x, y) ∈ X × X} . Thus W (M ) is generated by involutions (because of the 1’s on the ‘diagonal’ of M ). If X is non-empty the group is always non-trivial. Coxeter classified the M for which W (M ) is a finite group. These finite Coxeter groups have an uncanny way of appearing all over a large part of important mathematics: Lie Groups, Algebraic Groups, Buildings, to name a few areas. remarks: There are many times in which the choice of relations, R may very well imply that the presented group has order only 1. For example if 6 The real situation is basically topolgical. It is a fact that a (k, l, m)-group is finite if and only if 1/k + 1/l + 1/l > 1. The natural proof of this is in terms of manifolds and curvature. Far from discouraging anyone, such a connection shows the great and mysterious “interconnectedness” of mathematics. I find that joyful! 7 Of course if X is finite, it can be totally ordered so that the data provided by M can be encoded in an ordinary matrix. We just continue to say “matrix” in the infinite case – perhaps out of love of analogy. 223 G := hx|x3 = x232 = 1i then G has order 1. In fact, if we were to select at random an arbitrary set of relations on a set X of generators, the odds are that the presented group has order 1. From this point of view, it seems improbable that a presented group is not the trivial group. But this is illusory, since any non-trivial group bears some set of relations, and so the presented group with these relations is also non-trivial in these cases. So most of the time, the non-trivial-ness of a presented group is proved by finding an object on which the presented group can act in a non-trivial way.8 Thus a Coxeter group W (M ) is known not to be trivial for the reason that it can be faithfully represented as an orthogonal group over the real field, preserving a bilinear form B := BM : V × V → R defined by M . Thus Coxeter Groups are non-trivially acting groups, in which each generating involution acts as a reflection on some real vector space with some sort of inner product. That is the trivial part of the result. The Real result is that the group acts faithfully in this way. This means that there is no product of these reflections equal to the identity transformation unless it is already a consequence of the relations given in M itself. This is a remarkable theorem. Exercise 4 Suppose G is a (k, l, m)-group. Using only the directed Cayley graph C(G, X) where X is the set of two elements {x, y} of orders 2 and 4, Show that 1. any (2, 4, 4)-group has order 24 and so is Sym(4). 2. any (2, 3, 5)-group has order 120 6.4 The Brauer-Ree Theorem If G acts on a finite set Ω, and X is any subset of G, we let ν(X) be defined by the equation X ν(X) := Γ =hXi-orbit (|Γi | − 1), i where the sum ranges over the hXi-orbits Γi induced on Ω. Thus if 1 is the identity element, ν(1) = 0, and if t is an involution with exactly k orbits of length 2, then ν(t) = k. 8 It is the perfect analogue of Locke’s notion of “substance”: something on which to hang properties. 224 Theorem 131 (Brauer-Ree) Suppose G = hx1 , . . . , xm i acts on Ω and x1 · · · xm = 1. Then Xm ν(xi ) ≥ 2ν(G). i=1 proof: Let Ω = Ω1 + · · · + Ωk be a decomposition of Ω into G-orbits, and for any subset X of G, set νi (X) = so ν(X) = X Γij =hXi-orbits P in νi (X). By induction, if k > 1, X ν(xj ) = j X ν (x ) ≥ i,j i j X i Ωi (|Γij | − 1), νi (xj ) ≥ 2(|Ωi | − 1), so P 2(|Ωi | − 1) = 2ν(G). Thus we may assume k = 1, so G is transitive. Now each xi induces a collection of disjoint cycles on Ω. We propose to replace each such cycle of length d, by the product of d − 1 transpositions t1 · · · td−1 . Then the contribution of that particular cycle of xi to ν(xi ) is P ν(ti ) = d − 1 and xi itself is the product of the transpositions which were used to make up each of its constituent cycles. We denote this collection of transpositions T (xi ) so ν(xi ) = X ν(t) and xi = t∈T (x ) i Y t∈T (xi ) t, in an appropriate order. Since x1 · · · xm = 1, we have YY i t∈T (xi ) t = 1, Q or, letting T be the union of the T (xi ), we may say t∈T t = 1, with the transpositions in an appropriate order. But as i ranges over 1, . . . m, the cycles cover all points of Ω, since G is transitive. It follows that the graph G(T ); = (Ω, T ), where the transpositions are regarded as edges, is connected. So by Lemma 23 of Section 3.4.3 of Chapter 3. hT i = Sym(Ω). Q But Sym(Ω) is transitive and so ν(Sym(Ω)) = ν(G). Since T t = 1, P P hT i = Sym(Ω), and T ν(t) = i ν(xi ), it suffices to prove the asserted inequality for the case that G = Sym(Ω) and the xi are transpositions. 225 So now, G = Sym(Ω), x1 · · · xm = 1, where the xi are a generating set of transpositions. Among the xi we can select a minimal generating set {s1 , . . . , sk } taken in the order in which they are encountered in the product Q x1 · · · xm . We can move these si to the left in the product xi at the expense of conjugating the elements they pass over, to obtain s1 s2 · · · sk y1 y2 · · · yl = 1, k + l = m, where the yj ’s are conjugates of the xi ’s. Now by the Feit-Lyndon-Scott Theorem, the product s1 · · · sk is an n-cycle (where n = |Ω|). Thus k ≥ n − 1. But then y1 · · · yl is the inverse n-cycle, and so l ≥ n − 1, also. But P k + l = m = ν(xi ) and n − 1 = ν(G), so the theorem is proved. Exercise 5. Use the Brauer-Ree Theorem to show that Alt(7) is not a (2, 3, 7)-group. [Hint: Use the action of G = Alt(7) on the 35 3-subsets of the seven letters. With respect to this action show that an involution fixes exactly seven triplets, and so has ν-value 14. Any element of order 3 contributes ν = 22, and that an element of order 7 fixes no triplet, acts in five 7-cycles and contributes ν = 30.] 6.5 Appendix to Chapter 6: A Detective Story: An application of (k, l, m)-groups At 2030 hours it was brought in. It was an unidentified group. All we knew was 1. It was simple. 2. It had order 360. Not much to go on. I had a hunch that it was Alt(6), which already had a known record of being simple and order 360. We knew that if it was simple of at least this order that it could not have a subgroup H of index less than 6, for otherwise the standard homomorphism πH : G → Sym(G/H) 226 would yield a proper non-trivial kernel. Also if this G had a subgroup H of index exactly 6, simplicity would yield that πH is an embedding into Alt(G/H) ' Alt(6) and comparing group orders we would have “made” our group G as an isomorphic version of Alt(6). So, at the request of Prosecuting Attorney Martha Clark, we made, the working assumption 9 (H’) G is not Alt(6). G has no subgroup of index less than or equal to six. As usual, we methodically proceeded by a series of short steps, first exploiting the Sylow theory. STEP 1. G has ( or rather “had”) 36 5-Sylow subgroups. G contains 144 elements of order 5. What were the possibilites? We knew that n5 := |Syl5 (G)| was a divisor of 360 congruent to one modulo 5, and so was a divisor of 36 with the same properties. So the only possibilities were 1, 6, and 36. Since n5 is just the index of the normalizer of a 5-Sylow subgroup, the first two cases are ruled out by our working assumption (H 0 ). Since the 5-Sylow subgroups have prime order 5, they pairwise intersect trivially (they are TI subgroups) and so each contains 4 elements of order 5. In that way, we found that there were exactly 36 × 4 = 144 elements of order 5 in its system. STEP 2. The 3-sylow subgroups of G were TI-subgroups, that is, each intersected its other conjugates only at the identity. We knew that these 3-Sylow subgroups of G had order 9 and so were abelian. Thus if two of them intersected at a subgroup hxi of order 3, then CG (x) would contain at least two of them. This would mean that |CG (x)| = 9d where a divisor m of d is m = |Syl3 (CG (x))|, is not 1, and satisfies m ≡ 1(mod 3). If m = 40, CG (x) = G, against G simple. Now if m = 4, a 3-Sylow subgroup of CG (x) must induce a 3-cycle on X := Syl(CG (x)), and so CG (x) induces a 2-transitive group – either Alt(4) or Sym(X) – on X. Also, there would have to be a subgroup F , normal in CG (x), representing the kernel of 9 The department has been a lot more sensitive recently, about what the working assumptions the police are actually using. We are now required to display them openly in our reports. I do not mind this as much as I used to, for I understand this is now done in quite a number of professions. 227 this action on X, so |CG (x)| = |F | · 12 or |F | · 24. By (H’), the only possibility was |F | = 1, and CG (x) inducing Alt(4) on X. But then there would be a normal Klein 4-group, K, normal in CG (x). Since K was normal in some 2-Sylow subgroup, NG (K) has at least twice the order of CG (x), yielding a subgroup of index 5 or less, against (H’). There was no doubt about it: CG (x) could not contain two or more distinct 3-Sylow subgroups, and so the TI-condition must have held. STEP 3. Ω := Syl3 (G) has ten elements and any 3-Sylow subgroup P must act regularly by conjugation on Ω − {P }. Thus G must have been doubly transitive on Ω. The possibilities allowed us by Sylow’s Theorem were |Syl3 (G)| = 1, 4, 10, 40, with the first two eliminated by our assumption (H’). But if there were 40 3-Sylow subgroups, they would contribute 40 × 8 = 320 elements of orders 3 or 9 (since they are TI-subgroups), and would have caused the subject’s order to exceed 360. Thus |Ω| = 10 was the only possibility. Now if P ∈ Syl3 (G), P would act by conjugaton on Ω, leaving itself fixed and acting on Ω − {P } in orbits of length 3 or 9. But an orbit of length 3 would reveal a non-trivial intersection between P and one of its conjugates, against Step 2. Thus P acts with just one orbit regularly on Ω − {P }. Since the Sylow theory showed us that there was a small degree doubly transitive representation, my blood began to tingle. Since G was simple, involutions could only act as even permutations. Moreover this action would really show up since the simplicity of G would force the transitive action on Ω to be faithful. We staked it out. Soon along came an involution t. We saw that it had to act as an even permutation whether it wanted to or not, and so had to fix at least 2 letters, and thus be represented by an involution in a 3-Sylow normalizer, say, as an involution t in NG (P ). Then, of course t must fix the unique point {P } of Ω fixed by P . And then it must act on the remaining points Ω − {P } exactly as it does on P by conjugation. (One of my clercks will put up the relevant file “Lemma 8 (iii)” at this point.) Thus t had to fix 1 + |CP (t)| letters – that is 1 or 4 letters. But if it fixed 4 letters it would have been an odd permutation and would have been arrested at once. Thus we were left with 228 STEP 4. Every involution of G appears as an involution t in some 3-sylow normalizer, NG (P ). Moreover an involution t in NG (P ) acts on P as a fixedpoint-free automorphism of order 2, so t “inverts” P – that is, tyt = y −1 for each y ∈ P . The last statement is a standard ‘MO’ on involutions which act a little too much fixed-point-freely than they should (we have a standard file on these offenders). At this point we knew what we were expecting: STEP 5. 1. A 2-Sylow subgroup of a 3-Sylow normalizer was a cyclic group of order 4. 2. All involutions that had ever lived in G, had been conjugate in G 3. G had never possesed elements of order 6, 15, or 10. Suppose a 2-sylow subgroup T of a 3-Sylow normalizer NG (P ) contained two commuting involutions s and t. Then, according to Step 4, each of s, t and st must invert P . Clearly this is impossible. Thus T cannot be Z2 × Z2 and so must be cyclic. Now any involution does appear in some 3-Sylow normalizer and hence (by two applications of Sylow’s Theorem) is conjugate to the unique involution in the cyclic group T . So, since all involutions are conjugate to the same involution, they are all conjugate. Now since 3-Sylow subgroups are TI-subgroups, any element of order 6 or 15 would have to appear in a 3-Sylow normalizer, NG (P ), not inverting P : absurd by Step 4. If u were an element of order 10, it’s square, u2 would be an element of order 5 which could only act on the ten elements of Ω as two disjoint 5-cycles. Then u could only have acted as a single 10-cycle, an odd permutation that could not live within G. Things were happening quickly now. STEP 6. G had never contained an element of order 8 or 9. We knew this because first, if G had ever possesed an element of order 8, then G would have had a cyclic 2-Sylow subgroup. The file (Corallary 14) 229 clearly shows that in this case G cannot be a simple group10 . On the other hand, we know that if there had ever been an element of order 9, then any 3-Sylow subgroup of G would be cyclic. and so have an automorhism group of order 6. Thus an element of order 4 normalizing P (whose existence was certified by Step 5 (1)) would have it’s square centralizing P against Step 5 (3). Here, the trail went cold. We could add only a few immediate footnotes to what we already knew above: (1) there were 45 involutions, each normalizing exactly two 3-Sylow subgroups. (2) Each 3-Sylow subgroup was elementary abelian – i.e. of type Z3 × Z3 . Not much else. There just didn’t seem to be any way to turn. We didn’t have the slightest idea what the structure of a 2-Sylow subgroup of G might have been, without a case-by-case pathology scan on all five possible subgroups of order 8, and most of us were much too close to retirement to do that. We only knew that a 2-Sylow subgroup could not be cyclic of order 8. At 0830 hours the next morning, one of my deputies (Gee, a Navaho) knocked persistently on my door. He said, “Sir? I have been looking at the file on (k, l, m)-groups, and I think it may be of help.” Normally I would have said something sarcastic like “How would you know?” But I didn’t want to burden the Department with another one of those required training sessions on Diversity, so I turned the subject to the impersonal issue at hand. I said, “How could that be? The integers k, l and m could be anything.” “But Sir,” he said, bowing a little too deeply for my taste, “if we fixed an involution t and looked at an element u of order 3, Steps 5 and 6 show that there are only certain orders available to the element tu: it could only have orders 2,3, 4 or 5. But these produce groups ht, ui, which are homomorphic images of (k, l, m)-groups with k = 2, l = 3 and m = 2, 3, 4, or 5, whose order is divisible by the least common multiple k, l, and m, since it contains elements of these orders.” I really hadn’t time for this obfuscation so I said “So?” ( This always works.) 10 Perhaps the only useful information we ever received from questioning the regular representation of a group. 230 “Moreover”, he went on interminably, “the (2, 3, m)-groups for m = 2, 3, 4 and 5 are all known in the file.” “You forgot the case m = 1”. I deliberately said this a little crossly, a trick I had learned in management school, or one of those schools I went to or was sent to. “But in that case Sir, we would have an element of order 2 times an element of order 3 being an element of order 1.” It sounded familiar. I said, “Certainly, I was just testing you, Deputy.” “It is part of the same principle, Sir.” (Somehow I knew he was going to preach to me.) “If an element t of order 2 times an element u of order 3 has order m, then the group they generate must have order a multiple of 2, 3, and m. “I knew that“, I snapped (I had learned to “snap”, in management school; they said it maintains discipline). “ But which (k, l, m)-groups have proper homomorphic images of these orders?” “None. Sir. I looked it up. For k = 2, l = 3, we have this: If m = 2, then the group is Sym(3). If m = 3, it is a homomorphic image of a (2, 3, 3)-group – that is, Alt(4) – of order at least 6. Thus it is exactly Alt(4) in this case. If m = 4, it is a homomorphic image of Sym(4) of order at least 12, and hence all of Sym(4). If m = 5, it is a homorphic image of Alt(5) of order at least 30. Thus in each case the group generated by t and u is the full abstract (2, 3, m)-group.” “Good work, Deputy. What do you propose to do?”. “Well, sir, I suggest we fix an involution t (recall that all involutions are conjugate) as a sort of ‘sting’. Then we watch various Z3 -subgroups hui approach t and record the order tu.” (I just knew Gee had no feeling for Departmental expenditures.) “This order (he went on) determines the group ht, ui, the (2, 3, m) group. We know that m cannot be 5 for otherwise ht, ui ' Alt(5) has index 6 against (H’). So something is bound to happen.” I put down my notebook and excused the Deputy. In part, my job is to make these vague ideas of my deputies precise. We set everything up by fixing an involution t. The potential interactions of t with the set A of 40 Z3 ’s were to be sorted out. We had A2 A3 A4 A5 := := := := {A ∈ A|ht, Ai = Sym(3)} {A ∈ A|ht, Ai = Alt(4)} {A ∈ A|ht, Ai = Sym(4)} {A ∈ A|ht, Ai = Alt(5)}, 231 (6.10) (6.11) (6.12) (6.13) and the last set was empty by our working assumption. Then setting ai := |Ai |, i = 2, 3, 4, we knew that a2 + a3 + a4 = 40. The trap was set. I was worried; we still didn’t know anything about the 2-Sylow subgroups (except that they weren’t Z8 ’s). First came A2 . If A were in A2 , then A would be normalized (and inverted) by t and hence also t would invert the unique 3-Sylow subgroup P which contains A (Steps 2 and 4). Each such 3-Sylow contributes 4 elements of A2 and t normalizes just two 3-Sylow subgroups (letters of Ω), so a2 = 2 × 4 = 8. We carefully entered this in the report as STEP 7. |A2 | = 8. None of the deputies seemed to know what to do with the subgroups A ∈ A3 ∪ A4 . We only knew that there had to be some A coming along, and patience was our game. It was at this time that Deputy Gee insinuated himself once more. “Any ideas?”, I asked. “Are you still mad at me sir?” (At first I did not recall what he was talking about, perhaps some abruptness on my part.) “I don’t know what you are talking about. Besides, you responded to my question with a question. What have you found out, Deputy?” (Always take charge of the dialogue: the Management Manual (MM) p. 43.) “Well sir, if A is in one of the classes A3 or A4 , then A and t act on some Klein 4-group, K. In case A ∈ A3 , t actually lies in K; but if A ∈ A4 then t does not lie in K and does not normalize A (otherwise A would be in A2 ). Thus 4-groups of the latter sort contribute just two members of A4 while 4-groups of the former sort contribute four members of A3 .” I broke the silence of my scratching in my notebook with this review: “So, according to you, our prayers are answered if we know all of the 4-groups K which are normalized by t. But how in this bloody world are we to do this if we dont even know what the 2-Sylow subgroups look like?” I have never seen a more expressionless face when he ventured that if a 2Sylow subgroup S were abelian, the Burnside Fusion theorem together with the fact that all involutions were conjugate would eliminate most groups: If S = Z2 ×Z2 ×Z2 then 7 would divide 360 – not a popular idea; S = Z2 ×Z4 has 232 two species of involutions, one square in S and three non-square involutions, so they cannot be conjugate in the normalizer of S. “ Finally, sir, you told me yourself that Z8 had been eliminated. Thus S is one of the two non-abelian groups of order 8 and the quaternion group of order 8 is eliminated because of the presence of a 4-group.” I was really glad I had thought of this, and am now encouraged to reconsider my promised retirement. At any rate I have once again made Gee’s vague ideas more precise by writing STEP 8. A 2-Sylow subgroup of G is dihedral of order 8. Sometimes Gee can be a pest. He then insisted that we now enjoyed (“enjoyed” was his word) the hypotheses: (K1) G is a simple group. (K2) There exists an involution t such that CG (t) ' D8 . Naturally I knew that; I even knew the name of the file (I think it was called “A Geometry of Klein 4-groups.”) and I knew that he would cough it up. I hadn’t time to read it fully, but he said that it contained these statements: STEP 9. (From a file) If G is a simple group and if t is an involution in G with CG (t) := D ' D8 , the dihedral group of order 8, then 1. D contains exactly two 4-groups, K1 and K2 , both containing the unique central involution t of D. 2. K1 is not conjugate to K2 in G. 3. Ki is a TI-subgroup of G – that is, Ki ∩ Kkg = Ki or 1, for all g ∈ G, i = 1, 2. 4. G has just one class of involutions. 5. NG (Ki ) ' Sym(4), i = 1, 2. 6. And some other stuff about a partial linear space Γ := (KiG , K2G ) with incidence being mutual normalization. 233 Perhaps after all, this would give us a handle on A3 and A4 . We now knew all about normalizers of fours-groups. They were all conjugate to NG (K1 ) or NG (K2 ), and Gee’s lucky guess that all (2, 3, 3)- and (2, 3, 4)-subgroups must appear only inside such subgroups was going to pay off. In calculating a3 we needed only look at the two 4-groups K1 and K2 containing t, and record four Z3 ’s in A3 for each. Thus STEP 10. |A3 | = 8. Next we had to examine the collection K(t) of all 4-groups K normalized by t but not containing it. For each K in K(t), CK (t) is a group hsi ' Z2 and there are just four choices of involution s in CG (t) − {t}. For each such choice, K is the unique 4-group on s distinct from the 4-group hs, ti, so that t∈ / K. Thus |K(t)| = 4. with two members in K1G and two members in K2G . Then, as noted above, each member of K(t) yields two members of A4 . Thus STEP 11. |A4 | ≤ 8. We write “≤” in case we might have counted some Z3 two times. Chee, with the confidence of his youth, insists that that does not happen. At any rate we see that a2 + a3 + a4 ≤ 24 and so we have reached a contradiction. There is no possible way G is not Alt(6). The case is closed. Your Servant, Inspector van Naughty. 234 Chapter 7 Elementary Properties of Rings 7.1 Elementary Facts about Rings In the next several parts of this course we shall take up the following topics and their applications: 1. The Galois Theory. 2. The Arithmetic of Integral Domains. 3. Semisimple Rings. 4. Multilinear Algebra. All of these topics require some basic facility in the language of rings and modules. For the purposes of this survey course all rings possess a multiplicative identity. This is not true of all treatments of the theory of rings. But virtually every major application of rings is one which uses the presence of the identity element (for example, this is always part of the axiomatics of Integral Domains, Field Theory, and appears in Artinian Rings and Finitedimensional Associative Algebras). So in order to even have a language to discuss these topics, we must touch base on a few elementary definitions and constructions. 235 7.1.1 Definitions A ring is a set – say R – endowed with two distinct binary operations, called (ring) addition and (ring) multiplication – which we denote by “+” and by juxtaposition or “·”, respectively – such that Addition laws: (R, +) is an abelian group. Multiplicative Rules: (R, ·) is a monoid. Distributive Laws: For all triples (a, b, c) ∈ R × R × R, a(b + c) = ab + ac (a + b)c = ac + bc (7.1) (7.2) Just to make sure we know what is being asserted here, for any elements a, b and c of R: 1. a + (b + c) = (a + b) + c 2. a + b = b + a 3. There exists a “zero element” (which we denote by 0R ) with the property that 0R + a = a for all ring elements a. 4. Given a ∈ R, their exists an element (−a) in R such that a+(−a) = 0R . 5. a(bc) = (ab)c. 6. There exists an “identity element” (which we denote by 1R ) with the property that 1R a = a1R = a, for all ring elements a ∈ R. 7. The distributive laws which we have just stated. As we know from the group theory, the zero element 0R is unique in R. Also given element x in R, there is exactly one additive inverse −x, and always −(−x) = x. The zero element may not be distinct from the multiplicative identity 1R (although this is required for integral domains as seen below); this makes the class of rings closed under homomorphisms. When there is no confusion (and sometimes with multiple uses of the notation) we often write 0 for 0R and 1 for 1R . 236 Lemma 132 For any elements a, b in R: 1. (−a)b = −(ab) = a(−b), and 2. 0 · b = b · 0 = 0. Some notation: Suppose X and Y are subsets of the set R of elements of a ring. We write X + Y := {x + y|x ∈ X, y ∈ Y } XY := {xy|x ∈ X, y ∈ Y }, −X := {−x|x ∈ X} (7.3) (7.4) (7.5) There are various species of rings. • A ring is said to be commutative if and only multiplication is commutative in it. • A division ring is a ring in which (R∗ := R − {0}, ·) is a group. • A commutative division ring is a field. • A commutative ring in which (R∗ , ·) is a monoid – i.e. the non-zero elements are closed under multiplication and possesses an identity – is called an integral domain. There are many other species of rings – Artinian, Noetherian, Semiprime, Primitive, etc. which we shall meet later, but there is no immediate need for them at this point. Finally, if R is a ring with multiplicative identity element 1, then a subring of R is a subset S of R such that 1. S is a subgroup of (R, +). Thus S = −S and S + S = S. In particular 0R ∈ S. 2. SS ⊆ S, –i.e. S is closed under multiplication. 3. 1 ∈ S. Thus the ring of integers, Z, is a subring of any of the three fields Q, R and C, of rational, real and complex numbers, respectively. But the set of even numbers is not a subring of the ring of integers. 237 7.1.2 Units in rings Suppose R is a ring with multiplicative identity element 1R . We say that an elements x in R has a right inverse if and only if there exists an element x0 in R such that xx0 = 1R . Similarly, x has a left inverse if and only if there exists an element x00 in R such that x00 x = 1R . (If R is not an integral domain these right and left inverses might not be unique.) An element in R is called a unit if and only if it possesses both a right and a left inverse in R. The element 1R , of course, is a unit, so we always have at least one of these. We observe Lemma 133 1. An element x has a right inverse if and only if xR = R. Similarly, x has a left inverse if and only if Rx = R. 2. The right inverse and left inverse of a unit element are equal. 3. The set of units of any ring form a group under multiplication. proof: Part 1. If x has a right inverse xR contains xx0 = 1R , so xR contains R. The converse, that xR = R implies that xx0 = 1R , for some element x0 in R, is transparent. The left inverse story follows upon looking at this argument through a mirror. Part 2. Suppose xl and xr are respectively the left and right inverses of an element x. Then xr = (1R )xr = (xl x)xr = xl (xxr ) = xl (1R ) = xl . Part 3. Suppose x and y are units of the ring R with inverses equal to x and y −1 , respectively. Then xy has y −1 x−1 as both a left and a right inverse. Thus xy is a unit. Thus the set U (R) of all units of R is closed under an associative multiplication, has an identity element with respect to which every one of its elements has a 2-sided inverse. Thus it is a group. −1 A ring in which every non-zero element is a unit must be a division ring and conversely. Example 1.. Some examples of groups of units: 1. In the ring of integers, the group of units is {±1} , forming a cyclic group of order 2. 238 2. The integral domain Z ⊕ Zi of√all complex numbers of the form a + bi with a, b integers and i = −1, is called the ring of Gaussian integers. The units of this ring are {±1, ±i} forming the cyclic group of order four under multiplicaton. 3. The complex numbers of the form a + b(e2iπ/3 ), a and b integers, form an integral domain called the Eisenstein numbers. The units of this ring are {1, z, z 2 , z 3 = −1, z 4 , z 5 }, where z = −e2iπ/3 , forming a cyclic group of order 6. √ √ 4. The number λ := 2 + 5 is a unit in the domain D := {a + b 5|a, b ∈ Z}. Now λ is a real number, and since 1 < λ, we see that the successive powers of λ form a monotone increasing sequence, and so we may conclude that the group U (D) of units for this domain is an infinite group. 5. In the ring R := Z/Zn of residue classes of integers mod n, under addition and multiplication of these residue classes, the units are the residue classes of the form w + Zn where w is relatively prime to n, (Of course, if we wish, we may take 0 ≤ w ≤ n − 1.) They thus form a group of order φ(n), where φ denotes the Euler phi-function. Thus the units of R = Z/8Z are {1 + 8Z, 3 + 8Z, 5 + 8Z, 7 + 8Z}, and form the Klein fours group rather than the cyclic group of order four. 6. In the ring of n-by-n matrices with entries from a field F , the units are the invertible matrices and they form the general linear group under multiplication. 7.1.3 Homomorphisms Let R and S be rings. A mapping f : R → S is called a ring homomorphism if and only if 1. For all elements a and b of R, f (a + b) = f (a) + f (b). 2. For all elements a and b of R, f (ab) = f (a)f (b). A basic consequence of Part 1 of this definition is that f induces a homomorphism of the underlying additive groups, so, from our group theory we know that f (−a) = −f (a), and f (0R ) = 0S . In fact any polynomial 239 expression formed from a finite set of elements using only the operations of addition, multiplication and additive inverses gets mapped by f to a similar expression with all the original elements replaced by their f -images. Thus: f ([(a + b)c2 + (−c)bc]a) = [(f (a) + f (b))(f (c))2 + (−f (c))(f (b)f (c))]f (a). The homomorphism is onto or an epimorphism or surjection, injective, or an isomorphism if and only if these descriptions apply to the induced group homomorphism of the underlying additive groups. 7.1.4 Ideals and factor rings. Again let R be a ring. A left ideal is a subgroup (I, +) of the additive group (R, +), such that it is invariant under left multiplications from R, – that is RI ⊆ I. Similarly, a subgroup J ≤ (R, +) is a right ideal if JR ⊆ J. Finally, a (two-sided) ideal is a left ideal which is also a right ideal. The parentheses are to indicate that if we use the word “ideal” without adornment, a two-sided ideal is intended. A subset X of R is a two sided ideal if and only if X +X = X −X = X RX + XR ⊆ X. (7.6) (7.7) (7.8) If X is a two sided ideal in R, then it is a (normal) subgroup of the commutive additive group (R, +). Thus we may form the factor group R/X := (R, +)/(X, +) whose elements are the additive cosets x + X in R. Not only is R/X an additive group, but there is a natural multiplication on this set, induced from that fact that all products formed from two such additive cosets, x + X and y + X, lie in a unique further additive coset. This is evident from (x + X)(y + X) ⊆ xy + xX + Xy + XX ⊆ xy + X + X + X = xy + X. (7.9) With respect to addition and multiplication, R/X forms a ring which we call the factor ring. Moreover, by equation (7.9), the homomorphism of additive groups R → R/X is a ring epimorphism. 240 7.1.5 The fundamental theorems of ring homomorphisms Suppose f : R → S is a homomorphism of rings. Then the kernel of the homomorphism f is the set ker f := {x ∈ R|f (x) = 0S }. We have Theorem 134 (The first fundamental theorem of ring homomorphisms.) 1. The kernel of any ring homomorphism is a two-sided ideal. 2. If f : R → S is a homomorphism of rings, then there is a natural isomorphism between the factor ring R/ ker f and the image ring f (R), which takes x + ker f to f (x). proof: Part 1. Suppose f : R → S is a ring homomorphism and that x is an element of ker f . Then as ker f is the kernel of the group homomorphism of the underlying additive groups of the two rings, ker f is a subgroup of the additive group of R. Moreover, for any element y in R, f (rx) = f (r)f (x) = f (r))0S = 0S . Thus R(ker f ) + (ker f )R ⊆ R, and so ker f is a two-sided ideal. Part 2. The image ring f (R) is a subring of S formed from all elements of S which are of the form f (x) for some element x in R. We wish to define a mapping φ : R/(ker f ) → f (R) which will take the additive coset x + ker f to the image element f (x). We must show that this mapping is well-defined. Now if x + ker f = y + ker f , then x + (−y) ∈ ker f , so 0S = f (x + (−y)) = f (x) + (−f (y)), whence f (x) = f (y). So changing the coset representative does not change the image, and so the mapping is well-defined. (In fact the φ-image of x + ker f is found simply by applying f to the entire coset, that is, f (x + ker f ) = f (x).) The mapping φ is easily seen to be a ring homomorphism since φ((x + ker f ) + (y + ker f ) = = φ((x + ker f )(y + ker f ) = = φ(x + y + ker f ) = f (x + y) f (x) + f (y) = φ(x + ker f ) + φ(y + ker f ). φ(xy + ker f ) = f (xy) f (x)f (y) = φ(x + ker f ) · φ(y + ker f ). 241 Clearly φ is onto since by definition, f (x) = φ(x + ker f ) for any image element f (x). If φ(x+ker f ) = φ(y+ker f ) then f (x) = f (y) so f (x+(−y)) = 0S , and this gives x + ker f = y + ker f . Thus φ is an injective mapping. We have assembled all the requirments that f be an isomorphism. Theorem 135 Suppose f : R → S is an epimorphism of rings. Then there is a one-to-one correspondence between the left ideals of S and the left ideals of R which contain ker f . Under this correspondence, 2-sided ideals are matched only with 2-sided ideals. proof: If J is a left ideal of the ring S, set f −1 (J) := {r ∈ R|f (r) ∈ J}. Clearly if a and b are elements of f −1 (J), and r is an arbitrary element of R, then f (ra) = rf (a) ∈ J f (−a) = −f (a) ∈ J f (a + b) = f (a) + f (b) ∈ J So ra, −a, a+b all belong to f −1 (J). Thus f −1 (J) is a left ideal containing f (0S ) = ker f . If also J is a right ideal of S, we have JS ⊆ J. Then f (f −1 (J)R) ⊆ Jf (R) = JR ⊆ J, whence f −1 (J)R ⊆ f −1 (J). Thus f −1 (J) is a right ideal if and only if J is. Now if I is a left ideal of R, then so is f (R) a left ideal of F (R) = S. But if I contains ker f , then f −1 (f (I)) = I, because, for any element x ∈ f −1 (f (I)), there is an element i ∈ I such that f (x) = f (i). But then x + (−i) ∈ ker f ⊆ I, when x ∈ I. We thus see that the operator f −1 induces a mapping of the poset of left ideals of S onto the set of left ideals of R containing ker f . It remains only to show that this (containment-preserving) operator is one-to-one. Suppose I = f −1 (J1 ) = f −1 (J2 ). We claim J1 ⊆ J2 . If not there is an element y ∈ J1 − J2 . Since f is an epimorphism, y has the form y = f (x), for some element x in R. Then x is in f −1 (J1 ) but not f −1 (J2 ), a contradiction. Thus we have J1 ⊆ J2 . By symmetry, J2 ⊆ J1 and so J1 = J2 . This completes the proof. −1 a second proof: We can take advantage of the fact that the kernel of the surjective ring homomorphism f : R → S is also the kernel of the homomorphism f + : (R, +) → (S, +) of the underlying additive groups and exploit 242 the fact that the fundamental theorems of homomorphisms of group theory give us a poset isomorphism A(R : ker f ) → A(S : 0), where A(M, N ) denotes the poset of all subgroups of M which contain the subgroup N . The isomorphism is given by sending subgroup A of A(R, ker f ) to the subgroup f (A) of S. We need only show that if X is a subgroup of A(R : S). then 1. RX ⊆ X if and only if Sf (X) ⊆ f (X). 2. XR ⊆ X if and only if f (X)S ⊆ f (X). Both of these are immediate consequences of the fact that f is a ring homomorphism, and S = f (R). 7.1.6 The poset of ideals. We wish to make certain statements that hold for the set of all left ideals of R as well as for the set of all right ideals and all (two-sided) ideals of R. We do this by asserting “X holds for (left, right) ideals with property Y”. This means the statement is asserted three times: once about ideals with property Y, again about left ideals with property Y, and again about right ideals with property Y. It does save a little space, and helps point out a uniformity among the theorems. But on the other hand it makes statements difficult to scan, and sometimes renders the statement a little less indelible in the students memory. So I will keep this practice at a minimum. Lemma 136 Suppose A and B are (left, right) ideals in the ring R. Then 1. A + B := {a + b|(a, b) ∈ a × B} is itself a (left,right) ideal and is the unique minimal such which contains both A and B. 2. A ∩ B is a (left, right) ideal which is the maximal such contained in both A and B. 3. The set of all (left, right) ideals of R form a partially ordered set — that is a poset – with respect to the containment relation. In view of the preceeding two conclusions, this poset is a lattice. 243 As in the case of groups, when we say maximal (left, right) ideal, we mean a maximal member in the poset of all proper (left, right) ideals. Thus the ring R itself, though an ideal, is not a maximal ideal. Lemma 137 Maximal (left, right) ideals exist. proof: Let P be the full poset of all proper ideals, all proper left ideals or all proper right ideals of the ring R. Let J 1 ≤ J2 ≤ · · · be an ascending chain in the poset P . Let U be the set-theoretic union of the Ji . Then any element of U belongs to some Jj . Thus any (left, right) multiple of that element by an element of R belongs to U . Thus it is easy to see that RU R, RU , or U R is a subset of U in the three respective cases of P . Similarly, −U = U and using the total ordering of the chain, U + U ⊆ U . Thus U is an (left, right) ideal in R. If U = R, then 1R would belong to some Jj , whence RJ1 , J1 R and RJj R would all coincide with R, against Jj ∈ P . Thus U is a member of P and is an upper bound to the ascending chain. Since any ascending chain in P has an upper bound, it follows from Zorn’s Lemma that P contains maximal members. A simple ring is a ring with exactly two (two-sided) ideals: 0 := {0R } and R. (Note that the “zero-ring” in which 1R = 0R is not considered a simple ring here. Simple rings are non-trivial. Since every non-zero element of a division ring D is a unit, 0 is a maximal ideal of D. Thus any division ring is a simple ring. The really classical example of a simple ring, however, is the full matrix algebra over a division ring, which appears among the examples at the end of this section. Lemma 138 1. A (two-sided) ideal A of R is maximal if and only R/A is a simple ring. 2. An ideal A of a commutative ring R is maximal if and only if R/A is a field. proof: Part 1, is a direct consequence of Theorem 135 of the previous subsection. 244 Part 2. Set R̄ := R/A. If R̄ is a field, it is a simple ring, and A is maximal by Part 1. If A is a maximal ideal, the commutative ring R̄ has only 0 for a proper ideal. Thus for any non-zero element x̄ of R̄, x̄R̄ = R̄x̄ = R̄. Thus any non-zero element of R̄ is a unit. Thus R̄ is a commutative division ring, and hence is a field. Suppose now that A and B are ideals in a ring R. In a once-only departure from our standard notational convention –special for (two-sided) ideals – we let the symbol AB denote the additive subgroup of (R, +) generated by the set of elements {ab|(a, b) ∈ A × B} (the latter would ordinarily have been written as “AB” in our standard notation). That is, for ideals A and B, AB is the set of all finite sums of the form X ab, j∈I j j where |I| < ∞, aj ∈ A, bj ∈ B. If we multiply such an expression on the left by a ring element r, then by the left distributive laws, each aj is replaced by raj which is in A since A is a left ideal, yielding thus another such expression. Similarly, right multiplication of such an element by any element of R yields another such element. Thus AB is an ideal. 1 Such an ideal AB clearly lies in A ∩ B, but it might be even smaller. Indeed, if B lies in the right annihilator of A, that is the set of elements x of R such that Ax = 0, then AB = 0, while possibly A ∩ B is a non-zero ideal (with multiplication table all zeroes, of course). This leads us to still another construction of new ideals for old: Suppose A and B are right ideals in the ring R with B ≤ A. The right annililator of A modB is the set AnnR (A : B) := {r ∈ R|Ar ⊆ B}. Similarly, if B and A are left ideals of R, with B ≤ A, the same symbol AnnR (A : B) will denote the left annihilator of A mod B, the set of ring elements r such that rA ⊆ B. Lemma 139 If B ≤ B is a chain of right (left) ideals of the ring R, then AnnR (A : B) is a two-sided ideal of R. 1 Indeed, one can see that it would be a two-sided ideal even if A were only a left ideal and B were a right ideal. It is unfortunate to risk this duplication of notation: AB as ideal product, or set product, but I feel bound to hold to the standard notation of the literature – when there is a standard –even if there are small overlaps in that standard. 245 proof: We prove here only the right-hand version. Set W := AnnR (A : B). Clearly W + W ⊆ W , W = −W , and W R ⊆ W , since B is a right ideal. It remains to show RW ⊆ W . Since A is also a right ideal, A(RW ) = (AR)W ⊆ AW ⊆ B, so by definition, RW ⊆ W . A prime ideal of the ring R is an ideal – say P – such that whenever two ideals A and B have the property that AB ⊆ P , then either A ⊆ P , or B ⊆ P . (Note here AB is the product of two ideals as defined above. But of course, since P is an additive group, AB (as a product of two ideals) lies in P if and only if AB (the old set product) lies in P – so the multiple notation is not critical here. A rather familiar example of a prime ideal would be the set of all multiples of a prime number – say 137, to be specific –in the ring of integers. This set is both a prime ideal and a maximal ideal. The ideal {0} in the ring of integers would be an example of a prime ideal which is not a maximal ideal. Theorem 140 In any ring, a maximal ideal is a prime ideal. proof: Suppose M is a maximal ideal in R and A and B are ideals such that AB ⊆ M . Assume A is not contained in M . Then A + M = R, by the maximality of M , but RB = AB + M B ⊆ M , so B belongs to the two-sided ideal AnnR (R : M ). Since M is a left ideal, M also belongs to this right annihilator. Thus we see that M + B lies in this right annihilator, AnnR (R : M ). But the latter cannot be R since 1R does not annihilate R mod M on the right. Thus M + B is a proper ideal of R and maximality of M as a two-sided ideal forces B ⊆ M . Thus either A ⊆ M or (as just argued) B ⊆ M . Thus M is a prime ideal. 7.1.7 Examples Example 2. Constructions of fields. Perhaps the three most familiar fields are the rational numbers Q, the real numbers R, and the complex numbers C. In addition we have the field of integers modulo a prime. Later, when we begin the Galois theory and the theory of fields, we shall meet a number of ways to construct a field, namely 246 1. Forming the factor ring D/M where M is a maximal ideal in an integral domain D. 2. Forming the ring of quotients of an integral domain. 3. Forming the topological completion of a field. The first of these methods is contained in the theorems already encountered. Any commutive ring , modulo a maximal ideal is a field. It turns out that any commutative ring modulo a maximal two-sided ideal is a field. The other two must be deferred. Example 3. The monoid algebras. Recall that a monoid is a set with an associative operation (say, multiplication) and an identity element with respect to that operation. Here are some monoids: • The natural numbers (the positive integers) N under multiplication. The integer 1 is the multiplicative identity. • The multisets over a set X. The elements of this monoid consist of the integer zero together with formal positive-integer linear combinations a1 x1 + a2 x2 + · · · am xm , m, a1 , . . . am ∈ N, xi ∈ X, where addition is performed by adding coefficients of like elements of X. For example if X = {a, b, o}, where a = apples, b = bananas, and o = oranges, then (a+2b)+(b+2o) = a+3b+2o – that is, an apple and two bananas added to a banana and two oranges is one apple, three bananas and two oranges. Thus one can add apples and oranges in this system! We used just such a multiset monoid over the simple groups in proving the Jordan-Hölder Theorem for groups. • The Free Monoid over X. This was the monoid whose elements are the words of finite length over the alphabet X. These are sequences of juxtaposed elements of X, x1 x2 · · · xm where m ∈ N and the “letters” xi are elements of X, as well as the empty word which we denote by φ. The associative operation is concatenation of words, the act of writing one word right after another. The empty word is an identity element. We utilized this monoid in constructing free groups. The free monoid over the single letter alphabet X = {x}, consists of φ, x, xx, xxx, . . . which we denote as 1, x, x2 , x3 . . ., etc. Clearly this 247 monoid is isomorphic to the monoid of non-negative integers Z+ := N ∪ {0} under addition. • Of course any group is a monoid. Now let F be any field, and let M be a monoid. The monoid algebra F M is the set of all formal finite F -linear combinations of the elements of M . Two such expressions are added in the usual way, by adding coefficients of the same elements of M to obtain the coefficient of that element in the sum. Similarly multiplication is defined by X ( X a σ)( σ σ b τ) = τ τ X X κ ( a b )κ. σ,τ,στ =κ σ τ Essentially this “convolution formula” results from applying every possible distributive law in sight (assuming that scalars from F commute with monoid elements) and then collecting terms. • In the case that M is a group, F M is called a group algebra. • In the case that M is the free monoid on the single generator {x}, then F M is called the ring of polynomials over F and is denoted F [x]. • In the case that M is the monoid of multisets over the finite set X = {x1 , x2 , . . . , xn }, the monoid algebra F M is called the ring of polynomials in the indeterminates {x1 , . . . , xn } and is denoted F [x1 , . . . , xn ]. Example 4. Endomorphism rings. Let (A, +) be an additive abelian group. Let Hom(A, A) denote the set of all mappings from A into itself. We can define addition on this set by letting f + g be the mapping A → A, whose value at any element a in A is f (a) + g(a) — here f and g can be any mappings from A into A. With respect to this definition of addition, the reader should be able to show that Hom(A, A) is an abelian group with respect to addition. Notice that the zero mapping which sends every element of A to its additive identity, is the identity element of Hom(A, A). Now there is a second binary operation that can be performed on the set Hom(A, A), and that is the composition of mappings: If f and g are in Hom(A, A), then f ◦ g is the mapping which takes any element a of A to 248 f (g(a)). Compositions like this are always associative, and form a monoid which has the identity mapping a → a as its identity element. Is this composition distributive with respect to addition? The reader should be able to see that it is not. The trouble is that f (g(x) + h(x)) may not be f (g(x)) + f (h(x)), if f does not respect group addition. So instead let us restrict to a subgroup of Hom(A, A), namely HomZ (A, A) which denotes the set of all endomorphisms of the group A into itself. The reader can quickly verify the following: 1. HomZ (A, A) is indeed a subgroup of Hom(A, A) that is, it is closed under the addition operation and the taking of additive inverses. 2. HomZ (A, A) is closed under map compositions – that is, the composition of two group endomorphisms is an endomorphism. 