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Chapter 4 Algebras 4.1 Definition It is time to introduce the notion of an algebra over a commutative ring. So let R be a commutative ring. An R-algebra is a ring A (unital as always) that is an R-module (left, say) such that r(ab) = (ra)b = a(rb) for all r ∈ R, a, b ∈ A. You can say it the other way round: an R-algebra is an R-module equipped with an R-bilinear multiplication R × A → A making it into a ring. Thus, this multiplication satisfies the conditions (A1) 1R a = a; (A2) (r + s)a = ra + sa; (A3) r(a + b) = ra + rb; (A4) (rs)a = r(sa); (A5) r(ab) = (ra)b = a(rb) for all r, s ∈ R, a, b ∈ A. (Note strictly speaking I should call this an “associative, unital R-algebra” – there are other important sorts of algebra which are not associative – but since we won’t meet them in this course I’ll just stick to algebra.) Note a Z-algebra is just the old definition of ring: “Z-algebras = rings” just as “Z-modules = Abelian groups”. So you should view the passage from rings to R-algebras as analogous to the passage from Abelian groups to R-modules! This is the idea of studying objects (e.g. Abelian groups, rings) relative to a fixed commutative base ring R. There is an equivalent formulation of the definition of R-algebra: an R-algebra is a ring A together with a distinguished ring homomorphism (the “structure map”) s:R→A such that the image of s lies in the center Z(A) = {a ∈ A | ab = ba for all b ∈ A}. Indeed, given such a ring homomorphism, define a multiplication R × A → A by (r, a) $→ s(r)a. Now check this satisfies the above axioms (A1)–(A5) (the last one being because im s ⊆ Z(A)). Conversely, given an R-algebra as defined originally, one obtains a ring homomorphism s : R → A by defining s(r) = r1A , and the image lies in Z(A) by (A5). Let A be an R-algebra. Then, given an A-module M , we can in particular think of M as just an R-module, defining rm = s(r)m for r ∈ R, m ∈ M . So you can hope to exploit the additional structure of the base ring R in studying A-modules. In particular, if R = F is a field and A is 97 98 CHAPTER 4. ALGEBRAS an F -algebra, A and any A-module M is in particular a vector space over F . So we can talk about finite dimensional F -algebras and finite dimensional modules over an F -algebra, meaning their underlying dimension as vector spaces over F . Note given A-modules M, N , an A-module homomorphism between them is automatically an Rmodule homomorphism (for the underlying R-module structure). However, it is not necessarily the case that a ring homomorphism between two different R-algebras is an R-module homomorphism. So one defines an R-algebra homomorphism f : A → B between two R-algebras A and B to be a ring homomorphism in the old sense that is in addition R-linear (i.e. it is an R-module homomorphism too). Ring homomorphisms and R-algebra homomorphisms are different things! Now for examples. Actually, we already know plenty. Let R be a commutative ring. Then, the polynomial ring R[X1 , . . . , Xn ] is evidently an R-algebra: indeed, R[X1 , . . . , Xn ] contains a copy of R as the subring consisting of polynomials of degree zero. The ring Mn (R) of n × n matrices over R is an R-algebra: again, it contains a copy of R as the subring consisting of the scalar matrices. But note there is a big difference between this and the previous example: Mn (R) is finitely generated as an R-module (indeed, it is free of rank n2 ) whereas R[X1 , . . . , Xn ] is not. In case F is a field, Mn (F ) is a finite dimensional F -algebra, F [X1 , . . . , Xn ] is not. For the next example, let M be any (left) R-module. Consider the Abelian group EndR (M ). We make it into a ring by defining the product of two endomorphisms of M simply to be their composition. Now I claim that EndR (M ) is in fact an R-algebra: indeed, we define rθ for r ∈ R, θ ∈ EndR (M ) by setting (rθ)(m) = r(θ(m))(= θ(rm)) for all m ∈ M . Let us check that rθ really is an R-endomorphism of M . Take another s ∈ R. Then, s((rθ)(m)) = sr(θ(m)) = θ(srm) = θ(rsm) = (rθ)(sm). Note