3. Composition satisfies both distributive laws with respect to addition. Thus we obtain a ring, and we write HomZ (A, A) as End(A) when we wish to emphasize its ring structure. This is the endomorphism ring of an abelian group. Suppose now that A happens to be a right vector space over the division ring D. Then we can specialize even further. Instead of considering all endomorphisms, as in End(A), we may just take those which are linear transformations of A. This set, HomD (A, A), is also a subgroup of HomZ (A, A) which is closed under multiplication and contains both the “zero map” and the identity mapping. Again, it is a ring, called the D-endomorphism ring of the D-vector space A and is denoted EndD (A). When A has finite dimension n, then by choosing a basis, every linear transformation of V into itself can be rendered uniquely as an n × n matrix. Then addition and composition of linear transformations is duplicated as matrix addition and multiplication. In this guise, EndD (A) is seen to be isomorphic to the full matrix algebra over D – the ring Dn×n of all n × n matrices over D, with respect to matrix addition and multiplication. Each basis of A yields a ring isomorphism EndD (A) → Dn×n . The algebra Dn×n is a simple ring for every natural number n and division ring D. We will not prove this until much later; it is only offered here as a “sociological” statement about the population of mathematical objects which are known to exist. It would be nearly sadistic to propose definitions 249 to students which turn out to be inhabitated by nothing – still one more description of the empty set. So I perceive some value in pointing out (without proof) that things exist. Example 5. Opposite rings. Suppose we have a ring R with a given multiplication table. We can define a new multiplication (let’s call it “*”) by transposing this multiplication table. That is, a ∗ b := ba for all a and b in R. The new operation is still distributive with respect to addition with both distibutive laws, and it is easy for the student to verify that (R, +, ∗) is a ring. (Remember, we need a multiplicative identity, and the associative law should at least be checked.) We call this ring the opposite ring of R and denote it by the symbol Opp(R). Example 6. Direct Product of rings. Suppose {Rσ }I is a family of rings indexed by I. Let 1σ be the multiplicative identity element of Rσ . Now we can can form the direct product of the additive groups {(Rσ , +)}I : P := Y σ∈I (Rσ , +). Recall that the elements of the product are functions f : I → ∪I Mσ (the abstract disjoint union), with the property that f (σ) ∈ Rσ . We can then convert this additive group into a ring by defining the multiplication of two elements f and g of P by the rule that f g is the function I → ∪I Rσ defined by (f g)(σ) := f (σ)g(σ), where the juxtapostion on the right indicates multiplication in the ring Rσ . (This sort of multiplication is often called “coordinatewise multiplication”.) It is an easy exercise that this multiplication is associative and is both left and right distributive with respect to addition in P . Finally, one notes that the function 1P whose value at σ is 1σ , the unique multiplicative identity of the ring Rσ , comprises a multiplicative identity element for P , so P is indeed a ring. Note that if K is any subset of the index set I, then the subset PK of all functions f : I → ∪I Rσ which vanish on K forms a (two-sided) ideal of P . Now that we understand a direct product of rings, is there a corresponding direct sum of rings? The additive group (S, +) of such a ring should be a direct sum ⊕I (Rσ , +) of additive groups of ring. But in that case, the multiplicative identity element 1S of the ring S must be uniquely expressible 250 as a function which vanishes on all but a finite subset F of indices in I. It then follows that if multiplication is that defined by “pointwise multiplication” as for its ambient direct product, that in fact the direct sum of rings, ⊕I Rσ = ⊕F Rσ , is always a finite direct sum of rings. This is an expense one pays for in maintaining a multiplicative identy. Example 7. Integral domains. Let D be any integral domain. Recall that this means that D is a commutative non-zero ring in which the product of a non-zero element and any other is again a non-zero element. In other words, D∗ := D − {0D } is closed under multiplication (and hence, from the other ring axioms, D∗ is a multiplicative monoid. Obviously, the example par excellance is the ring of integers. ( Who knows? Perhaps this ring was the inspiration of the actual term, “integral domain”.) But perhaps that is also why it need not be the example par example. The integers certainly enjoy every property that the any integral domain does. But there are many further properties of the integers which are not shared by many integral domains. For example, every integer possesses an essentially unique factorization (up to the order of the factors or the replacing of any factor by a unit times that factor) in prime numbers. That is true of a large class of integral√domains, but it is not true of many further integral domains (such as Z ⊕ −7Z). Every ideal of Z has the form Zx where x is a particular element of Z. Any ideal of this form is called a principal ideal. (That is for an arbitrary ring, a principal ideal is a ring of the shape RxR for some element x. If all ideals of an integral domain have this shape, we say that D is a principle ideal domain or PID. Many integral domains don’t have this property. I am telling you this for your own good. The characteristic of an integral domain. Suppose now that D is an integral domain. We define the characteristic of D as the smallest natural number n such that nx = xn is zero for every element x in D. Of course we can make this concept very local by considering the natural numbers nx such that nx is the smallest natural number such that x + x + · · · x (with exactly nx summands) is 0R . 251 Then it should be obvious from the distributive laws that nx divides n1 , since 1 + 1 + · · · + 1( to n1 factors) implies that x + . . . + x = x(1 + · · · + 1) = xnx = x0R = 0R . Thus nx divides n1 . On the other hand, if x is nonzero, and if we consider x added to itelf with exactly nx summands, we obtain 0 = x + x + · · · , +x = x(1 + 1 + · · · + 1) = xnx . (7.10) Since D∗ is closed under multiplication either x = 0 or 1 + · · · + 1 (to nx summands ). The former alternitive is forbidden by the choice of x. Thus n1 divides nx . Thus nx = n1 , since these are natural numbers, each dividing the other. We have shown Lemma 141 For any integral domain D, one of the following two alternatives hold: 1. Either for every non-zero element x, no sum of x0 s can ever be zero, or 2. There exists a natural number c such that cx = xc = 0 for every every element x of R and for every smaller number c0 < c, Dc0 6= 0. In the former case we say that D has characteristic zero, while in the latter case we say that has characteristic c where c is the number described in the second part. Since, in an integral domain, the product of two non-zero elements must remain non-zero, it follows that the number c := n1 is a prime number . All subrings of an integral domain of characteristic c are integral domains of the same characteristic c. Thus if D is any sort of subring of the complex numbers C, then it is an integral domain of characteristic 0. Thus the Gaussian Integers, the Eisenstein numbers, or any other subring of the reals or complex numbers, must be an integral domain of characteristic 0. The field Z/pZ, where p is a prime number, clearly has characteristic p. Thus the characteristic is either 0 or a positive prime number p. Of course every field is also an integral domain, so they too possess a characteristic. The cancellation law for integral domains. Suppose D is an integral domain. If, for any three elements (a, b, c) ∈ D × D × D, b 6= 0, ab = cb, then 252 (7.11) b = c (7.12) As a check on one’s ability to construct proofs that everyone might understand, the student should attempt to lay out an unassailable proof of the above assertion. Finite integral domains are fields. Suppose now that D is a finite integral domain. Then for any non zero element d, right multiplication by d must be an injective mapping D → D. For if ad = bd then (a − b)d = 0R . But if d 6= 0 then by the integral domain hypothesis, a = b. Thus right multiplication by any non-zero element is injective. Therefore , as D∗ is a finite set, it must be surjectve (the famous “Pidgeon Hole Principle). Thus xR = Rx = R for every non-zero element x. Of course, this means that x is a unit. Thus D is a commutive ring in which every non-zero element is right invertible, hence left invertible, hence a unit – i.e. D is a field. 253 Chapter 8 Elementary properties of Modules 8.1 8.1.1 Basic theory Modules over rings. Fix a ring R. A left module over R or left R-module is an additive group M = (M, +), which admits a composition R×M →M (which we denote as a juxtaposion of arguments to indicate multiplication) such that for all elements (a, b, m, n) ∈ R × R × M × M , a(m + n) (a + b)m (ab)m 1·m = = = = am + an am + bm a(bm) m, (8.1) (8.2) (8.3) (8.4) where 1 denotes the multiplicative identity element of R. It follows from 0R + 0R = 0R that for any element m of M , 0R m + 0R m = 0R m. Thus 0R m = 0M for any m ∈ M . Here of course 0R denotes the additive identity of the additive group (R, +) while 0M denotes the additive identity of the additive group (M, +). A right R-module is an additive group M = (M, +) admitting a right composition M × R → M , such that, subject to the same conventions of 254 denoting this composition by juxtaposition, we have for all (a, b, m, n) ∈ R × R × M × M, (m + n)a m(a + b) m(ab) m1 = = = = ma + na ma + mb (ma)b m. (8.5) (8.6) (8.7) (8.8) Again we can easily deduce the identities m0R = 0M m(−a) = −(ma) (8.9) (8.10) where 0R and 0M are the additive identity elements of the groups (R, +) and (M, +), and m is an arbitrary element of M . If should be clear from the second equation of the definition of right Rmodule (Equation 8.5) that right multiplication of all of M by an element r ∈ R, induces an endomorphism f (r) of the additive group M . But if we denote the composition of endomorphisms in their chronological order, by left-to-right juxtaposition (that is, the one on the left is applied first) we see f (ab) = f (a)f (b), (8.11) right multiplication by a, followed by right multiplication by b, (a, b) ∈ R × R. Also, the first equation (8.6) tells us that addition of endomorphisms represents ring addition – i.e.f (a + b) = f (a) + f (b) for all (a, b) ∈ R × R. In short, we have a ring homomorphism R → End(M ). Conversely, if we are given such a ring homomorphism f , and if we regard endomorphisms as being right operators of the group (M, +), then M acquires the structure of a right R-module by setting mr := m · f (r), for all (m, r) ∈ M × R. We can take this point of view with left R-modules as well, but there are two approaches to it: (1) using opposite rings, or (2) using the “circle” convention for composition of endomorphisms. (1) Here, given a left R-module M and regarding the composition of endomorphisms to be indicated by juxtapostion in their chronological order (reading left to right), we obtain a ring homomorphism Opp(R) → End(M ). 255 (2) Here all endomorphisms of M are viewed as left operators. The “circle” convention for composing mappings, asserts that (α ◦ β)(m) := α(β(m)), for all m ∈ M. Thus in writing α ◦ β we mean that left operator β is chronologically applied first, and α is applied at some later date. Thus if left multiplication of the entire left R-module M by element r ∈ R induces the endomorphism f (r) of the additive group M , then we have f (ab) = f (a) ◦ f (b). But then, we obtain the ring homomorphism r → (End(M ), ◦), where “circle” indicates composition. In other words, the “opposite ring” construction is built into the definition “◦”, for we have a ring isomorphism (End(M ), ◦) ' Opp(End(M )). Finally, there is a third type of module, an (R, S)-bimodule. Here M is a left R-module and at the same time a right S-module, and the left and right compositions are connected by the law r(ms) = (rm)s, for all (r, m, s) ∈ R × M × S. (8.12) Although this resembles a sort of “associative law”, what it really says is that every endomorphism of M induced by right multiplication by an element of ring S commutes with every endomorphism induced by left multiplication by an element of ring R. Example 1.. Let T be a ring and let R and S be subrings of T . Then (T, +) is an additive group, which admits a composition T × R → T . It follows directly from the ring axioms, that with respect to this composition, T becomes a right R module. We denote this module by the symbol TR when we wish to view T in this way. Also, (T, +) admits a composition R × T → T with respect to which T becomes a left R-module which we denote by R T . Finally (T, +) admits a composition S×T ×R→T 256 inherited directly from ring multiplication. With respect to this composition T becomes an (S, R)-bimodule, which we denote by the symbol S TR . Notice that here, the special law for a bimodule (Equation 8.12) really is the associative law of T (restricted to certain triples, of course). 8.1.2 Submodules Fix a (right) R-module M . A submodule is an additive subgroup of (M, +) which is invariant under right multiplication by elements of R – that is, right multiplication of the subgroup by any element of R yields a subset of that subgroup. It is easy to see that a subset N is a submodule of the right R-module M , if and only if 1. N + N = N 2. −N = N , and 3. N R ⊆ N . It should be remarked that the last containment is actually the equality N R = N , since right multiplication by the identity element 1R induces the identity endomorphism of N . Suppose now N = {Nσ |σ ∈ I} is a family of submodules of the right R-module M . The intersection ∩N Nσ of all these submodules is clearly a submodule of M . This is the unique supremum of the poset of all submodules contained in all of the submodules in N . Now let X be an arbitrary subset of the right R-module M . The submodule generated by X is defined to be the intersection of all submodules of M which contain the subset X and is denoted hXiR . From the previous paragraph, this intersection is the unique smallest submodule in the poset of all submodules of M which contain X This object has another description. Let X̄ be the set of all elements of M which are finite R-linear combinations of elements from X – that is elements of the form x1 r1 + . . . + xn rn where n is any natural number, the xi belong to X and the rj belong to the ring R. Then X̄ is a subgroup (recall 257 that x(−1R ) = −x) and X̄R = X̄, so X̄ is a submodule. On the other hand, it must lie in any submodule which contains X and so must coincide with hXiR . In the case that X = {x1 , . . . , xn } is a finite set, then we see X̄ = x1 R + · · · + xn R = hXiR . A right R-module is said to be finitely generated if and only if it is generated by a finite set. Thus right R module N is finitely generated if and only if N = x1 R + · · · + xn R for some finite subset {x1 , . . . xn } of its elements. Now once again let N = {Nσ |σ ∈ I} be a family of submodules of the right R-module M . The sum over N is the submodule of M generated by the set-theoretic union ∪N of all the submodules in the family N . It is therefore the set of all finite R-linear combinations of elements which lie in at least one of the submodules in N . But of course, as each submodule is invariant under right multiplication by elements of R, this simply consists of all finite sums of elements lying in at least one module of the family. In any event, this is the unique member of the set of all submodules of M which contain every member of N . [Notation: if N = {N1 , . . . , Nn } is a finite family of submodules of M , we write N1 + · · · + Nn for the sum of these submodules. ] We summarize this information in Lemma 142 1. The intersection over any family of submodules of a given module is a submodule. 2. The sum over any family of submodules is the submodule consisting of all finite sums of elements each of which is in at least one member of the family of modules. 3. The poset of all submodules of a module is a lattice with unrestricted meets and joins, given by the intersection and sum operations on arbitrary families of submodules. Everything that we have said in this subsection about submodules of a right R module holds for submodules of a left R-module with the phrase N R = R replaced by RN = N . For this reason the above lemma has been 258 stated without any reference to left or right. It is assumed that we are either talking about the poset of all submodules of a right R-module or else the poset of all submodules of a left R-module. 8.1.3 Factor modules. Now suppose N is a submodule of the right R-module M . Since N is a subgroup of the additive group (M, +), we may form the factor group M/N whose elements are the additive cosets x+N, x ∈ M of N . Now a composition M/N × R → M/N is well-defined since right multiplication of a coset of N by a ring element r lies in a unique additive coset of N – as is seen from the containment (x + N )r ⊆ xr + N (since N R = N ). This composition obeys all the requisite axioms with respect to addition of cosets to endow the factor group M/N with the structure of a right R-module. This is called the factor module M/N Again, this construction can be duplicated for left R-modules: For each submodule N of a left R=module M , we obtain a left R-module M/N . 8.1.4 The fundamental theorems of homomorphisms for R-modules. First the definition of the relevant morphisms: Suppose M and N are right R modules (for the same ring R). An R-homomorphism M → N is a homomorphism f : (M, +) → (N, +) among the underlying additive groups, such that f (m)r = f (mr), (8.13) for all module elements m and ring elements r. In other words, these are group homomorphisms which commute with the right multiplications by each of the elements of the ring R. (The student should be able to draw the relevant commutative square diagram on the board.) The adjectives which accrue to an R-homomorphism are exactly those which are applicable to it as a homomorphism of additive groups: Thus the R-homomorphism f is • an R-epimorphism • an R-monomorphism 259 • an R-isomorphism if and only if f is an epimorphism, monomorphism, or isomorphism, respectively, of the additive groups. We define the kernel of an R-homomorphism f : M → N among right R-modules, simply to be the kernel of f as a homomorphism of additive groups. Explicitly, ker f is the set of all elements of M which are mapped to the additive identity 0N of the additive group N . But the critical equation 0N r = 0N shows that if x lies in ker f , then so must the entire set xR since f (xR) = f (x)R = 0N R = 0N . Thus the kernel of an R-homomorphism (such as f ) must be an R-submodule. The student should be able to restate the obvious analogues of the above definitions for left R-modules as well. Now we can state Theorem 143 (The First Fundamental Theorem of Module Homomorphisms) Suppose f : M → N is a homomorphism of right (left) R-modules. 1. ker f is a submodule of M . 2. The image f (M ) is a submodule of N . 3. There is a natural R-isomorphism φ : M/(ker f ) → f (M ) which takes coset m + ker f to f (m). proof: Part 1 was proved above. Part 2 is immediate from the definition of R-homomorphism. Part 3. By definition φ is already an isomorphism of additive groups. It remains only to verify equation (8.13), – that is, to observe that φ((x + ker f )r) = f (xr) = f (x)r = (φ(x + ker f ))r, for all (x, r) ∈ M × R. Theorem 144 (Second Fundamental Theorem of Module Homomorphisms.) 260 1. If f : M → N is an R-epimorphism of right (left) R-modules, then there is a poset isomorphism between S(M, ker f ), the poset of all submodules of M which contain ker f and the poset S(N, 0N ) of all submodules of N. 2. Suppose A ≤ B ≤ M is a chain in the poset of submodules of a right (left) R-module M . Then there is an R-isomorphism M/A → (M/B)/(B/A) Theorem 145 (The Third Fundamental Theorem of Module Homomorphisms.) Suppose A and B are submodules of the right (left) R-module M . Then there is an R-isomorphism (A + B)/A → B/(B ∩ A). sketch of the proof of the two theorems: All isomorphisms mentioned are exactly those defined in the case of abelian groups. One need only observe that when the arguments forming the poset or group isomorphisms are R-modules, the isomorphisms in question commute with the action of the endomorphisms induced by R. Thus sending an R-submodule C containing ker f to f (C) yields an R-invariant subroup of (N, +), and vice versa. The isomorphisms M/B → (M/A)/(B/A), and (A + B)/A → B/(B ∩ A) are all R-homomorphisms. 8.1.5 The Jordan-Hölder Theorem for R-modules. An R-module is said to be irreducible if and only if its only submodules are itself and the zero submodule – that is, it has no proper submodules. Let Irr(R) denote the collection of all isomorphism classes of non-zero irreducible R modules. Suppose M is a given irreducible R-module, and let P (M ) be the lattice of all its submodules. Recall that an interval [A, B] in P (M ) is the set {C ∈ P (M )|A ≤ C ≤ B} (this is considered to be undefined if A is not a submodule of B). A cover is an interval, [A, B] such that A is a maximal proper submodule of B – that is (A, B) is itself an unrefinable chain of submodules of length one. Finally, recall that a non-empty interval [A, B] is said to be algebraic if and only if there exists a finite unrefinable chain of submodules from A to B. 261 Because of the Second Fundamental Theorem of Homomorphisms, we see that (A, B) is a cover if and only if B/A is a non-trivial irreducible R-module. Thus we have a well-defined mapping µ : Cov(P (M )) → Irr(R), from the set of all covers in the lattice P (M ) to the collection of all nonzero irreducible R-modules. If A1 and A2 are distinct proper submodules of a submodule B, with each Ai maximal in B, then A1 + A2 = B, and by the Third Fundamental Theorem of Homomorphisms, A2 /(A1 ∩ A2 ) (being isomorphic to B/A1 ) is an irreducible R-module. Thus if {[A1 , B], [A2 , B]} ∈ Cov(P (M )) and A1 6= A2 , then [A1 ∩ A2 , Ai ], i = 1, 2 are also covers. This is the statement that • The poset P (M ) of all submodules of the R-module M , is a semimodular lattice. We can now conclude from the Jordan-Hölder Theorem for semi-modular lower semilattices, Theorem 146 (Jordan-Hölder Theorem for R-modules.) Let M be an arbitrary R-module, and let P (M ) be its poset of submodules. 1. Then the mapping µ : Cov(P (M )) → Irr(R), extends to an interval measure µ : Alg(P (M )) → M(Irr(R)), from the set of all algebraic intervals of P (M ) to the multiset monoid over the collection of all isomorphism classes of non-trivial irreducible R-modules. 2. Suppose M itself possesses a composition series – that is, there exists a finite unrefinable chain of submodules beginning at the zero submodule and terminating at M . The composition factors are the finitely many non-trivial irreducible submodules formed from the factors between successive members of the chain. Then the collection of irreducible modules so obtained is the same with the same multiplicities, no matter what composition series is used. In particular all composition series of M have the same length. 262 8.1.6 Direct sums and products of modules. Let I be any index set, and let {Mσ |σ ∈ I} be a family of right R-modules indexed by I. Recall that the direct product of the additive groups (Mσ , +) is the collection of all functions f : I → ∪I Mσ , from I into the disjoint union of the Mσ so that f (σ) ∈ Mσ . This had the structure of an additive Q group I Mσ with addition defined “coordinatewise ” so that f + g was the Q mapping σ → f (σ) + g(σ). Here, since each Mσ is an R-module, I Mσ can Q be endowed with the structure of a right R-module, by defining f r ∈ I Mσ by Y f r : σ → f (σ)r, for all (f, r, σ) ∈ ( I Mσ ) × R × I. (8.14) Q When all the Mσ are R-modules, I Mσ is understood to be this R-module and is called the direct product over the family of R-modules {Mσ |σ ∈ I}. P Clearly its additive subgroup: I Mσ , consisting of all functions f for which f σ 6= 0Mσ at only finitely many σ, is also a submodule. Thus again, P when the Mσ are all R-modules, I Mσ is understood to be this R-submodule Q of I Mσ . This is the direct sum of the Mσ . As was the case for groups, when I is a finite set {1, . . . , n}, the two notions of product and sum coincide, and we write Y I M σ = M 1 ⊕ M2 ⊕ · · · ⊕ M n . Lemma 147 (How To Recognize Direct Sums Internally) Let M1 , M2 , . . . be a countable sequence of submodules of an R-module M which generate M . P Then M ' N Mi if and only if, for any natural number k, (M1 + M2 + · · · + Mk−1 ) ∩ Mk = {0M }. P proof: For each mapping f in I Mσ , let supp(f ), be the set of σ for which f (σ) is not the identity element of Mσ . Then by definition of the sum, supp(f ) is always a finite subset of the index set N, the natural numbers. P We next define a mapping κ : I Mj → M , by the recipe: κ:f → X j∈supp(f ) f (j). Now it is easy to see that κ(f + g) = κ(f ) + κ(g) κ(−f ) = −κ(f ) 263 (8.15) (8.16) so κ is a homomorphism of groups. Moreover the definition of κ reveals κ(f r) = κ(f ) · r, for all r ∈ R. (8.17) Thus κ is actually an R-homomorphism. Since M is generated by the Mj , each element of M is a finite sum of elements each lying in at least one of the Mj ’s. Thus a typical element would be m = m1 + · · · mt , where mi lies in Mp(i) , for some index p(i) possibly depending on (but not necessarily uniquely determined by) i. By adding together summands in the same submodule, we may assume without loss that the indices p(i), 1 ≤ i ≤ t are all distinct. Now there is a function f in P i∈N Mi whose value at p(i) is mp(i) , for 1 ≤ i ≤ t, and is 0M at all other indices. Clearly f has been devised so that κ(f ) = m. Thus κ is a surjective mapping and so is an R-epimorphism. It remains only to show that κ is one-to-one (injective) for in that case κ is the desired isomorphism of R-modules. To that end, suppose f is an P element of i∈N Mi ; set J := supp(f ), a finite subset of natural numbers; and suppose κ(f ) = 0, the additive identity of M . Clearly J is empty if and only if f = 0. So assume J is non-empty and P set k = max(J). Then κ(f ) = j∈J f (j) = 0. This means −mk = X j∈J−{k} f (j) ∈ Mk ∩ (M1 + · · · + Mk−1 ) = {0} (8.18) This is a contradiction since mk = f (k) and k ∈ J, the support of f , and so mk cannot be 0. Thus J = ∅ and f = 0. We have shown then that ker κ (as a group homomorphism) is trivial. It follows that the mapping κ is one-to-one as desired. This completes the proof. 8.1.7 Free modules P A free right R-module over I is a direct sum σ∈I Mσ where each summand Mσ is isomorphic as right R-modules to the R-module RR . This object is uniquely determined up to R-isomorphism by the index set I and so is here P denoted by I RR . Analogously, a free left R-module is one that is a direct P sum of copies of R R. If I indexes the list of summands, we write this I R R. There is a special subset EI of the free right module over I which we call the special R-basis over I. It is the set {ei |i ∈ I}, where ei denotes the function from I into RR for which: 264 ei (j) = ( 1R if j = i 0R otherwise In keeping with our general notation, 1R is the multiplicative identity of R, and 0R is the additive identity of R. There is a canonical bijection β : EI → I in which ei is mapped to i. (Notice that there is nothing “natural” about the definition of ER or the bijection β. It all depends on how the free module is presented as a direct sum. Clearly there may be R-automorphisms of the free module in which the presented direct summand ei R may be sent to a submodule which is not one of the presented direct summands at all.) Despite the drawbacks of the preceeding remark, there is a useful property of a special basis. P Lemma 148 1. Every element of the direct sum i∈I RR has a unique expression as a finite R-linear combination of the elements of EI . P 2. Suppose EI is a special basis for the free module I RR . Then any arbitrary mapping f : EI → M into a right R-module M can be extended P to an R-homomorphism I RR → M . proof: Part 1. If their were two distinct ways to express some element of the direct sum in this way, then by substracting one from the other, we would obtain a nontrivial expression X er i∈J j j = 0, (8.19) where J is a finite non-empty subset of I, and each rj is non-zero. Now the right side of equation (8.19) is the function taking each member of I to 0R . But for each k ∈ J, this function has value rk 6= 0R which is a contradiction. P Part 2. First, every element of I RR is a finite R-linear combination of the elements of ER since we are dealing with a direct sum of the subP modules ej R. Given f as described we propose to define f ( i∈J ej rj ) to be P j∈J f (ej )rj . But this is well-defined by the uniqueness of the respesentation of the R-linear expression. From its definition, it is automatically defining an R homomorphism with the desired properties. The main result is this: 265 Theorem 149 Suppose M is a right (left) R-module generated by a subset X – that is M = hXiR (R hXi). 1. Then there is an R-epimorphism κ from the free module over X to M such that the restriction of κ to the special R-basis EX yields the bijection βX : EX → X. 2. Moreover, if f : N → M is any R-epimorphism, then there exists an P R-homomorphism hf : x∈X RR → N from the free module to N for which the composition f ◦ hf is κ. remark: The minimal use of “right” and “left” in this theorem is deliberate. We are either talking throughout about morphisms in the category of right Rmodules, or throughout about morphisms in the categrory of left R-modules. proof: The proof is simpler than the the statement of the theorem would suggest. The first part is an immediate consequence of the fact that the bijection βX : EX → X when composed with the (somewhat metaphysical) containment mapping X ,→ M , can be extended to an R-homomorphism. This is clearly an R-epimorphism since X generates M as an R-module. Suppose we are given the epimorphism f : N → N . In each fibre f −1 (x) ( a coset of ker f mapping onto x), x ∈ X, choose a representative x̂ so that f (x̂) = x, and set X̂ := {x̂|x ∈ X}. Then X̂ generates some submodule N 0 of N with the property that f (N 0 ) = M . Now by Part 1, the bijection EX → X̂, extends to an R-homomorphism hf which takes ex to x̂, as x ranges over X. Then hf has all the desired properties since X (f ◦ hf )( 8.1.8 X ex rx ) = f ( X hf (ex )rx = f ( = X X f (x̂)rx = = X κ(ex rx ) = κ( x̂rx ) xrx X ex rx ). (8.20) Vector spaces. Let D be a division ring. A right D-module M is called a right vector space over D. Similarly, a right F -module is a right vector space over F. The student should be able convert everything that we define for a right vector space to the analogous notion for a left vector space. There is thus no 266 need to go through the discussion twice, once for left vector spaces and once again for right vector spaces – so we will stick to right vector spaces for this discussion. The elements of the vector space are called vectors. We say that a vector v linearly depends on a set {x1 , . . . , xn } of vectors if there exist elements of the field F , α1 , . . . , αn , such that v = x1 α1 + · · · + xn αn , – that is, v is a member of the (right) submodule generated by x1 , . . . , xn . Clearly from this definition, the following properties hold: Reflexive Property x1 linearly depends on {x1 , . . . , xn }. Transitivity If v linearly depends on {x1 , . . . , xn } and each xi linearly depends on {y1 , . . . , ym }, then v linearly depends on {y1 , . . . , ym }. The Exchange Condition If vector v linearly depends on {x1 , . . . , xn } but not on {x1 , . . . , xn−1 }, then xn linearly depends on {x1 , . . . , xn−1 , v}. Thus linear dependance is a dependence relation as defined in the last section of Chapter 2. It follows from Theorem 40, page86 and Theorem 41, page86, that maximal linearly independant sets exist and any two maximal linearly independant sets in M have the same cardinality. Therefore this cardinality is an invariant of M and is called the dimension of M over F and is denoted dimF (V ). Thus in particular, if F is a finitely generated F -module, then it is finite dimensional. Recall from Chapter 2, Theorem 39, that a maximal independant set is at the same time, a minimal spanning set. Such a set is called an F -basis for M . In this case, every vector of M is uniquely expressible as a finite F -linear combination of the vectors in this basis. Suppose now v is any non-zero vector of the vector space V . Then v generates an F -submodule vF which is a one-dimensional subspace. Then it is easy to see that the mapping mv : FF → vF which sends field element α to vα gives an F -isomorphism of F -modules. (The argument runs like this: It is easily seen to be an F -homomorphism and its kernel is the two-sided ideal {α ∈ F |vα = 0M }. But there is a severe shortage of two-sided ideals in a field. The only possibilities are 0 and F itself. But the latter alternative for ker mv is not available since v1F = v, by the definition of module. Thus ker mv = 0F and so mv is injective.) Now we can say 267 Lemma 150 If B is any basis for the F -vector space M , then M is isoP morphic to the direct sum b∈B bF . Since each bF is isomorphic to the right F -module FF , M is a free F -module. proof: The unique expression of each element as a finite F -linear combination of elements of B defines the group isomorphism M→ X b∈B bF which is easily seen to behave as an F -isomorphism. A morphism V → W of right F -modules is called a linear transformation. Given a linear transformation T : V → V of a vector space into itself, we can convert V into a left F [x]-module by defining p(x) · v := p(T )(v) for any p(x) ∈ F [x] and v ∈ V . (Note that F [T ] is the subring of the F -endomorphism ring EndF (M ) genereated by {T }.) We will have occasion to examine finitely generated modules of this type – even to classify them up to isomorphism. remark: Note that we have defined dimension of a right vector space. Similarly we can define linear dependance and dimension for a left vector space. But if V is a (D, D) bimmodule we have both a left dimension and a right dimension? Must these dimensions be the same? Is a left-linearly independant set also a right-linearly independant set? More about this later. 8.1.9 Examples and exercises Example 2. (Ideals as modules.) A submodule of RR is simply a right ideal of R. Similarly the left ideals of R are just the submodules of the (free) left R-module R R. This is certainly a trivial consequence of the definitions, but it is a useful notion, for it allows one to apply module notions (such as being finitely generated or semisimple) to the ideals of a ring to get a richer language in which to discuss the parent ring itself. Exercise 1. ( Annihilators of modules as two-sided ideals.) Suppose that M is a right R-module. Show that Ann(M ) := {r ∈ R|M r = 0M } is a two-sided ideal in R. 268 Exercise 2. (Abelian groups are Z-modules.) Obviously Z-modules, like all R-modules, are abelian groups. Conversely suppose A is an Abelian group. A composition A × Z → A can be defined as follows: na := a + a + · · · + a.( with n summands ), n ∈ N (−n)a := (−a) + . . . + (−a)( with n summands), n ∈ N 0a := 0A . (8.21) (8.22) (8.23) Show that A bcomes a Z-module with respect to this composition. So the two notions, Z-module and Abelian group, are coextensive. Exercise 3. (The Jordan-Hölder theorem implies the fundamental theorem of Arithmetic.) This exercise is going to occur in four steps to be demonstrated by the student: 1. A Z-module M is irreducible if and only if it is a cyclic group of prime order. 2. If n is a natural number and p is a prime number, then nZ/pnZ is a cyclic group of prime order p, and hence is an irreducible Z-module 3. Suppose the integer n has a factorization n = pa11 · · · pakk for some set of prime numbers {p1 , . . . , pk } and natural numbers aj . Then there exists a composition series A0 := nZ < A1 < · · · < Am := Z, such that the composition factors involve Z/p1 Z (exactly a1 times), Zp2 Z (exactly a2 times), . . . Z/pk Z exactly ak times. [Hint: apply the previous step repeatedly.] 4. By the Jordan-Hölder Theorem any two composition series of Z/nZ yield the same multiset of composition factors. By the previous step these are recorded in any prime-factorization. Hence up to order and unit multiples (association classes) these are unique. Example 3. (Ideals defined by varieties and vice versa.) Let V be the n-dimensional vector space F (n) of all n-tuples over a field F . Let R := F [x1 , x2 , . . . , xn ] be the ring of all polynomials in n indeterminates {x1 , . . . , xn } having coefficients in the field F . Suppose we begin with an ideal J in R. Then the variety of zeroes of J is defined to be the subset V(J) := {(a1 , . . . , an ) ∈ F (n) |p(a1 , . . . , an ) = 0 ∀p ∈ J} 269 –that is, the vectors which are zeroes of each poynomial p(x1 , . . . , xn ) in J. This produces an order-reversing mapping V : the poset of ideals of R → the poset of subsets of V. Conversely, for every subset X of F (n) , we may consider the collection I(X) of all polynomials p(x1 , . . . , xn ) in F [x1 , x2 , . . . , xn ] such that p(a1 , . . . , an ) = 0 for all vectors (a1 , . . . , an ) ∈ X. Any polynomial r times such a polynomial p ∈ I(X) is again such a polynomial. Similarly, the sum or difference of two polynomials which vanish on X, also vanish on X. Thus I(X) is an ideal in R. The result is an order-reversing mapping of these posets in the opposite direction: I : poset of subsets of V → poset of ideals of F [x1 , x2 , . . . , xn ]. Moreover, we possess the monotone relations J¯ := I(V(J)) ≥ J X̄ := V(I(X)) ≤ X (8.24) (8.25) for every ideal J in R and subset X within vector space V . Thus the pair of mappings forms a Galois connection in the sense of Chapter 0, and the ¯ and subset X → X̄, are closure operators two “bar” operations, ideal J → J, (idempotent order preserving maps on a poset). Their images – the “closed” objects – are a special subposet which defines a topology. If we start with an ideal J, its closure J¯ is a particular ideal rad(J) := {x ∈ R|xn ∈ J, for some natural number n}. This is the content of the famous “zeroes Theorem” of David Hilbert. 1 An analogue for varieties would be a description of the closure X̄ of X in terms of the original subset X. These image sets X̄ are called (affine) algebraic varieties. That there is not a uniform description of affine variesties in terms of V alone should be expected. The world of “V ” allows only such a primitive language to describe things, that we cannot say words like “nilpotent”, “ideal”, etc. That is why the Galois connection is useful. The mystery of these ”topology-inducing” closed sets can be pulled back to a better understood algebraic world, the poset of ideals. 1 Note that rad(J), as defined, is not patently an ideal when F is replaced by a noncommutative division ring. 270 Is the new topology on V really new? Of course, it is not ordained in heaven that any vector space actually has a topology which is more interesting than the discrete one. In the case of finite vectors spaces, one cannot expect anything better, and indeed every subset of a finite vector space is an affine algebraic variety. Also, if F is the field of complex numbers, the two topologies coincide, although that takes a non-trivial argument. But there are cases where this topology on algebraic varieties differs from other standard topologies on V , or, in some cases is the only available topology known. Every time this happens, one has a new opportunity to do analysis another way, and to deduce new theorems. 8.2 8.2.1 Consequences of chain conditions on modules Definitions: Noetherian and Artinian modules Let M be a right R-module. We are interested in the poset P (M ) := (P, ≤), of all submodules of M , parially ordered by inclusion. We have mentioned in a previous section that P (M ) was a semimodular lower semi-lattice for the reason that that was all that was needed for activating the general JordanHölder theory. Actually, the fundamental homomorphism theorems imply a much stronger property: Theorem 151 The poset P (M ) = (P, ≤) of all submodules of a right Rmodule is a modular lattice with a minimal element 0 (the lattice “zero”) and a unique maximal element, M (the lattice “one”). Recall from Chapter 2 that a lattice L is modular if and only if (M) a ≥ b implies a ∧ (b ∨ c) = b ∨ (a ∧ c) for all a, b, c in L. It is manifestly equivalent to it’s dual: (M∗ ) a ≤ b implies a ∨ (b ∧ c) = b ∧ (a ∨ c) for all a, b, c in L. In the poset P (M ) the condition M reads: If A, B, and C are submodules with A contained in B, then A ∩ (B + C) = B + (A ∩ C). 271 This is an easy argument, acheived by considering an arbitrary element of each side. We say that module M is Noetherian if and only if the poset P (M ) satisfies the ascending chain condition (ACC). Similarly, we say that M is Artinian if and only if the poset P (M ) satisfies the descending chain condition (DCC). Obviously each of these properties inherits to submodules and factor modules. Finally, we say that the module M has a finite composition series if and only if there is a finite unrefineable chain of submodules in P (M ) proceeding from 0 to M . Now since P (M ) is a modular lattice, so are each of the intervals [A, B] := {X ∈ P (M )|A ≤ X ≤ B}. Moreover, from the modular property, we have that any interval [A, A + B] is poset isomorphic to [A ∩ B, B]. Using these facts the student should be able to prove Lemma 152 The following three conditions for an R-module M are equivalent: 1. M has a finite composition series. 2. M is both Noetherian and Artinian. 3. Every unrefinable chain of submodules if finite. Exercise 4. Prove the preceeding Lemma. Exercise 5. (Heredity of the Noetherian (Artinian) properties) 1. Show that any submodule or factor module of a Noetheriean (Artinian) module is Noetherian (Artinian) 2. Suppose M = N1 ⊕ · · · ⊕ Nk , a direct sum of finitely many submodules Ni . Show that if each submodule Ni is Noetherian (Artinian) then M is Noetherian (Artinian). 3. Show that if M/A and M/B are Noetherian (Artinian) right modules, then so is M/(A ∩ B). 4. Let R be a ring. Suppose the right R-module RR is Noetherian (Artinian) (in this case we say that the ring R is right Noetherian (Artinian)). Show that any finitely generated right R-module is Noetherian (Artinian). [Hint: Such a right module M is a homomorphic image 272 of a free module F = x1 R⊕· · ·⊕xk R over a finite number of generators. Then apply other parts of this exercise.] 8.2.2 Effects of the chain conditions on endomorphisms. The basic results Let f : M → M be an endomorphism of the R-module M . As usual, we regard f as operating from the left. Set f 0 to be the identity mapping on M , f 1 := f and inductively define f n = f ◦ f n−1 , for all positive integers n. Then f n is always an R-endomorphism, and so its image Mn := f n (M ) and its kernel Kn := ker(f n ) are submodules. We then have two series of submodules, one ascending and one descending: 0 = K 0 ≤ K1 ≤ K2 ≤ · · · M = M 0 ≥ M1 ≥ M 2 ≥ · · · Lemma 153 Let f be an endomorphisms of M and let Kn := ker(f n ) and Mn := f n (M ), as above. The following statements hold: 1. Mn = Mn+1 implies M = Mn + Kn . 2. Kn = Kn+1 implies Mn ∩ Kn = 0. 3. If M is Noetherian, then for sufficiently large n, Mn ∩ Kn = 0. 4. If M is Artinian, then M = Mn + Kn for sufficiently large n. 5. If M is Noetherian and f is surjective, then f is an automorphism of M. 6. If M is Artinian and f is injective, then again f is an automorphism of M . Proof: 1. Since Mn = f (Mn ) we have Mn = M2n . Suppose x is an arbitrary element of M . Then, f n (x) = f 2n (y), for some element y. But we can write 273 x = f n (y) + (x − f n (y)). Now the latter summand is an element of Kn while the former summand belongs to Mn . Thus x ∈ Mn + Kn and the result follows. 2. Suppose Kn = Kn+1 , so that Kn = K2n . Choose x ∈ Kn ∩ Mn . Then x = f n (y) for some element y, and so 0 = f n (x) = f 2n (y). But as K2n = Kn , f 2n (y) = 0 implies f n (y) = 0 = x. Thus Kn ∩ Mn = 0. 3. If M is Noetherian, the ascending series K0 ≤ K1 ≤ · · · must stabilize at some point, so there exists a natural number n such that Kn = Kn+1 . Now by Part 2., Mn ∩ Kn = 0. 4. If M is Artiniam, the decending series in the Mi eventually stabilizes and so we have Mn = Mn+1 for sufficiently large n. Now apply Part 1. 5. If f is surjective, M = M0 = Mn for all n. But M is Noetherian, so by Part 3., Mn ∩ Kn = 0 for some n. Thus Kn = 0. But since the Ki are ascending, K1 = 0, which makes f injective. Hence f is a bijective endomorphism, that is, an automorphism. 6. If f is injective and M is Artinian, Mn = Mn+1 for sufficiently large n, and so M = Mn + Kn . But Kn = 0 since K1 = 0 by hypothesis. Thus M = Mn , and since the Mi form a descending series, M = M1 and f is surjective. Hence f is a bijection once more. Corollary 154 Suppose M is both Artinian and Noetherian and f : M → M is an endomorphism. Then for some natural number n, M = M n ⊕ Kn . Fittings Lemma A module M is said to be indecomposable if and only if it is not the direct sum of two non-trivial submodules. Theorem 155 (Fitting’s Lemma.) Suppose M 6= 0 is an indecomposable right R-module which possesses a finite composition series. Then the following hold: 1. For any R-endomorphism f : M → M , f is either invertible (that is, f is a unit in the endomorphism ring S := EndR (M )) or f is nilpotent (which means that f n = 0, the trivial endomorphism, for some natural number n). 274 2. The endomorphism ring S = EndR (M ) possesses a unique maximal right ideal which is also the unique maximal left ideal, and which consists of all nilpotent elements of S. Proof: 1. Recall that since P (M ) is a modular lattice, M has a finite composition series if and only if it is both Artinian and Noetherian. It follows from the preceeding Corollary that for sufficiently large n, M = Mn + Kn and Mn ∩ Kn = 0, so, in fact M = Mn ⊕ Kn . Since M is indecomposable, either Mn = 0 or Kn = 0. In the former case f is nilpotent. In the latter case f is injective, and it is also surjective since M = Mn . Thus f is an automorphism of M and so possesses a two-sided inverse in S. 2. It remains to show that the nilpotent elements of S form a two-sided ideal, and to prove the uniqueness assertions about this ideal. Suppose f is nilpotent. Since M 6= 0, K1 = ker f 6= 0, and so for any endomorphism s ∈ S, neither sf nor f s are injective. So, by the dichotomy imposed by Part 1., both sf and f s must be nilpotent. Now assume f and g are nilpotent endomorphisms while h := f + g is not. Then by the dichotomy, h possesses a 2-sided inverse h−1 . Then 1M = hh−1 = f h−1 + gh−1 . It follows that the two endomorphisms, s := f h−1 and t := gh−1 , commute and, by the previous paragraph, are nilpotent. Then there is a natural number m large enough to force sm = tm = 0. Then 2m 1M = (1M ) 2m = (s + t) = X2m j=0 2m j ! s2m−j tj . Each monomial s2m−j sj = 0 since at least one of the two exponents exceeds m − 1. But that yields 1M = 0, an absurdity as M 6= 0. Thus h must also be nilpotent. Thus the set nil(S) of all nilpotent elements of S forms a two-sided ideal. That this ideal is the unique maximal member of the poset of right ideals of S follows from the fact that S − nil(S) consists entirely of units. A similar statement holds for the poset of left ideals of S. The proof is complete. The Krull-Remak-Schmidt Theorem. Lemma 156 Let f : A = A1 × A2 → B = B1 × B2 be an isomorphism of Artinian modules, and define mappings α : A1 → B1 and β : A1 → B2 by the 275 equation f (a1 , 0) = (α(a1 ), β(a1 )) for all a1 ∈ A1 . Clearly α and β are homomorphisms of R-modules. If α is an isomorphism, then there exists an isomorphism g : A2 → B2 . Proof: Clearly α = π1 ◦ f ◦ κ1 and β = π2 ◦ f ◦ κ1 , where πj is the canonical projection B → Bj with kernel B3−j , j = 1, 2, and κ1 is the canonical embedding A1 → A1 × A2 sending a1 to (a1 , 0). Suppose first that β = 0. Then f restricted to the first summand has its image equal to B1 × 0, and so induces an isomorphism A/(A1 × 0) → B/(B1 × 0) – that is, an isomorphism A2 → B2 . So our proof will be complete is we can construct from f another isomorphism h : A → B such that h(a1 , 0) = (α(a1 ), 0) for all a1 ∈ A1 . We define h in the following way: If f (a1 , a2 ) = (b1 , b2 ), set h(a1 , a2 ) := (b1 , b2 − βα−1 (b1 )). Clearly h is an R-homomorphism. Since B is Artinian it suffices to show that h is injective. Suppose h(a1 , a2 ) = 0. Then b1 = 0 = b2 so a1 = 0 = a2 since f is an isomorphism. Thus h has been constructed as desired. Lemma 157 Suppose A is a right R-module that is Artinian. Then A is a direct sum of finitely many indecomposable modules. Proof: First we argue that A even possesses an indecomposable direct summand. If A is already indecomposable then we are done. Otherwise there is a non-trivial decomposition B1 ⊕ A1 . If A1 is indecomposable we are done. Otherwise we write A1 = B2 ⊕ A2 . If A2 is indecomposable we have A = (B1 + B2 ) ⊕ A2 and we are done. As long as the process continues we obtain a properly descending series A > A1 > A2 > · · ·. By the Artinian condition the process must terminate at an indecomposable direct summand An := D1 . Now that we can write A = C1 ⊕ D1 , where D1 is non-zero indecomposable, we may, noting that C is Artinian, similarly write C1 = C2 ⊕ D2 , and Ci = Ci+1 ⊕ Di+1 with Di+1 indecomposable, as long as Ci remains 276 decomposable. But the properly descending chain C1 > C2 > · · · must eventually terminate in some finite number m of steps. Then one has a direct decomposition A = Cm ⊕ Dm ⊕ Dm−1 ⊕ · · · ⊕ D1 into indecomposables. Theorem 158 (The Krull-Remak-Schmidt Theorem) Let the Artinian and Noetherian module A = A1 × · · · × Am be isomorphic to B = B1 × · · · × Bn where the Ai and Bj are indecomposable. Then m = n and there is a permutation ρ of the index set I = {1, . . . , n} such that Ai is isomorphic to Bρ(i) for all i ∈ I. Proof: Without loss of generality, we may assume m ≥ n. We shall use induction on min(n, m) = n since the result holds trivially if n = 1. Let f : A → B be the assumed isomorphism. Let κi : Ai → A and 0 κj : Bj → B be the cannonical embeddings associated with the direct sums. Similarly, let πi : A → Ai and πj0 : B → Bj be the corresponding canonical projection epimorphisms associated with these direct sums. Then set αi : = π10 ◦ f ◦ κi : Ai → B1 , βi : = πi ◦ f −1 ◦ κ01 : B1 → Ai , each an R-homomorphism. Then Xm α i=1 i ◦ βi is the identity mapping on the indecomposable module B1 . By Fitting’s Lemma (Theorem 155) not all the αi ◦ βi can be nilpotent. Reindexing if necessary, we may assume α1 ◦ β1 is not nilpotent. Then it is in fact an automorphism of B1 . Then α1 : A1 → B1 is an isomorphism. Then we have f (a1 , 0, . . . , 0) = (α1 (a1 ), . . .). By the preceeding Lemma 156, there is an isomorphism A2 × · · · × Am ' B2 × · · · × Bn . Now we apply the induction hypothesis to deduce that m = n, and that for some permutation σ of J = {2, . . . , n}, Ai is isomorphic to Bσ (i), i ∈ J. The result now follows upon feeding in the isomorphism α1 . 277 8.2.3 Noetherian Modules and finite generation. We can come to the point at once: Theorem 159 Suppose M is a right R-module, for some ring R. Then the poset of right R-submodules of M satisfies the ascending chain condition (ACC) if and only if every right submodule of M is finitely generated. proof: 1. ACC implies every submodule is finitely generated (fg). Assume the poset of right R-submodules of M possesses the ACC. Suppose, by way of contradiction that N is a submodule which is not finitely generated. Then we claim that the collection N of all finitely generated submodules of N possesses no maximal member. For if S ∈ N then S = a1 R + a2 R + · · · am R for elements a1 , · · · am pf N , since S is a finitely generated submodule of N . But as N is not finitely generated there exists an element am+1 ∈ N − S. Then a1 R + · · · am+1 R is a module in N which properly contains S. Thus the maximal condition fails among the right R-submodules of M , and so by the previous section, the ACC fails. 2. Every submodule of the right R-module M finitely generated implies the ACC on submodules of M . Let M1 < M2 < · · · be a properly ascending chain of submodules of H. Then clearly the union M∞ := ∪i Mi is a submodule N of M and so, by hypothesis, is finitely generated. Then N = x1 R + · · · + xm R for some finite subset {xi } ⊆ M∞ . Now each xj lies in some Mf (j) . Thus setting N = max({f (j)|j ≤ m}, we see that MN is the terminal member of the properly ascending sequence. Thus ACC holds on the poset of right R-submodules of M Corollary 160 Suppose the right R-module M satisfies the ascending chain condition on submodules. Then every generating X – that is a set such that hXiR = M – contains a finite subset Y which generates M . The proof is an exercise that the reader should be able to do at this point. Although we have lost the “up-down” symmetry between a poset and its dual by locking the ACC to finite-generatedness of submodules, we have not lost right-left symmetry. Thus 278 A left R-module possesses the ascending chain condition on submodules if and only if every left R-submodule is finitely generated. A right R-module M which possesses either of the two equivalent conditions of the above theorem (the ACC or the maximum condition in the poset of submodules, and the fact that all submodules are finitely generated) has been called Noetherian. Similarly we call a left R module Noetherian if and only every left R-submodule is finitely generated. There is no need to invent two words (“left Noetherian” or “right Noetherian” as long as we know that M is either a right R-module or a left R-module, but not both. These words will be required for bimodules such as R RR , since conceivably one of these things could be Noetherian as a left module without being Noetherian as a right module. But that leads us into the definitions of the next subsection. 8.2.4 Right and left Noetherian rings. The reader shall soon discover that many properties of a ring are properties which are lifted (in name) from the R-modules RR or R R. We say that a ring R is a right Noetherian ring if and only if RR is a Noetherian R-module. In view of Theorem 159 we are saying that a ring R is right Noetherian if and only every right ideal is finitely generated. Similarly, R is a left Noetherian ring if and only R satisfies the ACC on left ideals – or, equivalently, every left ideal is finitely generated. Beleive it or not, there really do exist rings which are right Noetherian but not left Noetherian (Dieudonné). By taking opposite rings one may also have it the other way round. Of course on modules, there is quite a difference between saying, on the one hand, that a module is finitely generated, and on the other hand, that every submodule is finitely generated. That is, being finitely generated and Noetherian are in general two different things in the world of modules. That is where Noetherian rings come in: if M is a right (left) R-module over a right Noetherian ring , there is no difference between the two notions. Let’s make this precise by stating: Theorem 161 Suppose R is a ring which is right Noetherian – i.e. it satisfies the ascending chain condition on its right ideals. Then if M is a finitely 279 generated right R-module, then M is Noetherian – i.e. every submodule of M is also finitely generated. proof: Assume R is right Noetherian, and suppose N is a submodule of M = a1 R +· · · an R, a finitely generated right R-module. Then every element of N can be written as an “expression” a1 r 1 + a2 r 2 + · · · + an r n . (8.26) We next consider the totality of such expressions of the form in equation ( 8.26) which represent elements in N . We say that such an expression of the form ( 8.26) has length ≤ ` if and only if the last n − ` coefficients are all zero. Let I` be the set of coefficients rl of al appearing among expressions of the form ( 8.26) which have length at most `, and represent elements of N . Thus I` := {r` |a1 r1 + · · · + a`−1 r`−1 + a` r` ∈ N, ri ∈ R}. since N is a right R-module itself, it is clear that I` is a right ideal in R. (8.27) Moreover, from the definition of I` , every element b ∈ I` actually is realized as the `-th coefficient of at least one expression of the form ( 8.26), say B(b) := a1 r1 (b) + · · · a`−1 r`−1 (b) + a` b, to make the choice precise. Now by the hypothesis, R has the ascending chain condition on right ideals, and so the ideals I` are finitely generated. Thus we may write (l) (l) (`) I` = d1 R + d2 R + · · · dt(`) (`) (`) (`) (8.28) for some finite set of generators d1 , d2 , . . . dt(`) , depending on the choices of ` = 1, 2, · · · , n. P (`) We can now form the t = nl=1 t(`) expressions B(dj ), 1 ≤ j ≤ t(`), 1 ≤ ` ≤ n, and view these as the elements of N which they represent. We (`) now claim that the t elements B(dj ) generate N as an R-module. To see this, first consider an element of N represented by an expression of length n, say a1 c1 + a2 c2 + · · · + an cn , with cn 6= 0. (8.29) 280 Then cn ∈ In and so from equation (8.28) is an R-linear combination (n) (n) (n) d1 s1 + d2 s2 + · · · + dt(n) st(n) = cn . Then we see that (n) (n) (n) T = B(d1 )s1 + B(d2 )s2 + · · · + B(dt(n) )st(n) = Xt(n) Xn−1 i=1 ( (n−1) a r (d ))si j=1 j j i + an Xt(n) (n) d i=1 i = ( terms of length ≤ n − 1) + an cn is an element of N which is represented by an expression of the form ( 8.26) whose n-th coefficient is also cn . Thus it can be subtracted from the expression in (8.29) to transform it into an expression of length ≤ n − 1 still representing an element of N . This expression in turn can have it’s length reduced to n − 2 by subtracting off a suitable R-linear combination of (n−1) (n−1 B(d1 ), · · · , B(dt(n−1) ). Continuing in this way, a1 c1 + · · · an cn can eventually be reduced to an expression of length zero by succesively subtracting (j) off R-linear combinations of the B(di ). Since 0 is the only element represented by an expression of length zero, we see that any element of N is (j) an R-linearly combination of the t elements B(di ). Thus N is a finitely generated R-module. As N was an arbitrary submodule, the proof is complete. The student is encouraged to state the left-hand version of the statement of this theorem. Although the above theorem reveals a nice consequence of Noetherian rings, the applicability of the theorem appears limited until we know that such rings exist. One very important class are the principal ideal domains. These are the integral domains such that every ideal is generated by a single element. Since an integral domain is by definition commutative, it does not matter whether we are speaking of right ideals or left ideals here. Our most important canonical example is the ring Z of integers. Suppose K is an ideal in the ring of integers. Since K = −K, the ideal K contains positive integers, and so, by the “well-ordered property” contains a unique smallest positive integer, say k. Suppose y is any integer in the ideal K. Then the greatest common divisors of k and y consist of two integers d 281 and −d, where, without loss, d is positive. As is well known from Euclid’s algorithm, d can be expressed as an integer linear combination d = ak + by. Thus d lies in the ideal K. Now d is a positive integer in K dividing k, the smallest positive integer in K. Thus d = k, and so y is an integer multiple of k. Thus K = Zk. Hence Lemma 162 The ring of integers Z is a principal ideal domain. This argument used only certain properties shared by all Euclidean Domains (and there are many). All of them are principal ideal domains by a proof almost identical to the above. But principal ideal domains are a larger class of rings, and most of the important theorems (about their modules and the unique factorization of their elements into primes) encompass basically the analysis of finite-dimensional linear transformations and the elementary arithmetic of the integers. So this subject is given a seperate section. Nonetheless, just knowing that Z is a Noetherian ring, yields the important fact that algebraic integers form a ring. 8.2.5 Algebraic integers form a ring. An algebraic integer is any complex number α which √ is the root of a polynomial in Z[x] having lead coefficient one. Thus 3 is√an algebraic integer, but half of it is not. The complex number ω = (1 − −3)/2 is an algebraic integer because w2 + w + 1 = 0. An easy way to tell whether a complex number α is an algebraic integer is given in the following; Lemma 163 A complex number α is an algebraic integer if and only if Z[α] is a finitely generated Z-module. proof; (⇒). Suppose α is an algebraic integer. Then α satisfies a polynomial equation αn + a1 αn−1 + · · · + an−1 α + an = 0. Then αn is a Z-linear combination of lower powers of α (including the zero-th power). By an induction argument, one can easily establish that for every integer m larger than n, αm ∈ Z + Zα + · · · + Zαn−1 . 282 (8.30) First, it is true for m = 0, 1, . . . , n − 1. If the membership relation ( 8.30) holds for m = k, then αk = z1 + z2 α + · · · + zn αn−1 for integers z1 , . . . zn . Then αk+1 = ααk = α(z1 + · · · + zn αn−1 ) = αz1 + · · · + zn−1 αn−1 + zn αn . (8.31) (8.32) But since αn = −an − an−1 α − · · · − a1 αn−1 we see that αk+1 is a Z-linear combination of {αn−1 , . . . , α, 1}. Thus equation ( 8.30) holds. Since the powers of α generate the Z-module Z[α], ( 8.30) implies Z[α] = Z + Zα + · · · Zαn−1 and so Z[α] is finitely generated as a Z-module. (⇐) Suppose Z[α] is a finitely generated Z-module. Now Z is a principal ideal domain (as noted at the end of a previous section) and so has the ascending chain condition on ideals. Then by our theorem, the Z-module Z[α] satisfies the ascending chain condition on submodules. But the set X := {1, α, α2 , . . .} is a set of generators and so by Lemma 160 there exists a finite subset Y = {αn1 , . . . , αnm } of X which generates Z[α]. Thus if m := max{ni }, then we have αm+1 equal to a Z-linear combination of the αni , i = 1, . . . m, so αm+1 = b1 αn1 + b2 αn2 + · · · + bm αnm , bi ∈ Z so α is a root of the monic polynomial p(x) = xm+1 − bnx 1 − b2 xn2 − · · · − bm xnm in Z[x] of degree m+1 and having leading coefficient 1. Thus α is an algebraic integer. Theorem 164 The algebraic integers form a ring. 283 proof: Certainly if α is a root of a monic polynomial with integer coeficients, than so is −α. Let α and β be any two algebraic integers. It remains only to show that α + β and αβ are algebraic integers. By the previous Lemma 163, both Z[α] and Z[β] are finitely generated Z-modules, and so we may write: Z[α] = Zu1 + Zu2 + · · · Zun Z[β] = Zv1 + Zz2 + · · · + Zvm (8.33) (8.34) Now form the ring Z[α, β] of all polynomial expressions in both complex numbers α and β. This is generated as a Z-module by all monomials αi β j ; 0 ≤ i, j. Since αi and β j lie in Z[α] and Z[β], from equations (8.33) and (8.34) just above, the general monomial αi β j is a Z-linear combination of the set X := {us vt |1 ≤ s ≤ n, 1 ≤ t ≤ m}. Thus X is a set of generators of the Z-module Z[α, β]. Thus Z[α, β] is finitely generated. Since Z is Noetherian, every submodule of Z[α, β] is finitely generated. Thus the submodules Z[α + β] and Z[αβ] are finitely generated. But by the Lemma 163, this means that α + β and αβ are algebraic integers. In the opinion of the author, the spirit of abstract algebra is immortalized in this maxim: • A GOOD DEFINITION IS BETTER THAN MANY LINES OF PROOF. Perhaps no theorem illustrates this principle more than the one just given. In the straightforward way of proving this theorem one is obligated to produce monic polynomials with integer coefficients for which α +β and αβ are roots; not an easy task. Yet the above proof was not hard. That was because we had exactly the right definitions! 284 8.3 The Hilbert Basis Theorem This section gives us a second important application of Theorem 161. We begin with a few remarks about polynomial rings. Previously, in attempting to supply ourselves with examples of rings, we considered monoid rings M R where R was any ring and M was any monoid. The elements of M R were finite (right) R-linear combinations of elements of M , with component-wise addition and right multiplication by elements of R – i.e. in terms of language we have since learned – M R is a free right R-module. Multiplication in M R is defined by assuming two principles: 1. Elements of R commute with elements of M . 2. Every conceivable distributive law holds. Thus M R = RM is an (R, R)-bimodule. In particular, if M is the free monoid generated by an element x, then M R is all expressions r0 + xr1 + · · · xn rn , for all natural numbers n and coefficients rj in R, and is the ring of polylomials over R. The reason this needs to be spelled out is that usually the notion of a ring of polynomials in many books and contexts entails at least 1. that R is a commutative ring, or even 2. that R is a field. . Neither of these are assumed here: only that elements of the ring R commute with powers of x. Of course if M is the commutative multiset monoid generated by a finite set {x1 , . . . xn }, then RM is the ring of polynomials over R and is denoted R[x1 , . . . , xn ]. Theorem 165 (The Hilbert Basis Theorem) That is one of the two applications of the above theorem and is given two subsections ahead. Let R be any (not necessarily commutative ) ring and let R[x] be the ring of polynomials with coefficients in R. If R has the ascending chain condition on right ideals, then so does R[x]. proof: Let A be a right ideal in R[x]. We wish to show that A is finitely generated as a right R[x]-module. Let A+ be the collection of all “lead” 285 coefficients of all polynomials lying in A, where the “lead coefficients” of any polynomial is the coefficient of the highest power of x appearing (with nonzero coefficient) in the polynomial or is zero in the case of the zero polynomial. Since A is a right ideal in the polynomial ring, A+ is a right ideal in R. Then by Theorem 159 A+ is finitely generated, so we may write A+ = a1 R + a2 R + · · · + an R. From the definition of A+ , there exist polynomials pi (x) in A having lead coefficient ai . Let ni := degree pi (x) and set m = max{ni }. Now let Rm [x] be the polynomials of degree at most m in R[x] and set Am = A ∩ Rm [x]. We claim A = Am + p1 (x)R[x] + p2 (x)R[x] + · · · + pn (x)R[x]. (8.35) Let f (x) be a polynomial in A of degree N larger than m. Then f (x) = αxN + terms involving lower powers of x. Then, as α is the leading coefficient of f (x) we have α ∈ A+ . Thus α = a1 r 1 + a2 r 2 + · · · an r n . Then f1 (x) = f (x) − xN −n1 p1 (x)r1 − xN −n2 p2 (x)r2 − · · · − xN −nn pn (x)rn is a polynomial in A which has degree less than N . In turn, if the degree of f1 (x) exceeds m this procedure can be repeated. Thus eventually we obtain a polynomial h(x) in A of degree at most m, which is equal to the original f (x) minus an R[x]-linear combination of {p1 (x), p2 (x), . . . , pn (x)}. But since each pi (x) and f (x) all lie in A, we see that h(x) ∈ Am . Thus f (x) ∈ Am p1 (x)R[x] + · · · pn (x)R[x]. Since f (x) was an arbitrary member of A, equation ( 8.35) holds. In view of equation ( 8.35), it remains only to show that Am R[x] is a finitely generated R[x]-module, and this is true if Am is a finitely generated 286 R-module. But Am is an R-submodule of Rm [x] which is certainly a finitely generated R-module (for {1, x, . . . xm } is a set of generators). Thus by the main theorem of the previous section, since R has the ACC on right ideals, the submodule Am is finitely generated. This completes the proof. remark: Again there exists a left version of this theorem. The ring R[x] is the same. It asserts that if R is left Noetherian, than R[x] is also a left Noetherian ring. Corollary 166 If R is right Noetherian, than so is R[x1 , . . . , xn ]. 287 Chapter 9 The Arithmetic of Integral Domains. 9.1 Introduction. Let us recall that an integral domain is a species of commutative ring with the simple property (ID) If D is an integral domain then there are no zero divisors – that is, there is never a product of two non-zero elements equalling the zero element of D. Of course this is equivalent (in the realm of commutative rings) to the proposition (ID’) The set D∗ of non-zero elements of D, is closed under multiplication, and thus (D∗ , ·) is a commutative monoid under ring multiplication. An immediate consequence is Lemma 167 The collection of all non-zero ideals of an integral domain are closed under intersection, and so form a lattice under either the containment relation or its dual. Under the containment relation, the ‘meets’ are intersections and the ‘joins’ are sums of two ideals. proof: The submodules of DD form a lattice under the containment relation with ‘sum’ and ‘intersection’ playing the roles of and “join” and “meet”. 288 Since such submodules are precisely the ideals of D, it remains only to show that two nonzero ideals cannot intersect at the zero ideal in an integral domain. But if A and B are ideals carrying non-zero elements a and b, respectively, then ab is a non-zero element of A ∩ B. We had earlier introduced integral domains as a class of examples in the Chapter 6 on basic properties of rings. One of the unique and identifying properties of integral domains presented there was: (The Cancellation Law) If (a, b, c) ∈ D∗ × D × D, then ab = ac implies b = c. It is this law alone which defines the most interesting aspects of integral domains – their arithmetic – which concerns who divides who among the elements of D. (It does not seem to be an interesting question in general rings with zero divisors – such as full matrix algebras.) Of course for any commutative ring R we may extract the group of units, U (R). This gives rise to an equivalence relation among the elements of D: the associate relation, where one element is a (left or right) multiple of another by a unit: clearly an equivalence relation. Equivalence relations are fine, but how can we connect these up with “divisibility”? To make the notion precise, we shall say that element a divides element b if and only if b = ca for some element c in R. Notice that this concept is very sensitive to the ambient ring R which contains elements a and b, for that is the “well” from which a potential element c is to be drawn. By this definition, • Every element divides itself. • Zero divides only zero, while every element divides zero. • If element a divides b and element b divides c. then certainly a divides c. Thus the divisibility relationsihip is transitive and so we automatically inherit a preorder which is trying to tell us as much as possible about the question of who divides whom. As one may recall from Chapter 1, for every preorder, there is an equivalence relation (that of being less-than-or-equal in both directions) whose 289 eqivalence classes become the elements of a partially ordered set. Under the divisility preorder for an integral domain, D, equivalent elements a and b should satisfy a = sb, and b = ta. for certain elements s and t in D. If either one of a or b is zero, then so is the other, so zero can only be equivalent to zero. Otherwise, a and b are both non-zero and so, by the Cancellation Law, a = s(ta) = (st)a and st = 1. In this case, both s and t are units, and a and b are associates. Conversely, if a and b are associates, they divide one another. Thus the equivalence classes for the associate-relation in an integral domain D (which are the multiplicative cosets xU (D) of the group of units in the multiplicative monoid (D, ·) of the domain D) are also the equivalence classes of the divisibility preorder . Thus from the definition of the word“divides”, we obtain a poset which we denote Div(D) whose elements are the associate-classes of the integral domain D. Moreover Lemma 168 (Properties of the Divisibility Poset) 1. The additive identity element 0D , is an associate-class that is a global maximum of the poset Div(D). 2. The group of units U (D) is an associate-class comprising the global minimum of the poset Div(D). One can also render this poset in another way. We have met above the lattice of all ideals of D under inclusion – which we will denote here by the symbol L. Its dual lattice (ideals under reverse containment) is denoted L∗ . We are interested in a certain induced sub-poset of L∗ . Ideals of the form xD which are generated by a single element x are called principal ideals. Let P (D) be the poset of L induced on the collection of all principal ideals, including the zero ideal, 0 = 0D. We call this he principal ideal poset. To obtain a comparison with the poset Div(D), we must pass to the dual. Thus P (D)∗ is defined to be the subposet of L∗ induced on all principal ideals — that is, the poset of principal ideals ordered by reverse inclusion. We know the following: 290 Lemma 169 (Divisibility and principal ideals posets) In an intedgral domain D, the following holds: 1. Element a divides element b if and only if bD ⊆ aD. 2. Elements a and b are associates if and only if aD = bD. 3. So there is a bijection between the poset P (D)∗ of all principal ideals under reverse containment, and the poset Div(D) of all multiplicative cosets of the group of units (the association classes of D) under the divisiblity relation, given by the map f : xD → xU (D). 4. Moreover, the mapping f : P (D)∗ → Div(D), is an isomorphism of posets. One should be aware that the class of principal ideals of a domain D need not be closed under taking intersections and taking sums. This has a lot to do – as we shall soon see – with the questions of the existence of “greatest” common divisors, “least” common multiples and ultimately the success or failure of unique factorization. Notice that here, the “meet” of two cosets xU (D) and yU (D) would be a coset dU (D) such that dU (D) divides both xU (D) and yU (D) and is the unique coset maximal in Div(D) having this property. We pull this definition back to the realm of elements of D by saying: The element d is a greatest common divisor of two non-zero elements a and b of an integral domain D if and only 1. d divides both a and b. 2. If e divides both a and b, then e divides d. Clearly, then, if a greatest common divisor of two elements of a domain exists, it is unique up to taking associates (i.e. only a coset of U (D) is being specified) and is unchanged upon replacing any of the two original elements by any of their associates. In fact it is easy to see that if d is the greatest common divisor of a and b, then dD is the smallest principal ideal containing both aD and bD. Note 291 that this might not be the ideal aD + bD, the join in the lattice of all ideals of D. Similarly, lifting back the meaning of a “join” in the poset Div(D), we say The element m is a least common multiple of two elements a and b in an integral domain D if and only if 1. a and b both divide m (i.e. m is a multiple of both a and b). 2. If n is also a common multiple of a and b, then n is a multiple of m Again, we note that if m is the least common multiple of a and b, then mD is the largest principal ideal contained in both aD and bD. Again, this does not mean that mD is aD ∩ bD, the meet of aD and bD in the full lattice L of all ideals of D. The next two lemmas display further instances in which a question on divisibility in an integral domain, refers back to properties of the poset of principal ideals. A factorization of an element of the integral domain D is simply a way of representing it as a product of other elements of D. Of course one can write an element as a product of a unit and another associate in as many ways as there are elements in U (D). The more interesting factorizations are those in which none of the factors are units. We call these proper factorizations. Suppose we begin with an element x of D. If it is a unit, it has no proper factorization. Suppose element x is not a unit, and has a proper factorization x = x1 y1 into non-zero non-units. We attempt, then, to factor each of these factors into two further factors. If such a factorization is not possible, one proceeds with the other proper factor. In this way one may image an infinite schedule of such factorizations. In the graph of the divisility poset Div(D), this schedule would correspond to a downwardly growing binary tree. An endnode to this tree results if and only if one obtains a factor y which possesses no proper factorization. We call such an element “irreducible”. Precisely, an element y of D∗ is irreducible if and only if it is a non-zero non-unit with the property that if y = ab is a factorization, one of the factors a or b, is a unit. Applying the poset isomorphism f : Div(D) → P IP (D) given in Lemma 169 above, these comments force 292 Lemma 170 If D is an integral domain, an element y of D is an irreducible element if and only if the principal ideal yD is non-zero and does not properly lie in any other principal ideal except D itelf (that is, it is maximal in the induced poset P (D) − {D, 0} of proper principal ideals of D under containment). Another important central Lemma of this type which holds for arbitrary integral domains is this: Lemma 171 Let D be an integral domain possessing the ascending chain condition on its poset P (D) of principal ideals. Then every non-zero nonunit is the product of a unit and a finite number of irreducible elements. proof: Suppose there were an element x ∈ D∗ − U (D) which was not the product of a finite number of irreducible elements. Then the collection T of principal ideals xD where x is not a product of finitely many irreducible elements is non-empty. We can thus choose an ideal yD which is maximal in T because T inherits the ascending chain condition from the poset of principal ideals. Since y itself is not irreducible, there is a factorization y = ab where neither a nor b is a unit. Then aD properly contains yD, since otherwise aD = yD and y is an associate ua of a where u is a unit. Instantly, the cancellation law yields b = u contrary to hypothesis. Thus certainly the principal ideal aD does not belong to T . Then a is a product of finitely many irreducible elements of D. Similarly, b is a product of finitely many irreducible elements of D. Hence their product y must also be so, contrary to the choice of y. The proof is complete. What about the converse to this lemma? In any event, the divisibility structure of D is revealed by the divisibility poset Div(D) and is in part controlled by the group of units. It is then interesting to know that the group of units can be enlarged by a process described in the following section.1 1 When this happens, the student may wish to verify that the new divisibility poset so obtained is a homomorphic image of the former. 293 9.2 9.2.1 Localization in Commutative Rings. Localization by a multiplicative set. We say that a subset S of a ring is multiplicatively closed if and only if SS ⊆ S and S does not contain the zero element of the ring. If S is a non-empty multiplicatively closed subset of the ring R, we can define an equivalence relation “∼” on R × S by the rule that (a, s) = (b, t) if and only if, for some element u in S, u(at − bs) = 0. Let us show first that it truly is an equivalence relation. Obviously the relation “∼” is symmetric and reflexive. Suppose now (a, s) ∼ (b, t) ∼ (c, r) for {r, s, t} ⊆ S. Then there exists elements u and v in S such that u(at − bs) = 0 and v(br − tc) = 0. Multiplying the first equation by vr we get vr(uat) = vr(ubs). But vr(ubs) = us(vbr) = us(vtc), by the second equation. Hence vut(ar) = vut(sc), so (a, s) ∼ (c, r), since vut ∈ S. Thus ∼ is a transitive relation. We now let the symbol a/s denote the ∼-equivalence class containing the ordered pair (a, s). We call these equivalence classes fractions. Next we show that these equivalence classes enjoy a ring structure. First observe that if there is an element u in S such that u(as0 − sa0 ) = 0 – i.e. (a, s) ∼ (a0 , s0 ) – then u(abs0 t−sta0 b) = 0, so (ab, st) ∼ (a0 b, s0 t). Thus we can unambiguously define the product of two equivalence classes – or ‘fractions’ – by setting (a/s) · (b/t) = ab/st. Similarly, if (a, s) ∼ (a0 , s0 ), so that for some u ∈ S, u(as0 − sa0 ) = 0 , we see that u(at + sb)s0 t − u(a0 t + s0 b)st = 0, so (at + bs)/s = (a0 t + s0 b)/s0 t. Thus ‘addition’, defined by setting at + bs a b + := , s t st 294 is well defined, since this is also (a0 t + s0 b)/s0 t. The set of all ∼-classes on R × S is denoted RS . Now that multiplication and addition are defined on RS , we need only verify that 1. (RS , +) is an abelian group with identity element 0/s – that is, the ∼-class containing (0, s) for any s ∈ S. 2. (RS , ·) is a commutative monoid with multiplicative identity s/s, for any s ∈ S. 3. Multiplication is distributive with respect to addition in RS . All of these are left as Exercise 1. The conclusion, then, is that (Rs , +, ·) is a ring. This ring Rs is called the localization of R by S. 9.2.2 Special features of localization in domains. For each element s in the multiplicatively closed set S, there is a homomorphism of additive groups ψs : (R, +) → (RS , +) given by ψs (r) = r/s. This need not be an injective morphism. If a/s = b/s, it means that there is an element u ∈ S such that us(a − b) = 0. Now if the right annihilator of us is nontrivial, such an a − b exists with a 6= b, and ψs is not one-to-one. Conversely, we can say, however, Lemma 172 If s is an element of S such that no element of sS is a “zero divisor” – an element with a non-trivial annihilator – then ψs : R → RS is an embedding of additive groups. The converse is also true. In particular, we see that in an integral domain, where every non-zero element has only a trivial annihilator, the map ψs is injective for each s ∈ S. Lemma 173 Suppose S is a multiplicative closed subset of non-zero element of the commutative ring R and suppose no element of S is a zero divisor in R. 295 1. Suppose ac 6= 0 and b, d are elements of S. Then in the localized ring RS a c ( )( ) is non-zero in RS . b d 2. If a is not a zero divisor in R, then for any b ∈, a/b is not a zero divisor in RS . 3. If R is an integral domain, then so is RS . 4. If R is an integral domain, the mapping ψ1 : R → RS is an injective homomorphism of rings (that is an embeddig of rings). proof: : Part 1. Suppose ac 6= 0, but that for some {b, d, s} ⊆ S, a c 0 ( )( ) = . b d s (9.1) Then (ac, bd) ∼ (0, s). Then by the definition of the relation “∼”, there is an element u ∈ S such that u(acs − ab · 0) = 0 which mean uacs = 0, But since u and s are elements of S, they are not zero divisors, and so we obtain ac = 0, a contradiction. Thus the assumption ac 6= 0 forces equation (9.1) to be false. This proves Part 1. Parts 2. and 3. follow immediately from Part 1. Part 4. Assume R is an integral domain. From the second statement of Lemma 172, the mapping ψs : R → RS which takes element r to element r/s is an injective homomorphism of additive groups for each s ∈ S. Now put s = 1. We see that ψ1 (ab) = ψ1 (a)ψ1 (b) for all a, b ∈ R. Thus ψ1 is an injective ring homomorphism. 9.2.3 Localization at a prime ideal. Now let P be a prime ideal of the commutative ring R – that means P has this property: If, for two ideals A and B of R, one has AB ⊆ P , then at least one of A and B lie in P . For commutative rings, this is equivalent to asserting either of the following (see the exercise at the end of Chapter 7): (1) For elements a and b in R, ab ∈ P implies a ∈ P or b ∈ P . 296 (2) The set R − P is a multiplicatively closed set. It should be clear that a prime ideal contains the annihilator of every element outside it. Now let P be a prime ideal of the integral domain D, and set S := D − P , which, as noted is multiplicatively closed. Then the localization of D by S is called the localization of D at the prime ideal P . In the literature the prepositions are important: The localization by S = D − P is the localization of D at the prime ideal P . Now we may form the ideal M := {p/s|(p, s) ∈ P × S} in DS for which each element of DS − M , being of the form s0 /s, (s, s0 ) ∈ S × S, is a unit of DS . This forces M to be a maximal ideal of DS , and in fact, every proper ideal B of DS must lie in it. Thus we see DS has a unique maximal ideal M. (9.2) Any ring having a unique maximal ideal is called a local ring. This discussion together with Part 4. of Lemma reflem:localzero yields Theorem 174 Suppose P is a prime ideal of the integral domain D. Then the localization at P (that is, the ring DS where S = D − P ) is a local ring that is also an integral domain. Example 1. The zero ideal {0} of any integral domain is a prime ideal. Forming the localization at the zero ideal of an integral domain D thus produces an integral domain with the zero ideal as the unique maximal ideal and every non-zero element a unit. This localized domain is called the field of fractions of the domain D.2 9.3 Unique Factorization. Let D be any integral domain. Recall that a non-zero non-unit of D is said to be irreducible if it cannot be written as the product of two other non-units. This means that if r = ab is irreducible, then either a or b is a unit. 2 One finds “field of quotients” or even “quotient field” used in place of “‘field of fractions” here and there. Such usage is discouraged because we already have the term “quotient ring” commonly being used to mean “factor ring”. 297 A similar, but distinct notion is that of a prime element. A non-zero non-unit r in D is said to be a prime, if and only if, whenever r divides a product ab, then either r divides a or r divides b. In the case of the familiar ring of inegers, the two notions coincide, and indeed they are forced to coincide in a fairly large collection of domains. An integral domain D is said to be a unique factorization domain (or UFD) if and only if (UFD1) Every non-zero non-unit is a product of a unit and a finite number of irreducible elements. (UFD2) Every irreducible element is prime. We shall eventually show that a large class of domains – the principal ideal domains – are UFD’s. One of the very first observations to be made from these definitions is that the notion of being prime is stronger than the the notion of being irreducible, Thus: Lemma 175 In any integral domain, any prime element is irreducible. proof: Suppose r is a prime element of the integral domain D. Suppose r = ab. Then r divides ab, and so, as r is prime, r divides one of the two factors a or b – say a. Then r = rvb so, by the Cancellation laws, 1 = vb,whence b is a unit. Thus r is irreducible. Lemma 176 The following are equivalent: 1. D is a UFD. 2. In D every non-zero non-unit is a product of a unit and a finite number of primes. proof: 1. ⇒ 2. is obvious since 1. implies each irreducible element is a prime. 2. ⇒ 1. First we show that every irreducible element is prime. Let r be irreducible. Then by 2., r is a product of a unit and a finite number of primes. But since r is irreducible, r = up where u is a unit and p is a prime. 298 Now every non-zero non-unit is a product of a unit and finitely many irreducible elements, since each prime is irreducible by Lemma 175. Thus the two defining properties of a UFD follow from 2. and the proof is complete. Theorem 177 (Unique Factorization Theorem) Let D be an integral domain which is a UFD. Then every non-zero non-unit can be written as a product of a unit and a set of primes. Such an expression is unique up to the unit, the order of the factors, and the replacement of a prime by an associate – that is, a multiple of that prime by a unit. proof: Let r be a non-zero non-unit of the UFD D. Then by Lemma 176, r can be written as a unit times a product of finitely many primes. Suppose this could be done in more than one way, say, r = up1 p2 · · · ps = vq1 q2 · · · qt where u and v are units and the pi and qi are primes. If s = 0 or t = 0, then r is a unit, contrary to the choice of r. So without loss we may assume 0 < s ≤ t. Now as p1 is a prime, it divides one of the right hand factors, and this factor cannot be v. Rearranging the indexing if necessary, we may assume p1 divides q1 . Since q1 is irreducible, p1 and q1 are associates, so q1 = u1 p1 . Then r = up1 · · · ps = (vu1 )p1 q2 · · · qt so up2 · · · ps = (vu1 )q2 · · · qt . By induction on t, we have s = t and qi = ui pπ(i) for some unit ui , i = 2, . . . t and permutation π of these indices. Thus the two factorizations involve the same number of primes with the primes in one of the factorizations being associates of the primes in the other written in some possibly different order. We also have a local characterization of UFD’s. Theorem 178 (Local Characterization of UFD’s) Let D be an integral domain. Let S be the collection of all elements of D which can be written as a product of a unit and a finite number of prime elements. Then S is multiplicatively closed and does not contain zero; so the localization DS can be formed. D is a UFD if and only if DS is a field. 299 proof: (⇒) If D is a UFD, then, by Lemma 176, S comprises all nonzero non-units of D. One also notes that DS = DS 0 where S 0 is S with the set U (D) of all units of D adjoined. (By Exercize 2. of this Chapter, this holds for all domains.) Thus S 0 = S − {0}. Then DS 0 is the field of fractions of D. (⇐) It suffices to prove that S is all non-zero non-units of D (Lemma 176). In fact, as S ⊆ D∗ − U (D), it suffices to show that D∗ − U (D) ⊆ S. Let r be any non-zero non-unit – i.e. an element of D∗ − U (D), and suppose by way of contradiction that r ∈ / S. Since r is a non-unit, Dr is a proper ideal of D. Then (Dr)S ; = {a/s|a ∈ Dr, s ∈ S} is clearly a non-zero ideal of DS . Since DS is a field, it must be all of DS . But this means that 1/1 is an element of (Dr)S – i.e. one can find (b, s) ∈ D × S, such that br/s = 1/1. This means br = s so Dr ∩ S 6= ∅. Now from the definition of S, we see that there must be multiples of r which can be expressed as a unit times a non-zero number s of primes. Let us choose the multiple so that the number of primes that can appear in such a factorization attains a minimum m. Thus there exists an element r0 of D such that rr0 = up1 · · · pm where u is a unit and the pi are primes. We may suppose the indexing of the primes to be such that p1 , . . . , pd do not divide r0 , while pd+1 , . . . , do divide r0 . Then if d < m, pd+1 divides r0 so r0 = bpd+1 . Then we have rbpd+1 = up1 · · · pd pd+1 · · · pm so rb = up1 · · · pd pd+2 · · · pm (m − 1 prime factors) against the minimal choice of m. Thus d = m, and each pi does not divide r0 . Then each pi divides r. Thus r = a1 p1 so a1 r0 = up2 · · · pd , upon cancelling p1 . But again, as p2 does not divide r0 , p2 divides a1 . Thus a1 = a2 p2 so a2 r0 = up3 · · · pd . As each pi does not divide r0 , this argument can be 300 repeated, until finally one obtains ad r0 = u when the primes run out. But then r0 is a unit. Then r = ((r0 )−1 u)p1 · · · pm ∈ S. Thus as r was arbitrarily chosen in D∗ − U (D) ⊆ S, we are done. 9.3.1 A Special Case. Of course we are accustomed to the domain of the integers, which is a UFD. So it might be instructive to look√at a more pathological case. Consider the ring D = {a + b −5|a, b ∈ Z}. Being a subring of the field of complex numbers C, it is an integral domain. The mapping z → z̄, where z̄ is the complex conjugate of z, is an automorphism of C which leaves D invariant (as a set). and so induces an automorphism of D. We define the norm of a complex number ζ to be N (ζ) := ζ ζ̄. Then N : C → R+ , the positive and N (ζψ) = N (ζ)N (ψ) for √ real numbers, 2 2 all ζ, ψ ∈ C. Clearly, N (a + b −5) = a + 5b so the restriction of N to D has non-negative integer values. (Moreover, we obtain for free the fact that the integers of the form a2 + 5b2 are closed under multiplication.) Let us determine the units of√D. Clearly if ν ∈ D is a unit, then there is a µ in D so that νµ = 1 = 1 + 0 −5, the identity of D. Then N (ν)N (µ) = N (νµ) = N (1) = 12 + 5 · 02 = 1. But how can 1 be expressed as the product of two non-negative integers? Obviously, only if N (ν) = 1 = N (µ). But if a2 + 5b2 = 1, it is clear that b = 0 and a = ±1. Thus U (D) = {±1} = {d ∈ D|N (d) = 1}. We can also use norms to locate irreducible elements of D. For example, if ζ is an element of D of norm 9, then ζ is irreducible. For otherwise, one would have ζ = ψη where ψ and η are non-units. But that means N (ψ) and N (η) are both integers larger than one. Yet 9 = N (ψ)n(η) so N (ψ) = N (η) = 3, which is impossible since 3 is not of the form a2 + 5b2 . But now note that √ √ 9 = 3 · 3 = (2 + −5)(2 − −5) 301 are two factorizations of 9 into irreducible factors (irreducible, because the they have norm 9) and, as U (D) = {±1}, 3 is not an associate of either factor on the right hand side. √ Thus unique factorization fails in the domain D = Z⊕Z −5. The reason is that 3 is an irreducible √ element√which is not a prime. It is not a prime because 3 divides (2 + −5)(2 − −5), but does not divide either factor. 9.4 Euclidean Domains An integral domain D is said to be a Euclidean Domain if and only if there is a function g : D∗ → N into the non-negative integers3 satisfying (ED1) For a, b ∈ D∗ := D − {0}, g(ab) ≥ g(a). (ED2) If (a, b) ∈ D × D∗ , then there exist elements q and r such that a = bq + r, where either r = 0 or else g(r) < g(b). The notation in (ED2) is intentionally suggestive: q stands for “quotient” and r stands for “remainder”. Recall that in a ring R, an ideal I of the form xR is called a principal ideal. An integral domain in which every left ideal is a principal ideal is called a principal ideal domain or PID. We have Theorem 179 Every Euclidean domain is a principal ideal domain. proof: Let D be a Euclidean domain with grading function g : D∗ → N. Let J be any ideal in D. Among the non-zero elements of J we can find an element x with g(x) minimal. Let a be any other of J. Then a = qx + r as in (ED2) where r = 0 or g(r) < g(x). Since r = a − qx ∈ J, minimality of g(x) shows that the second alternative cannot occur, so r = 0 and a = qx. Thus J ⊆ xD. But x ∈ J, an ideal, already implies xD ⊆ J so J = xD is a principal ideal. Thus D is a PID. We have mentioned in the introduction to this chapter that the divisibility information about any integral domain is encoded in the divisibility poset Div(D) of the divisibility relation ‘|’ among the cosets of the group of units U (D) in the multiplicative monoid of D. 3 The term “natural numbers”includes zero. 302 In fact, the Lemma 169 of the introduction showed that there is an isomorphism between the divisibility poset Dvi(D), and the poset P (D)∗ of all principal ideals of D under reverse containment. Since, in any integral domain, the collection of all nonzero ideals are closed under intersections, the non-zero ideals of a principal ideal domain form a lattice. Thus in a principal ideal domain D the poset Dvi(D) becomes a lattice – that is, it has “meets” and “joins”. Following the developement in Section 8.1, we conclude Theorem 180 In a principal ideal domain D : 1. Any two non-zero elements a and b of D possess a greastest common divisor d which is a generator of the ideal aD + bD. 2. Any two non-zero elements a and b possess a least common multiple m which is generator of the ideal aD ∩ bD. The interesting feature of a Euclidean domain, is not only the algebraic property that it is a PID and hence has greatest common divisors, but the interesting computional-logical feature that the greatest common divisors can actually be computed! Behold the euclidean algorithm. Given a and b in D and D∗ , respectively, by (ED2) we have a = bq1 + r1 , and if r1 6= 0, b = r1 q2 + r2 , and if r2 6= 0, r 1 = r 2 q3 + r 3 , etc., until we finally obtain a remainder rk = 0 in rk−2 = rk−1 qk + rk . Such a termination of this process is inevitable for g(r1 ), g(r2 ), . . . is a strictly decreasing sequence of non-negative integers which can only terminate at g(rk−1 ) when rk = 0. We claim the number rk−1 ( the last non-zero remainder) is a greatest common divisor of a and b. First rk−2 = rk− qk and also rr−3 = rk−2 qk−1 +rk−1 303 are multiples of rk−1 . Inductively, if rk−2 divides both rj and rj+1 , it divides rj−1 . Thus all of the right hand sides of the equations above are multiples of rk−1 ; in particular, a and b are multiples of rk−1 . On the other hand if d0 is a common divisor of a and b, d0 divides r1 = a − bq1 and in general divides rj = rj−2 − rj−1 qj and so d0 divides rk−1 , eventually. Thus rk−1 is a greatest common divisor. Of course it follows from the poset isomorphism P → P r(D) → (all ideals of D, ⊇) that any greatest common divisor of aD and bD is aD + bD. However an explicit expression of rk−1 as a D-linear combination a and b can be obtained from the same sequence of equations. For d = rk−1 = rk−3 − rk2 qk−1 , and each rj similarly is a D-linear combination of rj−2 and rj−1 , j = 1, 2, . . .. 9.4.1 Examples of Euclidean Domains 1. The Gaussian integers and Eisenstein Integers. These are respectively the domains D1 := Z ⊕ Zi (where i2 = −1 6= −i) and D2 = Z ⊕ Zω (where ω = e2iπ/3 ). The complex number ω is a root of the irreducible polynomial x2 + x + 1, so it is an algebraic integer which is a primitve cube root of unity. Both of these domains are subrings of C, the complex number field, and each is invariant under complex conjugation. Thus both of them admit a multiplicative norm function N : C → R where N (z) := z · z̄ records the square of the Euclidean distance of z from zero in the complex plane. We shall demonstrate that these rings are Euclidean domains with the norm function in the role of g. That its values are taken in the non-negative integers follows from the formulae N (a + bi) = a2 + b2 N (a + bω) = a2 − ab + b2 . As already remarked, the norm function is multiplicative so (ED1) holds. To demonstrate (ED2) we consider two elements a and b of D with b 6= 0. 304 Now the elements of D1 form a square tessellation of the complex plane with {0, 1, i, i+1} as a fundamental square. Similarly, the elements of D2 form the equilateral triangle tessallation of the complex plane with {0, 1, ω 1/2 } at the corners of a fundamental triangle. We call the points where three or more tiles of the tessellation meet lattice points. When we superimpose the ideal bD on either one of these tessellations of the plane, we are depicting a tessellation of the same type on a subset of the lattice points with a possibly larger fundamental tile. Thus for the Gaussian integers, bD is a tesselation whose fundamental square is defined by the four “corner” points {0, b, ib, (1 + i)b}. The resulting lattice points are a subset of the Gaussian integers closed under addition (vector addition in the geometric picture). Similarly for the Eisenstein numbers, a fundamental triangle of the tessellation defined by the ideal bD is the one whose “corners” are in {0, b, bω 1/2 }. In either case the tiles of the tessellation bD cover the plane, and so the element a must fall in some tile T – either a square or an equilateral triangle – of the superimposed tessellation bD. Now let qb be a corner point of T which is nearest a. Since T is a square or an equilateral quadrangle, we have (Nearest Corner Principle) the distance from a point of T to its nearest corner is less than the length of a side of T q q Thus the distance |a − qb| = N (a − qb) is less than |b| = N (b). Using the fact that the square function is monotone increasing on non-negative real numbers, we have g(a − bq) = N (a − bq) < N (b) = g(b). Thus setting r = a − qb, we have r = 0 or g(r) < g(b). Thus (with respect to the function g), r serves to realize the condition (ED2).4 4 Usually boxing matches are held (oxymoronically) in square “boxing rings”. Even if (as well) they were held in equilateral triangles, it is a fact that when the referee commands a boxer to “go to a neutral corner”, the fighter really does not have far to go (even subject to the requirement of “neutrality” of the corner). But in some ethnological fantasia, one could well imagine a culture with such an intense emphasis on “gracious defeats” that their boxing rings approached a more ring-like N -gon, for larger N . Woe the fighter who achieves a knock-down near the center, for after the mandatory hike to a neutral corner, there is a good probably that the fighter can only return to the boxing match totally exhausted from the long trip. No Euclidean algorithm for these boxers. 