we really did use the commutativity of R! We can generalize the previous two examples. Suppose now that A is a (not necessarily commutative) R-algebra and M is a left A-module. Then, the ring D = EndA (M ) is also an R-algebra, defining (rd)(m) = r(d(m)) for all r ∈ R, d ∈ D, m ∈ M . For the final example of an algebra, let G be any group. Define the group algebra RG to be the free R-module on basis the elements of G. Thus an element of RG looks like ! rg g g∈G for coefficients rg ∈ R all but finitely many of which are zero. Multiplication is defined by the rule & ' ! ! ! rg g sh h = rg sh (gh). g∈G h∈G g,h∈G In other words, the multiplication in RG is induced by the multiplication in G and R-bilinearity. Note the RG contains a copy of R as a subring, namely, R1G , so is certainly an R-algebra. The construction of the group algebra RG allows the possibility of studying the abstract group G by studying the category of RG-modules – so module theory can be applied to group theory. Note finally in case G is a finite group and F is a field, the group algebra F G is a finite dimensional F -algebra. 99 4.2. SOME MULTILINEAR ALGEBRA 4.2 Some multilinear algebra Let F be a field (actually, everything here could be done more generally over an arbitrary commutative ring, e.g. Z – but let’s stick to the case of a field for simplicity). Given a vector space V over F , we have defined V ⊗ V (meaning V ⊗F V ). Hence, we have (V ⊗ V ) ⊗ V and V ⊗ (V ⊗ V ). But these two are canonically isomorphic, the isomorphism mapping a generator (u ⊗ v) ⊗ w of the first to u ⊗ (v ⊗ w) in the second. From now on we’ll identify the two, and write simply V ⊗ V ⊗ V for either. More generally, we’ll write T n (V ) for V ⊗ · · · ⊗ V (n times), where the tensor is put together in any order you like, all different ways giving canonically isomorphic vector spaces. Set ( T (V ) = T n (V ) n≥0 and call it the tensor algebra on V . Note T 0 (V ) = F by convention, just a copy of the ground field, and T 1 (V ) = V . Then, T (V ) is an F -vector space given, but we can define a multiplication on it making it into an F -algebra as follows: define a map T m (V ) × T n (V ) → T m+n (V ) = T m (V ) ⊗ T n (V ) by (x, y) $→ x ⊗ y. Then extend linearly to give a map from T (V ) × T (V ) → T (V ). Since ⊗ is associative, the resulting multiplication is associative, and T (V ) is an F -algebra. To be quite explicit, suppose that ei (i ∈ I) is a basis for V . Then, {ei1 ⊗ ei2 ⊗ · · · ⊗ ein | n ≥ 0, i1 , . . . , in ∈ I} gives a basis of “monomials” for T (V ). Multiplication of the basis elements is just by joining tensors together. In other words, T (V ) looks like “non-commutative polynomials” in the ei (i ∈ I) – an arbitrary element being a linear combination of finitely many non-commuting monomials. The importance of the tensor algebra is as follows: Universal property of the tensor algebra. Given any F -linear map f : V → A to an F algebra A, there is a unique F -linear ring homomorphism f¯ : T (V ) → A such that f = f¯ ◦ i, where i : V → T (V ) is the inclusion of V into T 1 (V ) ⊂ T (V ). (Note: this universal property could be taken as the definition of T (V )!) Proof. The vector space T n (V ) is defined by the following universal property: given a multilinear map fn : V × · · · × V (n copies) to a vector space W , there exists a unique F -linear map f¯n : T n (V ) → W such that fn = f¯n ◦ in , where in is the obvious map V × · · · × V → T n (V ). Now define the map fn so that fn (v1 , v2 , . . . , vn ) = f (v1 )f (v2 ) . . . f (vn ) (multiplication in the ring A). Check that it is multilinear in the vi , since A is an F -algebra. Hence it induces a unique f¯n : T n (V ) → A. Now define f¯ : T (V ) → A by glueing all the f¯n together. In other words, using the basis notation above, we have that f¯(ei1 ⊗ · · · ⊗ ein ) = f (ei1 )f (ei2 ) . . . f (ein ). This is an F -linear ring homomorphism, and is clearly the only option to extend f to such a thing. We can restate the theorem as follows: Let X be a basis of V . Then, given any set map f from