305 2. Polynomial domains over a field. Let F be a field and let F [x] be the ring of polynomials in indeterminate x and with coefficients from F . We let g be the degree function, F [x] → N. Then if f (x) and g(x) are non-zero polynomials, deg(f g) = deg f + deg g, so (ED1) holds. The condition (ED2) is the familiar long division algorithm of College Algebra and grade school. 9.5 Principal Ideal Domains are Unique Factorization Domains The following theorem utilizes the fact that if an integral domain possesses the ascending chain condition on principal ideals, then every element is a product of finitely many irreducible elements (Lemma 171). Theorem 181 Any PID is a UFD. proof: Assume D is a PID. Then D has the ascending chain condition on all ideals, and so by Lemma 171, the condition (UFD2) holds. It remains to show (UFD1), that every irreducible element is prime. Let x be irreducible. Then Dx is a maximal ideal ( Lemma 170). Now if x were not a prime, there would exist elements a and b such that x divides ab but x does not divide either a or b. Thus a ∈ / xD, b ∈ / xD, yet ab ∈ xD. Thus xD is not a prime ideal, contrary to the conclusion of the previous paragraph that xD is a maximal ideal, and hence a prime ideal. 9.6 If D is a UFD, then so is D[x]. Let D be a unique factorization domain (UFD), set S = D∗ = D − {0}, and left F := DS be the field of fractions of D. Then D[x], the domain of polynomials with coefficients from D, can be regarded as a subring of F [x] by the device of regarding coefficients d of D as fractions d/1. We wish to compare the factorization of elements of D[x] in D[x] and in F [x]. We begin with very elementary results. Lemma 182 U (D[x]) = U (D). 306 proof: Since D is an integral domain, degrees add in the products of nonzero polyomials, and so the units of D[x] must have degree zero. This shows that units have degree zero and so lie in D. That a unit of D is a unit of Dx depends only upon the fact that D is a subring of D[x]. Lemma 183 Let p be an irreducible element of D, regarded as a subring of D[x]. If p divides the polynomial q(x) in D[x], then every coefficient of q(x) is divisible by p. proof: If p divides q(x), then q(x) = pf (x) for some f (x) ∈ D[x]. Then the coefficients of q(x) are those of f (x) multiplied by p. Lemma 184 (Irreducible elements of D are prime elements of D[x].) Let p be an irreducible element of D. If p divides a product p(x)q(x) of two elements of D[x], then either p divides p(x) or p divides q(x). proof: Suppose the irreducible element p divides p(x)q(x) as hypothesized. Write Xm Xn p(x) = i=0 ai xi , and q(x) = j=0 bj xj , where ai , bj ∈ D. Suppose by way of contradiction that p does not divide either p(x) or q(x). Then by the previous Lemma 183, p divides each coefficient of p(x)q(x) while there is a first coefficient ar of p(x) not divisible by p, and a first coefficient bs of q(x) not divisible by p. Then the coefficient of xr+s in p(x)q(x) is cr+s = X ab, i+j=r+s i j subject to 0 ≤ i ≤ n and 0 ≤ j ≤ m (9.3) Now if (i, j) 6= (r, s), and i + j = r + s, then either i < r or j < s, and in either case ai bj is divisible by p. Thus all summands ai bj in the right side of equation ( 9.3) except for ar bs are divisible by p. But this would mean that p does not divide cr+s , against Lemma 183 and the fact that p divides p(x)q(x). Thus p must divide one of p(x) or q(x), completing the proof. We say that a polynomial p(x) ∈ D[x] is primitive if and only if, whenever d ∈ D divides p(x), then d is a unit. We now approach two lemmas that involve F [x] where F is the field of fractions of D. First, 307 Lemma 185 If p(x) = αq(x), where p(x) and q(x) are primitive polynomials in D[x], and α ∈ F , then α is a unit in D. proof: Since D is a UFD, greatest common divisors and least common multiples exist, and so there is a “lowest terms” representation of α as a fraction t/s where the greatest common divisor of s and t are just the units. If s is a unit, then t divides each coefficient of p(x), and since this is primitive, t must also be a unit. Otherwise, there is prime divisor p of s. Then p does not divide t, and so from sp(x) = tq(x), we see that every coefficient of q(x) is divisible by p, against q(x) being primitive. Lemma 186 If f (x) ∈ F [x], then f (x) has a factorization f (x) = rp(x) where r ∈ F and p(x) is a primitive polynomial in D[x]. This factorization is unique up to replacement of each factor by an associate. proof: We prove this in two steps. First we establish Step 1. There exists a scalar γ such that f (x) = γp(x), where p(x) is primitive in D[x]. Each non-zero coefficient of xi in f (x) has the form ai /bi with bi ∈ S. Multiplying through by a least common multiple m of the bi (recall that lcm’s exist in UFD’s), we obtain a factorization f (x) = ( 1 X ) ai (m/bi )xi , m whose second factor is clearly primitive in D[x]. Step 2. The factorization in Step 1 is unique up to associates. proof: Suppose f (x) = γ1 p1 (x) = γ2 p2 (x) with γi ∈ F , and pi (x) primitive in D[x]. Then p1 (x) and p2 (x) are associates in F [x] so p1 (x) = γp2 (x), for γ ∈ F. Then by Lemma 185, γ is a unit in D. But as γ = γ2 γ1−1 , the result follows. 308 Lemma 187 (Gauss’ Lemma) Suppose p(x) ∈ D[x] is reducible in the ring F [x]. Then p(x) is reducible in D[x]. proof: By way of contradiction, we may assume that p(x) is irreducible in D[x]. In particular, by Lemma 183 and Lemma 184, we may assume that p(x) is primitive. Suppose p(x) = f (x)g(x) where f (x) and g(x) are polynomials in F [x]. Then by Step 1, we may write f (x) = µf1 (x) g(x) = γf2 (x). where µ, γ ∈ F and f1 (x) and g1 (x) are primitive polynomials in D[x]. Now by Lemma 184, f1 (x)g1 (x) is a primitive polynomial in D[x] so p(x) = (µγ)(f1 (x)g1 (x), and Lemma 185 shows that µγ is a unit in D. Then p(x) = ((µγ)f1 (x)) · g1 (x) is a factorization of p(x) in D[x], with factors of positive degree. The conclusion thus follows. Theorem 188 If D is a UFD, then so is D[x]. proof: Let p(x) be any element of D[x]. We must show that p(x) has a factorization into irreducible elements which is unique up the replacement of factors by associates. Since D is a unique factorization domain, a consideration of degrees shows that it is sufficient to do this for the case that p(x) is primitive of positive degree. Let S = D−{0} and form the field of fractions F = DS , regarding D[x] as a subring of F [x] in the usual way. Now F [x] is a Euclidean domain, hence a PID, and hence a unique factorization domain. Thus we have a factorization p(x) = p1 (x)p2 (x) · · · pn (x) 309 where each pi (x) is irreducible in F [x]. Then by Lemma 186, there exist scalars γi , i = 1, . . . , n, such that pi (x) = γi qi (x), where qi (x) is a primitive polynomial in D[x]. Then p(x) = (γ1 γ2 · · · γn )q1 (x) · · · qn (x). Since the qi (x) are primitive, so is their product (Lemma 184). Then by Lemma 185 the product of the γi is a unit u of D. Thus p(x) = uq1 (x) · · · qn (x) is a factorization in D[x] into irreducible elements. If p(x) = vr1 (x) · · · rm (x) were another such factorization, the fact that F [x] is a UFD shows that m = n and the indexing can be chosen so that ri (x) is an associate of qi (x) in F [x] – i.e. there exist scalars ρi in F such that ri (x) = ρi qi (x), for all i. Then as ri (x) and qi (x) are irreducible in D[x], they are primitive, and so by Lemma 185, each γi is a unit in D. Thus the two factorizations of p(x) are alike up to replacement of the factors by associates in D[x]. The proof is complete. Corollary 189 If D is a UFD, then so is the ring D[x1 , . . . , xn ]. proof: Repeated application of the theorem to D[x1 , . . . , xj+1 ] ' (D[x1 , . . . , xj ])[xj+1 ]. EXERCISES 9.1 The first few exercises concern localizations. Exercise 1. Using the definitions of multiplication and addition for fractions, fill in all the steps in the proof of the following: 1. that (DS , +) is an abelian groups, 310 2. that multiplication of fractions is a monoid, and 3. that multiplication is distributive with respect to addition. Exercise 2. Let D be an integral domain with group of units U (D). Show that if S is a multiplicatively closed subset and S 0 is either U (D)S or U (D) ∪ U (D)S, then the localizations DS and DS 0 are isomorphic. Exercise 3. Below are given examples of a domain D and a multiplicatively closed subset of non-zero elements of D. Describe as best you can the local ring DS . 1. D = Z; S is all powers of 3. Is DS a local ring? 2. D = Z; S consists of all numbers of the form 3a 5b where a and b are natural numbers and a + b > 0. Is the condition a + b > 0 necessary? 3. D = Z; S is all positive integers which are the sum of two squares. 4. D = Z[x]; S is all primitive polynomials of D. Let D be a fixed unique-factorization-domain (UFD), and let p be a prime element in D. Then the principal ideal (p) := pD is a maximal ideal and D/(p) is a field. The next few excercises concern the following surjection between integral domains: mp : D[x] → (D/(p))[x], which reduces mod p the coefficients of any polynomial of D[x]. Exercise 4. Show that mp is a ring homomorphism. Conclude that the principal ideal in D[x] generated by p is a prime ideal and from this, that p is a prime element in the ring D[x]. Exercise 5. Show that every maximal ideal M of Z[x] is generated by two elements p and f (x), where p is a rational prime and f := f (x) is a polynomial in Z[x] for which mp (f ) is irreducible in (Z/(p))[x]. [Hint: First show that if M is a maximal ideal, then the field Z[x]/M cannot have characteristic zero. This forces the presence of a unique rational prime p in M . Then use the fact that (Z/(p))[x] is a Euclidean domain.] 311 Exercise 6. (Another proof of Gauss’ Lemma.) Let D be a UFD. A polynomial h(x) in D[x] is said to be primitive if and only there is no prime element p of D dividing each of the D-coefficients of h(x). Using the morphism mp of the first exercise, show that if f and g are primitive polynomials in D[x], then the product f g is also primitive. [Hint: by assumption, there is no prime in D dividing f or g in D[x]. If a prime p of D divides f g in D[x] then mp (f g) = 0 = mp (x) · mp (g). Then use the fact that (D/(p)[x] is an integral domain.] Exercise 7. (Eisenstein’s Criterion) Let D be a UFD. Suppose p is a prime in D and f (x) = an xn + · · · a1 x + a0 is a polynomial in D[x] with n > 1 and an 6= 0. Further, assume that 1. p does not divide the lead coefficient an ; 2. p divides each remaining coefficient ai for i = 0, . . . , n − 1; and 3. p2 does not divide a0 . Show that f cannot factor in D[x] into polynomial factors of positive degree. In particular f is irreducible in F [x], where F is the field of fractions of D. [Hint: Suppose by way of contradiction that there was a factorization f = gh in D[x] with f and g of degree at least one. Then apply the ring morphism mp to get mp (f ) = mp (g) · mp (h) = mp (an )xn 6= 0. Now by hypothesis 1) the polynomials f and mp (f ) have the same degree, while the degrees of g and h are at least as large as the respective degrees of their mp images. Since f = gh, and deg mp (f ) = deg mp (g) + deg mp (h), we must have the last two summands of positive degree. Since (D/(p)[x] is a UFD, x must divide both mp (g) and mp (h). This means the constant coefficients of g and h are both divisible p. This contradicts hypothesis 3.] Exercise 8. Show that under the hypotheses of the previous exercise, one cannot conclude that f is an irreducible element of D[x]. [Hint: In Z[x], the polynomial 2x + 6 satisfies the Eisenstein hypotheses with respect to the prime 3. Yet it has a proper factorization in Z[x].] Exercise 9. Let p be a prime element of D. Using the morphism mp show that the subset 312 B := {p(x) = a0 + a1 x + · · · + an xn |ai ≡ 0mod 0, for i > 0} of D[x] is a subdomain. 9.7 9.7.1 Appendix to Chapter 8: The Arithmetic of Quadratic Domains. Introduction We understand the term quadratic domain to indicate the ring of algebraic integers of a field K of degree two over the rational number field Q. The arithmetic properties of interest are those which define Euclidean domains, principal ideal domains (or PID’s) and unique factorization domains (or UFD’s, also called “factorial domains” in the literature). Quadratic Fields. Throughout Q will denote the field of rational numbers. Suppose K is a subfield of the complex numbers. Then it must contain 1, 2 = 1 + 1, all integers and all fractions of such integers. Thus K contains Q as a subfield and so is a vector space over Q. Then K is a quadratic field if and only of dimQ (K) = 2. Suppose now that K is a quadratic field, so as a Q-space, K = Q ⊕ ωQ, for any chosen ω ∈ K − Q. Then ω 2 = −bω − c for rational numbers b and c, so ω is a root of the monic polynomial p(x) = x2 + bx + c in Q[x]. Evidently, p(x) is irreducible, for otherwise, it would factor into two linear factors, one of which would have ω for a rational root, against our choice of ω. Thus p(x) has two roots ω and ω̄ given by √ (−1 ± b2 − 4c)/2. 313 We see that K is a field generated by Q and either ω or ω̄ — a fact which can be expressed by writing K = Q(ω) = Q(ω̄). Evidently √ K = Q( d), where d = b2 − 4c. We anticipate a few elementary results on fields reviewed in the next chapter, by observing that substitution of ω for the indeterminate x in each polynomial of Q[x] produces an onto ring homomorphism, vω : Q[x] → K, whose kernel is the principal ideal p(x)Q[x]. But as vω̄ : Q[x] → K has the same kernel, we see that the Q-linear transformation σ : Q ⊕ ωQ → Q ⊕ ω̄Q, which take each vector a + bω to a + bω̄, a, b ∈ Q, is an automorphism of the field K. Thus σ(ω) = ω̄, and σ 2 = 1K (why?). Associated with the group hσi = {1K , σ} of automorphisms of K are two further mappings. First is the trace mapping, T r : K → Q, which takes each element ζ of K to ζ + ζ σ . Clearly this mapping is Q-linear. The second mapping is the norm mapping, N : K → Q, which takes each element ζ of K to ζζ σ . This mapping is not Q-linear, but it follows from its definition (and the fact that multiplication in K is commutative) that it is multiplicative – that is N (ζψ) = N (ζ)N (ψ), for all ζ, ψ in K. Just to make sure everything is in place, consider the roots ω and ω̄ of the irreducible polynomial p(x) = x2 +bx+c. Since p(x) factors as (x−ω)(x− ω̄) in K[x], we see that the trace is T (ω) = T (ω̄) = −b ∈ Q, and that the norm is N (ω) = N (ω̄) = c, also in Q. 314 Quadratic Domains We wish to identify the ring A of algebraic integers of K. These are the elements ω ∈ K which are roots of at least one monic irreducible polynomial in Z[x]. At this stage the reader should be able to prove: Lemma 190 An element ζ ∈ K − Q is an algebraic integer if and only if its trace T (ζ) and its norm N (ζ) are both rational integers. √ √ Now putting ζ = d, where d is square-free and K = Q( d), the lemma implies: Lemma 191 √ If d is a square-free integer, then the ring of algebraic integers of K = Q( d) is the Z-module spanned by √ {1, d} if d ≡ 2, 3 mod 4, √ {1, (1 + d)/2} if d ≡ 1 mod 4. √ √ Thus the ring of√integers for Q( 7) is the set {a + b 7|a, b ∈ Z}. But if d = −3, √ then (1 + −3)/2 is a cube root of unity, and the ring of integers for Q( −3) is the domain of Eisenstein numbers. 9.7.2 Which quadratic domains are Euclidean, principal ideal or factorial domains? Fix a square free integer d and let A(d) denote the full ring of algebraic √ integers of K = Q( d). A great deal of effort has been devoted to the following questions: 1. When is A(d) a Euclidean domain? 2. More generally, when is A(d) a principal ideal domain? In this chapter we showed that the Gaussian integers, A(−1) and the Eisenstein numbers, A(−3) were Euclidean by showing that the norm function could act as the function g : A(d) → Z with respect to which the Euclidean algorithm is defined. In the case that the norm can act in this role, we say that the norm is algorithmic. The two questions have been completely answered when d is negative. 315 Proposition 192 The norm is algorithmic for A(d) when d = −1, −2, −3, −7 and −11. For all remaining negative values of d, the ring A(d) is not even Euclidean. However, A(d) is a principal ideal domain for d = −1, −2, −3, −7, −11, −19, −43, −67, −163, and for no further negative values of d. There are ways of showing that A(d) is a principal ideal domain, even when the norm function is not algorithmic. For example: Lemma 193 Suppose d is a negative square-free integer and the (positively valued) norm function of A(d) has this property: • If N (x) ≥ N (y) then either y divides x or else there exist elements u and v (depending on x and y) such that 0 < N (ux − yv) < N (y). Then A(d) is a principal ideal domain. This lemma is used, for example, to show that A(−19) is a PID.5 √ For positive d (where Q( d) is a real field) the status of these questions is not so clear. We can only report the following: Proposition 194 Suppose d is a positive square-free integer. Then the norm function is algorithmic precisely for d = 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57, 73. The domain A(d) is known to be a principal ideal domain for many further values of d. But we dont even know if this can occur for infinitely many values of d! Moreover, among these PID’s, how many are actually Euclidean by some function g other than the norm function? Are there infinitely many? 5 See an account of this in Pollard’s The Theory of Algebraic Numbers, Carus Monograph no. 9, Math. Assoc. America, page 100, and in Wilson, J. C., A principal ideal domain that is not Euclidean, Mathematics Magazine, vol 46(1973), pp. 34-48. 316 9.7.3 Subdomains of Euclidean domains which are not UFD’s Here it is shown that standard Euclidean domains such as the Gaussian integers contain subdomains for which unique factorization fails. These nice examples are due to P. J. Arpaia (“A note on quadratic Euclidean Domains”, Amer. Math. Monthly, vol. 75(1968), pp 864-865). Let A0 (d) be the√subdomain of A(d) consisting of all integral linear combinations of 1 and d, where d is a square-free integer. Thus A0 (d) = A(d) if d ≡ 2, 3 mod 4, and is a proper subdomain of A(d) only if d ≡ 1 mod 4. Let p be a rational prime. Let A0 /(p) be the ring whose elements belong to the vector space √ (Z/(p))1 ⊕ (Z/(p)) d, √ where multiplication is defined by setting ( d)2 to be the residue class modulo p containing d. Precisely, A0 /(p) is the ring Z[x]/I where I is the ideal generated by the prime p and the polynomial x2 − d. It should be clear that A0 /(p) is a field if d is not zero or is not a quadratic residue mod p, and is the direct (ring) sum of two fields if d is a quadratic residue mod p. If p divides d, it has a 1-dimensional nilpotent ideal. But in all cases, it contains Z/(p) as a subfield containing the multiplicative identity element. There is a straightforward ring homomorphism np : A0 → A0 /(p), √ which takes a + b d to ā + b̄ d where “bar” denotes the taking of residue classes of integers mod p. Now set √ B(d, p) := n−1 p (Z/(p)). Then B(d, p) is a ring, since it is the preimage under np of a subfield of the image A0 /(p). In fact, as a set, √ B(d, p) := {a + b d ∈ A0 |b ≡ 0 mod p}. Clearly it contains the ring of integers and so is a subdomain of A(p). Note that the domains A0 and B(d, p) are both σ-invariant. In the subdomain B(d, p), we will show that under a mild condition, there are elements which can be factored both as a product of two irreducible elements and as a product of three irreducible elements. Since d and p are fixed, we write B for B(d, p) henceforward. 317 Lemma 195 An element of B of norm one is a unit in B. Proof: Since B = B σ , N (ζ) = ζζ σ = 1 for ζ ∈ B implies ζ σ is an inverse in B for ζ. Lemma 196 Suppose ζ is an element A0 whose norm N (ζ) is the rational prime p. The ζ does not belong to B. Proof: By hypothesis, N (ζ) = a2 − b2 d = p. If p divided integer b, it would divide integer a, so p2 would in fact divide N (ζ), an absurdity. Thus p does not divide b and so ζ is not an element of B. Corollary 197 If an element of B has norm p2 , then it is irreducible in B. Proof: By Lemma 195, elements of norm one in B are units in B. So if an element of norm p2 in B were to have a proper factorization, it must factor into two elements of norm p. But the preceeding lemma shows that B contains no such factors. Theorem 198 Suppose there exist integers a and b such that a2 − db2 = p (that is, A0 contains an element of norm p). Then the number p3 admits these two factorizations into irreducible elements of B: p3 = p · p · p √ √ p3 = (pa + pb d) · (pa − pb d). Proof: The factors are clearly irreducible by Corollary 197. We reproduce Arpaia’s table displaying p = a2 − db2 for all d for which the norm function on A(d) is algorithmic. d −11 −7 −3 −2 −1 2 3 p = a2 − db2 47 = 62 − (−11)12 11 = 22 − (−3)12 7 = 22 − (−3)12 3 = 12 − (−2)12 2 = 1 − (−1)12 −7 = 12 − 2 · 22 −2 = 12 − 3 · 12 d 5 6 7 11 13 17 19 p = a2 − db2 11 = 42 − 5 · 12 −5 = 12 − (6)12 −3 = 22 − 3 · 12 −7 = 22 − 11 · 12 3 = 42 − 13 · 12 −13 = 22 − 17 · 12 −3 = 42 − 19 · 12 318 d 21 29 33 37 41 57 73 p = a2 − db2 −17 = 22 − 21 · 12 −13 = 42 − (26)12 −29 = 22 − 33 · 12 107 = 122 − 37 · 12 −37 = 22 − 41 · 12 −53 = 22 − 57 · 12 −37 = 62 − 73 · 12 A Student Project In the preceding subsection, a non-factorial subdomain B was produced for √ each case that the domain A(d) = alg. int.(Q( d)) was Euclidean. But there are only finitely many of these domains A(d). This raises the question whether one can find such a subdomain B in more standard generic Euclidean domains? We know that no such subdomain can exist in the domain of integers Z due to the paucity of it subdomains. But are there non-factorial subdomains B of the classical Euclidean domain of polynomials over a field? This might be done by duplicating the development of the previous section replacing the ring of integers Z, by an arbitrary unique factorization domain D which is distinct from its field of fractions k. Suppose further that K is a field of degree two over k generated by a root ω of an irreducible quadratic polynomial x2 − bx − c in D[x] which has distinct roots {ω, ω̄}. (One must assume that the characteristic of D is not 2.) Then the student can show that, as before, K = k ⊕ ωk as a vector space over k and that the mapping σ defined by σ(u + ωv) = u + ω̄v, for all u, v ∈ k, is a field automorphism of K of order two so that the trace and norm mappings T and N can be defined as before. Now let A be the set of elements in K whose norm and trace are in D. The student should be able to prove that if D is Noetherian, A is a ring. In any event, A0 := {a + ωb|a, b ∈ D} is a subring. Now since D is a UFD not equal to its quotient field, there is certainly a prime p in D. It is easy to see that Bp = {a + ωb|a, b ∈ D, b ∈ pD} = D ⊕ ωpD is a σ-invariant integral domain. Then the proofs of Lemmas 195, and 196 and Corollary 197 go through as before (note that the fact that D is a UFD is used here). The analogue of the Theorem will follow: If A0 contains an element of norm p, B is not factorial. Can we do this so that between A and B lies the Euclidean ring F [x]? The answer is “yes”, if the characteristic of the field F is not 2. In fact we can pass directly from F [x] to the subdomain B and fill in the roles of D, k, K, A, and A0 later. 319 Let p(x) be an irreducible polynomial in F [x] where the characteristic of F is not 2. Then (using an obvious ring isomorphism) p(x2 ) is an irreducible element of the subdomain F [x2 ]. Now the student may show that the set Bp : F [x2 ] ⊕ xp(x2 )F [x2 ] is a subdomain of F [x]. Now D = F [x2 ] is a UFD with quotient field k = F (x2 ) – the so called “field of rational functions” over F . The quotient field K = F (x) of F [x], is degree two over k, with the indeterminate x playing the role of ω. Thus F (x) = F (x2 ) ⊕ xF (x2 ) since ω = x is a root of the polynomial y 2 − x2 which is irreducible in F [x2 ][y] = D[y]. Since the characteristic is not 2, the roots x and −x are distinct. Thus we obtain our field automorphism σ which takes any element ζ = r(x2 ) + xs(x2 ) of F (x), to r(x2 ) − xs(x2 ) (r and s are quotients of polynomials in F [x2 ]). (Note that this is just the substitution of −x for x in F (x).) Now the student can write out the norm and trace of such a generic ζ. For example N (ζ) = r2 − x2 s2 . (9.4) At this point the ring Bp given above is not factorial if there exists a polynomial ζ = a(x2 ) + xb(x2 ) in F [x] of norm p(x2 ). That is, a(x2 )2 − x2 b(x2 )2 = p(x2 ) is irreducible in F [x2 ]. Here is an example: Take a(x2 ) = 1 = b(x2 ), so ζ = x + 1. Then N (1+x) = x2 −1 is irreducible in F [x2 ]. Then in Bp = F [x2 ]⊕x(x2 −1)F [x2 ], we have the factorizations (1 − x2 )3 = (1 + x − x2 − x3 )(1 − x − x2 + x3 ) into irreducible elements. 320 Chapter 10 Principal Ideal Domains and Their Modules 10.1 Introduction Thoughout this Chapter, D (and sometimes R) will be a principal ideal domain (PID). Thus it is a commutative ring for which a product of two elements is zero only if one of the factors is already zero – that is, the set of non-zero elements is closed under multiplication. Moreover every ideal is generated by a single element – and so has the form Dg for some element g determined by the ideal. Of course, as rings go, principal ideal domains are hardly typical. But two of their examples, the ring of integers Z, and the ring F [x], of all polynomials in a single indeterminate x over a field F , are so pervasive thoughout mathematics that their atypical properties deserve special attention. In the previous chapter we observed that principal ideal domains were in fact unique factorization domains. In this chapter, the focus is no longer on the internal arithmetic of such rings, but rather on the structure of all finitely-generated submodules over these domains. As one will soon see, the structure is rather precise. 10.2 Quick Review of PID’s Principal Ideal Domains (PID’s) are integral domains D for which each ideal I has the form I = aD for some element a of D. One may then infer 321 from the fact that every ideal is finitely generated, that the Ascending Chain Condition (ACC) holds for the poset of all ideals. Recall that a divides b if and only if aD ⊇ bD. It follows — as we have seen in the arguments that a principal ideal domain (PID) is a unique factorization domain (UFD) – that a is irreducible if and only if aD is a maximal ideal, and that this happens if and only if aD is a prime element. We also recall that for ideals, xD = yD if and only if x and y are associates. For this means x = yr and y = xs for some r and s in D, so x = xrs, so sr = 1 by the cancellation law. Thus r and s, being elements of D, are both units. The least common multiple of two elements a, b ∈ D is a generator m of the ideal aD ∩ bD = mD where m is unique up to associates. The greatest common divisor of two non-zero elements a and b is a generator d of aD + bD = dD, again up to the replacement of d by an associate. Two elements a and b are relatively prime if and only if aD + bD = D. Our aim in this chapter will be to find the structure of finitely-generated modules over a PID. Basically, our main theorem will say that if M is a finitely-generated module over a principal ideal domain D, then M ' (D/De1 ) ⊕ (D/De2 ) ⊕ · · · ⊕ (D/Dek ) as right D-modules, where k is a natural number and for 1 ≤ i ≤ k − 1, ei divides ei+1 . This is a very useful theorem. Its applications are (1) the classification of finite abelian groups, and (2) the complete analysis of a single linear transformation T : V → V of a vector space into itself. It will turn out that the set of elements ei which are not units is uniquely determined up to associates. But all of that still lies ahead of us. 322 10.3 Free modules over a PID. Recall that we have earlier defined a free module over the ring R to be a direct sum of submodule Mσ each of which is isomorphic to RR as right R-modules. We say that an R-module M has a basis if and only if there is a subset B of M (called a basis) such that every element m of M has a unique expression X bi ri , bi ∈ B, ri ∈ R, as a finite R-linear combination of elements of B. The uniqueness is the key item here. Lemma 199 An R-module M is a free R-module if and only if it has a basis. Proof: If M is free, it is isomorphic to a direct sum of submodules Mσ each of which is isomorphic to RR as right R-module. Let mσ be the element of Mσ which corresponds to the element 1 ∈ R under this R-isomorphism. The reader can verify that the set B := {mσ |σ ∈ I}, where I indexes the direct summands Mσ , is a basis for M . Now suppose the R-module M has a basis B. Suppose b ∈ B. Then the uniqueness of expression means that for any r, s ∈ R, br = bs implies r = s. Thus the map which sends br to r is a well-defined isomorphism bR → RR of right R-modules. The directness of the sum M = ⊕b∈B bR follows from the uniquness of expression. Thus M is a free R-module. In general, there is no result asserting that two bases of a free module have the same cardinality. However, the result is true when R is an integral domain D. The reader may recall from Chapter 7 that this was true of any K-module, where K was a field. Recall that we only needed to invoke our general dependence theory (modeled by linear dependance) to achieve this result: namely that vector spaces have a dimension. Since any K-module is free, and any field K is certainly a PID, the theorem proposed below is clearly a vast generalization of the vector space result. On the other hand it readily stems from it. Theorem 200 Let D be an integral domain which is not neccessarily a principal ideal domain. Let F be a free D-module with basis X := {xi }i∈I . Then the cardinality of I is uniquely determined. 323 proof: Let P be a maximal ideal of D (recall that at least one must exist). Then the factor ring, D/P := K is a field. Then F P := hf r|f ∈ F, r ∈ P i is a submodule of F . Then the factor module F̄ := F/F P is annihilated by P and so becomes a K-vector space. We are done once we show that the D-basis X for F maps to a K-basis of F̄ under the natural modulehomomorphism F → F̄ . That it maps to a spanning set is no problem: it is the K-linear independance of the x̄i which is at issue. P P P P Suppose r̄i x̄i = 0 ∈ K. Then ri xi ∈ F P . This means ri x1 = aj xj , where each coefficient aj lies in P . Since X is an D-basis the uniqueness of the expression for the right side forces the ri to lie in P . But then each r¯i = 0, proving the K-independance. We call the unique cardinality of a basis for a free D-module F , the D-rank of F and denote it rkD (F ) or just rk(F ) if D is understood. In general, a submodule of a free R-module is not free. However the result holds if R is an integral domain. This is proved in a later Chapter where projective modules are introduced. 10.4 A technical result Theorem 201 Let A be an n × m matrix with coefficients from a principal ideal domain D. Then there exists an n × n matrix P invertible in (D)n×n , and an m × m matrix Q, invertible in (D)m×m , such that P AQ = d1 0 0 d2 . . . . ... ... ... ... 0 0 . . where di divides di+1 . The elements {d1 , d2 , . . . , dk }, where k = min(n, m), are uniquely determined up to associates. proof: For each non-zero element d of D let `(d) be the number of prime divisors occurring in a factorization of d into primes. (Recalling that a PID is always a UFD, the function ` is seen to be well-defined.) Our goal is to convert A into a diagonal matrix with successively dividing diagonal elements by a series of right and left multiplications by invertible 324 matrices. These will allow us to perform several so-called elementary row and column operations namely: (I) Add b times the i-th row (column) to the j-th row (column), i 6= j. This is left (right) multiplication by a suitably sized matrix having b in the (i, j)-th (or (j, i)-th) position while having all diagonal elements equal to 1D , and all futher entries equal to zero. (II) Multiply a row (or column) throughout by a unit u of D. This is left (right) multiplication by a diagonal matrix diag(1, . . . , 1u, 1, . . . , 1) of suitable size. (III) Permute rows (columns). This is left (right) multiplication by a suitable permutation matrix. First observe that there is nothing to do if A is the n × m matrix with all entries zero. In this case each di = 0D , and they are unique. Thus if we consider the entire collection of matrices S = {P AQ|P, Q units in D(n×n) and D(m×m) , respectively}, among their non-zero entries, there can be found a matrix element bij in some matrix B ∈ S, with `(bij ) minimal. Then by rearranging rows and columns of B if necessay, we may assume that this bij is in the (1, 1)-position of B, – that is, (i, j) = (1, 1). We now claim b11 divides b1k for k = 2, . . . , m. (10.1) To prove the claim, assume b11 does not divide b1k . Permuting a few columns if necessary, we may assume k = 2. Since D is a principal ideal domain, we may write b11 D + b12 D = Dd. Then as d divides b11 and b12 , we see that b12 = ds and − b11 = dt, with s, t ∈ D. Then, as d = b11 x + b12 y, for some x, y ∈ D, we see that d = (−dt)x + (ds)y so 1 = sy − tx, 325 by the cancellation law. Thus −t s y −x ! x s y t ! = 1 0 0 1 ! . So, after multplying B on the right by the invertible Q1 = x y 0 . 0 s 0 ... 0 t 0 ... 0 0 1 0 ... . . ... . 0 ... 0 1 , The first row of BQ1 is (d, 0, b13 , . . . , b1m ). Thus, as d now occurs as a nonzero element of D in a matrix in S, we see that `(b11 ) ≤ `(d) ≤ `(b11 ) so that b11 is an associate of d. Thus b11 divides b12 as claimed. Similarly, we see that b11 divides each entry bk1 in the first column of B. Thus, subtracting suitable multiples of the first row from the remaining rows of B, we obtain a matrix in S with first column, (b11 , 0, . . . , 0)T . Then subtacting suitable multiples of the column from the remaining columns of this matrix, we arrive at a matrix B0 := b11 0 0 . . . 0 0 · , , · B1 0 where B0 ∈ S. We set B1 = (cij ), where 1 ≤ i ≤ n − 1 and 1 ≤ j ≤ m − 1. We now expand on this: we even claim that b11 divides each entry cij of matrix B1 . This is simply because addition of the j-th row to the first row of the matrix B0 just listed above, yields a matrix B 0 in S whose first row is (b11 , cj2 , cj3 , . . . .cjm ). But we have just seen that in any matrix B0 in S containing an element of minimal possible length in the (1, 1)-position, that this element divides every element of the first row and column. Since B 0 is such a matrix, b11 divides cij . 326 Now by induction on n + m, there exist invertible matrices P1 and Q1 such that P1 B1 Q = diag(d2 , d3 , . . . , ) with d2 dividing d3 dividing . . . dk .1 Moreover, by induction, the numbers d2 , . . . dk , are uniquely determined up to associates by the matrix B1 . But in turn, the element b11 with minimal length determined the matrix B1 , so that the pair (b, B1 ) is determined up to the multiplication of each entry by a unit of D. Since the d2 , . . . dk are in turn determined by B1 and hence b11 , they are uniquely determined quantities. 10.5 Finitely generated modules over a PID Theorem 202 (The Invariant Factor Theorem) Let D be a principal ideal domain and let M be a finitely generated D-module. Then as right Dmodules, we have M ' (D/Dd1 ) ⊕ (D/Dd2 ) ⊕ . . . ⊕ (D/Ddn ), where di divides di+1 , i = 1, ..., n − 1. proof: Let M be a finitely generated D-module. Then there is a finitely generated free module F and a D-epimorphism f : D → M. Then F has a D-basis {x1 , . . . , xn }, and as D is Noetherian, its submodule ker f is finitely generated. Thus we have M = x1 D ⊕ x2 D ⊕ . . . ⊕ xn D ker f = y1 D + y2 D + . . . + ym D. (10.2) (10.3) We then have m unique expressions yi = x1 ai1 + x2 ai2 + · · · + xn ain , i = 1, . . . , m, 1 I apologize for the notation here: diag(d2 , d3 , . . . , ) should not be presumed to be a square matrix. Of course it is still an n × m-matrix, with all entries zero except those that bear a di at the (i, i)-position. If n < m the main diagonal hits the “floor” before it hits the “east wall”, and the other way round if n > m. 327 and an m × n matrix A = (aij ) with entries in D. By the technical result of the previous subsection, there exist matrices P = (pij ) and Q = (qij ) invertible in Dm×m and Dn×n , respectively, such that d1 0 . . . 0 P AQ = 0 d2 . . . 0 ( where di |di+1 ), .. an m × n matrix whose only non-zero entries are on the principal diagonal and proceed until they cant go any further (that is, they are defined up to subscript min(m, n)). If we set Q−1 = (qij∗ ), and yi0 = X yj pij and x0i X xj qij∗ = then P AQ expresses the yi0 in terms of the x0i , which now comprise a new D-basis for F . Thus we have y10 = x01 d1 , y20 = x02 d2 , . . . etc., (where, by convention the expression dk is taken to be zero for min(m, n) < k ≤ m). Also, since P and Q−1 are invertible, the yi0 also span ker f . So, bearing in mind that some of the yi0 = x0i di may be zero, and extending the definition of the dk by setting dk := 0 if m < k ≤ n, we may write F = x01 D ⊕ x02 D ⊕ · · · ⊕ x0n D ker f = x01 d1 D ⊕ x02 d2 D ⊕ · · · ⊕ x0n dn D. Then M = f (F ) ' F/ ker f ' (D/d1 D) ⊕ (D/d2 D) ⊕ · · · ⊕ (D/dm D) and the proof of the theorem is complete. Remark The essential uniqueness of the elements {di } appearing in the above theorem will soon emerge in Section 9.6 below. These elements are called the invariant factors (or sometimes the elementary divisors) of the finitely generated D-module M . 328 At this stage their uniqueness does not immediately follow from the fact that they are uniquely determined by the matrix A since this matrix itself depended on the choice of a finite spanning set for ker f and not all such finite spanning sets can be moved to one another by the action of an invertible transformation Q in homD {ker f, ker f }. Indeed they might have different cardinalities. An element m of a D-module M is called a torsion element if and only if there exists a non-zero element d ∈ D such that md = 0. Recall that Ann(m) = {d ∈ D|md = 0} is an ideal and that we have just said that m is a torsion element if and only if Ann(m) 6= 0. In an integral domain, the intersection of two non-zero ideals is a non-zero ideal; it follows from this that the sum of two torsion elements is a torsion element. In fact the set of all torsion elements of M forms a submodule of M called torM . Corollary 203 A finitely generated D-module over a principal ideal domain D, is a direct sum of its torsion submodule torM and a free module. proof: By the Invariant Factor Theorem, M = (D/d1 D) ⊕ · · · ⊕ (D/dn D) where di divides di+1 . Choose index t so that i ≤ t implies di 6= 0, and, for i > t we have di = 0. Then torM = (D/d1 D) ⊕ · · · ⊕ (D/dt D) and so M = torM ⊕ DD ⊕ DD ⊕ · · · ⊕ DD , with n − t copies of DD . If M = torM , we say that M is a torsion module. Let M be a module over a principal ideal domain, D. Let p be a prime element in D. We define the p-primary part of M as the set of elements of M annihilated by some power of p, that is Mp = {m ∈ M |mpk = 0 for some natural number k = k(m)}. It is easy to check that Mp is a submodule of M . We now have 329 Theorem 204 (Primary Decomposition Theorem) If M is a finitely generated torsion D-module, where D is a principal ideal domain, then there exists a finite number of primes p1 , p2 , . . . , pN such that M ' Mp1 ⊕ · · · ⊕ MpN . proof: Since M is finitely generated, we have M = x1 D+. . . xm D for some finite generating set {xi }. Since M is a torsion module, each of the xi is a torsion element, and so all of the ideals Ai := Ann(xi ) are non-zero; and since D is a principal ideal domain, Ai = Ann(xi ) = di D, where di 6= 0, i = 1, . . . , m. Also, as D is a unique factorization domain as well, each di is expressible as a product of primes: a di = pai1i1 pai2i2 · · · pifif(i)(i) . We let {p1 , . . . , pN } be a complete re-listing of all the primes pij which appear in these factorizations. If f (i) = 1, then di is a prime power and so xi D ⊆ Mpj for some j. If, on the other hand, f (i) 6= 1, then, forming the elea ments vij = di /pijij , we see that the greatesr common divisor of the elements {vi1 , . . . , vif (i) } is 1, and this means vi1 D + vi2 D + · · · + vif (i) D = D. Thus there exist elements b1 , · · · , bf (i) such that vi1 b1 + vi2 b2 + · · · + vif (i) bf (i) = 1. Note that a bj vij xi ∈ Mpij since pijij vij bj xi = 0, so xi = xi · 1 = X bj vij xi ∈ Mpi1 + · · · + Mpif (i) . Thus in all cases xi D ⊆ Mp1 + · · · + MpN for i = 1, . . . , m, and the sum on right side is M . Now it remains only to show that this sum is direct and we do this by the criterion for direct sums of D-modules. Consider an element a ∈ (Mp1 + · · · + Mpk ) ∩ Mpk+1 . 330 Then the ideal Ann(a) contains an element m which on the one hand is a product of primes in {p1 , . . . , pk }, and on the other hand is a power of the prime pk+1 . Since gcd(m, pk+1 ) = 1, we have Ann(a) ⊇ mD + p`k+1 D = D. thus 1 · a = a = 0. Thus we have shown that (Mp1 + · · · + Mpk ) ∩ Mpk+1 = (0). This proves that M = Mp1 ⊕ Mp2 ⊕ · · · ⊕ MpN . 10.6 The uniqueness of the Invariant Factors. We have seen that a finitely generated module M over a principal ideal domain has a direct decomposition as a torsion module and a finitely generated free module (corresponding to the number of invariant factors which are zero). This free module is isomorphic to M/torM and so its rank is the unique rank of M/torM (Theorem 200), and consequently is an invariant of M . The torsion submodule, in turn, has a unique decomposition into primary parts whose invariant factors determine those of tor(M ). Thus, in order to show that the so-called invariant factors earn their name as genuine invariants of M , we need only show this for a module M which is primary – i.e. M = Mp for some prime p in D. In this case, applying the invariant factor theorem, we have M ' D/ps1 D ⊕ · · · ⊕ D/pst D (10.4) where s1 ≤ s2 ≤ · · · ≤ st , and we must show that the sequence S = (s1 , . . . , st ) is uniquely determined by M . We shall accomplish this by producing another sequence Ω = {ωi }, whose terms are manifestly invariants determined by the isomorphism type of M , and then show that S is determined by Ω. For any p-primary D-module A, set Ω1 (A) := {a ∈ A|ap = 0}. Then Ω1 (A) is a D-submodule of A which can be regarded as a vector space over the field F := D/pD. We denote the dimension of this vector space by 331 the symbol rk(Ω1 (A)). Clearly, Ω1 (A) and its rank are uniquely determined by the isomorphism type of A. Now let us apply these notions to the module M with invariants S = (s1 , . . . , st ) as in equation ( 10.4). First we set ω1 := rk(Ω1 (M )), and K1 := Ω1 (M ). In general we set Ki+1 /Ki := Ω1 (M/Ki ), and ωi+1 := rk(Ki+1 /Ki ). The Ki form an ascending chain of submodules of M which ascends properly until it stabilizes at i = st , the maximal element of S. (Note that Ann(M ) = Dpst , so that pst can be thought of as the “exponent” of M .) Moreover, each of the D-modules Ki , is completely determined by the isomorphism type of M/Ki−1 , which is determined by Ki−1 , and ultimately by M alone. Thus we obtain a sequence of numbers Ω = (ω1 , ω2 , . . . , ωst ), which are completely determined by the isomorphism type of M . One can easily see (*) ωj is the number of elements in the sequence S which are at least as large as j. Now the sequence S can be recovered from Ω as follows: (**) For j = 0, 1, . . . , t − 1, st−j is the number of elements ωr of Ω of cardinality at least j + 1. example: Suppose p is a prime in the principal ideal domain D and A is a primary D-module with invariant factors: p, p, p2 , p3 , p3 , p4 , p7 , p7 , p10 , so S = (1, 1, 2, 3, 3, 4, 7, 7, 10). Then, setting F = D/pD, K1 is an F module of rank 9, K2 /K1 is an F -module of rank 7, and in general Ω = (9, 7, 6, 4, 3, 3, 3, 1, 1, 1). Then S = (1, 1, 2, 3, 3, 4, 7, 7, 10) is recovered by the recipe in (**). We conclude: Theorem 205 The non-unit invariant factors of a finitely-generated module over a principal ideal domain are completely determined by the isomorphism type of the module – i.e., they are really invariants. 332 10.7 Applications of the Theorem on PID Modules. 10.7.1 Classification of Finite Abelian Groups. As remarked before, an abelian group is simply a Z-module. Since Z is a PID, any finitely generated Z-module A has the form A ' Zd1 × Zd2 × · · · × Zdm –that is, A is uniquely expressible as the direct product of cyclic groups, the order of each dividing the next. The numbers (d1 , . . . dn ) are thus invariants of A and the decomposition just presented is called the Jordan decomposition of A. The number of distinct isomorphism types of abelian groups of order n is thus the number of ordered sets (d1 , . . . , dm ) such that di divides di+1 and d1 d2 · · · dm = n. example: The number of abelian groups of order 36 is 4. The four possible ordered sets are: (36), (2, 18), (3, 12), (6, 6). This number is usually much easier to compute if we first perform the primary decomposition. If we assume A is a finite abelian group of order n, and if p is a prime dividing n, then Ap = {a ∈ A|apk = 0, k ≥ 0} is just the set of elements of A of p-power order. Thus Ap is the unique p-Sylow subgroup of A. Then the primary decomposition A = Ap1 ⊕ · · · ⊕ Apt is simply the additive version of writing A as a direct product of its Sylow subgroups. Since the Api are unique, the isomorphism type of A is uniquely determined by the isomorphism types of the Api . Thus if Api is an abelian group of order ps , we can ask for its Jordan decomposition: Zps1 × Zps2 × · · · × Zpsm where psi divides psi+1 and ps1 ps2 · · · psm = ps . But we can write this condition s: si ≤ si+1 and s1 + s2 + · · · + sm = s. Thus each unordered set renders s as a sum of natuaral numbers without regard to the order of the summands – called an unordered partition of the integer s. The number of these is usually denoted p(s), and p(s) is called the partition function. For example 333 p(2) = 2, p(3) = 3, p(4) = 5, p(5) = 7, etc. Here, we see that the number of abelian groups of order ps is just p(s), and that: Corollary 206 The number of isomorphism classes of abelian groups of order n = pt11 · · · ptrr is p(t1 )p(t2 ) · · · p(tr ). Again, we can compute the number of abelian groups of order 36 = 22 32 as p(2)p(2) = 22 = 4. 10.7.2 The Rational canonical form of a linear transformation Let F be any field and let V be an n-dimensional vector space over F so V = F ⊕ · · · ⊕ F (n copies) as an F -module. Now suppose T : V → V is a linear transformation on V , which we regard, as a right operator on V . Now in homF (V, V ), the ring of linear transformations of V into V , multiplication is composition of transformations. Thus for any such transformation T , T 2 := T ◦ T and any polynomial in T is a well defined linear transformation of V . As is customary, F [x] will denote the ordinary ring of polynomials in an indeterminate x. We wish to convert V into an F [x]-module by defining v · p(x) := vp(T ) for any v ∈ V and p(x) ∈ F [x]. This module is finitely generated since an F -basis of V will certainly serve as a set of generators of V as an F [x]-module. It now follows from our main theorem that there are polynomials p1 (x), p2 (x),. . . , pm (x) with pi (x) dividing pi+1 (x) in F [x] such that VF [x] ' F [x]/(p1 (x)) ⊕ · · · ⊕ F [x]/(pm (x)), where as usual (pi (x)) denotes the principal ideal pi (x)F [x]. Note that none of the pi (x) is zero since otherwise VF [x] would have a (free) direct summand F [x]F [x] which is infinite dimensional over F . Also, as usual, those pi (x) which are units – that is, of degree zero – contribute nothing to the direct decomposition of V . In fact we may assume that the polynomials pi (x) are monic polynomials of degree at least one. Since they are now uniquely determined polynomials, they have a special name: the invariant factors of the transformation T . 334 Now since each pi (x) divides pm (x), we must have that V pm (x) = 0 and so pm (x)F [x] = AnnF [x] (V ). For this reason, pm (x) being a polynomial of smallest degree such that pm (T ) : V → 0, is called the minimal polynomial of T . Such a polynomial is determined up to associates, but bearing in mind that we are taking pm (x) to be monic, the word “the” preceeding “minimal polynomial” is justified. The product p1 (x)p2 (x) · · · pm (x) is called the characteristic polynomial of T and we shall soon see that the product of the constant terms of these polynomials is, up to sign, the determinant detT , familiar from linear algebra courses. How do we compute the polynomials pi (x)? To begin with, we must imagine the transformation T : V → V given to us in such a way that it can be explicitly determined, that is, by a matrix A describing the action of T on V with respect to a basis B = (v1 , . . . , vn ). Specifically, as T is regarded as a right operator, 2 [T ]B = A = (aij ) so T (vi ) = Xn va . j=1 j ij Then in forming the F [x]-module VF [x] , we see that application of the transformation T to V corresponds to right multiplication of every element of V by the indeterminate x. Thus vi x = X vj aij so 0 = v1 ai1 + · · · + vi−1 ai,i−1 + vi (aii − x) + vi+1 ai,i+1 · · · vn ain . Now as in the proof of the Invariant Factor Theorem, we form the free module F r = x1 F [x] ⊕ · · · ⊕ xn F [x] and the epimorphism f : F r → V defined by f (xi ) = vi . Then the elements ri := x1 ai1 + · · · + xi−1 ai,i−1 + xi (aii − x) + xi+1 ai,i+1 · · · xn ain . 2 Note that the rows of A record the fate of each basis vector under T . Similarly, if T had been a left operator rather than a right one, the columns of A would have been recording the fate of each basis vector. The difference is needed only for the purpose of making matrix multiplication represent composition of transformations. It has no real effect upon the invariant factors. 335 all lie in ker f . Now for any subscript i, x times the element xi of F r is X x i x = ri + xj aij , i = 1, . . . , n. Thus for any polynomial p(x) ∈ F [x, we can write xi p(x) = X r h (x) + k k ik X x j bj for some polynomials hik (x) ∈ F [x] and coefficients bj ∈ F . Now if w = x1 g1 (x) + x2 g2 (x) + · · · + xn gn (x) were an arbitrary element of ker f , then itwould have the form w = Xn x g (x) i=1 i i Xn = ( r h (x)) i=1 i i (10.5) + Xn xc i=1 i i (10.6) for appropriate hi (x) ∈ F [x] and ci ∈ F . Then since f (w) = 0 = f (ri ) and f (xi ) = vi we see that X f (w) = 0 = f ( = 0+ ri hi (x)) + X Xn vc) i=1 i i (10.7) vi ci . (10.8) So each ci = 0. This means w ∈ r1 F [x] + · · · + rn F [x]. Thus the elements r1 , . . . , rn are a set of generators for ker f . Thus we have: Corollary 207 To find the invariant factors of the linear transformation represented by the square matrix A = (aij ), it suffices to diagonalize the following matrix of F [x](n×n) , A − Ix = a11 − x a12 a13 a21 a22 − x a23 a31 a32 a33 − x .. .. .. an1 ... ... 336 ... a1n ... a2n ... a3n .. .. . . . ann − x by the elementary row and column operations to obtain the matrix diag(p1 (x), p2 (x), . . . , pn (x)) with each pi (x) monic, and with pi (x) dividing pi+1 (x). Then the pi (x) which are not units are the invariant factors of A. Now, relisting the invariant factors of T (the non-unit invariant factors) as p1 (x), . . . pr (x), r ≤ n the invariant factor theorem says that T acts on V exactly as right multiplication by x acts on V 0 := F [x]/(p1 (x)) ⊕ . . . ⊕ F [x]/(pr (x)). The space V 0 has a very nice basis. Each pi (x) is a monic polynomial of degree di > 0, say pi (x) = b0i + b1i x + b2i x2 + · · · + bdi −1,i xdi −1 + xdi . Then the summand F [x]/(p1 (x)) of V 0 is non-trivial, and has the F -basis 1 + (pi (x)), x + ((pi (x)), x2 + (pi (x)), . . . , xdi −1 + (pi (x)), and right multiplication by x effects the action: xj + (pi (x)) → xj+1 + (pi (x)), 1 ≤ j < di − 1 (10.9) di −1 x + (pi (x)) → −b0i (1 + (pi (x))) − b1i (x + (pi (x))) − . . . .(10.10) Thus, with respect to this basis, right multiplication of F [x]/(pi (x)) by x is represented by the matrix Cpi (x) := 0 1 0 ... 0 0 0 1 0... 0 .. .. .. ... .. 0 0 ... 0 1 −b0i −b1i −b2i . . . −bdi −1,i (10.11) called the companion matrix of pi (x). If pi (x) is a unit in F [x] – that is, it is just a nonzero scalar, then our convention is to regard the companion matrix as the “empty” (0 × 0) matrix. That convention is incorporated in the following 337 Corollary 208 If p1 (x), . . . , pn (x) are the invariant factors of A, then A is conjugate to the matrix P −1 AP = Cp1 (x) 0 0 Cp2 (x) .. .. 0 ... ... 0 0... 0 ... .. 0 Cpn (x) . This is called the rational canonical form of A. Remark: Note that each companion matrix Cpi (x) is square with di := deg(pi (x) rows. Using decomposition along the first column, one calculates that Cpi (x) in equation (10.11) has determinant (−1)di −1 (−b0i ) = (−1)di b0i . Thus the determinant of A is completely determined by the invariant factors. The determinant is just the product detA = Yr (−1)di b0i = (−1)n i=1 Yr b , i=1 0i which is the product of the constant terms of the invariant factors times the P parity of n = di , the dimension of the original vector space V . It may be of interest to know that the calculation of the determinant of matrix A by ordinary means uses on the order of n! steps. Calculating the determinant by finding the invariant factors (diagonalizing A − xI) uses on the order of n3 steps. The abstract way is ultimately faster. 10.7.3 The Jordan Form This section involves a special form that is available when the ground field of the vector space V contains all the roots of the minimal polynomial of transformation T . Suppose now F is a field containing all roots of the minimal polynomial of a linear transformation T : V → V . Then the minimal polynomial, as well as all of the invariant factors, completely factors into prime powers (x −λi )ai . Applying the primary decomposition to the right F [x]-module V , we see that a “primary part” V(x−λ) of V is a direct sum of modules, each isomorphic to F [x] modulo a principal ideal I generated by a power of x − λ. This means that any matrix representing the linear transformation T is equivalent to a matrix which is a diagonal matrix of submatrices which are companion 338 matrices of powers of linear factors. So one is reduced to considering F [x]modules of the form F [x]/I = F [x]/((x − λ)m ), for some root λ. Now F [x]/I has this F -basis: 1 + I, (x − λ) + I, (x − λ)2 I, . . . (x − λ)m−1 + I. Now right multiplying F [x]/I by x = λ + (x − λ) takes each basis element to λ times itself plus its successor basis element – except for the last basis element, which is merely multiplied by λ. That basis element, (x−λ)m−1 +I, and its scalar multiples comprise a 1-space of all “eigenvectors” of F [x]/I – that is, vectors v such that vT = λv. The resulting matrix, with respect to this basis has the form: λ 0 .. 0 0 1 λ .. 0 0 0 1 .. 0 0 ... 0 0 ... . . . .. ... λ ... 0 0 0 .. 1 λ . The matrix which results from assembling blocks according to the primary decomposition, and rewriting the resulting companion matrices in the above form, is called the Jordan form of the transformation T . example: Suppose the invariant factors of a matrix A were given as follows: x, x(x − 1), x(x − 1)3 (x − 2)2 . Here, F is any field in which the numbers 0, 1 and 2 are distinct. What is the Jordan form? First we separate the invariant factors into their elementary divisors: prime x: {x, x, x}. prime x − 2: {(x − 2)2 }. prime x − 1: {x − 1, (x − 1)3 }. Next we write out each Jordan block as above: the result being 339 0 0 0 2 1 2 . 1 1 1 1 1 1 (All unmarked entries in the above matrix are zero.) 10.7.4 Information Carried by the Invariant Factors What are some things one would like to know about a linear transformation T : V → V , and how can the invariant factors provide this information? 1. We have already seen that the last invariant factor pr (x) is the minimal polynomial of T , that is – the smallest degree monic polynomial p(x) such that p(T ) is the zero transformation V → 0V . (We recall that the ideal generated by p(x) is Ann(VF [x] ) when V is converted to an F [x]-module via T .) 2. Sometimes it is of interest to know the characteristic polynomial of T , which is defined to be the product p1 (x) · · · pr (x) of the invariant factors of T . One must bear in mind that by definition the invariant factors of T are monic polynomials of positive degree. The characteristic polynomial must have degree n, the F -dimension of V . This is because the latter is the sum of the dimensions of the F [x]/(pi (x)), that is, the sum of the numbers deg pi (x) = di . 3. Since V has finite dimension, T is onto if an only if it is one-to-one . 3 . 3 This is a well-known consequence of the so-called “nullity-rank” theorem for transformations but follows easily from the fundamental theorems of homomorphisms for R-modules applied when R = F is a field. Specifically, there is an isomorphism between the poset of subspaces of T (V ) and the poset of subspaces of V which contain ker T . Since dimension is the length of an unrefinable chain in such a poset one obtains codimV (ker T ) = dim(T (V )) 340 The dimension of ker T is called the nullity of T , and is given by the number of invariant factors pi (x) which are divisible by x. 4. The rank of T is the F -dimension of T (V ). It is just n minus the nullity of T (see the footnote just above). 5. V is said to be a cyclic F [x]-module if and only if there exists a module element v such that the set vF [x] spans V – i.e. v generates V as an F [x]-module. This is true if and only if the only non-unit invariant factor is pr (x) the minimal polynomial (equivalently: the minimal polynomial is equal to the characteristic polynomial). 6. Recall that an R-module is said to be irreducible if and only if there are no proper submodules. (We met these in the Jordan-Hölder theorem for R-modules.) Here, we say that T acts irreducibly on V if and only if the associated F [x]-module is irreducible. This means there exists no proper sub-vector space W of V which is T -invariant in the sense that T (W ) ⊆ W . One can then easily see that T acts irreducibly on V if and only the characteristic polynomial is irreducible in F [x].4 7. An eigenvector of T associated with the root λ or λ-eigenvector is a vector v of V such that vT = λv. A scalar is called an eigen-root of T if and only if there exists a non-zero eigenvector associated with this scalar. Now one can see that the eigenvectors associated with λ for right multiplication of VF [x] by x, are precisely the module elements killed by the ideal (x−λ)F [x]. Clearly they form a T -invariant subspace of V whose dimension is the number of invariant factors of T which are divisible by (x − λ). 8. A matrix is said to be nilpotent if and only some power of it is the zero matrix. Similarly a linear transformation T : V → V is called a nilpotent transformation if and only if some power (number of iterations) of T annihilates V . Quite obviously the following are equivalent: • T is a nilpotent transformation. • T can be represented by a nilpotent matrix with respect to some basis of V . 4 This condition forces the minimal polynomial to equal the characteristic polynomial (so that the module is cyclic) and both are irreducible polynomials. 341 • The characteristic polynomial of T is a power of x. • The minimal polynomial of T is a power of x. 342 Chapter 11 Theory of Fields. 11.1 Introduction In this chapter we are about to enter one of the most fascinating and historically interesting regions of abstract algebra. I can think of at least three reasons why this is so: 1. Fields underlie nearly every part of mathematics. Of course we could not have vector spaces without fields (if only as the center of the coefficient division ring). Independantly of vector spaces, fields figure in number theory, algebraic varieties and sesquilinear forms. So it is a very basic subject. 2. There are presently many open problems involving fields – each motivated by some application of fields to another part of mathematics. 3. Third, there is the strange drama that so many field-theoretic questions bear on old historic questions, and indeed, its own history has a dramatic story of its own. Recall that a field is an integral domain in which every non-zero element is a unit. Examples are: 1. The field of rational numbers – in fact the “field of fractions” of an arbitrary integral domain D (that is, the localization DS where S = D − {0}). A special case is the field of fractions of the polynomial domain F [x]. This field is called the field of rational functions 343 over F in the indeterminate x (even though they are not functions at all). 2. The field of real numbers. 3. The field of complex numbers. 4. The field of integers modulo a prime – more generally, the field D/M where M is a maximal ideal of an integral domain D. Of course there are many further examples. In the past few chapters, we have been passing from genus to species to subspecies. We considered integral domains, then principal ideal domains, and now fields. As we get more specialized by adding more axioms, one might suppose that we are seeing the same theory, with the proofs getting easier and easier. But what actually happens is that each theorem rises to its appropriate level of generality, so that with each new speciation of rings, an entirely new generation of questions becomes feasible. Thus for arbitrary integral domains, divisibility is an important question, and hence the UFD condition, which fails for general domains but works for PID’s, is of interest. But for fields, divisibility questions are almost meaningless since every nonzero element divides every other such element. The new generation of feasible questions for fields involves some of the oldest and longest-standing problems in mathematics – they all involve the existence or non-existence of roots of polynomial equations in one indeterminate. 11.2 Elementary Properties of Field Extensions. Let K be a field. A subfield is a subset F of K which contains a nonzero element, and is closed under addition, multiplication, and the taking of multiplicative inverses. It follows at once that F is a field with respect to these operations inherited from K, and that F contains the multiplicative identity of K as its own multiplicative identity. Obviously, the intersection of any collection of subfields of F is a subfield, and so one may speak of the subfield generated by a subset X of a field F as the intersection of all 344 subfields of F containing set X. The subfield of K generated by F and X is denoted F (X).1 The subfield generated by zero is called the prime subfield of F . (By definition, it contains the multiplicative identity as well as zero.) Since a field is a species of integral domain, it has a characteristic. If the multiplicative identity element 1, as an element of the additive group (F, +) has finite order, then its order is a prime number p which we designate as the characteristic of F in that case. Otherwise 1 has infinite order as an element of (F, +) and we say that F has characteristic zero in that case. We write char(F ) = p or 0 in the respective cases. In the former case, it is clear that the prime subfield of F is just Z/pZ; in the latter case the prime subfield is isomorphic to the field Q of rational numbers. Since we shall often be dealing in this chapter with cases in which the subfield is known but the over-field is not, it is handy to reverse the point of view of the previous paragraph. We say that K is an extension of F if and only if F is a subfield of K. A chain of fields F1 ≤ F2 ≤ · · · ≤ Fn , where each Fi+1 is a field extension of Fi , is called a tower of fields. If the field K is an extension of the field F , then K is a right module over F , that is, a vector space over F . As such, it possesses a dimension as an F -vector space, which we call the degree of the extension F ≤ K, and denote by the symbol [K : F ]. 11.2.1 Algebraic and Transcendental Extensions. Let K be a field extension of the field F , and let p(x) be a polynomial in F [x]. We say that an element α of K is a zero of p(x) (or a root of the equation p(x) = 0) if and only it p(x) 6= 0 and yet p(α) = 0.2 If α is a 1 Of course this notation suffers the defect of leaving K out of the picture. Accordingly we shall typically be using it when the ambient overfield K of F is clearly understood from the context. 2 Two remarks are appropriate at this point. First note that the definition makes it “ungrammatic” to speak of a zero of the zero polynomial. Had the non-zero-ness of p(x) not been inserted in the definition of root, we should be saying that every element of the field is a zero of the zero polynomial. That is such a bizarre difference from the situation with non-zero polynomials over infinite fields that we should otherwise always be apologizing for that exceptional case in the statements of many theorems. It thus seems better to utilize the definition to weed out this akwardness in advance. 345 zero in K of a polynomial in F [x], then α is said to be algebraic over F . Otherwise α is said to be transcendental over F . If K is an extension of F , and if α is an element of K, then the symbol F [α] denotes the subring of K generated by F ∪ {α}. This would be the set of all elements of K which are F -linear combinations of powers of α. Note that as this subring contains the field F , it is a vector space over F whose dimension is again denoted [F [α] : F ]. There is then a ring homomorphism ψ : F [x] → F [α] ⊆ K which takes the polynomial p(x) to the field element p(α) in K. Clearly ψ(F [x]) = F [α]. Suppose now ker ψ = 0. Then F [x] ' F [α] and α is transcendental since any non-trivial polynomial of which it is a zero would be a polynomial of positive degree in ker ψ. On the other hand, if ker ψ 6= 0, then, as F [x] is a principal ideal domain, we have ker ψ = F [x]p(x), for some polynomial p(x). Then F [α] = ψ(F [x]) ' F [x]/kerψ = F [x]/(p(x)). Now, as F [α] is an integral domain, the principal ideal (p(x)) is a prime ideal, and so p(x) is a prime element of F [x]. But since F [x] is a UFD (not to mention even simpler reasons) p(x) is an irreducible element of F [x]. Consequently the ideal (p(x)) = ker ψ is actually a maximal ideal of F [x], and so the subring F [α] is a field. The irreducible polynomial p(x) determined by α is unique up to multiplication by a non-zero scalar. The associate of p(x) which is monic is denoted IrrF (α). It is the the unique monic polynomial of smallest possible degree which has α for a zero. Note that in this case, the F -dimension of F [α] is the F - dimension of F [x]/(p(x)), which, by the previous chapter can be recognized to be the degree of the polynomial p(x). Second, there are natural objections to speaking of a “root” of a polynomial. One of my colleagues (an editor of the Intelligencer whose opinion I regard most highly) asks whether it might not be better to follow the German example and write “zero” (Nullstelle) for “root” (Wurzel), despite usage of the latter term in English versions of Galois Theory. I will follow tradition in speaking of a “root” of an equation, p(x) = 0 but the zero of a polynomial. 346 Now conversely, assume that [F [α] : F ] is finite. Then there exists a non-trivial finite F -linear combination of elements in the infinite list {1, α, α2 , α3 , . . .} which is zero. It follows that α is algebraic in this case. Summarizing, we have Theorem 209 Let K be an extension of the field F and let α be an element of K. 1. If α is transcendental over F , then the subring F [α] which it generates is isomorphic to the integral domain F [x]. Moreover, the subfield F (α) which it generates is isomorphic to the field of fractions of F [x], the field of rational functions over F . 2. If α is algebraic over F , then the subring F [α] is a subfield intermediate between K and F . As an extension of F its degree is [F [α] : F ] = deg IrrF (α). 3. Thus dimF F [α] is finite or infinite according as α is algebraic or transcendental over F . 11.2.2 Indices Multiply: Ruler and Compass Problems. Lemma 210 If F ≤ K ≤ L is a tower of fields, then the indices “multiply” in the sense that [L : F ] = [L : K][K : F ]. proof: Let A and B be bases of L as a vector space over K, and K as a vector space over F , respectively. Let AB be the set of all products {ab|a ∈ A, b ∈ B}. We shall show that AB is an F -basis of L. If α ∈ L, then α is a finite K-linear combination of elements of A, say α = a1 α1 + · · · + am αm . 347 Now each coefficient αi , being an element of K, is a finite linear combination αj = bj1 βj1 + · · · + bjmj βjmj of elements bij of B (with the βij in F ). Substituting these expressions for αj in the L-linear combination for α, we have expressed α as an F -linear combination of elements of AB. Thus AB is an F -spanning set for L. It remains to show that AB is an F -linearly independant set. So suppose for some finite subset S, that X (a,b)∈S abβa,b = 0. Then as each βa,b is in F . and each b is in K, the left side of the presented formula may be regarded as a K-linear combination of a’s equal to zero, and so, by the K-linear independance of A, each coefficient X b bβa,b of each a is equal to zero. Hence each βa,b = 0, since B is F -linearly independent. Thus AB is F -linearly independent and hence is an F -basis for L. Since this entails that all the elements of AB are distinct, we see that |AB| = |I × J| = |I||J| = |A||B|, which proves the theorem. Corollary 211 If F1 ≤ F2 ≤ · · · ≤ Fn is a finite tower of fields, then [Fn : F1 ] = [Fn : Fn−1 ] · [Fn−1 : Fn−2 ] · · · [F2 : F1 ]. We can use this observation to sketch a proof of the impossibility of certain ruler and compass constructions. Given a unit length 1, we can replicate it m times, or divide it into n equal parts, and so can form all lengths which are rational numbers. We can also form right-angled triangles inscribed in a circle and so with ruler and compass, we can extract the square root of the differences of squares, and hence any square root because of the formula c+1 c= 2 2 c−1 − 2 348 2 . If α is such a square root, where c is a rational length, then by ruler and compass we can form all the lengths in the ring Q[α], which (since α is a zero of x2 − c) is a field extension of Q of degree 1 or 2. Iterating these constructions a finite number of times, the possible lengths that we could encounter all lie in the uppermost field of a tower of fields Q = F1 < F2 < · · · < Fn = K with each Fi+1 an extension of degree two over Fi . By the Corollary above, the index [K : Q] is a power of 2. Now we come to the arm-waving part of the proof: One needs to know that once a field L of constructable distances has been achieved, the only new number not in K contructed entirely from old numbers already in K is in fact obtained by the construction of a missing side of a right triangle, two of whose sides have lengths in K. (A really severe proof of that fact would require a formalization of exactly what “ruler and compass” constructions are – a logical problem beyond the field-theoretic applications whose interests are being advanced here.) In this sketch, we assume that that has been worked out. This means, for example, that we cannot find by ruler and compass alone, a length α which is a zero of an irreducible polynomial of degree n containing an odd prime factor. For if so, Q[α] would be a subfield of a field K of degree 2m over Q. On the other hand, the index n = [F [α] : F ] must divide [K : Q] = 2m , by the Theorem 209. But the odd prime in n cannot divide 2m . That’s it! That’s the whole amazingly simple argument! Thus one cannot solve the problem posed by the oracle of Delos, to “duplicate” (in volume) a cubic altar – i.e., find a length α such that α3 − 2 = 0 – at least not with ruler and compass. Similarly, given angle α, trisect it with ruler and compass. If one could, one could construct the length cos β where 3β = α. But cos(3β) = 4cos3 β − 3 cos β = λ = cos α. This means we could always find a zero of 4x3 − 3x − λ and when α = 60o , so λ = 1/2, setting y = 2x yields a constructed zero of y 3 −3y −1, which is irreducible over Q. As observed above, this is impossible. 11.2.3 The Number of Zeros of a Polynomial Lemma 212 Let K be an extension of the field F and suppose the element α of K is a zero of the polynomial p(x) in F [x]. Then in the polynomial ring F [α][x], x − α divides p(x). 349 proof: Since p(α) = 0 we may write p(x) = = = = a0 + a1 x + · · · + an x n p(x) − p(α) a1 (x − α) + a2 (x2 − α2 ) + · · · + an (xn − αn ) (x − α)[a1 + a2 (x + α) + a3 (x2 + αx + α2 ) + · · · · · · + an (xn−1 + · · · αn−1 )]. Theorem 213 Let K be an extension of the field F and let p(x) be a polynomial in F [x]. Then K has at most deg p(x) zeros of p(x) (even counting multiplicities). proof: There is no loss in assuming that F = K and that p(x) is monic. Since K[x] is a UFD, p(x) has a factorization into irreducible factors as p(x) = (x − α1 )n1 · · · (x − αr )nr g1 (x) · · · gk (x), where the gi (x) are each irreducible in K[x] of degree at least 2. Then each of the αi are zeros. Conversely by Lemma 212, each zero α produces a linear (degree one) factor (x − α) of p(x), and so α must be one of the αi by uniqueness of the factorization. The number ni is called the multiplicity of the zeros αi , and clearly P ni ≤ deg p(x), proving the Theorem. For the next result, we require this definition: a group G is said to be locally cyclic if and only if every finitely-generated subgroup of G is cyclic. Clearly such a group is abelian. Corollary 214 Let F be any field and let F ∗ be its multiplicative group of non-zero elements. Then the torsion subgroup of F ∗ is locally cyclic. proof: Any finitely-generated subgroup of the torsion subgroup of F ∗ is a finite abelian group A. If A were not cyclic, it would contain a subgroup isomorphic to Zp × Zp , for some prime p. In that case, the polynomial xp − 1 would have at least p2 zeros in F , contrary to Theorem 213. 350 Corollary 215 (Polynomial Identities) Suppose a(x), b(x), c(x) and d(x) are polynomials in F [x]. Suppose a(b(α)) = c(d(α)) for all α ∈ F . If F contains infinitely many elements, then a(b(x)) = c(d(x)) is a polynomial identity — that is, both sides are the same element of F [x]. proof: : We need only show that the polynomial J(x) := a(b(x)) − c(d(x)) has degree zero. If it has positive degree, then by Theorem 213 it possesses only finitely many zeros. But then by hypothesis, the infinitely many elements of F would all be zeros of J(x), a contradiction. 11.3 Splitting Fields and Their Automorphisms 11.3.1 Extending Isomorphisms Between Fields Suppose E is a field extension of F and that α is an element of E which is algebraic over F . Then the development in the previous section showed that there is a unique irreducible monic polynomial, g(x) = IrrF (α) ∈ F [x], having α as a zero. The subfield F (α) generated by F ∪ {α} is the subring F [α] isomorphic to the factor ring F [x]/(g(x)). Now suppose f : F1 → F2 is an isomorphism of fields. Then f can be extended to a ring isomorphism f ∗ : F1 [x] → F2 [x], which takes polynomial p := a0 + a1 x + · · · + an xn to f (p) := f (a0 ) + f (a1 )x + · · · + f (an )xn . Then of course f ∗ takes irreducible polynomials of F1 [x] to irreducible polynomials of F2 [x]. With this notation in mind we record 351 Lemma 216 (Fundamental Lemma on Extending Isomorphisms) Let E1 be a field extension of F1 , and suppose f : F1 :→ E be an isomorphism which embeds F1 as a subfield F2 of some other field E. Let f ∗ be the extension of the field isomorphism f : F1 → F2 to a ring isomorphism F1 [x] → F2 [x], as described above. Finally, let α1 be an element of E1 which is algebraic over F1 , and let g1 be its monic irreducible polynomial in F1 [x]. Then the following assertions hold: 1. The embedding (injective morphism) f : F1 → E can be extended to an embedding fˆ : F1 (α1 ) → E, if and only if the polynomial f ∗ (g1 ) := g2 contains a zero in E. 2. Moreover, for each zero α2 of g2 in E, there is a unique embedding fˆ : F1 (α1 ) → E, taking α1 to α2 and extending f . 3. The number of extensions fˆ : F1 (α1 ) → E of the embedding f is equal to the number of distinct zeros of g2 to be found in E. proof: 1. If there is an embedding fˆ : F (α1 ) → E extending the embedding f , then fˆ(α1 ) is a zero of g2 in E. Conversely, if α2 is a zero of g2 in E, then F2 (α2 ) is a subfield of E isomorphic to F2 [x]/(g2 (x)). Similarly, F1 (α1 ) is isomorphic to F1 [x]/(g1 (x)). But the ring isomorphism f ∗ takes the maximal ideal (g1 (x)) of F1 [x] to the maximal ideal (g2 (x)) of F2 [x], and so induces an isomorphism of the corresponding factor rings: f 0 : F1 [x]/(g1 (x)) → F2 [x]/(g2 (x)). Linking up these three isomorphisms f0 F1 (α1 ) → F1 [x]/(g1 (x)) → F2 [x]/(g2 (x)) → F2 (α2 ) ⊆ E, yields the desired embedding. 2. If their were two embeddings F1 (α1 ) → E, taking α1 to α2 and extending f , then one composed with the inverse of the other would fix F1 ∪ {α1 } and hence F1 (α1 ) element-wise. Similarly, composition of the inverse of one with the other would be the identity mapping on F2 (α2 ). It follows that the two embeddings are equal. 3. This statement now follows from 1. and 2. 352 11.3.2 Splitting Fields. Suppose E is an extension field of F . Then, of course, any polynomial in F [x] can be regarded as a polynomial in E[x], and any factorization of it in F [x] is a factorization in E[x]. Put another way, F [x] is a subdomain of E[x]. A polynomial p(x) in F [x] of positive degree is said to split over E if and only if p(x) factors completely into linear factors (that is, factors of degree 1) in E[x]. Such a factorization has the form p(x) = a(x − a1 )n1 (x − a2 )n2 · · · (x − ar )nr (11.1) where a is a nonzero element of F , the exponents ni are positive integers and the collection {a1 , . . . , ar } of zeros of p(x) in E (also called roots of p(x) = 0) is contained in E. Obviously, if E 0 is a field containing E and p(x) splits over E, then it splits over E 0 . Also, if E0 were a subfield of E containing F together with all of the zeros {a1 , . . . , ar }, then p(x) splits over E0 since the factorization in equation (11.1) can take place in E0 [x]. In particular, this is true if E0 = F (a1 , . . . , ar ), the subfield of E generated by F and the zeros ai , i = 1, . . . , r. An extension field E of F is called a splitting field of the polynomial p(x) ∈ F [x] over F if and only if 1. F ⊆ E. 2. f (x) factors completely into linear factors in E[x] as f (x) = a(x − a1 )(x − a2 ) · · · (x − an ). 3. E = F (a1 , . . . , an ), that is, E is generated by F and the zeros ai . The following observations follow directly from the definition just given and the proofs are left as an exercise. Lemma 217 The following hold: 1. If E is a splitting field for p(x) ∈ F [x] over F , and L is a subfield of E containing F , then E is a splitting field for p(x) over L. 2. Suppose p(x) and q(x) are polynomials in F [x]. Suppose E is a splitting field for p(x) over F and K is a splitting field for q(x) over E. Then K is a splitting field for p(x)q(x) over F . 353 Our immediate goals are to show that splitting fields exist, to show that they are unique up to isomorphism, and to show that between any two such splitting fields, the number of isomorphisms is bounded by a function of number of distinct zeros of f (x). Theorem 218 (Existence of Splitting Fields.) If f (x) ∈ F [x], then a splitting field for f (x) over F exists. Its degree over F is at most d!, where d is the sum of the degrees of the non-linear irreducible polynomial factors of f (x). proof: Let f (x) = f1 (x)f2 (x) · · · fm (x) be a factorization of f (x) into irreducible factors in F [x]. Set n = deg f (x), and proceed by induction on k = n − m. If k = 0, then n = m, so each factor fi (x) is linear and clearly E = F satisfies all the conclusions of the theorem. So assume k > 0. Then some factor, say f1 (x), has degree greater than 1. Form field L = F [x]/(f1 (x)), (this is a field since f1 (x) is irreducible and is an extension of F since it is a vector-space over F ), and observe that the coset a := x + F [x]f1 (x) is a zero of f1 (x) ( and hence f (x)) in L. Thus in L[x], we have the factorizations: f1 (x) = (x − a)h1 (x), and f (x) = (x − a)h1 (x)f2 (x) · · · fm (x), with ` > m irreducible factors. Thus n − ` < n − m and so by induction, there is a splitting field E of f (x) over L. Moreover, this degree is at most (d − 1)! since the sum of the degrees of the non-linear irreducible factors of f (x) over L[x] has been reduced by at least one because of appearance the new linear factor x − a. Now we claim that E is a splitting field of f (x) over F . Property (1.) holds since L ⊆ E implies F ⊆ E. We already have (2.) from the definition of E. It remains to see that E is generated by the zeros of f . Since (x − a) is one of the factors of f (x) in the factorization f (x) = (x − a1 )(x − a2 ) · · · (x − an ), 354 we may assume without loss of generality that a = a1 . We have E = L(a1 , . . . , an ) from the definition of E. But L = F (a), so E = L(a1 , . . . , am ) = F (a1 )(a1 , . . . , an ) = F (a1 , . . . , an ). Thus E is indeed a splitting field for f (x) over F . Finally, since [L : F ] = deg f1 (x) ≤ d, and [E : L] ≤ (d − 1)! we obtain [E : F ] = [E : L][L : F ] ≤ (d − 1)! · d = d! as required. Theorem 219 Let η : F → F̄ be an isomorphism of fields, which we extend to a ring isomorpism η ∗ : F [x] → F̄ [x]. and we write f¯(x) for η ∗ (f (x)) for each f (x) ∈ F [x]. Suppose E and Ē are splitting fields of f (x) over F , and and f¯(x) over F̄ , respectively. Then η can be extended to an isomorphism η̂ of E onto Ē, and the number of ways of doing this is at most the index [E : F ]. There are exactly [E : F ] such isomorphisms if f¯(x) has distinct zeros in Ē; or, equivalently, f (x) has distinct zeros in E. proof: We use induction on [E : F ]. If [E : F ] = 1, f (x) factors into linear factors, and there is precisely one isomorphism extending η, namely η itself. Assume [E : F ] > 1. Then f (x) contains an irreducible factor g(x) of degree greater than 1. Since η ∗ is an isomorphism of polynomial rings, ḡ(x) is an irreducible factor of f¯(x) with the same degree as g(x). Thus Ē contains a zero of ḡ(x), and E contains a zero r1 of g(x). Form K = F (r1 ). Then by the Fundamental Lemma on Extensions (Lemma 216), the isomorphism η has k extensions ζ1 , . . . , ζk : F (r1 ) → Ē, where k is the number of distinct zeros of ḡ(x) in Ē. Now if m = deg g(x), [K : F ] = m and k ≤ m. If the zeros of ḡ(x) are distinct, k = m. Now clearly E is a splitting field of f (x) over K, and Ē is a splitting field ¯ of f (x) over each ζi (K). Since [E : K] < [E : F ], induction implies each ζi can be extended to an isomorphism E → Ē in at most [E : K] ways, and in exactly k ways if the non-linear factors of f (x) have distinct zeros in E. This yields at most k[E : K] = [K : F ][E : K] = [E : F ] isomorphisms in general, and exactly [E : F ] isomorphisms if f (x) has distinct zeros in E. Since we possess at least one field isomorphism from E to Ē, the last statement is equivalent to f¯(x) having distinct zeros in Ē. This proves the theorem. 355 Corollary 220 If E1 and E2 are two splitting fields for f (x) ∈ F [x] over F , there exists an F -linear field isomorphism: σ : E1 → E2 . Proof: This follows immediately from Theorem 219 for the case σ = 1F , the identity mapping on F = F1 = F2 . EXAMPLE. Suppose F = Q, the rational field, and p(x) = x4 − √5, irreducible by Eisenstein’s criterion. Then if α is a zero of p(x), α2 is ± 5, √ so the √ splitting field E of p(x) over Q must contain the subfield Q( 5). In Q( 5)[x] we have a factorization √ √ x4 − 5 = (x2 + 5)(x2 − 5). q √ √ Noting that the complex number − 5 is i 4 5, the full factorization over E[x] is √ √ √ √ 4 4 4 4 x4 − 5 = (x + i 5)(x − i 5)(x + 5)(x − 5), √ where i2 = −1, and 4 5 denotes the positive real fourth zero of 5. We see that √ 4 E = Q( 5, i) and √ √ 4 4 [E : Q] = [E : Q( 5)] · [Q( 5) : Q] = 2 · 4 = 8. This shows that the bound [E : F ] ≤ d! need not be obtained. We close this subsection with a useful observation: Lemma 221 Suppose K is an extension of a field F , and K contains a subfield E which is a splitting field for a polynomial p(x) ∈ F [x] over F . Then any automorphism of K which fixes F pointwise, must stabilize E. Remark: The Lemma just says that the splitting field E is “characteristic” among fields in K which contain F – that is, E is invariant under all F -linear automorphisms of K. Proof: If σ is an F -linear automorphism of K, then for any zero α of p(x) in K, σ(α) is also a zero of p(x) in K. 356 11.3.3 Normal Extensions. So far the notion of splitting field is geared to a particular polynomial. The purpose of this section is to show that it can be seen as a property of a field extension itself, independant of any particular polynomial. Before proceeding further, let us stream-line our language concerning field automorphisms. Definition: Let K and L be fields containing a common subfield F . An isomorphism K → L of K onto L is called an F -isomorphism if and only if it fixes the subfield F elementwise. (Heretofore, we have been calling these F -linear isomorphisms.) Of course, if K = L, an F -isomorphism:K → L is called an F -automorphism. Finally, an F -isomorphism of K onto a subfield of L is called an F -embedding. We say that a finite extension E of F is normal over F if and only if E is the splitting field over F of some polynomial of F [x]. Just as a reminder, recall that this means that there is a polynomial p(x) which splits completely into linear factors in E[x] and that E is generated by F and all the zeros of p(x) that lie in E. Note that if E is a normal extension and K is an intermediate field, – that is, F ≤ K ≤ E – then E is a normal extension of K. We have a criterion for normality. Theorem 222 (A characterization of normal field extensions3 ) A finite extension E of F is normal if and only if, every irreducible polynomial g(x) in F [x] which has one zero in E, in fact splits completely over E. proof: Suppose E has the described property concerning irreducible polynomials of F [x]. Since E is a finite extension of F , it has an F -basis, say u1 , . . . , un . Set gi (x) = IrrF (ui ), and let p(x) be the product of the gi (x). Then by hypothesis, every zero of p(x) is in E and p(x) splits into linear factors in E[x]. since the {ui } are among these zeros and the ui and F generate E, E is a splitting field of p(x) over F . 3 In many books, the characterizing property given in this theorem is taken to be the definition of “normal extension”. This does not change the fact that the equivalence of two distinct notions must be proved. 357 Suppose, on the other hand, that E is the splitting field of a polynomial f (x) ∈ F [x]. Suppose, by way of contradiction that g(x) is an irreducible polynomial in F [x] with one zero a in E, and one zero b not in E. We may regard b as a zero in a field E(b), properly extending E. Now since g(x) is irreducible, there is an isomorphism σ : F (a) → F (b) which is the identity mapping when restricted to F . Now E is the splitting field for a polynomial f (x) ∈ F [x]. Thus E is a splitting field for f (x) over F (a). But also E(b) is a splitting field for f (x) over F (b) since E(b) is generated by F (b) and all zeros of f (x). By Theorem 219, the isomorphism σ extends to an F -linear isomorphism τ : E → E(b). It follows that E and E(b) have the same degree over F . But as E is a subfield of E(b), we have E = E(b), so b ∈ E, contrary to our choice of b. Thus all zeros of g(x) lie in E, as required. Corollary 223 If E is a (finite) normal extension of F and K is any intermediate field, then any F -embedding σ : K → E. can be extended to an automorphism of E fixing F . proof: E is a splitting field over F for a polynomial f ∈ F [x]. Thus E is a splitting for f (x) over K as well as over σ(K). The result then follows from Theorem 219. Suppose K/F is a finite extension. Then K = F (a1 , . . . , an ) for some finite set of elements {ai } of K (for example, an F -basis of K). Now let p(x) be the product of the gi (x) := IrrF (ai ). Now any normal extension E/F capturing K as an intermediate field, contains every zero of gi (x) and hence every zero of p(x). Thus between E and K is a splitting field S of p(x) over F . The splitting field S is the “smallest” normal extension of F containing K in the sense that any other finite normal extension E 0 of F which contains K, contains an isomorphic copy of S – that is, there is an injective morphism S → E 0 whose image contains K. Now this global description of the field S is independent of the polynomial p(x), and so we have 358 Corollary 224 (Normal Closure) If K is a finite extension of F , then there is a normal extension S of F , containing K with the property that for every normal extension E of F containing K, there is an F -isomorphism of S to a subfield of E which is normal over F and contains K. Clearly S is unique up to isomorphism. The extension S of F so defined is called the normal closure of K over F . The reader should bear in mind that this notion depends critically on F . For example it may happen that K is not normal over F , but is normal over some intermediate field L. Then the normal closure of K over L is just K itself while the normal closure of K over F could be larger than K. Another consequence of Corollary 223 is this: Corollary 225 (Normality of invariant subfields) Suppose K is a normal extension of the field F of finite degree, and let G be the full group of F automorphisms of K. If L is a subfield of K containing F , then L is normal over F if and only if it is G-invariant. proof: If L is normal over F , then L is the splitting field for some polynomial f (x) ∈ F [x]. But the roots of f (x) in K are permuted among themselves by G, and so the subfield over F that they generate, namely L, is G-invariant. On the other hand, assume L is G-invariant. Suppose p(x) is an irreducible polynomial in F [x] with at least one zero, α in L. Suppose β were another zero of p(x) in K. Then there is an F -isomorphism F [α] → F [β] of subfields of K, which, by Corollary 223, can be extended to an element of G. Since L is G invariant, β ∈ L. Thus p(x) has all its zeros in L and so splits completely over L. Now by Theorem 222, L is normal over F . 11.4 Some Applications to Finite Fields 11.4.1 Prime Subfields For any field F there is a unique subfield P generated by the multiplicative identity element 1. It is called the prime subfield of F and is uniquely defined by the field F . Of course it must contain the following set of field elements N = {0, 1, 1 + 1, 1 + 1 + 1, . . .} 359 as well as the additive inverses of elements of this set, −N . There are two cases: (1) N is a finite set, and (2) N is an infinite set. In the former case, there exists a smallest positive integer n for which 1 + 1 + · · · + 1 with n summands is the zero element of F . In the latter case, there exists no such integer n. (In language that we have understood earlier,we are discussing the order of the element 1 in the additive group (F, +) of the field F . In the one case 1 has finite order n and in the second case, “1 has infinite order”. The reader should recognise that F is an integral domain, then the two cases being discussed simply describe the characteristic of F . In case (1) n must be a prime number. But that means that the set N is itself a field, so N = P ' Z/pZ, where, p is a prime number. In case (2) M := N ∪ −N is a copy of the ring of integers contained in P . It follows that F contains the set M −1 , of multiplicative inverses of non-zero elements of M , and that M M −1 is isomorphic to the field of rational numbers. Lemma 226 In any field, the prime subfield is either the field Z/pZ or is the field of rational numbers. The algebraic definition of the prime subfield as the subfield generated by 1, forces an invariance under any isomorphism: Corollary 227 Any field isomorphism φ : K → L induces an isomorphism P (K) → P (L) between their prime subfields. 11.4.2 Finite fields Suppose now that F is a finite field – that is, one which contains finitely many elements. Then, of course, its prime subfield P is the field Z/pZ, where p is a prime. Then, of course, F is a finite-dimensional vector space over P – say, of dimension n. Then clearly, |F | = pn = q, a prime power. It now follows from Corollary 214 that the finite multiplicative group F ∗ of non-zero elements of F is a cyclic group of order q − 1. This means that F contains all the q − 1 roots of the equation xq−1 − 1 = 0 in P [x], and since F is generated by its set of non-xero elements, we have 360 Lemma 228 F is a splitting field of the polynomial xq − x in P [x]. It follows immediately, that F is uniquely determined up to P -isomorphism by the prime-power q alone. Corollary 229 For any given prime-power q, there is, up to isomorphism, exactly one field with q elements. One denotes any member of this isomorphism class by the symbol GF(q). 11.4.3 Automorphisms of finite fields Suppose F is a finite field with exactly q elements, that is, F ' GF(q). We have seen that char (F ) = p, a prime number, where, for some positive integer n, q = pn . We know from Theorem 237 of Subsection 10.4.1, that the “p-th power mapping” σ : F → F is a ring endomorphism. Its kernel is trivial since F contains no non-zero nilpotent elements. Thus σ is injective, and so, by the pigeon-hole principle, is bijective – that is, it is an automorphism of k F . Now suppose σ had order k. Then, αp = α for all elements α of F . But k as there are at most pk roots of xp − x = 0, we see that k = n. Thus σ generates a cyclic subgroup of Aut(F ) of order n. Now let τ be an arbitrary automorphism of F . Now the multiplicative group C of non-zero elements of F is cyclic of order q−1. Let θ be a generator of C. Then there exists an integer t , 0 < t < q − 1, such that τ (θ) = θt . Then, as τ is an automorphism, τ (θi ) = (θi )t . Since 0t = 0 = τ (0), it is true that the automorphism τ is a power mapping α → αt on F . Since αt + 1 = (α + 1)t = αt + tαt−1 + . . . + 1t we have that if t > 1, all of the q −1 elements α in C are roots of the equation Xt−1 k=1 t k ! xk = 0. (11.2) If the polynomial on the left were not identically zero, its degree t − 1 would be at least as large as the number of its distinct zeros which is at least q1 . That is outside the range of t. Thus each of the binomial coefficients in the equation is zero. Thus either p divides t or t = 1. 361 Now if t = pk ·s, where s is not divisible by p, we could apply the argument k of the previous paragraph to ρ := τ p , to conclude that s = 1. Thus t is a prime power and so τ is a power σ. We have shown the following: Corollary 230 If F is a the finite field GF(q), q = pn , then the full automorphism group of F is cyclic of order n and is generated by the p-th power mapping. remarks: The argument above, that Aut(F ) = hσi, could be achieved in one stroke by a principle to be introduced later, called the “independence of F -characters”. The above argument uses only the most elementary properties introduced so far. We see at this point that [F : P ] = |Aut(F )|. In the language of a later section, we would say that F is a “Galois extension” of P . It will have very strong consequences for us. For one thing, it will eventually mean that no irreducible polynomial of postive degree in GF(q)[x] can have repeated zeros in some extension field. 11.4.4 Polynomial equations over finite fields: the ChevalleyWarning Theorem The main result of this section is a side-issue in so far as it doesnt really play a role in the developement of the Galois theory as do the chain of sections before and after this one. However, it is represents a property of finite fields which is too important not to mention before leaving a section devoted to these fields. The student wishing to race on to the theory of Galois extensions and solvability of equations in radicals may safely skip this section, hopefully for a later revisitation. For this subsection, fix a finite field K = GF(q) of characteristic p and order q = p` . Now for any natural number k, let S(k, q) := X a∈K ak . the sum of the k-th powers of the elements of the finite field K of q elements. Clearly, if k = 0, the sum is S(0, q) = q = 0, since the field has characteristic 362 p dividing q. If k is a multiple of q − 1 distinct from 0, then as K ∗ is a cyclic group of order q − 1, S(k, q) = X K−{0} 1 = q − 1 = −1. Now suppose k is not a multiple of q − 1. Then there exists an element b in K ∗ with bk 6= 1. Since multiplying on the left by b simply permutes the elements of K, we see that S(k, q) = X ak = a∈K X a∈K (ba)k = bk S(k, q). So (bk − 1)S(k, q) = 0. Since the first factor is not zero, we have S(k, q) = 0 in this case. We summarize this in Lemma 231 0 if k = 0 S(k, q) = −1 if q − 1 divides k 0 otherwise Now fix a positive integer n. Let V = K (n) , the vector space of n-tuples with entries in K. For any vector v = (a1 , . . . , an ) ∈ V and polynomial p = p(x1 , . . . xn ) ∈ K[x1 , . . . , xn ], the ring of polynomials in the indeterminates x1 , . . . xn , we let the symbol p(v) denote the result of substituting ai for xi in the polynomial, that is, p(v) := p(a1 , . . . , an ). Now suppose p = xk11 xk22 · · · xknn is a monomial of degree Then X Yn = S(ki , q) = 0, v∈V i=1 ki < n(q − 1). P since at least one of the exponents ki is less than q − 1 and so by Lemma 231 introduces a factor of zero in the product. Since every polynomial p in K[x1 , . . . , xn ] is a sum of such monomials of degree less than n(q − 1) , we have Lemma 232 If p ∈ K[x1 , . . . , xn ] has degree less than n(q − 1), then S(p) := X v∈V 363 p(v) = 0. Now we can prove the following Theorem 233 (Chevalley-Warning) Let K be a finite field of order q and characteristic p. Suppose p1 , . . . , pm is a family of polynomials of K[x1 , . . . , xn ], the sum of whose degrees is less than n, the number of variables. Let X be the collection {v ∈ K (n) |pi (v) = 0, i = 1, . . . , m} of common zeroes of these polynomials. Then |X| ≡ 0 mod p. proof: Let R := i=1 (1 − pq−1 ). Then R is a polynomial whose degree i P (q − 1) deg pi is less than n(q − 1). Now if v ∈ X, then v is a common zero of the polynomials, so R(v) = 1. But if v ∈ / X, then for some i, pi (v) 6= 0, so pi (v)q−1 = 1, introducing a zero factor in the definition of R(v). Thus we see that the polynomial R induces the characteristic function of X. It follows that X S(R) := v∈V R(v) ≡ |X| modp. (11.3) Qn But since deg R < n(q − 1), Lemma 232 forces S(R) = 0, which converts (11.3) into the conclusion. Corollary 234 Suppose p1 , . . . pm is a collection of polynomials over K = GF(q) in the indeterminates x1 , . . . , xn , each with a zero constant term. Then there exists a common non-trivial zero for these polynomials. (Non-trivial means one that is not the zero vector of V .) In particular, if p is a homogeneous polynomial over K in more indeterminates than it’s degree, then it must have a non-trivial zero. 11.5 Separable Elements and Field extensions. 11.5.1 Separability. A polynomial p(x) ∈ F [x] is said to be separable if and only if its irreducible factors in F [x] each have all their roots distinct in a splitting field for p(x). Clearly a splitting field for p(x) contains a splitting field for any one of its factors (just take the subfield generated by the roots of that factor and note that the factor splits there). There is a simple way to tell whether a polynomial f (x) in F [x] is separable. 364 Lemma 235 The polynomial f (x) has distinct roots in its splitting field if and only if F [x]f (x) + F [x]f 0 (x) = F [x], where f 0 (x) represents the formal derivative of f (x). Equivalently, f (x) is separable if and only if f and its derivative f 0 are relatively prime in F [x]. proof: If f (x) = a0 + a1 x + · · · + an xn , then its formal derivative f 0 (x) is defined to be the polynomial a1 + 2a2 x + · · · + nan xn−1 . For any field K, the mapping ∂ : K[x] → K[x] which takes each polynomial to its formal derivative, is a K-linear mapping which satisfies the “product rule”: ∂(f · g) = f · ∂(g) + (∂(f )) · g. Now if E is a splitting field for f (x) in F [x], we have a factorization of f (x) in E[x] into linear factors: f (x) = (x − a1 )b1 (x − a2 )b2 · · · (x − am )bm , P where m is the number of distinct roots and bi = n, the degree of f (x). Then the product rule (iterated a number of times) renders the derivative as f 0 (x) = Xm b (x − ai )b1 −1 ( i=1 i Y j6=i (x − aj )bj ). Now if the roots of f (x) are not distinct, some exponent bi exceeds 1, and so (x − bi ) divides both f (x) and f 0 (x). Conversely, if there is a common non-unit factor to f (x) and f 0 (x), then (by the UFD property) there is a common linear factor (x − ai ), forcing bi > 1 so that the root ai is repeated in f (x). This lemma has a very nice application when f (x) is irreducible. A finite extension E of F is said to be separable if and only if, for each element a ∈ E, the irreducible monic polynomial ga (x) := Irr F (a) of F [x] which vanishes at a is separable. 365 11.5.2 p-th Power Mappings in Characteristic p Before beginning with a discussion of simple and repeated roots of polynomials we need to familiarize ourselves with certain properties of the p-th power mapping in domains and fields of characteristic p. So, for the moment, assume D is an integral domain of prime characteristic p. The p-th power mapping on D is the mapping which takes each element x to its p-th power. We immediately observe Lemma 236 If D is an integral domain of characteristic p, then the p-th power mapping is an isomorphism of D onto a subdomain Dp of D. In case D is a field, the image Dp is a subfield of D. Proof: For any two elements x and y of D, the “binomial theorem” implies (x + y)p = xp + y p , thanks to the fact that the intermediate binomial coefficients are all divisible by the prime p. Since we have (xy)p = xp y p for any x and y, we see that the p-th power mapping is a ring endomorphism of D. Since D, being a domain, has no non-zero elements whose p-th power is zero, the kernel of this endomorphism is zero. Thus the p-th power mapping is an injective endomorphism of D. But if D is a field, the inverse of xp is (x−1 )p . Thus the set Dp of p-th powers in D forms a subfield. We shall have need of the following technical result: Theorem 237 Suppose K is an extension of finite degree of the field F . Suppose further that char K = p, and that K = F K p – that is, K is the join of it’s two subfields F and K p . Then the p-th power mapping on K preserves F -linear independence of subsets. Proof: Since any F -linearly independant subset of K can be completed to a F -basis for K, it suffices to prove that the p-th power mapping preserves the linear independence of any F -basis for K. Suppose, then, that {a1 , . . . , an } is a F -basis of K. Let c be a generic element of K. Then c = α1 a1 + · · · + αn an , where αi ∈ F, for all i. Then cp = α1p ap1 + · · · + αnp apn , 366 so K p = F p ap1 + · · · + F p apn . Then K = F K p = F ap1 + · · · + F apn . Thus the p-th powers of the basis elements ai form a F -spanning set of K of size n. Since n = dimF K, these n spanning elements must be F -linearly independent. Armed with these results, we can now carry out our discussion of simple and repeated roots. 11.5.3 Simple Roots Let f (x) be any polynomial in F [x]. We have learned from the previous subsection that there exists a splitting field for f (x) and that any two such fields are F -isomorphic. This means that there is a field K, distinct elements a1 , . . . , ar , and positive integers n1 , . . . , nr such that 1. K = F (a1 , . . . , ar ), and 2. there is a complete factorization f (x) = (x − a1 )n1 (x − a2 )n2 · · · (x − ar )nr in K[x]. . The elements ai are called the roots of the equation f (x) = 0 in K and are uniquely determined in any extension field of K. The number ni is called the multiplicity of the root ai in f (x). Obviously the collection of multiplicities {n1 , . . . , nr }, is uniquely determined by the polynomial f (x). But it is also true that the multiplicity ni attached to the root ai is unique up to exchanging ai by some zero bi of the same irreducible monic polynomial in F [x] that ai satisfies. Suppose K1 and K2 are two splitting fields for f (x) over F . Then there is a F -isomorphism σ : K1 → K2 which extends to a ring isomorphism σ ∗ : K1 [x] → K2 [x]. 367 This carries the factorization f (x) = to f (x) = Yr i=1 Yr i=1 (x − ai )ni , (x − σ(ai ))ni in K2 [x]. Then ai and σ(ai ) are both zeros of the same irreducible polynomial p(x) over F and possess the same multiplicity, although they may lie in distinct splitting fields. In particular we see that if f (x) is irreducible in F [x], then any two of its zeros possess the same multiplicity in it. Note that “multiplicity” of a root has been defined relative to the polynomial of which it is a zero. Thus 1 has multiplicity one in x−1 but multiplicity two in x2 − 2x + 1. Most of the time we will be interested in this notion of multiplicity of a zero relative to an irreducible polynomial. A root ai of f (x) is said to be a simple zero of f (x) iff and only if it has multiplicity ni equal to one. Any root of f (x) = 0 of multiplicity greater than one is called a repeated zero of f (x). The following very elementary properties concerning roots and their multiplicities are useful: Lemma 238 Let K contain the subfield F . 1. Suppose f (x) and g(x) are polynomials in F [x] which share a common root a in K. Then they possess a non-unit greatest common divisor in F [x] – that is, there exists a polynomial of degree at least one in F [x] dividing both polynomials f and g. 2. If f (x) is irreducible in F [x] and f shares a root with g ∈ F [x], then f divides g. 3. If f (x) is irreducible in F [x], then all of its roots have the same multiplicity in f (x). 4. Suppose f (x) and g(x) are polynomials in F [x] with the property that in some splitting field for f (x), every root of f (x) is simple and moreover is a root of g(x). Then f (x) divides g(x) in F [x]. 368 Proof: 1. If the conclusion were false there would exist b(x) and c(x) in F [x] such that b(x)f (x) + c(x)g(x) = 1. This is impossible when one applies the homomorphism F [x] → K obtained by substituting a for x. 2. This is immediate from 1. 3. This was discussed above. 4. Let E be a splitting field for f (x)g(x). Then in E[x], f (x) = (x − a1 ) · · · (x − ad ) where the roots ai are all simple and distinct. By hypothesis each ai is a root of g(x), so g(x) = (x − a1 ) · · · (x − ad )h(x), a factorization in E[x]. Thus f (x) divides g(x) in E[x]. We claim that this means f (x) divides g(x) in F [x]. For if not, the Euclidean algorithm yields g(x) = q(x)f (x) + r(x), where 1 ≤ deg r(x) < deg f (x). But then each ai , i = 1, . . . , d is a root of r(x). That is impossible since the ai are pairwise distinct. In any ring of polynomials F [x], there is a mapping ∂ : F [x] → F [x] which takes each polynomial f (x) = a0 + a1 x + · · · + an xn , to it’s formal derivative ∂(f ) = f 0 (x) := a1 + 2a2 x + · · · + nxn−1 . Without any recourse to analytic arguments it can directly be proved that 1. The mapping ∂ is F -linear, 2. For any two polynomials f (x) and g(x) in F [x], ∂(f g) = f 0 (x)g(x) + f (x)g 0 (x). We now can prove this useful theorem: 369 Theorem 239 Let f (x) be an irreducible polynomial in F [x]. Then f (x) has a repeated root if and only if its formal derivative is the zero polynomial – that is, f 0 (x) = 0. Proof: Suppose first that a is a repeated root of f (x). This means that in K[x], where K is a splitting field for f (x) over F , one obtains a factorization f (x) = (x − a)m g(x), where g(a) 6= 0, and m > 1. Then f 0 (x) = m(x − a)m−1 g(x) + (x − a)m g 0 (x) has a for a root, and so by Lemma 238, part 3, f (x) divides f 0 (x). But since f 0 (x) has smaller degree than f (x), this implies f 0 (x) = 0. Now assume f (x) is irreducible in F [x] and that f 0 (x) = 0. Then it is easy to see F must have characteristic p, a rational prime, and that f (x) = g(xp ) for some irreducible polynomial g(x) in F [x]. Now in K, the p-th power of a root of f (x) is a root of g(x). Since the p-th power mapping is injective (see Lemma 236), f (x) cannot have more roots than g(x). But g(x) has smaller degree than f (x), and so the number of roots of f (x) is smaller than its degree. It follows that some must be repeated. 11.5.4 Separable Elements and Separable Field Extensions Again K is an extension of field F . We say that an element a in K is separable over F if and only if a is a simple root of IrrF (a), the monic irreducible polynomial of F [x] of which it is a root. Note that if a is already an element of F then its monic irreducible polynomial is IrrF (a) = x − a, of degree one. Clearly a is a simple root of this polynomial, so elements of F are automatically separable over F . There is a second observation: If L is a subfield of K containing F and a is separable over F then a is separable over L. The argument is this: Set f (x) := IrrF (a) and g(x) := IrrL (a). Then there is a factorization f (x) = g(x)h(x) in L[x]. Let E be the splitting field of f (x) over K. and let EL and EF be the splitting fields of f (x) over L and F , respectively. All the roots of f (x) in either EK or EL lie in EF and are distinct. But all the roots of g(x) are roots of f (x) and are pairwise distinct. Thus a is seperable over L. We record these observations: Lemma 240 Suppose K is an extension of the field F . 370 1. Every element of F is separable over F . 2. If L is a subfield such that F ⊆ L ⊆ K, and a ∈ K is separable over F , then a is also separable over L. We say that K is separable over F if and only if every element of K is separable over F . This definition also has several immediate consequences: Lemma 241 Suppose K is an extension of F . 1. If char F = 0, then K is separable over F . 2. Suppose L is a subfield with F ⊆ L ⊆ K. (a) If K is separable over F then L is separable over F . (b) If K is separable over F then K is also separable over L. Proof: 1. Suppose char F = 0. Choose a ∈ K and set f (x) := IrrF (a). Since f (x) is monic of degree at least one, and the characteristic is zero we cannot have f 0 (x) = 0. Then by Theorem 239, f (x) has no repeated roots, so a is a simple root of f (x). This means a is separable over F , and since a was arbitrary, K is separable over F . 2. Part (a) is obvious and part (b) follows from Lemma 240, part 2. We shall eventually show that if F ⊆ L ⊆ K is a tower of fields for which K is separable over L and L is separable over F , then K is separable over F . For the moment even the relation between separable elements and separable field extensions is extremely one-sided. We do not even know that if a is an element of K separable over F , that the subfield F (a) is separable over F . As one can see from the above lemma, most of the trouble occurs in prime characteristic. In fact it occurs precisely under the initial hypotheses of the following theorem: Theorem 242 Suppose F ⊆ K are fields of characteristic p, a prime. Suppose further that there exists an element a in K and an integer e such that e ap = b ∈ F . e r 1. If p(x) = IrrF (a), then xp − b = (p(x))p for some integer r ≥ 0. e 2. If b is not in F p then xp − b is irreducible and so is equal to IrrF (a). 371 3. If b is not in F p and moreover a is separable, then a is already an element of F . e Proof: Set f (x) = xp − b and let p(x) be any irreducible monic factor of f (x) in F [x]. Let c be a root of p(x) and set L = F (c). Then f (c) = 0, and e so b = cp . Using the additive property of the p-th power map in L[x], we obtain e f (x) = (x − c)p , as a factorization in L[x]. Now suppose q(x) is any other monic irreducible divisor of f (x) in F [x]. Then factoring q(x) in L[x] we see that q(x) is some positive power of (x − c) and so q(c) = 0. Thus we see that p(x) and q(x) are irreducible in F [x] and share a common root in an extension field of F . It follows from Lemma 238 that they divide each other. Since they are monic and irreducible, they are equal. It thus follows that f (x) = p(x)m . Since m divides deg f (x) = pe , we see that m = pr , for some non-negative integer r. Thus part 1. holds. Now suppose further, that b is not in F p . Let a0 = p(0), the constant r term of the polynomial p(x). Clearly a0 ∈ F . Then from Part 1., ap0 = −b. Since b is not the p-th power of an element of F , we must have r = 0. Thus f (x) = p(x) is irreducible in F [x] and so Part 2. has been established. Now assume not only that b is not in F p but that the element a is separable over F . Then by Theorem 239, f 0 (x) 6= 0 where now f (x) = IrrF (a) by Part e e 2. But f (x) = xp − b, so f 0 (x) = pe xp −1 6= 0. This forces e = 0. Thus e a = ap = b ∈ F as desired. The proof is complete. We now can deduce a useful result: Theorem 243 (Separability criterion for fields of prime characteristic) Let K be any algebraic field extension of the field F , where F (and hence K) have prime characteristic p. 1. If K is a separable extension of F , then K = F K p . (Note that the degree [K : F ] need not be finite here.) 2. If K is a finite extension of F such that K = F K p , then K is separable over F . 372 Proof: Part 1. Assume K is separable over F . Then K is separable over its subfield L := F K p (Lemma 241). Suppose there exists an element a in K that is not in L. Then its p-th power b = ap would be an element of L. If b were not the p-th power of an element of L, then (noting that a is separable over L) Theorem 242, Part 3.(with L in the role of F in the Theorem) would imply a ∈ L, a contradiction. Thus we must assume b = cp for some c ∈ L. Then a is a root of xp − b = (x − c)p ∈ L[x], and so IrrL (a) = x − c, whence a = c ∈ L, a contradiction once again. We must conclude that no such element a exists, and so L = K as we were to prove. Part 2. Now assume [K : F ] is finite, and that K = F K p . Suppose, by way of contradiction that a is an element of K which is not separable over F . Then if f (x) := IrrF (a), Theorem 239 implies f 0 (x) = 0. It follows that m f (x) = a0 + a1 xp + · · · + am xp = g(xp ), (11.4) for g(x) irreducible in F [x]. Now from (11.4), the elements {1, ap , a2p , . . . , amp } are linearly dependent over F . On the other hand, if {1, a, a2 . . . . , am } were F -linearly dependant, then there would exist a polynomial p(x) in F [x] of degree at most m having element a for a zero. But in that case f (x) = IrrF (a) would divide p(x), which is impossible as deg p(x) ≤ m < pm = deg f (x). Thus we must conclude that the elements {1, a, . . . , am } are linearly independant over F . But this too is impossible, for in that case the p-th power mapping takes this linearly independent set to one that is F -linearly dependent, against Theorem 237 and the hypothesis that K = F K p . Thus no non-seperable element exists in K, so K is separable over F . Corollary 244 (Separability of simple extensions) Fix a field F of prime characteristic p. 1. Suppose a is separable over F . Then F (a) is a separable field extension of F , and F (a) = F (ap ). 373 2. Conversely, if a is algebraic over F , and F (ap ) = F (a) where p 6= 0 is the characteristic of F , then a is separable over F . Proof: Part 1. Suppose first that a is separable over F and set L := F (ap ). We first show that F (a) = F (ap ). Suppose a is not in L, so that f (x) := IrrL (a) has degree greater than one. r Then clearly ap ∈ L = F (ap ), so by Theorem 242, Part 1, xp − ap = (f (x))p has degree p and deg f (x) is a non-identity divisor of this prime. Thus f (x) = xp − ap , and b = ap is not a p-th power of an element of L. But by hypothesis a is separable over F , and so by Theorem 242, Part 3., a is an element of L, a contradiction. Thus we must have a ∈ L and so F (a) = F (ap ). Now as every element of F (a) is a polynomial of F [x] evaluated at a, we have (F (a))p ⊆ F · F (ap ) and F · (F (a))p = F · F (a) = F (a). Then F · (F (a))p = F · F (ap ) = F · F (a) = F (a). (11.5) By Theorem 243, Part 2., F (a) is separable over F . Part 2. Conversely, F (a) = F (ap ) implies the full equation (11.5), and so again F (a) is separable over F as before. In particular, element a is separable over F . Corollary 245 (Transitivity of separability among finite extensions) Suppose K is a finite separable extension of L and that L is a finite separable extension of F . Then K is a finite separable extension of F . Proof: That [K : F ] is finite is known from the first section of this chapter, so we only need to prove the separability of K over F . By Lemma 241, Part 1., such a conclusion follows if F has characteristic zero. So we assume char F = p. Now by Theorem 243, the separability hypotheses yield K = LK p and L = F Lp . Since Lp ⊆ K p , we obtain K = (F Lp )K p = F K p . Then by part 2. of Theorem 243, (noting that [K : F ] is finite) K is separable over K. Corollary 246 Suppose K is an extension of F containing elements a1 , . . . , an each of which is separable over F . Then the subfield F (a1 , . . . , an ) is separable over F . 374 Proof: This is an easy induction using Corollaries 244 and 245. Corollary 247 Let K be an arbitrary extension of F . Then Ks = {a ∈ K|a is separable over F } is a subfield of K containing F . Proof: By Lemma 240 F ⊆ Ks . It remains to show that Ks is a subfield of K. Suppose a and b are arbitrary elements of Ks . Then by Corollary 246, F (a, b) is a separable extension of F so F (a, b) ⊆ Ks . But this means a ± b, ab, and, if b 6= 0, ab−1 , all lie in Ks . Thus Ks is a subdomain of K closed under taking multiplicative inverses of non-zero elements and so is a subfield of K. This field Ks is called the separable closure of F in K. 11.6 Galois Theory 11.6.1 Galois field extensions Let K be any field, and let k be a subfield of K. The k-isomorphisms of K with itself are called k-automorphisms of K. Under composition they form a group which we denote by Gal(K/k). The group of all automorphisms of K is denoted Aut(K), as usual. Suppose now that G is any group of automorphisms of the field K. The elements fixed by every automorphism of G form a subfield called the fixed subfield of G and denoted Fix(G). If G ≤ Gal(G/k) then clearly k is a subfield of Fix(G). Now let S(G) be the poset of all subgroups of G, and let S(K/k) be the poset of all subfields of K which contain k. In both cases the partial ordering is containment. Now, the mapping φ : S(Gal(K/k)) → S(K/k), which takes each subgroup G of Gal(K/k) to Fix(G), reverses inclusion. That is, if H ≤ G, then Fix(G) ≤ Fix(H). Similarly, the mapping gal : S(K/k) → S(Gal(K/k)), 375 which takes each field L, with k ≤ L ≤ K, to Gal(L/k), also reverses inclusion. It follows that the composition φ ◦ gal : S(K/k) → S(K/k), is a closure operator on the poset of fields between k and K. The Galois theory is concerned with the subposet of “closed” subfields. Accordingly, we say that K is a Galois extension of L, (or that K is Galois over L), if and only if L is the fixed subfield of Gal(K/L). Now we can always force this situation by “starting from the top down”. That is, for any subgroup G of Gal(K/k), we can form the fixed field L = Fix(G), and observe the following: 1. k ≤ L, 2. G ≤ Gal(K/L), and 3. K is Galois over L. 11.6.2 Independence of linear characters Let K be any field and let G be any group. A linear K-character of the group G is a group homomorphism λ : G → K ∗ of the group G into the multiplicative group K ∗ of non-zero elements of the field K. Thus a K-character is a very special vector in the left K-vector space Hom (G, K). Theorem 248 (Independence of the linear K-characters of a group) For any group G and field K, the linear K-characters of G are K-linearly independent in Hom (G, K). Proof: It suffices to show that any finite set of linear K-characters is linearly independent. Suppose, by way of contradiction that {σ1 , . . . σn } is a collection of linear K-characters of G of smallest cardinality n, forming a K-linearly dependent set. Since no linear K-character is the zero map, a single such character comprises a linearly independent set. So n > 1. Then there exist scalars a1 , . . . , an in K, not all zero, such that Xn ( a σ )(g) i=1 i i = 0, for all g ∈ G. 376 (11.6) Note that by the minimality of n, each scalar ai is non-zero. Since σ1 and σn are distinct characters, there exists a group element h at which they differ — i.e., σ1 (h) 6= σn (h). Now, from the definition of linear K-character, we must have σi (gh) = σi (g)σi (h), for all i. Then 0= Xn a σ (gh) = i=1 i i Xn a σ (g)σi (h), i=1 i i (11.7) for all g ∈ G. But also, just multiplying equation (11.6) throughout by σ1 (h) on the right, one obtains 0= Xn a σ (g)σ1 (h). i=1 i i (11.8) Now, subtracting equation (11.8) from equation (11.7), we obtain 0= Xn a σ (g)(σi (h) i=2 i i − σ1 (h)), for all g ∈ G. (11.9) Note that the linear combination on the right side of equation (11.9) is not zero, since the last coefficient an (σn (h) − σ1 (h)) is non-zero. But then (11.9) reveals that {σ2 , . . . , σn } is a linearly dependent set, against the minimal choice of n. We can parley this into a lower bound of the degree of an extension in terms of the number of k-isomorphisms of one extension of k into another. To be more precise, let ` and k be fields and let σ : ` → k be an isomorphism of fields. Next suppose L and K are fields containing ` and k, respectively. An injective ring morphism L → K which extends σ is called a `-isomorphism extending σ, or just a `-isomorphism if σ is fixed by the context. Theorem 249 Let L and K be fields containing the fields ` and k, respectively. Suppose σ : ` → k is a field isomorphism and that {σi : L → K|i = 1, . . . , n} is a finite collection of n distinct `-isomorphisms of L into K, each extending σ. Then [L : `] ≥ n. Proof: Suppose, by way of contradiction that {b1 , . . . , bm } is a `-basis for L, where m < n. Consider now the n-by-m matrix B= σ1 (b1 ) σ2 (b1 ) ... σn (b1 ) . . . σ1 (bm ) . . . σ2 (bm ) ... . . . σn (bm ) 377 . The rows of B are n vectors in the m-dimensional space of row vectors K , and, since m < n, they form a K-linearly dependent set of vectors. Thus there are scalars a1 , . . . , an of K, not all zero, forming a linear combination of these column vectors equal to zero – that is (a1 , . . . , an ) × B = 0 ∈ M 1×m . Thus for each j = 1, . . . , m indexing the basis elements, 1×m Xn a σ (b ) i=1 i i j = 0. Since the `-span of the bj is L, we see that for every element v in L, Xn ( a σ )(v) i=1 i i = 0 ∈ K. That means ni=1 ai σi is the zero mapping L → K – that is, the set {σ1 , . . . , σn } is a K-linearly dependent set of elements of Hom (L, K). Now observing that the σi induce distinct linear K-characters of the multiplicative group L∗ , of non-zero elements of L, we have contradicted Theorem 248. Thus we must have [L : `] ≥ n. P Setting L = K, ` = k and σ = 1k in this theorem, we have Corollary 250 Suppose K is an extension of k. Then |Gal(K/k)| ≤ [K : k]. 11.6.3 Group orders and degrees in Galois extensions Theorem 251 Let K be a field and let G be a finite subgroup of Aut(K). If k is the fixed subfield of G, then the following is true: 1. K is a normal separable extension of k. 2. |G| = [K : k]. 3. G = Gal(K/k). Proof: First we show that K is normal and algebraic over k. Let a ∈ K. Since G is finite, we can form the polynomial f (x) := Y g∈G 378 (x − ag ), all of whose zeroes are in K. The coefficent of each power of x in f (x) is clearly G-invariant, and so belongs to k. Thus f (x) is a polynomial in k[x] having a as one of its zeroes, It follows that a is algebraic over k, and that if p(x) = Irrk (a), then p(x) divides f (x). Thus the zeroes of p(x), being among the zeroes of f (x), are all found in K. In short, K is algebraic over k, and if an irreducible polynomial has at least one zero in K, then it has all its zeroes in K. Thus K is a normal algebraic extension of k. We leave the separability till later. Now we list the elements of G as {σ1 , . . . , σn }. Suppose by way of contradiction that [K : k] ≥ n+1 and let {a1 , . . . , an+1 } be a k-linearly independent subset of K. We now form the n-by-(n + 1) matrix B= σ1 (a1 ) σ2 (a1 ) ... σn (a1 ) σ1 (a2 ) σ2 (a2 ) ... σn (a2 ) . . . σ1 (an+1 ) . . . σ2 (an+1 ) ... . . . σn (an+1 ) . We let V be the set of all column vectors {x1 , . . . , xn+1 }T such that B · {x1 , . . . , xn+1 }T = 0. Clearly V is a vector space. In fact one may regard V as the subspace of K (n+1)×1 consisting of all vectors whose (“dot”) product with each row vector of B is zero. Since there are only n row vectors, the space V is non-trivial. Now for each vector v = {x1 , . . . , xn+1 }T in V , and element g ∈ G, set gv := {g(x1 ), . . . , g(xn+1 )}T . Then if rj ∈ K 1×(n+1) is the j-th row of B, we have rj · v = 0, for each j – that is, Xn+1 i=1 σj (ai )xi = 0. Now left multiplication by g −1 permutes the elements of G, so, given j, there is an index π(j) such that g −1 σj = σπ(j) . But then for all j, the “dot product” rj · gv is Xn+1 Xn+1 σ (a ) · g(xi ) = g( i=1 j i i=1 Xn+1 = g( i=1 g −1 σj (ai ) · xi ) σπ(j) (ai ) · xi ) = g(rπ(j) · v) = g(0) = 0. Thus, if v ∈ V , then gv ∈ V . 379 Now for each vector w in K (n+1)×1 let the weight of w be the number of non-zero entries in w, and denote it as wt(w). In the subspace V choose a non-zero vector u of minimal possible weight. Multiplying through by a scalar from K, and rearranging the indices of the rows if necessary, we may assume u = (1, c2 , c3 , . . . cr , 0, . . . , 0)T , where each ci is non-zero. If all ci were in k, then we should have Xn+1 σj ( ac)= i=1 i i Xn+1 σ (a )c = rj · u = 0 ⇒ i=1 j i i X ai ci = 0, against the fact that the ai were k-linearly independent. Thus we may assume that r > 1 and that cr is not in k. Since k is the fixed field of G, this means there is a σ ∈ G such that σ(cr ) 6= cr . Now consider σ(u) − u. This vector lies in V , has zero in its first coordinate, and in all coordinates beyond the r-th. Thus wt(σ(u) − u) < wt(u). So, from the fact that u was chosen non-zero with minimal weight, we see that σ(u) − u is zero. But that is impossible since its r-th coordinate is non-zero. This contradiction proves that [K : k] ≤ |G|. But now Theorem 249 forces |G| ≤ [K : k], so equality holds. To show that K is separable over k, we must show that in case k has characteristic p, a prime, that kK p = K (see Theorem 29). Set L = kK p . Clearly L is a G-invariant subfield of K. The group G thus induces a group of automorphisms of of L, through a group homomorphism f : G → Gal(L/k). The kernel of f is GL := G ∩ Gal(K/L). We claim that Gal(K/L) = {1K }, the identity subgroup. Let g be any element of Gal(K/L) and let a be an element of K. If a ∈ L, then g fixes a. If a ∈ K − L, then ap = b ∈ L. Thus a is a zero of xp − b ∈ L[x]. But in K[x], xp − b = (x − a)p . That means a is the unique zero of xp − a ∈ L[x] that the field K has to offer. For that reason a is fixed by g. Thus our claim that Gal(K/L) is trivial, is true, and so f is an injection. Thus G can be regarded as a subgroup of Gal(L/k). But since L is G-invariant, it is normal over k, and so Theorem 249 applies to yield, |G| ≤ [L : k]. But now the equality in part 2 forces [K : k] = |G| ≤ [L : k] ≤ [K : k], 380 so [L : k] = [K : k] and K = L. The proof is complete. Corollary 252 Suppose K is a separable normal extension of k of finite degree. Then K is a Galois extension of k, and |Gal(K/k) = [K : k]. Proof: Let G = Gal(K/k). In view of Theorem 251, it suffices to show that Fix(G) = k. But suppose θ is any element of K − k. Then as [K : k] is finite, θ is algebraic over k, and so f (x) := Irrk (θ) exists and has degree at least two. Since K is separable over k, there are at least two distinct zeros, θ = a1 and a2 of f (x) in K. Thus there is a k-isomorphism: k(a1 ) → k(a2 ) which, by Corollary 223, can be extended to an automorphism g of the splitting field K. Thus g is an element of Gal(K/k) which moves θ to a2 6= θ. It follows that Fix(G) = k, and the proof is complete. Theorem 253 (The Fundamental Theorem of Galois Theory) Suppose K is a finite separable normal extension of the the field k. Let G := Gal(K/k). Then the following hold: 1. (the Galois Connection) There is a one-to-one order-reversing correspondence between the poset S(G) of all subgroups of G under inclusion and the poset S(K/k) of all subfields L with k ≤ L ≤ K ordered by inclusion, which takes every subgroup H of G to the subfield Fix(H). We write this Fix : S(G) → S(K/k), The inverse mapping is Gal : S(K/k) → S(G), which takes each intermediate field L to Gal(K/L). Thus the two poset order-reversing bijections, Fix and Gal, form a Galois connection. 2. (The connection between upper field indices and group orders) For each intermediate field L of S(K/k), one has [K : L] = |Gal(K/L)|. 3. (The connection between groups indices and lower field indices) If H ≤ G, then [Fix(H) : k] = [G : H]. 4. (The correspondence of normal fields and normal subgroups). 381 (a) Suppose L ∈ S(K/k) is normal over k. Then Gal(K/L) is a normal subgroup of G = Gal(K/k). (b) Suppose N is a normal subgroup of G. Then Fix(N ) is a normal extension of k. 5. (Induced groups and normal factors) Suppose L is a normal extension of k contained in K. Then Gal(L/k) is isomorphic to the factor group Gal(K/k)/Gal(K/L). Proof: Thanks to our preparation, the statement of this Theorem is almost longer than its proof. Lets start with part 1. Since, by Lemma 241 part 2(b), K is finite normal separable over every intermediate field L, k ≤ L ≤ K, then by Corollary 252 and Theorem 251 we always have that |Gal(K/L)| = [K : L]. By hypothesis both sides of this equation are natural numbers. A great deal follows from this. First, it implies that the mapping Gal : S(K/k) → S(G) is injective. For if H = Gal(K/L1 ) = Gal(K/L2 ), then H fixes the subfield L3 generated by L1 and L2 , and certainly H = Gal(K/L3 ). Then we see that |H| = |Gal(K/L3 )| = |Gal(K/L1 )| = [K : L3 ] = [K : L1 ]. Since L1 ≤ L3 we are forced to conclude that L3 = L1 so L1 = L2 . Thus Gal is an injective mapping. On the other hand, for each intermediate field L, |Gal(K/k)| = [K : k] = [K : L][L : k] = |Gal(K/L)| · [L : k]. So [Gal(K/k) : Gal(K/L)] = [L : k]. Similarly, suppose L := Fix(H1 ) = Fix(H2 ). Then if H is the subgroup of G generated by H1 and H2 , we have L = Fix(H). Now K is finite normal and seperable over L, so, applying Theorem 251 to each group, |H1 | = |H2 | = |H| = [K : L]. Since H1 ≤ H, we have H = H1 = H2 . Thus Fix : S(Gal(K/k)) → S(K/k) is injective. 382 Now, the compositions Gal ◦ Fix : S(Gal(K/k) → S(Gal(K/k)) and Fix ◦ Gal : S(K/k) → S(K/k) are injective poset endomorphisms of finite posets and so by the Pigeon-Hole Principle, are isomorphisms. At this point we have proved all of parts 1, 2, and 3. Now suppose L is a subfield of K which is normal over k. Then L is invariant under the action of the Galois group, and Gal(K/L) is the kernel of the homomorphism Gal(K/k) → Gal(L/k) which for every k-automorphism of K, gives the k-automorphism induced on L. But by Corollary 223, every k-automorphism of L extends to one of K, so this homomorphism is onto. So Parts 4(a). and 5. hold. Finally, suppose N is a normal subgroup of Gal(K/k). Then Gal(K/k) leaves Fix(N ) invariant, from which it follows that Fix(N ) is a normal extension of k. The proof is complete. 11.7 Solvability of equations by radicals 11.7.1 Introduction Certainly one remembers that point in one’s education that one first encountered the quadratic equation. If we are given the polynomial equation ax2 + bx + c = 0, where a 6= 0 (to ensure that the polynomial is indeed “quadratic”), then the roots (or zeros) of this equation are given by the formula √ (−b ± b2 − 4ac)/2a. Later, perhaps, the formula is justified by the procedure known as “completing the square”. One adds some constant to both sides of the equation so that the left side is the square of a linear polynomial, and then one takes square roots. It is fascinating to realize that this idea of completing the square goes back at least two thousand years to the Near East and India. It means that at this early stage, there is the suggestion that there could be a realm where square roots could always be taken, and the subtlety that there are cases in which square roots can never be taken because the radican is negative. A second method of solving the equation involves renaming the variable x in a suitable way. One first divides the equation through by a so that it 383 has the form x2 + b0 x + c0 = 0. Then setting y := x − b0 /2, the equation becomes y 2 + c0 − (b0 )2 /4 = 0, from which extraction of the square root (if that can be done) yields y = q (4c0 − (b0 )2 )/2. The original zero named by x, is obtained by x = y + b0 /2, and the equations b0 = b/a and c0 = c/a. Certainly the polynomials in these equations were seen as generic expressions representing relations between some sort of possible numbers. But these numbers were known not to be rational numbers in many cases. One could say that solutions don’t exist in those√cases, or that there is a larger realm of numbers (for example fields like Q( 2) or the reals), and the latter view only began to emerge in the last three centuries. Of course the quadratic formula is not valid over fields of characteristic 2 since dividing by 2 was used in deriving and expressing the formula. The same occurs for the solutions of the cubic and quartic equations: fields of characteristic 2 and 3 must be excluded. Dividing through by the coefficient of u3 , the general cubic equation has the form u3 +bu2 +cu+d = 0, where u is the generic unkown root. Replacing u by x − b/3 yields the simpler equation, x3 + qx + r = 0. Next, one puts x = y + z to obtain y 3 + z 3 + (3yz + q)x + r = 0. (11.10) There is still a degree of freedom allowing one to demand that yz = −q/3, thus eliminating the coefficient of x in (11.10). Then z 3 and y 3 are connected by what amounts to the norm and trace of a quadratic equation: z 3 + y 3 = −r, and z 3 y 3 = −q 3 /27. One can then solve for z 3 and y 3 by the quadratic formula. Taking cube roots (assuming that is possible) one gets a value for z and y. The possible zeroes of x3 + qx + r are: y + z, ωy + ω 2 z, ω 2 y + ωz, 384 where ω is a complex primitive cube root of unity. The formula seems to have been discovered in 1515 by Scipio del Ferro and independently by Tartaglia. Just thirty years later, when a manuscript of Cardan published the cubic formula, Ferrari discovered the formula for solving the general quartic equation. Here one seeks the roots of a monic fourth-degree polynomial, whose cubic term can be eliminated by a linear substitution to yield an equation x4 + qbx2 + rx + s = 0. We would like to re-express the left side as (x2 − kx + u)(x2 + kx + v), for suitable numbers k u and v. From the three equations arising from equating coefficients, the first two allow u and v to be expressed in terms of rk, and a substitution of these expressions for u and v into the third equation produces a cubic in k 2 , which can be solved by the cubic formula. Then u and v are determined and so the roots can be obtained from two applications of the quadratic formula. It is hardly surprising, with all of this success concentrated in the early sixteenth century, that the race would be on to solve the general quintic equation. But this effort met with complete frustration for the next 270 years. Looking back we can make a little better sense of this. First of all where are we finding these square roots and cube roots that appear in these formulae? That can be answered easily from the concepts of this chapter. If we we wish to find the n-th root of a number w in a field F , we are seeking a root of the polynomial xn − w ∈ F [x]. And of course we need only form the field F [α] ' F [x]/p(x)F [x] where p(x) is any irreducible factor of xn − w. But in that explanation, why should we start with a general field F ? Why not the rational numbers? The answer is that in the formula for the general quartic equation, one has to take square roots of numbers which themselves are roots of a cubic equation. So in general, if one wishes to consider the possibility of having a formula for an n-th degree equation, he or she must be prepared to extract p-th roots of numbers which are already in an uncontrollably large class of field extensions of the rational numbers, or whatever field we wish to begin with. Then there is this question of actually having an explicit formula that is good for every n-th degree equation. To be sure, the sort of formula one has in mind would involve only the operations of root extraction, multiplication 385 and addition and taking multiplicative inverses. But is it not possible that although there is not one formula good for all equations, it might still be possible that roots of a polynomial f (x) ∈ F [x] can be found in a splitting field K which is obtained by a series of root extractions – put more precisely, K lies in some field Fr which is the terminal member of a tower of fields F = F0 < F1 < · · · < Fr , where Fi+1 = Fi (αi ) and αini ∈ Fi , for positive integers ni , i = 0, . . . , r − 1. In this case we call Fr a radical extension of F . Certainly, if there is a formula for the roots of the general n-th degree equation, there must be a radical extension Fr /F which contains a copy of the the splitting field Kf (x) for every polynomial f (x) of degree n. Yet something more general might happen: Conceivably, for each polynomial f (x), there is a radical extension Fr /F , containing a copy of the splitting field of f (x), which is special for that polynomial, without there being one universal radical extension which does the job for everybody. Thus we say that f (x) is solvable by radicals if and only if its splitting field lies in some radical extension. Then, if there is a polynomial not solvable by radicals, there can be no universal formula. Indeed that turned out to be the case for polynomials of degree at least five. This field-theoretic view-point was basically the invention of Evariste Galois. The next sections will attempt to describe this theory. 11.7.2 Roots of unity Any zero of xn − 1 ∈ F [x] lying in some extension field of F is called an n-th root of unity. If F has characteristic p, a prime, we may write n = m · pa where p does not divide m and observe that a xn − 1 = (xm − 1)p . Thus all n-th roots of unity are in fact m-th roots of unity in this case. Of course if α and β are n-th roots of unity lying in some extension field of K of F , then (αβ)n = αn β n = 1 so αβ is also an n-th root of unity. Thus the n-th roots of unity lying in K always form a finite multiplicative subgroup of K ∗ , which, by Corollary 214 is necessarily cyclic. Suppose now that n is relatively prime to the characteristic p of F , or that F has characteristic zero. Let K be the splitting field of xn − 1 over F . Then xn − 1 is relatively prime to its formal derivative nxn−1 6= 0. It follows that 386 K is a separable extension of F and that the n-th roots of unity in K form a cyclic group of order n. A generator of this group is called a primitive n-th root of unity. Since all other n-th roots are powers of some primitive n-th root of unity α, we see that K = F [α]. Clearly Gal(K/F ) induces a group of automorphisms of this cyclic group C. The subgroup inducing the identity automorphism fixes all n-th roots, whose F -span is K, and so is the identity group. Thus Gal(K/F ) is isomorphic to a subgroup of Aut(C), and so is abelian. We summarize these observations in Lemma 254 Let N be any positive integer and let F be any field. Write N = n · pk if F has prime characteristic p, or N = n, if F has characteristic zero. Let K be the splitting field of xN − 1 over F . Then the following hold: 1. K is the splitting field of xn − 1 over F . 2. K is separable over F and is a simple extension K = F [α], where α is a primitive n-th root of unity. 3. Gal(K/F ) is an abelian group. 11.7.3 Radical extensions Suppose r is a prime. We say that a field F contains all r-th roots of unity if and only if the polynomial xr −1 splits completely into linear factors in F [x]. Now if F has prime characteristic equal to r, we have xr −1 = (x−1)r and so it is true that F contains all r-th roots of unity in that case. Lemma 255 Let F be a field, and let r be a prime. Suppose K = F [α] where, αr ∈ F . Suppose F contains all r-th roots of unity. Then 1. the extension K is normal over F , and 2. the Galois group Gal(K/F ) is abelian. Proof: First suppose char F = r. Then α is the unique zero of xr − b where b = αr . Thus K is a splitting field for this polynomial, and Gal(K/F ) is the identity group. Now suppose char F 6= r. Then K is separable over F and by assumption, F contains all r-th roots of unity.. We may assume Gal(K/F ) is non-trivial, 387 so α ∈ K − F . As before, set αr = b. Let ζ be an r-th root of unity in F distinct from 1. Then {ζ i α|i = 0, . . . , r − 1}, is the complete set of all r zeros of p(x) := xr − b. Thus K is the splitting field for p(x) over F and so is normal over F . Now let σ and τ be any two elements of Gal(K/F ). Then σ(α) = ζ i α and τ (α) = ζ j α for some i and j. Then obviously both στ and τ σ take α to ζ i+j α. Thus the commutator [σ, τ ] fixes α and hence fixes K point-wise. So any two elements of Gal(K/F ) commute. At this point the reader is reminded that if K is a finite extension of F , then there exists a normal closure M of K over F , and that M is also a finite extension of F and can be assumed to contain K [The reader may recall that M can be formed by selecting an F -basis B = {α1 , . . . , αn } of K, Q and forming the splitting field M for the polynomial fB (x) := fi (x) where fi (x) := IrrF (α1 ). Then, as there is an F -isomorphism, K → M we may assume that M contains K without any loss of generality.] Lemma 256 Suppose K = F [α] where, for some prime r, αr = b ∈ F . Let M be a normal closure of K over F . Then Gal(M/F ) is solvable. Proof: If r = char F , F aleady contains all rth roots of unity and K is already its own normal closure M . Thus Gal(M/F ) = Gal(K/F ) = h1M i. So we may assume that r 6= char F and that α is not in F . Then IrrF (α) = f (x) is an irreducible factor of xr − b in F [x] of degree at least two. Moreover f (x) is separable over F , and so K is separable over F , so M contains two distinct roots of f (x), say α and α0 . Then ζ = α/α0 is a non-trivial r-th root of unity. Since r is a prime number, L = F [ζ] is normal over F . Now Gal(M/F ) induces an abelian group of F -automorphisms of L (Lemma 254) with the kernel of the action being Gal(M/L). But also, noting that M = F (ζ, α) = L(α), Lemma 255 implies that Gal(M/L) is also abelian. It follows that Gal(M/F ) is a metabelian group, and so is certainly solvable. Now we introduce several concepts motivated by the introduction to this section. Earlier we said that a field K was a radical extension of F if and only there exists a tower of fields F = F0 < F1 < · · · < Fm = K for which Fi+1 = F [αi ] where αini ∈ Fi for some positive integer ni . Consider, for example one of these generic intervals Fi+1 /Fi . Basically, it is an extension L[β] of a field L where β m ∈ L for some integer m. Suppose and m = pm0 where p is a prime and set γ = β p . Then we have a refined 388 chain of fields L < L[γ] < L[β] where β p ∈ L[γ]. This process may be repeated with γ in place of β. The end result would be a chain of subfields from L to L[α] each obtained by adjoining a p-th root of an element in its preceding field, where p is a prime depending on the position of the field Lin the chain. This means that in considering the chain {Fi } used to define K as a radical extension as above, by refining the chain of fields, we can assume that each exponent ni in that definition is a prime number. We shall always do so. For each polynomial p(x) in F [x], we say that the equation p(x) = 0 is solvable by radicals if and only if the splitting field E for p(x) over F lies in some radical extension of F . Our goal in this subsection is to prove that if p(x) = 0 is solvable by radicals then Gal(E/F ) is a solvable group. In the subsection following this, we shall prove the converse of this statement. Theorem 257 Suppose K is a radical extension of F , and let M be the normal closure of K over F . Then Gal(M/F ) is solvable. Proof: For any radical extension K of F there exists a chain of subfields F = F0 < F1 < · · · < F` = K where Fi+1 = Fi [αi ] where αpi ∈ Fi for some prime number pi , i = 0, . . . , ` − 1. The length of a shortest such chain will be christened the radical height of K over F and denoted rh (K). Note that if rh(K) = 0, then F = K = M and Gal(M/K) = 1 and is solvable in this case. The proof will proceed by induction on the radical height. To fix notation, we assume here that {Fi } is such a shortest chain so that ` = rh(K). Note that this means rh(F`−1 ) = ` − 1. Next Let M be a normal closure of K over F and set G := Gal(M/F ). Let Mi be the normal closure of Fi in M with respect to F . The fields Mi are all normal over F . They actually have a simple description. For each αi let αiG be the full G-orbit {σ(αi )|σ ∈ G}. Then G Mi = F (α1G ∪ · · · ∪ αi−1 ). Now F`−1 is a radical extension of F of radical height ` − 1. By induction, Gal(M`−1 /F ) is solvable. But since M and M`−1 are both normal over F , Gal(M/F )/Gal(M/M`−1 ) ' Gal(M`−1 /F ). 389 Thus to show that G is solvable, it remains only to show that the kernel G N = Gal(M/M`−1 ), is a solvable group. We have that M = M`−1 (α`−1 ). G Next α`−1 decomposes into N -orbits with representatives {β1 , . . . , βt }. Then Di := M`−1 (βiN ) is a normal extension of M`−1 , i + 1, . . . , t, and M = D1 D2 · · · Dt . Now βip` is in M`−1 , p` is a prime, and Di is the normal closure over M`−1 of M`−1 [βi ]. By Lemma 256, Gal(Di /M`−1 ) is solvable. Since both M and Di are normal over M`−1 , there are epimorphisms φi : Gal(M/M`−1 ) → Gal(Di /M`−1 ) with solvable images. The intersection of their kernels is ∩ti=1 Gal(M/Di ) which pointwise fixes D1 · · · Dt = M , and so is trivial. Thus Gal(M/M`−1 ) is a subdirect product of the solvable groups Gal(Di /M`−1 ) and so is itself solvable. The proof is complete. Corollary 258 Suppose p(x) is a polynomial in F [x] and that p(x) = 0 is solvable by radicals. Then if E is a splitting field for p(x) over F , Gal(E/F ) is a solvable group. Proof: By assumption there is a radical extension K of F containing E. Let M be the normal closure of K over F . By Theorem 257, Gal(M/F ) is solvable. Since E and M are both normal over F , there is an epimorphism of groups φ : Gal(M/F ) → Gal(E/F ). The conclusion follows. 11.7.4 Galois’ criterion for solvability by radicals In this subsection we will prove a converse to Corollary 258 for polynomial equations over a ground field in characteristic zero. (In characteristic p the converse statement is not even true.) In order to accomplish this we require two initial results, each bearing an unusual name. 390 Lemma 259 (Lemma on Accesory Irrationalities) Suppose K and E are both normal extensions of a field F , and let K1 be a larger field generated by K and E. Then K1 is normal over E and the group Gal(K1 /E) is isomorphic to a subgroup of Gal(K/F ). Proof: First K is the splitting field over F of some polynomial p(x) ∈ F [x]. Let K 0 be the splitting field of f (x) over E. As there is a k-isomorphism K → F , we may assume K 0 contains K. Then K 0 ' E(K) ' K1 – these are k-isomorphisms – so we see that K1 can be expressed as a splitting field over E. Thus K1 is normal over E. Since K is normal over F , it is invariant under Gal(K1 /F ). Under the action of its subgroup Gal(K1 /E) we induce k-automorphisms of K, so there is a group homomorphism ψ : Gal(K1 /E) → Gal(K/F ). But any k-automorphism of ker ψ fixes pointwise both E and K, and hence is the identity automorphism of K1 . Thus ψ is injective. The proof is complete. Now assume K is a Galois extension of F . Set G = Gal(K/F ). Then G is finite and Fix(G) = F . Then, for any element α ∈ K we define the norm of α as the right side of NK/F (α) := Y σ∈G σ(α). The “norm” function has several important properties listed in this Lemma 260 For any elements α and β in K, 1. NK/F (α) is fixed by G and so lies in F , 2. NK/F (α−1 ) = NK/F (α)−1 , and 3. NK/F (αβ) = NK/F (α)NK/F (β). Lemma 261 (Hilbert’s Theorem 90) Suppose K is a Galois extension of F and G = Gal(K/F ) is a cyclic group of order n with generator σ. An element α in K can be written in the form βσ(β −1 ) for some element β ∈ K if and only if NK/F (α) = 1. 391 Proof: First, if α = βσ(β −1 ), then NK/F (α) = 1 follows easily from the properties of the norm listed in the previous Lemma. Suppose now NK/F (α) = 1. We form a series of elements in K as follows: δ0 := α, δ1 := ασ(α), δ2 := ασ(α)σ 2 (α), . . . , and in general δj := Yj i=0 σ(α), j = 0, . . . n − 1, so that δn−1 = ασ(α) · · · σ n−1 (α) = NK/F (α) = 1. Note that from these definitions, ασ(δi ) = δi+1 , i = 0, . . . , n − 2. (11.11) Since the elements {1, σ, σ 2 , . . . , σ n−1 } are a set of K-linearly independent linear K ∗ -characters, the function Xn−2 σ n−1 + δσ i=0 i i , from K to K is not the zero function. Thus there is an element γ at which this function does not vanish, and we let β be the nonzero value assumed at γ. Thus: Xn−2 β := σ n−1 (γ) + i=0 δi σ i (γ) 6= 0. (11.12) Now using equation (11.11), one has σ(β) = α−1 [δ1 σ(γ) + · · · + δn−1 σ n−1 (γ)] + σ n (γ). But since σ n = 1, the last term on the right can be written as σ n (γ) = γ = α−1 δ0 · γ. Factoring out the α−1 , we recognize the equation just above to read σ(β) = α−1 β. So we see α = β · σ(β −1 ), as desired. Corollary 262 In this corollary the ground field F has characteristic 0. Suppose K is a Galois extension of F and that Gal(K/F ) is a cyclic group of order n. Assume further that F contains all n-th roots of unity. Then K = F (α) where αn ∈ F . In particular, K is a radical extension of F . 392 Remark: The condition that the fields in question have characteristic zero is absolutely necessary at this point. Suppose F = GF (2) and K = GF (4). Then K is a Galois extension of F and Gal(K/F ) is cyclic of order two. But K is not a radical extension of F . In fact F is closed under the operation of taking square roots. Proof: Let σ be a generator of G := Gal(K/F ). Since F contains n-th roots of unity and has characteristic zero, it contains a primitive n-th root of unity, say ζ. Since |G| = n, the norm of ζ is ζ n = 1. By Hilbert’s Theorem 90 (Lemma 261) there exists an element β such that ζ = β −1 σ(β), so σ(β) = ζβ. Then σ i (β) = ζ i β. for all integers i, and (since ζ is a primitive n-th root) are pairwise distinct elements for i = 0, 1, . . . , n − 1. The formula for i = 1 yields σ(β n ) = (σ(β))n = ζ n β n = β n . Thus β n is fixed by G = hσi and so is an element c in F . Then β is a root of xn − c = 0. Of course, the n elements ζ i β, 0 ≤ n − 1, comprise the full set of roots of this equation. Since G transitively permutes them, the polynomial xn − c is irreducible in F [x]. Thus [F [β] : F ] = n. But since K is Galois over F , [K : F ] = |G| = n, so K = F [β], as in the stated conclusion. Theorem 263 Suppose F is a field of characteristic zero and that f (x) is a polynomial in F [x]. Let K be the splitting field of f (x) over F . If Gal(K/F ) is a solvable group then the equation f (x) = 0 is solvable in radicals. Proof: Since F has characteristic zero, K is normal and separable over F and so is a Galois extension of F of degree n, say. Then G := Gal(K/F ) is a solvable group of order n. We first prove the result under the assumption that F contains all n-th roots of unity. As we are in characteristic zero, that means that F contains a primitive n-th root of unity. Since G is solvable, there is a composition series G = G0 > G1 > · · · Gr = 1. This means Gi+1 is a normal subgroup of Gi and the quotient Gi /Gi+1 is a cyclic group of prime order pi . Let Fi := Fix(Gi ). Then by the Fundamental Theorem of Galois Theory (Theorem 253) we have a tower of fields F = F0 < F1 < · · · < Fr = K, 393 where Fi+1 is normal over Fi of degree pi . Moreover Gal(Fi+1 /Fi ) ' Gi /Gi+1 is cyclic of order pi . Clearly pi divides n = [K : F ]. Thus Fi contains a primitive pi -th root of unity. Then by Corollary 262, Fi+1 = Fi [αi ] where αipi is an element of Fi . Thus Fi+1 is a radical extension of Fi . Since this assertion holds at each index i , K is a radical extension F , and f (x) = 0 is solvable by radicals. Now we drop the assumption that F contains all n-th roots of unity. Let J be the splitting field of xn − 1 over K. Then as the characteristic is zero, J contains a primitive n-th root ζ. Set E = F (ζ). Then E is a splitting field of xn − 1 over F . In particular E is a radical extension of F ; that is important. Since J = K[ζ], J is generated by E and K. Since both E and K are normal over F , the Lemma on accessory irrationalities (Lemma 259) informs us that Gal(J/E) is isomorphic to a subgroup of the solvable group Gal(K/F ). This tells us two things: (1) m := [J : E] = |Gal(J/E)| divides |Gal(K/F )| = n, and (2) Gal(J/E) is itself solvable. From (1), E contains m-th roots, and so, by the first part of this proof, (2) implies that J is a radical extension of E. But we have already emphasized that E is a radical extension of K. So J must be a radical extension of F . Since the splitting field K lies in this extension J, f (x) is solvable by radicals. The proof is complete. Corollary 264 (Galois’ solvability criterion) Assume f (x) ∈ F [x] where F has characteristic zero. Then f (x) = 0 is solvable by radicals if and only if the Galois group Gal(K/F ) of the splitting field K for f (x) is a solvable group. Proof: This is just a combination of Theorems 257 and 263. 11.8 Transcendental extensions 11.8.1 Simple transcendental extensions Suppose K ia some extension field of F . Recall that an element α of K is said to be algebraic over F if and only if there exists a polynomial p(x) ∈ F [x] such that p(α) = 0. Otherwise, if there is no such polynomial we say that α is transcendental over F and that F (α) is a simple transcendental extension of F . 394 Recall further that a rational function over F is any element in the field of quotients F (x) of the integral domain F [x]. Thus every rational function r(x) is a quotient f (x)/g(x) of two polynomials in F [x], where g(x) is not the zero polynomial, and f (x) and g(x) are relatively prime. For each element α ∈ K, we understand r(α) to be the element r(α) := f (α) · (g(α))−1 , provided α is not a zero of g(x). If α is transcendental over F , then r(α) is always defined. Lemma 265 If F (α) is a transcendental extension of F , then there is an isomorphism F (α) → F (x) taking α to x. Proof: Define the mapping : F (x) → F (α) by setting (r(x)) := r(α), for every r(x) ∈ F (x). Clearly is a ring homomorphism. Suppose r1 (x) = c(x) a(x) , r2 (x) = , b(x) d(x) are rational functions with r1 (α) = r2 (α). We may assume a(x), b(x), c(x) and d(x) are all polynomials with b(x)d(x) 6= 0. Then a(α)d(α) − b(α)c(α) = 0. Since α is transcendental, a(x)d(x) = b(x)c(x) so r1 (x) = r2 (x). Thus is injective and its image is a subfield of F (α) containing (x) = α. Thus is bijective, and so −1 is the desired isomorphism. Now suppose F (α) is a simple transcendental extension of F , and let β be any further element of F (α). By Lemma 265 there exist a coprime pair of polynomials (f (x), g(x)), with g(x) 6= 0 ∈ F [x], such that β = f (α)(g(α))−1 . Suppose f (x) = a0 + a1 x + · · · + an xn g(x) = bo + b1 x + · · · + bn xn 395 where n is the maximum of the degrees of f (x) and g(x), so at least one of an or bn is non-zero. Then f (α) − βg(α) = 0, so (a0 − βb0 ) + (a1 − βb1 )α + · · · + (an − βbn )αn = 0. Thus α is a zero of the polynomial hβ (x) := (a0 − βb0 ) + (a1 − βb1 )x + · · · + (an − βbn )xn in F (β)[x]. Claim: The polynomial hβ (x) is the zero polynomial if and only if β ∈ F . First, if β ∈ F , then hβ (x) = f (x) − βg(x) is in F [x]. If it were a non-zero polynomial, then its zero, α would be algebraic over F . Thus hβ (x) is the zero polynomial. On the other hand, if hβ (x) is the zero polynomial, each coefficent ai −βbi is zero. Now, since g(x) is not the zero polynomial, for some index j, bj 6= 0. Then β = aj · b−1 j , an element of F . The claim is proved Now suppose β is not in F . Then the coefficient (ai − βbi ) is zero if and only if ai = 0 = bi , and this cannot happen for i = n as n was chosen. Thus hβ (x) has degree n. So certainly α is algebraic over the intermediate field F (β). Since [F (α) : F is infinite, and F (α) : F (β)] is finite, [F (β) : F ] is infinite, and so β is transcendental over F . We shall now show that hβ (x) is irreducible in F (β)[x]. Let D := F [y] be the domain of polynomials in the indeterminate y, with coefficients from F . (This statement is here only to explain the appearance of the new symbol y.) Then F (y) is the quotient field of D. Since β is transcendental over F , Lemma 265, yields an F -isomorphism : F (β) → F (y) of fields which can be extended to a ring isomorphism ∗ : F (β)[x] → F (y)[x]. Then h(x) := ∗ (hβ (x)) = f (x) − yg(x) is a polynomial in D[x] = F [x, y]. Moreover, hβ (x) is irreducible in F (β)[x] if and only if h(x) is irreducible in 396 F (y)[x]. Now since D = F [y] is a UFD, by Gauss’ Lemma, h(x) ∈ D[x] is irreducible in F (y)[x] if and only if it is irreducible in D[x]. Now, D[x] = F [y, x] is a UFD and h(x) is degree one in y. So, if h(x) had a non-trivial factorization in D[x], one of the factors k(x) would be of degree zero in y and hence a polynomial of positive degree in F [x]. The factorization is then h(x) = f (x) − yg(x) = k(x)(a(x) + yb(x)), where k(x),a(x) and b(x) all lie in F [x]. Then f (x) and g(x) possess the common factor k(x) of positive degree, against the fact that f (x) and g(x) were coprime in F [x]. We conclude that h(x) is irreducible in F (y)[x], and so hβ (x) is irreducible of degree n in F (β)[x] as promised. We now have the following Theorem 266 (Lüroth’s Theorem) Suppose F (α) is a transcendental extension of F . Let β be an element of F (α) − F . Then we can write β = f (α)/g(α) where f (x) and g(x) are a pair of coprime polynomials in F [x], unique up to multiplying the pair through by a scalar from F . Let n be the maximum of the degrees of f (x) and g(x). Then [F (α) : F (β)] = n. In particular, α is algebraic over F (β). Remark: A general consequence is that if E is an intermediate field distinct from F – that is, F < E ≤ F (α) — then [E : F ] is infinite. In particular, F (α) − F contains no elements which are algebraic over F (a fact that has an obvious direct proof from first principles.) Lüroth’s theorem gives us a handle on the Galois group Gal(F (α)/F ) of a transcendental extension. Suppose now that σ is an F -automorphism F (α) → F (α). Then σ is determined by what it does to the generator α. Thus, setting β = σ(α), for any rational function t(x) of F , we have σ(t(α)) = t(β). Now in particular β = r(α) for a particular rational function r(x) = f (x)/g(x) where (f (x), g(x)) is a coprime pair of polynomials in F [x]. Since σ is onto, F (α) = F (β). On the other hand, by Lüroth’s Theorem, [F (α) : F (β)] is the maximum of the degrees of f (x) and g(x). Thus we can write aα + b , (11.13) β= cα + d where a,b,c and d are in F , a and c are not both zero, and ad−bd 6= 0 to keep the polynomials ax + b and cx + d coprime. Actually the latter condition implies the former, that a and c are not both zero. 397 Now let G = Gal(F (α)/F ) and set O := αG , the orbit of α under G. Our observations so far are that there exists a bijection between any two of the following three sets: 1. the elements of the group Gal(F (α)/F ), 2. the orbit O, and 3. the set LF (F ) of so-called “linear fractional” transformations, ω→ aω + b , a, b, c, d ∈ F, ad − cb non-zero, cω + d viewed as a set of permutations of the elements of O. A perusal of how these linear factional transformations compose shows that LF (F ) is a group, and that in fact, there is a surjective group homomorphism GL(2, F ) → LF (2, F ), defined by a b c d ! → [ the transformation z → az + b on set O ], cz + d where the matrix has non-zero determinant – i.e. is invertible. The kernel of this homomorphism is the subgoup of two-by-two scalar matrices γI, comprising the center of GL(2, F ). As the reader will recall, the factor group, GL(2, F )/Z(GL(2, F )) is called the “one-dimensional projective group over F ” and is denoted P GL(2, F ). Our observation, then, is that Gal(F (α)/F ) ' LF (F ) ' P GL(2, F ). Notice how the fundamental theorem of Galois Theory falls to pieces in the case of transcendental extensions. Suppose F is the field Z/pZ, the ring of integers mod p, a prime number. If α is transcendental over F , then [F (α) : F ] is infinite, while Gal(F (α)/F ) is P GL(2, p), a group of order p(p2 − 1) which, when p ≥ 5, contains at index two, a simple subgroup. 398 11.8.2 Transcendence Degree Let K be an extension of the field F 4 . We define a notion of algebraic dependence over F on the elements of K as follows: We say that element α algebraically depends (over F ) on the finite subset {β1 , . . . , βn } if and only if • α is algebraic over the subfield F (β1 , . . . , βn ). Theorem 267 Algebraic dependence over F is an abstract dependance relation on K in the sense of Chapter 1. proof: We must show that the relation of algebraic depence over F satisfies the three required properties of a dependance relation: reflexivity, transitivity and the exchange condition. Clearly if β ∈ {β1 , . . . , βn }, then β lies in F (β1 , . . . , βn ) and so is algebraic over it. So the relation is reflexive. Now suppose γ is algebraic over F (α1 , . . . , αm ), and each αi is algebraic over F (β1 , . . . , βn ). Then certainly, αi is algebraic over the field F (β1 , . . . , βn , α1 , . . . , αi−1 ). So, setting B := {β1 , . . . , βn }, we obtain a tower of fields, F (B) ≤ F (B, α1 ) ≤ F (B, α1 , α2 ) ≤ · · · ≤ F (B, α1 , · · · , αn ) ≤ F (β1 , . . . , αn , γ) with each field in the tower of finite degree over its predecessor. Thus the top field in the tower is a finite extension of the bottom field, and so γ is algebraic over F (B). We now address the exchange condition: we are to show that if γ is algebraic over F (β1 , . . . , βn } but is not algebraic over F (β1 , . . . , βn−1 }, then βn is algebraic over the field F (β1 , . . . , βn−1 , γ}.5 4 This subsection is drawn from notes of my colleague David Surowski, for which I am most grateful. 5 It is interesting to observe that in this context the Exchange Condition is essentially an “implicit function theorem”. The hypothesis of the condition, transferred to the context of algebraic geometry, says that the “function” γ is expressible locally in terms of {β1 , . . . , βn } ∂γ but not in terms of {β1 , . . . , βn−1 }. Intuitively, this means ∂β 6= 0 , from which we n anticipate the desired conclusion. 399 By hypothesis, there is a polynomial relation 0 = ak γ k + ak−1 γ k−1 + · · · + a1 γ + a0 (11.14) where ak is non-zero, and each aj is in F (β1 , . . . , βn ), and so, at the very least, can be expressed as a rational funtion, nj /dj , where the numerators nj and denominators dj are polynomials in F [β1 , . . . , βn ]. We can then multiply equation (11.14) through by the product of the di , from which we see that without loss of generality, we may assume the coefficients ai are polynomials – explicitly: ai = X b βj , j ij n bij ∈ F [β1 , . . . , βn−1 ], ak 6= 0.. (11.15) Then substituting the formulae of equation (11.15) into equation (11.14), and collecting the powers of βn , we obtain the polynomial relation 0 = pm βnm + pm−1 βnm−1 + · · · p1 βn + p0 , with each coefficient pt = Xk b γ s=0 st s , (11.16) (11.17) a polynomial in the ring F [β1 , . . . , βn−1 , γ]. Now we get the required algebraic dependence of βn on {β1 , . . . , βn−1 , γ} from equation (11.16), provided not all of the coefficients pt are zero. Suppose, by way of crontadiction, that they were all zero. Then as γ does not depend on {β1 , . . . , βn−1 }, having each of the expressions in equation (11.17) equal to zero forces each coefficient bst to be zero, so by equation (11.15), each ai is zero. But that is impossible since certainly ak is non-zero. Thus the Exchange Condition holds and the proof is complete. So, as we have just seen, if K is an extension of the field F , then the relation of being algebraically dependant over F is an abstract dependance relation, and we call any subset X of K which is independant with respect to this relation algebraically independant (over F ). By the basic development of Chapter 1, Section 3, maximal algebraically independant sets exist, they “span” K, and any two such sets have the same cardinality. We call such subsets of K transcendence bases of K over F , and their common cardinality is called the transcendence degree of K over F , and is denoted tr.deg(K/F ). 400 If X is a transcendence basis for K over F , then the “spanning” property means that every element of K is algebraic over the subfield F (X) generated by X. One calls K(X) a “purely transcendental” extension of F , although this notion conveys little since it is complely relative to X. For example, if x is a real number that is transcendental over Q, the field of rational numbers, then both Q(x) and Q(x2 ) are purely transcendental extensions of Q, while the former is algebraic over the latter.6 Next we observe that transcendence degrees add under iterated field extension just as ordinary degrees multiply. Specificically: Corollary 268 Suppose we have a tower of fields F ≤ K ≤ L. Then tr.deg[K : F ] + tr.deg[L : K] = tr.deq[L : F ]. proof: Let X be a transcendence basis of K over F and let Y similarly be a transcendence basis of L over K. Clearly then, X ∩ Y = ∅, every element of K is algebraic over F (Y ), and every element of L is algebraic over K(Y ). The corollary will be proved if we can show that X ∪ Y is a transcendence basis of L over F . First, we claim that X ∪ Y spans L – that is, every element of L is algebraic over F (X ∪ Y ). Let λ be an arbitrary element of L. Then there is an albebraic relation λm + am λm−1 + · · · + a1 λ + a0 = 0, where m ≥ 1, and the coefficients ai are elements of K(Y ). This means that each of them can be expressed as a quotient of two polynomials in K[Y ]. Let B be the set of all coefficients of the monomials in Y appearing in the numerators and denominators of these quotients representing ai , as i ranges over 0, 1, . . . , m. Then B is a finite subset of K, and λ is now algebraic over 6 In this connection, there is a notion that is very useful in looking at the subject of Riemann Surfaces from an algebraic point of view: A field extension K of F of finite transcendence degree over F is called an algebraic function field if and only if, for any transcendence basis X, K is a finite extension of F (X). 401 F (X ∪ Y ∪ B). Since each element of B is algebraic over F (X), F (X ∪ B) is a finite extension of F (X), and so F (X ∪ Y ∪ B) is a finite extension of F (X ∪ Y ). It then follows that F (X ∪ Y ∪ B, λ) is a finite extension of F (X ∪ Y ), and so λ is algebraic over F (X ∪ Y ) as desired. It remains to show that X ∪ Y is algebraically independant over Y . But if not, there is a polynomial p ∈ F [z1 , . . . zn+m ] and finite subsets {x1 , . . . , xm } and {y1 , . . . , yn } of X and Y , respectively, such that upon substitution of the m xi for the first m zi ’s, and the yi for the remaining n zi ’s, one obtains 0 = p(x1 , . . . , xm , y1 , . . . ym ). Now this can be rewritten as a polynomial whose monomial terms are monomials in the yi , with coefficients which are polynomials ck in F [z1 , . . . , zn ] evaluated at x1 , . . . , xn . Since the yi are algebraically independant over K, the coefficient polymomials ck are each equal to zero when evaluated at x1 , . . . , xn . But again, since the xj are algebraically independent over F , we see that each polynomial ck is identically 0. This means that the original polynomial p was the zero polynomial of F [z1 , . . . , zm+n ], eastablishing the algebraic independence of X ∪ Y . The proof is complete. . Corollary 269 An algebraic extension of an algebraic extension is algebraic. Precisely, if F ≤ K ≤ L is a tower of fields with K an algebraic extension of F and L an algebraic extension of K, then L is an algebraic extension of F . proof: This follows from the previous theorem and the fact that zero plus zero is zero. remark: Before, we knew this corollary only with “finite degree” replacing the word “algebraic” in the statement. 11.9 Appendix to Chapter 10: Fields with a Valuation 11.9.1 Introduction Let R denote the field of real numbers and let F be any field. A mapping φ : F → R from F into the field of real numbers is a valuation on F if and only if, for each α and β in F : 402 1. φ(α) ≥ 0, and φ(α) = 0 if and only if α = 0. 2. φ(αβ) = φ(α)φ(β), and 3. φ(α + β) ≤ φ(α) + φ(β). This concept is a generalization of certain “distance” notions which are familiar to us in the case that F is the real or complex field. For example, in the case of the real field R, the absolute value function |r| = ( r −r if r ≥ 0 if r < 0 is easily seen to be a valuation. Similarly, for the complex numbers, the function “k k”, defined by √ ka + bik = a2 = b2 , where a, b ∈ R and the square root is non-negative, is also a valuation. In this case, upon representing complex numbers by points in the complex plane, kzk measures the distance of z from 0. The “subadditive property” (3.) then becomes the triangle inequality. A valuation φ : F → R is said to be archimedean if and only if there is a natural number n such that if s(n) := 1 + 1 + · · · + 1 { with exactly n summands}, in F , then φ(s(n)) > 1. These valuations of the real and complex numbers are clearly archimedean. An example of a non-archemedian valuation is the trivial valuation for which φ(0) = 0, and φ(α) = 1 for all non-zero elements α in F . Here is another important example. Let F be the field of quotients of some unique factorization domain D which contains a prime element p. Evidently, F 6= D. Then any nonzero element α of F can be written in the form α= a k ·p , b where a, b ∈ D, p does not divide either a or b, and k is an integer. The integer k and the element a/b ∈ F are uniquely determined by the element α. Then set φ(α) = e−k . 403 where e is any fixed number larger than 1, such as 2.71 . . .. Otherwise we set φ(0) = 0. The student may verify that φ is a genuine valuation of F . (Only the subadditive property really needs to be verified.) From this construction, it is clear that φ is non-archimedean. Lemma 270 If φ is a valuation of F , then the following statements hold: 1. φ(1) = 1. 2. If ζ is a root of unity in F , then φ(ζ) = 1. In particular, φ(−1) = 1 and so φ(−α) = φ(α), for all α in F . 3. The only valuation possible on a finite field is the trivial valuation 4. For any α, β ∈ F , |φ(α) − φ(β)| ≤ φ(α − β). The proof of this lemma is left as an exercise for the student. Lemma 271 If φ is a non-archimedean valuation of F , then for any α, β ∈ F, φ(α + β) ≤ max(φ(α), φ(β)). proof: From the subadditivity, and that fact that for any number of summands, φ(1 + 1 + · · · + 1) ≤ 1, we have n φ((α + β) ) = φ ≤ Xn j=0 Xn j=0 n j ! α n−j β j ! φ(α)n−j φ(β)j ≤ (n + 1)max (φ(α)n , φ(β)n ). Since the taking of positive n-th roots in the reals is monotone, we have φ(α + β) ≤ (n + 1)1/n max (φ(α), φ(β)). Now since n is an arbitrary positive integer, and (n + 1)1/n can be made arbitrarily close to 1, the result follows. 404 11.9.2 The Language of Convergence We require a few definitions. Throughout them, φ is a valuation of the field F. First of all, a sequence is a mapping N → F , from the natural numbers to the field F , but as usual it can be denoted by a displayed indexed set, for example {α0 , α1 , . . .} = {αk } or {βk }. A sequence {αk } is said to be a Cauchy sequence if and only if, for every positive real number , there exists a natural number N (), such that φ(αm − αn ) < , for all natural numbers n and m larger than N (). A sequence {αk } is said to converge relative to φ if and only if there exists an element α ∈ F such that for any real > 0, there exists a natural number N (), such that φ(α − αk ) < , (11.18) for all natural numbers k exceeding N (). It is an easy exercise to show that an element α satisfying (11.18) for each > 0 is unique. [Note that in proving uniqueness, the choice function N : R+ → N taking to N () is posited for the assertion that α is a limit. To assert that β is a limit, an entirely different choice function M : → M () may be posited.] In that case we say that α is the limit of the convergent sequence {αk } or that the sequence {αk } converges to α. A third easy exercise is to prove that every convergent sequence is a Cauchy sequence. A sequence which converges to 0 is called a null sequence. Thus {αk } is a null sequence if and only if, for every real > 0, however small, there exists a natural number N (), such that φ(αk ) < , for all k > N (). A field F with a valuation φ is said to be complete with respect to φ if and only if every Cauchy sequence converges (with respect to φ) to an element of F . Now, finally, a completion of F (with respect to φ) is a field F̄ with a valuation φ̄ having these three properties: 1. The field F̄ is an extension of the field F . 2. F̄ is complete with respect to φ̄. 405 3. Every element of F̄ is the limit of a φ-convergent sequence of elements of F . (We capture this condition with the phrase “F is dense in F̄ ”. So far our glossary entails • Convergent sequence • Limit • Null sequence • Cauchy sequence • Completeness with respect to a valuation φ • A completion of a field with a valuation 11.9.3 Completions exist The stage is set. In this subsection our goal is to prove that a completion (F̄ , φ̄) of a field with a valuation (F, φ) always exists. In retrospect, one may now realize that the field of real numbers R is the completion of the field of rational numbers Q with respect to the “absolute” valuation, | |. The construction of such a completion should not be construed as a “construction” of the real numbers. That would be entirely circular since we used the existence of the real numbers just to define a valuation and to deduce some of the elementary properties of it. One may observe that the set of all sequences over F possesses the structure of a ring, namely the direct product of countably many copies of the field F . Thus S := F N = Y k∈N Fk where each Fk = F . Using our conventional sequence notation, addition and multiplication in S are defined by the equations {αk } + {βk } = {αk + βk } {αk } · {βk } = {αk βk}. For each element α ∈ F , the sequence {αk } for which αk = α, for every natural number k is called a constant sequence and is denoted ᾱ. The zero 406 element of the ring S is clearly the constant sequence 0̄. The multiplicative identity of the ring S is the constant sequence 1̄ (where “1” denotes the multiplicative identity of F ). Obviously, the constant sequences form a subring of S which is isomorphic to the field F . It should also be clear that every constant sequence is also a Cauchy sequence. We shall show that the collection Ch of all Cauchy sequences also forms a subring of S. In order to do this, we first make a simple observation. A sequence {αk } is said to be bounded if and only if there exists a real number b such that αk < b for all natural numbers k. Such a real number b is called an upper bound of the sequence {αk }. The observation is Lemma 272 Every Cauchy sequence is bounded. proof: Suppose {αk } is a Cauchy sequence. Then there is a natural number N = N (1) such that φ(αN − αm ) < 1, for all natural numbers m exceeding N . Now set b := 1 + max {α0 , α1 , . . . , αN }. If k ≤ N , αk < b, by the definition of b. If k > N then αk < b by the definition of N and the fact that αN ≤ b − 1. So b is an upper bound. Lemma 273 The collection Ch of all Cauchy sequences of S is closed under addition and multiplication of sequences. Thus Ch is a subring of S, the ring of all sequences over F . proof: Recall that if {αk } is a Cauchy sequence, there must exist an auxilliary function N : R+ → N such that if ∈ R+ , then φ(αn − αm ) < for all numbers n and m larger than N (). This auxillialy function N is not unique; it is just that at least one such function is available for each Cauchy sequence. If we wish to indicate an available auxilliary function N , we simply write ({αk }, N ) instead of {αk }. Now suppose ({αk }, N ) and ({βk }, M ) are two Cauchy sequences. For each positive real number set K() := max(N (/2), M (/2)). Then for any natural numbers n and m exceeding K(), we have φ(αn + βn − αm − βm ) ≤ φ(αn − αm ) + φ(βn − βm ) ≤ /2 + /2 = . Thus ({αk + βk }, K) is a Cauchy sequence. 407 Again suppose ({αk }, N ) and ({βk }, M ) are two Cauchy sequences. By the preceeding Lemma 272, we may assume that these sequences possess positive upper bounds b and c, respectively. For each positive real set K() := max (N (/2c), M (/2b)). Now suppose n and m are natural numbers exceeding K(). Then φ(αm βm − αn βn ) = φ(αm βm − αm βn + αm βn − αn βn ) ≤ φ(αm )(φ(βm − βn ) + φ(βn )φ(αn − αm ) < b(/2b) + c(/2c) = . Thus ({αk βk }, K) is a Cauchy sequence. Finally, it should be obvious that {−αk } is the additive inverse in S of {αk } and, from an elementary property of φ, that the former is Cauchy if and only if the latter is. It follows that the Cauchy sequences form a subring of S. The null sequences of S are simply the sequences that converge to zero. We have already remarked that every convergent sequence is a Cauchy sequence. Thus the set Z of all null sequences lies in Ch. Note that any sequence whose coordinates are nonzero at only finitely many places (that P is, the direct sum N F ) form a subset Z0 of Z. Clearly the sum or difference of two null sequences is null. Suppose {αk } is a null sequence so that for every positive real , there is a natural number N () such that αm , for all m > N (). Next let {γk } be any Cauchy sequence with upper bound b. Then, for every real > 0, and n > N (/b), we see that φ(αn γn ) ≤ b · (/2) = . Thus the product {αk }{γk } is a null sequence. So we have Lemma 274 Z is an ideal of the ring Ch. . A sequence {αk } is said to possess a lower bound ` if and only if for some real ` > 0, one has αn > ` for every natural number n. The next lemma, though mildly technical, is important. Lemma 275 The following assertions hold: 408 1. Suppose a sequence {γk } in S has the property that for every real > 0, γk > for only finitely many natural numbers k. Then {γk } is a null sequence. 2. Suppose {βk } is a Cauchy sequence which is not a null sequence. Then there exists a real number λ > 0, and a natural number N such that βk > λ for all k > N . 3. If {αk } is a Cauchy sequence having a lower bound `, then it is a unit in the ring Ch. proof: 1. The reader may want to do this as an exercise. The hint is that the hypotheses allow us to define a mapping K : R+ → N where K() := max {k ∈ N|γk > }. Then K serves the desired role in the definition of a null sequence. 2. Using the contrapositive of the statement in part 1., we see that if {βk } is a Cauchy sequence which is not null, there must exist some > 0 such that βk > infinitely often. Also, since the sequence is Cauchy, there exists a natural number N (/2) such that φ(βn − βm ) ≤ /2, for all natural numbers n and m exceeding N (/2). So there is an natural number N exceeding N (/2) such that βM > . Then for m > N , φ(βm ) = φ(βM − (βM − βm ) ≥ φ(βM ) − φ(βM − βn ) > − (/2) = /2. Thus λ := /2 and N satisfy the conclusion of part 2. 3. For each > 0, let K() := N ( · `2 ). Then for any natural numbers n and m exceeding K(), we have 1 1 − φ αn αm 1 = φ φ(αm − αn ) αn αm ≤ `−2 · (`2 ) = . Thus the sequence of reciprocals {αk −1 } is a Cauchy sequence and is a multiplicative inverse to {αk } in Ch. The proof is complete. 409 Corollary 276 The factor ring F̄ := Ch/Z is a field. proof: It suffices to show that every element of Ch − Z is congruent to a unit modulo the ideal Z. Suppose {βk } is a non-null Cauchy sequence. Then by Lemma 275, there is a positive real number λ and a natural number N such that βn > λ for all natural numbers n greater than N . Now let {ζk } be a sequence for which ( 1 if k ≤ N ζk = 0 if k > N . Then {ζk + βk } is bounded below by ` := min (1, λ) and so by Lemma 275, part 3., is a unit. Since {ζk } belongs to Z, we have shown the sufficient condition announced at the beginning of this proof. Note that we have a natural embedding F → F̄ taking each element α to the constant sequence ᾱ, all of whose terms are equal to α. Now we need to extend the valuation φ on (the embedded copy of) F to a valuation φ̄ of F̄ . The key observation is that whenever we consider a Cauchy sequence {αk }, then {φ(αk )} is itself a Cauchy sequence of real numbers with respect to the absolute value. For, given real > 0, there is a number N () such that φ(αn − αm ) < for all n and m exceeding N (). But |φ(αm ) − φ(αm )| ≤ φ(αn − αm ) < with the same conditions on n and m. Now Cauchy sequences {rn } of real numbers tend to a real limit which we denote by the symbol lim rk . k→∞ Thus, for any Cauchy sequence {αk } in Ch, we may define φ̄({αk }) := lim φ(αk ). k→∞ Now note that if {αk } happens to be a sequence that already converges to α in F , then, by the definition of convergence, limk→∞ φ(αk ) = φ(α). In short, on convergent sequences, φ̄ can be evaluated simply by takin φ of the existing limit. In particular this works for constant sequences, whereby we we have arrived at the fact that φ̄ extends the mapping φ defined on the embedded subfield F . Now observe 410 Lemma 277 If {αk } and {βk } are two Cauchy sequences with the property that φ(αk ) ≤ φ(βk ), for all natural numbers k larger than some number N , then φ̄({αk }) ≤ φ̄({βk }). proof: The student should be able to prove this, either by assuming the result for real Cauchy sequences, or from first principles by examining what it would mean for the conclusion to be false. Corollary 278 If {αk } and {βk } are two Cauchy sequences, φ̄({αk + βk }) ≤ φ̄({αk }) + φ̄({βk }). proof: Immediate from Lemma 277. Corollary 279 The following statements are true. 1. {αk } is a null sequence if and only if φ̄({αk }) = 0 ∈ R. 2. φ̄ assumes a constant value on each coset of Z in Ch. proof: Another exercise in the meaning of the definitions. Now for any coset γ := {βk } + Z of Z in Ch we can let φ̄(γ) be the constant value of φ̄ on all Cauchy sequences in this coset. In effect φ̄ is now defined on the factor ring F̄ = Ch/Z, which, as we have noted, is a field. Moreover, from the preceeding Corollary φ̄ assumes the value 0 only on the zero element 0̄ = 0 + Z = Z of F̄ . At this stage – and with a certain abuse of notation – φ̄ is a non-negativereal-valued function defined on two distinct domains, Ch and F̄ . However the ring epimorphism f : Ch → F̄ preserves the φ̄-values. That is important for three reasons. First it means that if φ̄ is multiplicative on Ch, then it is on F̄ . But it is indeed the case that φ̄ is multiplicative on Ch, since, for {αk } and {βk } in Ch, lim φ(αk βk ) = k→∞ lim φ(αk )φ(βk ) k→∞ = ( lim φ(αk )) · ( lim φ(βk )). k→∞ 411 k→∞ Second, the fact that the homomorphism f preserves φ̄, means that if φ̄ is subadditive on Ch, then it is also subadditive on F̄ . But the subadditivity on Ch is asserted in Corollary 278, part 1. Third, since F ∩ Z = 0 ∈ Ch (that is, the only constant sequence which is null is the zero constant sequence), the homomorphism f embedds F (in its transparent guise as the field of constant sequences) into F̄ without any change in the valuation. These observations, taken together, yield Corollary 280 The function φ̄ : F̄ → R+ is a valuation of F̄ , which extends the valuation φ on its subfield F . It remains to show that F̄ is complete with respect to the valuation φ̄ and that F dense in F̄ . Recall that the symbol ᾱ denotes the constant sequence {α, α, . . .}. Its image in F̄ is then f (ᾱ) = ᾱ+Z. Now suppose {αk } is any Cauchy sequence, and let γ = {αk } + Z be its image in F̄ . We claim that the sequence {f (ᾱ1 ), f (ᾱ2 ), . . .} (of constant sequences mod Z) is a sequence of elements in F̄ converging to γ relative to φ̄. Choose a positive real number . Then there exists a natural number N () such that φ(αN () − αm ) < for all m > N (). Then as f preserves φ̄, φ̄ f (ᾱN () ) − γ̄ = φ̄ ᾱN () − {αk } = lim φ αN () − αn < . n→∞ So our claim is true. F is indeed dense in F̄ . Now suppose {γ` } is a Cauchy sequence of elements of F̄ with respect to φ̄. For each index `, we can write γ` = {β`k }, a Cauchy sequence over F indexed by k. So there exists a natural number N (`) such that φ(β`N (`) − β`m ) < 2−` (11.19) 412 for all m exceeding N (`). This allows us to define the following set of elements of F : δ` := β`N (`) , for ` = 0, 1, . . . . Then taking limits in equation (11.19), we have φ̄(f (δ̄` ) − γ` ) < 2−` , where, as usual, f (δ̄k ) is the image of a constant sequence, namely the coset {δk , δk , δk , . . .} + Z. Now for any natural numbers n and m, we have φ(δn − δm ) = φ̄ f (δ̄m ) − f (δ̄n ) = φ̄ f (δ̄n ) − γm + γm − γn + γn − f (δ̄m ) ≤ φ̄(f (δ̄n − γm ) + φ̄(γn − γm ) + φ̄(γm − f (δ̄m )) ≤ φ̄(γn − γm ) + 2−m + 2−n But since {γ` } is a Cauchy sequence, the above equations tell us that δ := {δk } is a Cauchy sequence in Ch. It is now clear that φ̄(δ − γ` ) = lim (δk − β`k ) < 2−` . k→∞ Thus {γ` } converges to δ with respect to φ̄. It follows that F̄ is a complete field. Summarizing all of the results of this subsection, one has the following Theorem 281 The field F̄ = Ch/Z with valuation φ̄ is a completion of (F, φ). 11.9.4 The non-archimedean p-adic valuations Our classical example of a non-archimedean valuation was derived from the choice of a prime p in a unique factorization domain D. The valuation φ is defined on the field F of quotients of D by the rule that φ(r) = e− k, whenever we write r ∈ F in the canonical form r = (a/b)pk where a and b are elements of D which are not divisible by p, and k is an integer. Of course to make this function subadditive, we must take e to be a fixed real number 413 greater than 1. (Many authors choose p itself. That way, the values never overlap from one prime to another.) Now notice that from Lemma 271, the collection of all elements α of F for which φ(α) ≤ 1 is closed under addition as well as multiplication, and so form a subring O of F called the valuation ring of (F, φ). From the multiplicative properties of φ, it is clear that every non-zero element of F either lies in O, or has its multiplicative inverse in O. A somewhat smaller set is P := {α ∈ F |φ(α) < 1}. It is easy to see that this is an ideal in O. In fact the set O − P = {u ∈ F |φ(u) = 1} is closed under taking multiplicative inverses in F and so consists entirely of units of O. Conversely, since φ(1) = 1, every unit of O must have φ-value 1, so in fact O − P = U (O), (11.20) the (multiplicate) group of units of O. Thus P is the unique maximal ideal of O. Thus O is a local ring. Note that O and P have been completely determined by F and φ. The factor ring O/P is a field called the residue field of F with respect to φ. The same sort of argument used to show (11.20) can be used to show that there exists an element π ∈ P , such that for any natural number k, P k − P k+1 = π k U (O). Taking the disjoint union of these set differences, one concludes that P is the principal ideal P = Oπ. Suppose p is a prime integer,D = Z, so F = Q, the field of rational numbers. Then O is the ring of all fractions ab where a and b are integers and p does not divide b. Terribly odd things happen! For example {1, p, p2 , . . .} is a null sequence (since φ(pk ) = e−k , for all positive integers k), and 1 + p + p2 + · · · is a “convergent series” – that is its sequence of partial sums is a Cauchy sequence. Let us return to our classic example. There F was the field of fractions of a unique factorization domain D, p is a prime element of D and φ(r) = e−k when r ∈ F is written in the form (a/b)pk , a, b ∈ D, k an integer. Let O be the valuation ring and let P be its unique prime ideal. Similarly, let F̄ , φ̄, Ō and P̄ be the completion of F , its extended valuation, valuation ring and uniqe maximal ideal, respectively. 414 Theorem 282 The following statements hold: 1. D ⊆ O, and D ∩ P = pD. 2. D + P = O. 3. F ∩ Ō = O, O ∩ P̄ = P . 4. O + P̄ = Ō. 5. The residue class fields of F and F̄ are both D/pD. proof: 1. Clearly every element of D has valuation at most 1 so the first statement holds. Those elements of D of value less than 1 are divisible by p, and vice versa. So the second statement holds. 2. Suppose r ∈ O − P . Then r has the form a/b where a and b are elements of D which are not divisible by p. Since p is a prime and D is a UFD, this means there elements c and d of D such that cp + db = a. Then a c r= = · p + d. b b (11.21) Now p does not divide d, otherwise it would divide a = cp + db, against our assumption about r. Thus d is an element of D also in O−P ,and by equation (11.21), r ≡ (a/b) modulo P . This proves 2. 3. Both statements here just follow from that fact the φ̄ extends φ. 4. Now let ᾱ be any element in Ō − P̄ so φ̄(ᾱ) = 1. Since F is dense in F̄ , there is an element r ∈ F such that φ(ᾱ − r) < 1. This means r − ᾱ is an element of P . Since O is an additive subgroup of F containing P , r belongs to Ō ∩ F = O, and we can say r ≡ ᾱ mod P̄ . This proves 4. 5. First using 3. and 4., and then 1. and 2., Ō/P̄ = ' O/P = ' (O + P̄ )/P̄ O/(O ∩ P̄ ) = O/P, while (D + P )/P D/(D ∩ P ) = D/pD. Now fix a system Y of coset representatives of D/pD and suppose 0 ∈ Y . For example, if D = Z, then p is a rational prime, the residue class field is Z/pZ, and we may tke Y to be the set of integers, {0, 1, 2, . . . , p − 1}. 415 For another example, suppose D = K[x] where K is some field, and p = x + a is a monic polynomial of degree one. Then the residue class field is K[x]/(x + a) ' K. But as K is a subring of D, it makes perfect sense to take Y = K (the polynomials of degree zero). Now in the notation of Theorem 282, D + P̄ = (D + P ) + P̄ = O + P̄ = Ō. So, for a fixed ᾱ ∈ Ō, there is a unique element a0 ∈ Y such that ᾱ − a0 ∈ P̄ . (If ᾱ is already in P̄ , this element is 0.) Now P̄ = pŌ, so 1 α¯1 = (ᾱ − a0 ) ∈ Ō. p Then we repeat this procedure to obtain a unique element a1 ∈ Y such that α¯1 − a1 ∈ pŌ. Then ᾱ ≡ a0 + pa1 mod p2 Ō. Again, one can find a2 ∈ Y such that if α¯2 := (1/p)(α¯1 − a1 ), then α¯2 − a2 ∈ pŌ. So, after k repetitions of this procedure, we obtain a sequence of k + 1 unique elements of Y , (a0 , a1 , . . . , ak ) such that ᾱ ≡ a0 + a1 p + a2 p2 + · · · + ak pk mod P̄ k+1 . Thus we can write any ᾱ ∈ F̄ in the “power series” form ᾱ = pk [a0 + a1 p + a2 p2 + · · ·] where the ai are uniquely determined elements of Y , and a0 6= 0, so that k is determined by e−k = φ̄(ᾱ). For example, if p = x ∈ K[x] = D, then, taking Y = K, we see that F̄ is the field of Laurent series over K – that is, the ring of power series of the form b−k x−k + · · · + b−1 x−1 + b0 + b1 x + b2 x2 + · · · , k ∈ N where the bi are in K and one follows the usual rules of adding and multiplying power series. The ring of Laurent series is very important in the study of power series generating functions. Be aware that the field K here is still arbitrary. It can even be a finite field. Note that if D = Z and p is a rational prime, the system of coset representatives Y is not closed under either addition or multiplication and so, in order to restore the coefficients of the powers of p to their proper location in 416 Y , some local adjustments are needed after addition or multiplication. For example if p = 5, then we must rewrite (1 + 3(5))(1 + 4(5)) = 1 + 7(5) + 12(52 ) as 1 + 2(5) + 3(52 ) + 2(53 ). 417 11.10 Appendix to Chapter 10: Finite Division Rings In 1905 J. H. M. Wedderburn proved that every finite division ring is in fact a field. In van Der Waerden’s book (1930-31) [??. pp 104-105] one finds a seemingly short proof that depends only on the easy exercise which proves that no finite group can be the union of the conjugates of a proper subgroup. But the proof is short only in the sense that the number of words required to render it is not large. Measured from actual elementary first principles, the proof spans a considerable “logical distance”. It uses the notion of a splitting field for a division ring, and a certain structure theorem involving a tensor product of algebras. Both of these notions are beyond what has been developed in this and the preceeding chapters. Several other proofs are available. The first proof given below is due to T. J. Kaczynski [??] and has several points of interest. We begin with a small lemma. Lemma 283 Every element in a finite field is the sum of two squares. In particular, when p is a prime number, there exist numbers t and r such that t2 + r2 ≡ −1, mod p, with t 6≡ 0. proof: If q is even, all elements are squares and hence is the sum of 02 and a square. So suppose q is odd and let S be the set of non-zero squares. Then, for any non-square q ∈ GF(q), we have a partition GF(q)∗ = S + Sg . Now using 0 as one of the summand squares, we see that every element of {0} ∪ S is the sum of two squares, and if (S + S) ∩ gS 6= ∅ (where now “S + S” indicates all possible sums of two squares), then this is true of gS as well and the result is proved. Otherwise S + S ⊆ {0} ∪ S and so {0} ∪ S is a subfield of GF(q). That is impossible since |{0} ∪ S| = (q + 1)/2 is not a prime power dividing q. In particular, −1 can be represented as the sum of two squares, and one of them at least (say t) is non-zero. Theorem 284 (Wedderburn) Every finite division ring is a field. proof: (Kaczynski, 1964) Let D be a finite division ring and let P be the prime subfield in its center. Also let D∗ be the multiplicative group of nonzero elements of D. For some prime r let S be an r-Sylow subgroup of D∗ , of 418 order ra , say. Now S cannot contain an abelian subgroup A of type Zr × Zr otherwise the subfield generated by P ∪ A would have too many roots of xr − 1 = 0. Thus S is either cyclic or else r = 2 and S is a generalized quaternion group. We shall show that the latter case is also forbidden. Suppose S is generalized quaternion. Then the characteristic p is not 2 and S contains a subgroup S0 which is quaternion of order 8, generated by two elements a and b or order 4 for which bab = a−1 and z = a2 = b2 is its central involution. Then the extension P (z) is a subfield containing the two roots of x2 − 1 = 0, namely 1 and −1. It follows that a2 = −1 = b2 so a3 = −a. Thus ba = a−1 b = −ab. Now let p = |P |, the characteristic of D. Then by the lemma we can find elements t and r in P (and hence central in D) such that t2 + r2 = −1. Then ((ta + b) + r) ((ta + b) − r) = (ta + b)2 − r2 = t2 a2 + r(ab + ba) + b2 − r2 = −t2 + b2 − r2 = 0. So one of the factors on the left is zero. But then b = −ta ± r and so as t 6= 0, b commutes with a, a contradiction. Thus, regardless of the parity of p, every r-Sylow subgroup of D∗ must be cyclic. If follows that D∗ must be solvable. Now let Z ∗ = Z(D∗ ) be the center of the group D∗ and suppose Z ∗ 6= D∗ . Then we can find a normal subgroup of D∗ containing Z ∗ such that A/Z ∗ is a minimal normal subgroup of D∗ . Since the latter is solvable with all Sylow subgroups cyclic, A/Z ∗ is cyclic. Now Z ∗ is in the center of A, and since no group has a non-trivial cyclic factor over its center, A is abelian. Now we shall show that in general, D∗ cannot contain a normal abelian subgroup A which is not in the center Z ∗ . Without loss of generality we can take (as above) Z ∗ < A D∗ . Choose y ∈ A − Z ∗ . Then there exists an element x in D∗ which fails to commute with y. Then also x + 1 is not in A as it fails to commute with y as well. Then, as A is normal, (1 + x)y = a(1 + x), 419 for some element a ∈ A. Then y − a = ax − xy = (a − yxy −1 )x. (11.22) Now if y − a = 0, then a = y and a = xyx−1 by (11.22), so y = xyx−1 against the fact that x and y do not commute. Thus (a−xyx−1 ) is invertible, and from (11.22), x = (a − xyx−1 )−1 (y − a). (11.23) But a, y and xyx−1 all belong to A and so commute with y. Thus every factor on the right side of (11.23) commutes with y, and so x commutes with y, the same contradiction as before. the proof is complete. Remark: As noted, the last part of Kaczynski’s proof did not use finiteness at all. As a result, one can infer Corollary 285 (Kaczynski) Suppose D is any division ring with center Z. Then any abelian normal subgroup of D∗ lies in Z. One might prefer a proof that did not use the fact that a finite group with all sylow subgroups cyclic is solvable. This is fairly elementary to a finite group theorist (one uses a transfer argument on the smallest prime and applies induction) but it would not be as transparent to the average man or woman on the street. Another line of argument due to to I. N. Herstein [??] develops some truly beautiful arguments from first principles. Consider: Lemma 286 Suppose D is a division ring of characteristic p, a prime, and let P be it prime subfield, which, of course, is contained in the center Z. Suppose a is any element of D − Z which is algebraic over P . Then there exists an element y ∈ D such that yay −1 = ai 6= a for some integer i. proof: Fix element a as described. The commutator mapping δ : D → D, which takes an arbitrary element x to xa − ax, is clearly an endomorphism 420 of the additive group (D, +). Also, for any element λ of the finite subfield P ((a) obtained by adjoining a to P , the fact that λ commutes with a yields δ(λx) = λxa − aλx = λ(xa − ax) = λδ(x). Thus, regarding D as a left vector space over the field P (a), we see that δ is a linear transformation of V . Now, for any x ∈ D, and positive integer k, an easy computation reveals k δ (x) = Xk j j=0 (−1) k j ! aj xak−j . m Now P (a) is a finite field of order pm , say. Then ap characteristic p, putting k = pm in (11.24) gives m m (11.24) = a. Since D has m δ p (x) = xap − ap = xa − ax = δ(x). Thus the linear transformation δ of V satisfies m δ p − δ. Thus, in the commutative subalgebra of Hom (V, V ) generated by δ, we have m 0 = δp − δ = Y λ∈P (a) (δ − λI), where the λI are the scalar transformations V → V , λ ∈ P (a). Now ker δ is the subalgebra CD (a) of all elements of D which commute with a. (This is a proper subspace of V since a is not in the center Z.) Thus Q δ acts as a non-singular transformation of W := V /CD (a) while (δ − λI) (the product taken over all non-zero λ ∈ P (a)) vanishes on it. It follows that for some λ ∈ P (a) − {0}, δ − λI is singular on W . This means there is a coset b0 + CD (a) 6= CD (a) such that δ(b0 ) = λb0 + c, c ∈ CD (a). Setting y = b0 + λ−1 c, we have δ(y) = λy 6= 0. This means yay −1 = a + λ ∈ P (a). Now if m is the order of a in the multiplicative group P (a)∗ , then yay −1 is one of the roots of X m − 1 = 0 lying in P (a). But since these roots are just 421 the powers of a, we have yay −1 = ai 6= a, for some i, 1 < m ≤ m − 1, as desired. Now we proceed with a proof of Wedderburn’s theorem. We may assume that D is a non-commutative finite division ring in which all proper subalgebras are fields. In particular, for any element v ∈ D − Z, the centralizing subalgebra CD (v) is a field. It follows that the commuting relation on D − Z is an equivalence relation. For each commuting equivalence class Ai , and element ai ∈ Ai , we have Fi := Z ∪ Ai = CD (ai ). (11.25) Let D∗ be the multiplicative group of non-zero elements of D, so Z ∗ := Z − {0} be its center. Choose a coset aZ ∗ so that its order in D∗ /Z ∗ is the smallest prime s dividing the order of D∗ /Z ∗ . The actual representative element a can then be chosen to have s-power order. Now by Herstein’s lemma, there is an element y ∈ D such that yay −1 = at 6= a. From (11.25) it follows that conjugation by y leaves the subfield F := CD (a) invariant, and induces on it an automorphism whose fixed subfield is CD (y) ∩ CD (a) = Z. But the same conclusion must hold when y is replaced by a power y k which is not in Z. It follows that Gal(F/Z) is cyclic of prime order r and conjugation by y induces an automorphism of order r. Thus y r ∈ CD (a)∩CD (a) = Z, so r is the order of yZ ∗ in the group D∗ /Z ∗ . We can then choose the representative y to be in the r-Sylow subgroup of Z(y)∗ , so that it has r-power order. Now, by the “(very) little Fermat Theorem” is−1 ≡ 1 mod s. Thus s−1 y (s−1) ay −(s−1) = ai ≡ a mod Z ∗ . So, setting b := y s−1 , we have bab−1 = az, for some element z in Z which is also a power of a. Now from this equation, z = 1 if and only if b ∈ CD (y) ∩ CD (a) = Z. 422 Case 1: z = 1 and b is in the center. Then y s−1 ∈ Z, so r, the order of yZ in D∗ /Z ∗ , divides s − 1. That is impossible since s was chosen to be the smallest prime divisor of |D∗ /Z ∗ |. Case 2. z 6= 1 and b has order r mod Z ∗ . Then br a−r = a = az r . But since z is an element of s-power order, we now have r = s is the order of z. Now at this stage, the multiplicative subgroup H := ha, bi of D∗ generated by a and b is a nilpotent group generated by two elements of r-power order, and so is an r group. Since it is non-abelian without a subgroup of type Zr × Zr , it is generalized quaternion and so contains the quaternion subgroup of order 8. Herstein goes on to eliminate the presence of Q8 by an argument that essentially anticipates by three years the argument given in the first part of Kaczynski’s proof above7 There is another fairly standard proof which comes closer to being a proof from elementary principles. It uses an argument due to Witt [??] concerning cyclotomic polynomials and the partition of the group D∗ into conjugacy classes. One may consult Jacobson’s book (cited below), pages 431-2, for an example of this proof. ∗ References Herstein, I.Wedderburn’s theorem and a theorem of Jacobson, Amer. Math. Monthly, 68 (1961), 249-251. Jacobson, Nathan, Basic Algebra I, W. H. Freeman Co., San Fransisco, 1974. Kaczynski, T. J. Another proof of Wedderburn’s theorem. Amer. Math. Monthly, 71 (1964), 652-653. Witt, Ernst, Über die Kommutativität endlicher Schiefköpern. Abhandlung. Math. Sem. Univ. Hamburg, 8 (1931), p. 413. Wedderburn, J. H. M., A theorem on finite algebras, Trans. Amer. Math. Soc. 6 (1905), 349-352. van der Waerden, B. L. Algebra vol II Springer-Verlag, Berlin-New York, 1991 7 Actually Herstein inadvertantly omitted Case 1 by taking s = r from a misreading of the definition of s. As we see, this presents no real problem since his lemma applies when s is the smallest possible prime. 423 Chapter 12 Semiprime Rings 12.1 Introduction In this chapter we develope the material necessary to reach the famous ArtinWedderburn theorem which – in the version given here – classifies the Socle of a prime ring. We begin with the important concept of complete reducibility of modules. When transferred to rings, this notion and its equivalent incarnations give us quick access to the Socle of a ring and its homogeneous components. Most of the remaining development takes place in a general class of rings called the semiprime rings. It has several important subspecies, the prime, primitive and completely reducible rings, where the important role of idempotents unfolds. 12.2 Complete Reducibility One defines the radical radM of a right R-module M as the intersection of all maximal submodules of M . If there are no maximal submodules — as in the Z-module 2 Z∞ p = h1/p, 1/p , . . .i ≤ Q/Z – then radM = M , by definition. There is a dual notion: One defines the socle Soc(R) as the submodule generated by all minimal submodules (that is, all irreducible submodules) of M . Similarly, if there are no irreducible submodules – as in the case of 424 the free Z-module ZZ – then by convention one sets Soc(M ) = 0, the zero submodule. Theorem 287 If M is a right R-module then 1. Soc(M ) = ⊕i∈J Mi , where {Mi }i∈J is a family of irreducible submodules of M , and the sum is direct. 2. In addition, Soc(M ) is invariant under the endomorphism ring, E = homR (M, M ). proof: We prove the second statement first. Recall that for a right Rmodule M , homR (M, M ) is regarded as acting on the left, so that the fact that application of such endomorphisms commutes with the right action of R can be expressed as an “associative law”, that is, ψ(mr) = (ψm)r, for all (ψ, m, r) ∈ homR (M, M ) × M × R. Now if Mj is an irreducible submodule of M , and ψ is an R-endomorphism of M , then ψ(Mj ) = 0 or is irreducible because of its extremely limited P lattice of submodules. Thus for any element m lying in a finite sum Mi P of irreducible submodules of M , ψ(m) ∈ ψ(Mi ), where ψ(Mi ) is zero or irreducible. Since the hypothesis on m in the previous sentence characterizes the elements of Soc(M ), we see that ψ(Soc(M )) ≤ Soc(M ) for all ψ ∈ homR (M, M ), as desired. Let the set I index the family of all irreducible modules of M . Trivially, we may assume I is non-empty. We shall call a subset J of I a direct set if P and only if j∈J Mj = ⊕j∈J Mj is a direct sum. [Recall that this means that P if a finite sum k∈K ak = 0 where ak ∈ Mk and K ⊆ J, then each ak = 0. We note also that singleton subsets of I are direct, and so, as irreducible modules are assumed to exist, direct subsets of I really exist.] Suppose I0 ⊂ L1 ⊂ · · · ⊂ · · · is a properly ascending tower of direct sets, and let J be the union of the sets P in this tower. If J were not direct, there would be a finite sum k∈K ak = 0 425 with not all ak equal to zero. But as each index k of the sum lies in some If (k) , the finite index set K for this sum is contained in some Im in the tower where m = max{f (k)|k ∈ K}. But then the existence of such a sum contradicts the assumption that each member Ij of the tower was a direct set. Thus in the poset of all direct subsets of I, partially ordered by inclusion, all simply ordered chains possess an upper bound. Thus by Zorn’s Lemma, this poset contains a maximal member, say H. We now claim that MH := ⊕j∈H Mj = Soc(M ). Clearly, MH ≤ Soc(M ). Suppose MH was properly contained in Soc(M ). Then there would exist an irreducible submodule M0 not in MH . Then as 0 is the only proper submodule P of M0 , we see MH ∩ M0 = 0. Then the sum ( H Mj ) ⊕ M0 is direct, for if not, there would be a finite subset K of H ∪ {0}, and elements ak , k ∈ K, P not all zero, such that k∈K ak = 0. Since H is direct, K is not a subset of H, and so 0 ∈ K and X a0 = − a k∈K−{0} k 6= 0. (12.1) But then both sides of equation (12.1) represent a non-zero element of MH ∩ M0 , an impossibility. Thus H ∪ {0} is direct, and that contradicts the maximality of H. Thus Soc(M ) = ⊕j∈H Mj , and the proof is complete. Corollary 288 The following conditions on an R-module are equivalent: 1. M = Soc(M ). 2. M is a sum of minimal submodules. 3. M is isomorphic to a direct sum of irreducible submodules. An R-module M satisfying any of the above three conditions is said to be completely reducible. A submodule L of a right R-module M is called large if it has a nonzero intersection with every non-zero submodule. (Note that if M 6= 0, M is a large submodule of itself.) We can produce (but not actually construct) many large submodules using the following Lemma 289 If B is a submodule of M and C is maximal among submodules of M meeting B at zero, then B + C is large. 426 proof: Assume B and C chosen as above and let D 6= 0 be any non-trivial submodule of M . Suppose (B + C) ∩ D = 0. If b, c, and d are elements of B, C and D, respectively, and b = c + d, then d = b − c and this is zero by our supposition. Thus b = c ∈ B ∩ C = 0, so b = c = 0. Thus B ∩ (C + D) = 0. By maximality of C, D ⊆ C. Then D = (B + C) ∩ D = 0, a contradiction. Thus always (B + C) ∩ D is non-zero, and so B + C is large. That such a submodule C exists which is maximal with respect to meeting B trivially, as in Lemma 289, is an easy application of Zorn’s Lemma. If M is a right R-module, let L(M ) be the lattice of all submodules of M (sum is the lattice “join”, intersection is the lattice “meet”). We say L(M ) is complemented if every submodule B of M has a complement in L(M ) – i.e. there exists a submodule B 0 such that B ∩ B 0 = 0 and B + B 0 = M (this means M = B ⊕ B 0 ). Lemma 290 If L(M ) is complemented, then so is L(N ) for any submodule N of M . proof: Let C be a submodule of N . By hypothesis there is a complement C 0 to C in L(M ). Then setting C 00 := C 0 ∩ N , we see C 00 ∩ C ⊆ C 0 ∩ C = 0, and C 00 + C = (C 0 ∩ N ) + C = (C + C 0 ) ∩ N, (for C ≤ N ) = M ∩ N = N. Thus C 00 is a complement of C in N . remark: Of course the above is actually a lemma about any modular lattice: if it is complemented, then so are any of its intervals above zero. Lemma 291 If M is a right R-module for which L(M ) is complemented, then rad(M ) = 0. proof: Let m be a non-zero element of M and let A be maximal among the submodules which do not contain m. (A exists by Zorn’s Lemma.) Suppose 427 now A < B < M so A is not maximal in L(M ). Let B 0 be a complement to B in L(M ). Then by the modular law (Dedekind’s Lemma) A = (B ∩ B 0 ) + A = B ∩ (B 0 + A), as A < B. Now as B < M implies B 0 6= 0 and A < B implies B 0 ∩ A = 0, we see that A is properly contained in B 0 + A. Then the maximality of A implies m ∈ B 0 + A and also m ∈ B. Thus m lies in B ∩ (B 0 + A) = A, contrary to the choice of A. Thus no such B exists and A is in fact a maximal submodule of M . Then we see that every non-zero module element is avoided by some maximal submodule of M , and hence is outside rad(M ). It follows that rad(M ) = 0. Theorem 292 The following conditions on an R-module M are equivalent: 1. M is completely reducible. 2. M contains no proper large submodule. 3. L(M ) is complemented. proof: 1. implies 2. Assume M = Soc(M ). If A is a proper large submodule of M , then there exists an irreducible submodule M0 not contained in A, since the irreducibles generate Soc(M ). But then as M0 is irreducible, A ∩ M0 = 0. But this contradicts the property that A is large. 2. implies 3. Let B be any proper submodule of M . Choose C maximal so that C ∩ B = 0 (Zorn’s Lemma gives us this). Then by Lemma 289, C + B is large and so by part 2., C + B = M . This means C is a complement to B in L(M ). 3. implies 1. Since L(M ) is complemented, Soc(M ) has a complement C in L(M ). Since we are trying to prove that Soc(M ) = M , let us assume by way of contradiction that Soc(M ) is a proper submodule of M . Then its complement C is non-zero. Now by Lemma 290, the sublattice L(C) is also complemented. Then by Lemma 291, we must have rad(C) = 0. Thus, for each nonzero element c of C, there is a maximal submodule Cc of C not containing c. Then Cc has a complement C 0 in L(C), so C = Cc ⊕ C 0 . Since Cc is a maximal submodule, C 0 ' C/Cc is irreducible. Thus C 0 is a minimal submodule of M not lying in Soc(M ), which is a contradiction. Thus M = Soc(M ) so M is completely reducible. The proof is complete. 428 12.3 Homogeneous Components. 12.3.1 The action of the R-endomorphism ring. Let M be a right R-module and let E = homR (M.M ) be its R-endomorphism ring. As remarked earlier, we may regard M as a left E-module, E M , and because of the “associative law” e(mr) = (em)r for all (e, m, r) ∈ E × M × R, we say that in this case that M is naturally an (E, R)-bimodule, denoted E MR . (Obviously, by replacing E by a ring S with morphism S → E, we obtain a general definition of an (S, R)-bimodule.) Now let Mi be any irreducible submodule of M and let e be an endomorphism of M (i.e. e ∈ E). Then, as remarked in the proof of Lemma 287 of the previous section, either eMi = 0 or else eMi is irreducible. But in the latter case we should also notice that E, restricted to Mi , is an isomorphism Mi → e(Mi ). This motivates the following definition: If Mi is an irreducible P submodule of M , we call the sum Mj over all irreducible submodules Mj of M that are isomorphic to Mi , the homogeneous component of M containing Mi . Clearly the homogeneous components of M are submodules of Soc(M ). Lemma 293 If M is a completely reducible R-module and E = homR (M, M ) and Mi is any irreducible submodule of M , then EMi is the homogeneous component of M containing Mi . proof: If e ∈ E, then eMi is either the zero submodule, or is an irreducuble module isomorphic to Mi . In either case eMi lies in the homogeneous component containing Mi . Conversely, if Mk is a submodule isomorphic to Mi , then there is a complement Mi0 to Mi in the lattice L(M ), so M = Mi ⊕ Mi0 . Then there is an endomorphism e : M → Mk with kernel Mi0 obtained by composing the projection onto Mi with the isomorphism Mi → Mk . 12.3.2 The Socle of a Module is a Direct Sum of the Homogeneous Components. For the moment, we don’t even know that the homogeneous components do not intersect non-trivially, or that they are ever proper submodules. This is 429 clarified by the next Lemma 294 Suppose C1 is the homogeneous component obtained as a sum of all irreducible submodules of M isomorphic to the irreducible submodule M1 . Suppose M0 is an irreducible submodule of M lying within C1 . Then M0 ' M1 . proof: First, C1 is a submodule of Soc(M ) which is completely reducible. Thus L(C1 ) is complemented. So there is a complementing submodule M00 to M0 in L(C1 ). Let F be the collection of all irreducible submodules of M which are isomorphic to M1 . Since these modules generate C1 , there exists one –say N1 – not in M00 . Then from irreducibility, N1 ∩ M00 = 0 so N1 ' (N1 + M00 )/M00 ≤ M/M00 = (M0 ⊕ M00 )/M00 ' M0 . Since M0 is irreducible, this shows N1 ' M0 and so M0 ∈ F. Corollary 295 Soc(M ) is the direct sum of the homogeneous components of M. proof: Let I index the isomorphism types of the irreducible submodules of M and for σ ∈ I let Cσ be the corresponding homogeneous component of M . As remarked above ECσ = Cσ ≤ Soc(M ). P Obviously Soc(M ) = σ∈I Cσ since every irreducible module lies in some Cσ . We claim that for any finite subset K of the index set I, and σ ∈ I − K, that X ( k∈K Ck ) ∩ Cσ = 0. P This will imply that the sum σ∈I Cσ = Soc(M ) is direct, completing the proof. Suppose the contrary, that for some σ and K as above, C := ( X k∈K Ck ) ∩ Cσ 6= 0. Since C is completely reducible, it contains an irreducible submodule M0 . Since it is also an irreducible submodule of the completely reducible module P P 0 K Ck , it contains a complement M0 in it. Since K Ck is generated by its 430 irreducible modules, there exists an index s ∈ K and irreducible submodule Ms of Cs such that Ms does not lie in M00 . Then considering that M0 is P irreducible and M00 is maximal in K Mk , the latter can be expressed as both M0 ⊕ M00 and as Ms ⊕ M00 . Thus M0 ' Ms . Now by Lemma 294, all the irreducible submodules of Cσ and Cs are of the same isomorphism type, against σ ∈ / K. So the claim holds. The next sections concern rings rather than modules. But we can already apply a few of the results of this section to the module RR . Note that because of the associative law in R, left multiplication of RR by any element of R is an R-endomorphism of RR , and so by Lemma 293 leaves each homogeneous component of RR invariant. Thus we see Corollary 296 Every homogeneous component of Soc(RR ) is a 2-sided ideal of R. Also Soc(RR ) is itself a 2-sided ideal of R (which we write as Soc(R)). 12.4 Semiprime rings. 12.4.1 Introduction We now pass to rings. The basic class of rings that we shall deal with are the semiprime rings and certain subspecies of these. In order to state the defining properties, we should touch base by reviewing three notions about 2-sided ideals that may or may not have been covered before. If A and B are 2-sided ideals of the ring R, the product AB of ideals is the additive group generated by the set of all products {ab|(a, b) ∈ A × B}. This additive group is itself already a 2-sided ideal, which is denoted AB, and which lies in the ideal A ∩ B. (Taking products among ideals is associative.) We say that a (2-sided) ideal A is nilpotent if and only if there exists some positive integer n such that An := A · A · · · · A(to n factors ) = 0 := {0R }, the zero ideal. An ideal P is said to be a prime ideal if and only if, whenever A and B are ideals such that AB ⊆ P , then either A ⊆ P or B ⊆ P . 431 12.4.2 The semiprime condition Theorem 297 The following conditions on a ring R are equivalent: 1. 0 is the only nilpotent ideal. 2. 0 is the intersection of all prime ideals. 3. For any ideals A and B of R, AB = 0 implies A ∩ B = 0. proof: 1. implies 2. Let a0 = a 6= 0. Then Ra0 R is not nilpotent. So its square, Ra0 Ra0 R is non-zero, so there is a non-zero element a1 in a0 Ra0 . In general we can iterate this process to find an element ai+1 non-zero in ai Rai . Set T = {an |n ∈ {0} ∪ N}. Before proceeding further, we note an important property of this sequence: (S) If element ai is in an ideal J, then so are all of its successors in T . Let P be an ideal maximal with respect to not containing any element of T . (It exists by Zorn’s Lemma.) Now suppose A and B are ideals of R neither of which is contained in P but AB ⊆ P . By the maximality of P the ideals A + P and B + P both meet T non-trivially. So there are subscripts i and j such that ai ∈ A + P , and aj ∈ B + P . Set m = max(i, j). Then by the property (S), am lies in both P + A and P + B, so am+1 ∈ am Ram ⊆ (A + P )(B + P ) ⊆ AB + P ⊆ P against P ∩ T = ∅. Thus no such ideals A and B exist, so P is a prime ideal. Thus, we have just argued, for each non-zero element a of R, there is a prime ideal P = Pa not containing it. It follows that the intersection of all prime ideals of R is the zero ideal 0 := {0R }. 2. implies 3. Assume A and B are ideals for which AB = 0. Then AB ⊆ P for any prime ideal P so either A lies in P or B lies in P . In any event A ∩ B ⊆ P for each prime ideal P . Thus A ∩ B is contained in the intersection of all prime ideals, and so is zero. 3. implies 1. Suppose A is a nilpotent ideal. Then for some positive integer n, An = 0. Then by 3., 0 = A ∩ An−1 = An−1 = A ∩ An−2 = · · · = A. Thus 1. holds. 432 A ring satisfying any of the conditions above is called semiprime. An easy way to characterize semiprime rings is given in Lemma 298 The following conditions are equivalent: 1. R is semiprime. 2. For each non-zero element x in R, xRx 6= 0. proof: 1. implies 2. Suppose 1., but suppose that 2. fails. Then there is a non-zero x such that xRx = 0. Then as x ∈ RxR, we see that RxR is a non-trivial ideal which is nilpotent since (RxR)2 = RxRxR = R · 0 · R = 0. This contradicts 1. 2. implies 1. Now, by way of contradiction, assume that 2. holds, but that 1. fails. Then there is a non-trivial nilpotent ideal A. Let n be the smallest positive integer such that An = 0 (here, n ≥ 2). Then N := An−1 is a non-trivial ideal and N 2 = 0. Thus if x is any non-zero element of N , we have xRx ⊆ N 2 = 0, contrary to our assumption 2. 12.4.3 Completely Reducible and Semiprimitive Rings are Species of Semiprime Rings At the beginning of this chapter we have defined the radical of a right Rmodule M to be the intersection of all maximal submodules of M , or, if there are no maximal submodules, to be M itself. Similarly, we define the radical, Rad(R), of the ring R to be rad(RR ) – the radical of the right module RR . Accordingly, Rad(R) is the intersection of all maximal right ideals of R and is called the Jacobson Radical of R. Although we do not require this fact, it turns out that the Jacobson radical, despite its being defined in terms of right ideals, is actually a 2-sided ideal. (For a neat proof of this fact see Lambeck:Rings and Modules, page 57.) One can then “mod out” the radical and observe that the factor ring has trivial Jacobson radical. But these are side issues. A ring is said to be semiprimitive if and only its Jacobson radical is trivial. Finally we say that a ring R is completely reducible if and only if the right module RR is a completely reducible module. Now we can state 433 Theorem 299 The following implications hold among properties of a ring: completely reducible ⇒ semiprimitive ⇒ semiprime proof: The first implication is Lemma 291 applied to the module M = RR . For the second implication suppose R is a ring which is semiprimitive but not semiprime. Then Rad(R) = 0 and there exists a non-zero nilpotent ideal N . By replacing N by an appropriate power if necessary, we may assume N 2 = 0 6= N . Let M be a maximal right ideal and suppose N is not contained in M . Then R = N + M by maximality of M . Then 1 = n + m, where n ∈ N and m ∈ M . Then n = 1 · n = n2 + mn = 0 + mn ∈ M. Thus 1 = n + m ∈ M which is impossible. Thus we see that N must lie in every maximal right ideal, so N ⊆ Rad(R) = 0. This contradicts N 6= 0. Thus no such ring R exists, so the implication holds. 12.4.4 Idempotents and Minimal Right Ideals of Semiprime Rings. The importance of semiprimness for us is that it affects the structure of the minimal right ideals of R. But first we need a lemma which holds for rings generally. Lemma 300 (Brauer’s lemma) Let K be a minimal right ideal of a ring R, Then either (i) K 2 = 0, or (ii) K = eR where e2 = e. proof: Suppose K 2 6= 0. Then by the minimality of K, K = K 2 = kK for some element k ∈ K. Now observe that AnnK (k) := {x ∈ K|kx = 0}, the right annihilator of k in K, is a right ideal of R lying properly in K. Hence it is the zero ideal. Then there is a unique element e such that ke = k. Then k(e2 − e) = (ke)e − ke = ke − ke = 0, 434 so e2 − e ∈ AnnK (k) = 0, so e2 = e. Then 0 6= eR ⊆ K, so eR = K as K was minimal. Corollary 301 If R is semiprime, then every minimal right ideal of R has the form eR where e2 = e. proof: Let K be a minimal right ideal of R. Then by Brauer’s Lemma, the conclusion holds or else K 2 = 0. But in the latter case, RK is a 2-sided ideal containing K which is nilpotent, since (RK)2 = R(KR)K ⊆ RK 2 = 0. This contradicts the fact that R is semiprime. 12.4.5 Idempotents and Left-Right Symmetries An element e in a ring R with the property that e2 = e is called an idempotent. Clearly the defining property of an idempotent is left-right symmetric; yet it makes its first appearance here in the two results of the previous subsection in connection with right ideals. We wish to exploit the symmetry of idempotents to acheive results relating minimal right ideals to minimal left ideals, right Socles to left Socles and so on. First we require a technical observation: Lemma 302 (Technical Lemma on Idempotents) Let e be an idempotent of R and let f be any element of R. Then there is a bijective morphism of additive groups homR (eR, f R) → f Re. If, moreover f = e, this an isomorphism of rings. proof: We define a mapping µ : f Re → homR (eR, f R), where µ(f re) is left multiplication of eR by f re. Thus µ(f re) sends element er0 of eR to (f re)er0 = f rer0 ∈ f R. Clearly associativity of multiplication 435 in R reveals that this left multiplication is an R-homomorphism of right Rmodules: eR → f R. Clearly, eRf is a subgroup of (R, +), and homR (eR, f R) is an additive group in the usual way. Since µ(f r1 e + f r2 e)(er0 ) = = = = µ(f (r1 + r2 )e)(er0 ) f (r1 + r2 )er0 = f r1 er0 + f r2 er0 µ(f r1 e)(er0 ) + µ(f r2 e)(er0 ) (µ(f r1 e) + µ(f r2 e))(er0 ) for all er0 ∈ eR, we see that µ is a group homomorphism. If µ(f re) is the zero R-homomorphism, then f reR = 0R so f re = 0. Thus ker µ = 0R , and µ is injective. To show that µ is a surjective (“onto”) mapping we need only show that any R-homomorphism φ ∈ homR (eR, f R) has the form of an appropriate left multiplication. Now φ(e) = f r for some element r ∈ R. Then for any r0 ∈ R, φ(er0 ) = φ(e2 r0 ) = φ(e) · er0 = f rer0 = (f re)(er0 ) as required. Thus µ is an isomorphism of the two additive groups. For the last statement, we need only observe that when f = e, then f Re = eRe is a subring of R, homR (eR, eR) is a ring with composition of morphisms as its ring multiplication, and that µ preserves multiplication. To demonstrate the latter, observe that for arbitrary elements r1 and r2 , the action of µ((er1 e)(er2 e)) = µ(er1 er2 e) is left multiplication of eR by the element er1 er2 . But as R is associative, the latter is left multiplication by er2 e followed by left multiplication by er1 e, the composition µ(er1 e)◦µ(er2 e). The proof is complete. Lemma 303 If R is semiprime and e2 = e ∈ R, then eR is a minimal right ideal if and only eRe is a division ring. proof: If eR is an irreducible right R module (minimal right ideal), then by Lemma 302 above, eRe ' hom(eR, eR), and the latter is a division ring by Schur’s Lemma (which was Corollary 7.16, to Fitting’s Lemma, Theorem 7.15). Conversely, suppose eRe is a division ring. Let r be such that er 6= 0. Then as R is semiprime, erRer 6= 0 (see Lemma 298). Thus erse 6= 0 for 436 some s ∈ R. But as erse is in the division ring eRe, it has an inverse ete. Then erse(ete) = ersete = e. thus erR = eR. Now if there were a right ideal A such that 0 < A < eR, we could find a non-zero er ∈ A. But by the sentence before last, eR = erR ⊆ AR ⊆ A, a contradiction. Thus no such A may exist and eR is a minimal right R-module. Corollary 304 If R is semiprime and e2 = e ∈ R, then eR is a minimal right ideal if and only if Re is a minimal left ideal. proof: In the preceeding Lemma 303, the condition that eRe is a division ring has complete left-right symmetry. Thus if it is equivalent to eR being a minimal right ideal, it is also equivalent to Re being a minimal left ideal. We require one last technical result. Lemma 305 Let e and f be idempotents of R. Then eR and f R are isomorphic as right R-modules if and only if there exist elements u and v in R such that vu = e and uv = f . proof: First assume there is an isomorphism of eR onto f R. Now by the Technical Lemma 302, this isomorphism corresponds to left multiplication of eR by an element u := f u0 e in f Re. Since f and e are idempotents, f ue = u, so without loss we may take u0 = u. Similarly, the inverse isomorphism corresponds to left multiplication of f R by an element v = evf . Then, becauses these isomophisms are inverses of each other, vue = vu = e, and similarly, uv = f . Conversely, assume that elements u and v exist such that vu = e and uv = f . Then ue = u(vu) = (uv)u = f u and similarly vf = ev. Then left multiplication of eR by u, acting as: er → uer = f ur, is an isomorphism: eR → f R. Its inverse is left multiplication of f R by v, acting as: f r0 → vf r0 = evr0 . We obtain, then, a Corollary relating left ideals to right ideals: 437 Corollary 306 If e and f are idempotents of a ring R, then the right ideals eR and f R are isomorphic as right R-modules if and only if the left ideals Re and Rf are isomorphic as left R-modules. proof: This follows from the left-right symmetry of the pair of conditions uv = f and vu = e. Theorem 307 If R is semiprime, then R R and RR have the same Socles. Also, they have the same homogeneous components, and these components are the minimal 2-sided ideals of R. P proof: S := Soc(RR ) = eR where e ranges over all idempotents e for P which eRe is a division ring. Similarly, S 0 := Soc(R R) = Re where the sum extends over the same set of idempotents e. Now by Corollary 296 S is a 2-sided ideal of R, so S 0 ⊆ S. By symmetry S = S 0 Next let H be any homogeneous component of S = Soc(RR ). Then P H = eR where now e ranges over those idempotents e such that eR is isomorphic as right R-module to some fixed minimal right ideal gR, where g 2 = g. Now by Corollary 306 above, eR is isomorphic to gR as (irreducible) right R-modules if and only Re is isomorphic to Rg as left R-modules. By Corollary 304 Rg is a minimal left ideal. Thus RH = H contains all Re such that Re is isomorphic to Rf , whose sum is a homogeneous component H 0 of 0 0 R R. Thus H ⊆ H. By symmetry again H = H . Now we show that H is a minimal 2-sided ideal. Suppose K is a nonzero ideal contained in H. Since H is a sum of minimal right ideals, it is a completely reducible right R-module, and so, therefore, is K. Thus K contains a minimal right ideal eR. Then by Lemma 294, all irreducible submodules of H are isomorphic to eR. Each of these has the form f R where f 2 = f , and since f R ' eR there must exist elements u and v such that f = uv = uev (Lemma 305). Thus f R = u(eR) ⊆ uK ⊆ K. Thus H, being a sum of all such f R, lies in K, whence H = K. This shows that H = H 0 is a minimal ideal of R. We apply these ideas to completely reducible rings in the next section. 438 12.5 Completely Reducible Rings The left-right symmetry we developed in the previous section can be applied to characterization of completely reducible rings. Theorem 308 The following four statements are equivalent for any ring R. 1. Every right R-module is completely reducible. 2. RR is completely reducible. 3. Every left R-module is completely reducible. 4. RR is completely reducible. proof: Clearly 1. implies 2. P Assume 2., so R = I Ai where the Ai are some minimal right ideals. Let M be any right R-module. Then clearly M= X mR = m∈M X m∈M X i∈I mAi . (12.2) Now if N were a non-zero submodule of one of the summands mAi , then the set B := {r ∈ Ai |mr ∈ N } is a right ideal of R contained in Ai , itself a minimal right ideal. Now elements of N have the form ma,a ∈ Ai . Then such an a is in B. Thus B 6= 0, so B = Ai and so N = mAi . Since N was arbitrary in mAi , we see that each summand in the right hand side of equation (12.2) is zero or is an irreducible module. Then by Corollary 288 at the beginning of this Chapter, M is completely reducible. So 1. and 2. are equivalent. Similarly, 3. and 4. are equivalent (just place the proof in the preceeding paragraphs in front of a mirror). We next show that 2. implies 4. Assume 2. Then R is semiprime, since by Theorem 299, completely reducibility implies semiprime-ness of rings. But then, by Theorem 307, R R and RR have the same Socles. Since R = Soc(RR ) we see that Soc(R R) = R so R R is completely reducible. By symmetry 4. implies 2. We are done. There is an immediate application: Corollary 309 Let D be a division ring. Then any right vector space V over D is completely reducible. In particular V is a direct sum of copies of DD . 439 proof:DD is an irreducible module so any right D module V (for that is what a vector space is!) is completely reducible. We need, for a moment, a very special notion. A ring is said to be a prime ring if and only if 0 is a prime ideal. This means that if A and B are ideals such that AB = 0 than either A = 0 or B = 0. Note that a prime ring is semiprime, for in a prime ring no non zero ideal K satisfies K 2 = 0. Lemma 310 Let R be a prime ring. Assume S = Soc(RR ) 6= 0. Then if eR is a minimal right ideal of R, then homR (S, S) is isomorphic to the ring of all linear transformations homeRe (Re, Re) . proof: Since S = Soc(RR ) is a 2-sided ideal and is the direct sum of its homogeneous components each of which is a 2-sided ideal, the prime hypothesis implies that S is itself a homogeneous component. Let S be P written as a direct sum ei R of minimal right ideals with the ei idempotent, and fix one of these, say e = e1 . Then each ei R ' eR so by Lemma 305, there exist elements ui and vi such that vi ui = e and ui vi = ei . (Note that as e = e1 , u1 = v1 = 1.) Then ui evi = ui vi ui vi = e2i = ei . Now consider any φ in homR (S, S). We define τ (φ) : Re → Re by the rule τ (φ)(re) = φ(re) = φ(re)e, for all r ∈ R. Clearly τ (φ) is right eRe-linear. Also, for φ1 , φ2 ∈ homR (S, S) , we have τ (φ1 + φ2 ) = τ (φ1 ) + τ (φ2 ). Also, τ (φ1 ◦ φ2 )(re) = (φ1 ◦ φ2 )(re)) · e. = φ1 (φ2 (re)e) = φ1 (φ2 (re)) · e So τ is a ring homomorphism. 440 Clearly, τ (φ) = 0 if and only if φ(re) · e = 0 for all r ∈ R. So φ(ei ) = φ(ui evi ) = φ(ui e)vi = 0 · vi = 0. for all indices i. Thus φ(S) = 0, the zero mapping. It follows that τ is one-to-one. We shall show that τ is onto by producing a mapping σ : homeRe (Re, Re) → homR (S, S) such that τ ◦ σ = 1, the identity map on homeRe (Re, Re). Now fix ψ ∈ homeRe (Re, Re). Define σ(ψ) : S → S as follows: If s ∈ S, we can uniquely write X X s = ei ri = ui evi ri . Then set σ(ψ)(s) := X ψ(ui e)vi ri , which is completely determined by ψ and the unique ri . Now, in order to evaluate τ ◦ σ, we express an arbirary element re of Re P in the unique form ui evi ri . (This is a finite sum.) Now τ (σ(ψ))(re) = σ(ψ)(re) · e X = ( ψ(ui e) · vi ri ) · e = X ψ(ui e2 ) · vi ri · e = X ψ(ui e) · euvi ri e X = ψ( ui evi ri e) = ψ(re · e) = ψ(re). Notice that the last three lines of equations use the fact that ψ is an eRemorphism in two places. Thus τ is onto and so we may conclude that τ : homR (S, S) → homeRe (Re, Re) is a ring isomorphism. Theorem 311 (Wedderburn-Artin) 1. A ring R is completely reducible if and only if it is isomorphic to a finite direct product of completely reducible simple rings. 2. A ring R is completely reducible and simple if and only if it is the ring of all linear transformations of a finite dimensional vector space. 441 proof: Part 1. Let R be completely reducible, so R is a direct sum of minimal right ideals. Now the identity element 1 is a finite sum of elements in these direct summands so 1 = e1 + · · · + en and 12 = 1 and the directness of the sum implies e2i = ei and R = ei R ⊕ · · · ⊕ en R, ei R a minimal right ideal. Collecting R = SocRR as a sum of homogeneans components Cσ , again expresssing 1 as a finite sum of elements from these components, we see that 1 = c1 + c2 + · · · + ck where ci ∈ Ci . Again 12 = 1 implies c2i = ci . But since Ci is a 2-sided ideal of R, the directness of the sum R = ⊕k1 Ci allows us to infer from the equation r1 = 1r, the equations rci = ci r, i = 1, · · · , k, for all r ∈ R. Thus the ci are central idempotents, and each component ci R = Rci = Ci is itself a ring with identity element ci . Let Ci = H be any homogeneous component. Since Ci · Cj = 0 for i 6= j, it is easy to see that the ideals and right ideals of R which lie in H are precisely the ideals and right ideals of the ring H itself. Thus H is a sum of minimal right ideals of H and so H is completely reducible. Why is H simple? Because by Theorem 307 H is a minimal ideal of R. Conversely, if R is a sum of ideals H which are sums of irreducible right R-modules, RR is completely reducible. Part 2. Let R be completely reducible and simple. Let eR be a minimal right ideal of R, where e2 = e in R. Now RR is its own socle, hence R ' HomR (R, R) ' HomeRe (Re, Re). (12.3) The second isomorphism is the previous Lemma 310. The first isomorphism sends element r to left multiplication of R by the element r. By the right distributive law and the associative law, the latter is an R-homomorphism. By the left distributive law, this morphism R → homR (R, R) is a homomorphism of rings. Finally, rR = 0 if and only if r = 0, so it is injective. On the other hand, if φ is an arbitrary R-endomorphism of RR , we may set φ(1) := fφ . Then for any element r in R, φ(r) = φ(1 · r) = φ(1)r = fφ · r, – that is the left action of φ on RR is just left multiplication of all the elements of R by an element, fφ of R. Thus the first mapping in equation (12.3) is also 442 surjective, and so depicts an isomorphism of rings. (In fact this part of the argument assumed nothing about R; the first isomorphism always holds.) Clearly homeRe (Re, Re) is a full ring of linear transformations of the right vector space Re over eRe. It remains only to show that Re has finite dimension over the division ring eRe. It suffices to show that (Re)eRe is Noetherian. To prove this we note that as RR is a direct sum of minimal right Rmodules, Mj , we can write the identity element as a finite sum 1 = m1 + m2 + · · · + mn where mi ∈ Mi − {0}. Then clearly RR = 1 · R = ⊕mi R, and so the increasing partial sums form a composition series of RR of length n. By the Jordan-Hölder theorem, and the fact that the lattice of submodules of RR is modular, any chain of submodules of RR has length at most n, and so RR is Noetherian. Let K = {Kτ } be the family of all subspaces (that is, eRe-submodules) of Re. For each Kτ ∈ K, Kτ R is a right R-submodule of (ReR)R . But as RR is Noetherian, so is its submodule (ReR)R . Thus among the submodules Kτ R, Kτ ∈ K, we can find one – say Kµ R which is maximal. Then if Kµ ≤ Kσ , then Kµ R = Kσ R. But, noting that as e is the identity of eRe, Kµ e = Kµ , we have Kµ = Kµ eRe = Kµ Re = Kσ Re = Kσ eRe = Kσ Thus Kµ is a maximal element of K. Thus all ascending chains K1 < K 2 < · · · < · · · in K convert to chains K1 R < · · · which stabilize at some Km R from which it follows that Km is maximal in K. Thus Re is a Noetherian eRe-module, and so has finite dimension. 12.6 Exercises for Chapter 12 12.6.1 Exercises relating to Sections 12.1 and 12.2 All of the results of this subsection possess a strong lattice-theoretic component. What I propose here is really a project – that means that the proposer 443 does not know the answer but beleives it answerable. So the basic thrust of these exercises is for the student to investigate to what extent it is possible to recast the theory of this section in the context of abstract lattices. It would seem that minimally we must assume that L is a complete lattice. This means that L has a “zero” element, 0 bounding below all elements of L, a “universal one” which bounds above all elements, and unrestricted joins and meets, which means that any subset of L contains a least upper bound and greatest lower bound in L. There are then two pertinent subsets of L: max(L) , the set of all elements which are maximal with respect to being distinct from 1, and min(L), the set of elements of L which are minimal with respect to being distinct from the lattice zero (usually called the atoms of a poset with 0). Either of these sets might be empty. We can then define : rad(L) = ( V Soc(L) = ( W if max(L) 6= ∅, if max(L) = ∅ A∈min(L) A L A∈min(L) A 0 if min(L) 6= ∅, if min(L) = ∅ Also, with these lattices, we may define the analogue of a direct sum. We say that d is the direct join of the set {ai |i ∈ I} if and only if 1. d = W i∈I ai 2. For each finite subset J ⊆ I, and σ ∈ I − J, we have aσ ∧ _ a j∈J j = 0. Exercise 1. With only these hypotheses on L, show that Soc(L) is the direct join of some set of minimal elements. [One defines direct subsets of a set I indexing min(L) and uses the Zornification to obtain a maximal direct set as in the proof of Theorem 287.] Exercise 2. Of course there is the usual notion of duality of posets under which (P, ≤) becomes (P, ≥). Then complete lattices as defined just above is a self-dual notion. Define a notion of direct meet which would be the dual of “direct sum” as above. Then one can define “codirect” subsets of the set K indexing max(L). State the dual form of the result of Exercise 1. 444 More remarks: One notices that the later theorems of this section link facts about rad(M ) with Soc(M ). There are two mediating concepts: L(M ) complemented and the notion of large submodules. So let us say that a complete lattice L is complemented if and only if for every element c of L, there exists an element c0 such that 1. c ∨ c0 = 1 2. c ∧ c0 = 0. The second concept is the notion of a “large“ submodule. So we say that an element l is large in L if and only if for every non-zero element d in L, one has l ∧ d 6= 0. These two notions are easy to define, but do they work in the same way? One of the Lemmas employing large submodules in it’s proof relied on recognizing that a certain module C + B where C was maximal with respect to meeting B trivially, was large. Even to get such modules to exist, a Zornification was used. Even later, a Zornification argument chose a submodule maximal with respect to not containing a module element m. I propose two analogues to these arguments in a general lattice L: the Zorn Axioms: (M1) For every non-zero element d of L, there exists an element m of L maximal with respect to d not less than m. (M2) For every non-zero element d of L, there exists an element m of L maximal with respect to d ∧ m = 0. Finally, there is one further notion used throughout the arguments of this section. The lattice of submodules of a module is modular. Recall from Chapter 1, that a lattice (complete or not) is modular if and only if The Modular Law (Dedekind’s Lemma) Whenever we have a ≤ b, then for any c in L, b ∧ (a ∨ c) = a ∨ (c ∧ b). Suppose L(M ) is complete modular lattice satisfying hypotheses (M1) and (M2). Investigate these questions: Exercise 3. If L is complemented, does L = Soc(L)? 445 Exercise 4. Is it true that when rad(L) = 0, that there are no large elements? (You may assume (M1).) Exercise 5. Is there a way to get around the arguments involving individual elements of M to adapt Lemma 289 to modular complete lattices satisfying any desired Zorn-axiom, (M1) or (M2)? Exercise 6. Write a report adapting the theory of section to complete modular modular lattices satifying the Zorn Conditions (M1) and (M2). Explain why it works or doesn’t work. 12.6.2 Warm up Exercises 7. An element x of a ring R is nilpotent if and only if, for some positive integer n, xn = 0. Show that no nilpotent element is right or left invertible. 8. Show that if x is a nilpotent element, then 1 − x is a unit of R. 9. A right ideal A of R is said to be nilpotent if and only if, for some positive integer n (depending on A) An = 0. Show that if A and B are nilpotent right ideals of R, then so is A + B. 10. (An interesting example) Let F be any field. Let F hx, yi denote the ring of all polynomials in two “non-commuting” variables x and y. Precisely F hx, yi is the monoid-ring F M where M is the free monoid generated by the two-letter alphabet {x, y}. (We shall see later on that this ring is the tensor algebra T (V ) where V is a two-dimensional vector space over F .) Let I be the 2-sided ideal generated by {x2 , y 2 }, and form the factor ring R := F hx, yi/I. (a) For each positive integer k, let (x)k := xyx · · · ( k factors ) + I, (y)k := yxy · · · ( k factors ) + I. Show that the set {1, (x)k , (y)k |0 < k ∈ Z} is a basis for R as a vector space over F . Thus each element of R is a finite sum r = α0 · 1 + X i∈J αi (x)i + βi (y)i , where J is a finite set of positive integers. 446 (b) Show that the sum (x+I)R +(y +I)R of two right ideals is direct, and is a maximal two-sided ideal of R. (c) An element (x)k is said to be x-palendromic if it begins and ends in x – i.e., if k is odd. Similarly (y)k is y-palendromic if and only if k is odd: Show that any linear combination of x-palendromic elements is nilpotent. (d) Let Px ( Py ) denote the vecor subspace spanned by all x-palendromic (y-palendromic) elements. Then the group of units of R contains (1+Px )∪(1+Py ) and hence all possible products of these elements. (e) Show that for any positive integer k, (x+y)k = (x)k +(y)k . (here is a sum of two nilpotent elements which is certainly not nilpotent.) 11. A right ideal is said to be a nil right ideal if and only all of its elements are nilpotent. (If the nil right ideal is 2-sided, we simply call it a nil ideal.) (a) For any family F of nilpotent right ideals, show that the sum of all ideals in F is a nil ideal. P F (b) Show that if A is a nilpotent right ideal, then so is rA. (c) Define N (R) to be the sum of all nilpotent right ideals of R. (This invariant of R can be thought of as a certain kind of “radical” of R.) Show that N (R) is a 2-sided nil ideal. 12.6.3 Exercises concerning the Jacobson radical Recall that Jacobson radical of a ring R (with identity) was defined to be the intersection of all the maximimal right ideals of R, and was denoted by the symbol Rad(R). 12. Show that an element r of the ring r is right invertible if and only if it belongs to no maximal right ideal. 13. Let “1” denote the multiplicative identity element of R. Show that the following four conditions on an element r ∈ R are equivalent: (a) r belongs to the Jacobson radical of R. (b) For each maximal right ideal M , 1 does not belong to M + rR. 447 (c) For every element s in R, 1 − rs belongs to no maximal right ideal of M . (d) For every element s in R, 1 − rs is right invertible. 14. Show that if an element 1 − rt is right invertible in R, then so is 1 − tr. [Hint: By hypothesis there exists an element u such that (1 − rt)u = 1. So we have : u = 1 + rtu. Use this to show that the product (1 − tr)(1 + tur) = 1 + tur − t(1 + rtu)r, is just 1.] 15. Demonstrate that the Jacobson radical Rad(R) is a two-sided ideal – that is, if r ∈ Rad(R), then so is tr. 16. Show that Rad(R) contains every nilpotent 2-sided ideal.[Hint: If false, there is a nilpotent 2-sided ideal A and a maximal ideal M such that A2 ⊆ M while R = A + M . Then 1 = a + m where a ∈ A and m ∈ M . Thus 1 − a is on the one hand an element of M , and on the other hand it is a unit by Exercise 8.] 12.6.4 The Jacobson radical of Artinian rings A ring whose poset of right ideals satisfies the descending chain condition is called an Artinian ring. The next group of exercizes will lead the student to a proof that for any Artinian ring, the Jacobson radical is a nilpotent