X to an F -algebra A, there exists a unique F -linear homomorphism f¯ : T (V ) → A extending f (proof: extend the map X → A to a linear map V → A in the unique way then apply the theorem). In other words, T (V ) plays the role of the “free F -algebra on the set X”. Then you can define algebras by “generators and relations” but we won’t go into that... Next, we come to the symmetric algebra on V . Continue with V being a vector space. We’ll be working with V × · · · × V (n times). Call a map f : V × ··· × V → W 100 CHAPTER 4. ALGEBRAS a symmetric multilinear map if its multilinear, i.e. f (v1 , . . . , cvi + c# vi# , . . . , vn ) = cf (v1 , . . . , vi , . . . , vn ) + c# f (v1 , . . . , vi# , . . . , vn ) for each i, and moreover f (v1 , . . . , vi , vi+1 , . . . , vn ) = f (v1 , . . . , vi+1 , vi , . . . , vn ) for each i (so its symmetric). The symmetric power S n (V ), together with a distinguished map i : V × · · · × V → S n (V ), is characterized by the following universal property: given any symmetric multilinear map f : V × · · · × V → W to a vector space W , there exists a unique linear map f¯ : S n (V ) → W such that f = f¯ ◦ i. Note this is exactly the same universal property as defined T n (V ) in the proof of the universal property of the tensor algebra, but we’ve added the word symmetric too. As usual with universal properties, S n (V ), if it exists, is unique up to canonical isomorphism (so we call it “the” symmetric algebra). Still, we need to prove existence with some sort of construction. So now define S n (V ) = T n (V )/In where In = *v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn , is the subspace spanned by all such terms for all v1 , . . . , vn ∈ V . The distinguished map i : V × · · · × V → S n (V ) is the obvious map V × · · · × V → T n (V ) followed by the quotient map T n (V ) → S n (V ). Now take a symmetric multilinear map f : V × · · · × V → W . By the universal property defining T n (V ), there is a unique F -linear map f¯1 : T n (V ) → W extending f . Now since f is symmetric, we have that f¯1 (v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn ) = 0. Hence, f¯1 annihilates all generators of In , hence all of In . So f¯1 factors through the quotient S n (V ) of T n (V ) to induce a unique map f¯ : S n (V ) → W with f = f¯ ◦ i. Note we have really just used the univeral property of T n followed by the universal property of quotients! So now we’ve defined the nth symmetric power of a vector space. Note its customary to denote the image in S n (V ) of (v1 , . . . , vn ) ∈ V × · · · × V under the map i by v1 · · · · · vn . So v1 · · · · · vn is the image of v1 ⊗ · · · ⊗ vn under the quotient map T n (V ) → S n (V ). We can glue all S n (V ) together to define the symmetric algebra on V : ( S(V ) = S n (V ) n≥0 where again S 0 (V ) = F, S 1 (V ) = V . Since each S n (V ) is by construction a quotient of T n (V ), we see that S(V ) is a quotient of T (V ), i.e. S(V ) = T (V )/I ) where I = n≥0 In . I claim in fact that I is an ideal of T (V ), so that S(V ) is actually a quotient algebra of T (V ). Actually, I claim even more, namely: 4.2.1. Lemma. I is the two-sided ideal of T (V ) generated by the elements v ⊗ w − w ⊗ v for all v, w ∈ V . Proof. Let J be the two-sided ideal generated by the given elements. By definition, I is spanned as an F -vector space by terms like v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn = v1 ⊗ · · · ⊗ vi−1 ⊗ (vi ⊗ vi+1 − vi+1 ⊗ vi ) ⊗ vi+2 ⊗ · · · ⊗ vn . 4.2. SOME MULTILINEAR ALGEBRA 101 This clearly lies in J, hence I ⊆ J. On the other hand, the generators of the ideal J lie in I, so it just remains to show that I is a two-sided ideal of T (V ). Since T (V ) is spanned by monomials of the form u = u1 ⊗ · · · ⊗ um , it suffices to check that for any generator v = v1 ⊗ · · · ⊗ vi ⊗ vi+1 ⊗ · · · ⊗ vn − v1 ⊗ · · · ⊗ vi+1 ⊗ vi ⊗ · · · ⊗ vn of I, we have that uv and vu both lie in I. But that’s obvious! The lemma shows that S(V ) really is an F -algebra, as a quotient of T (V ) by an ideal. Multiplication of two monomials v1 · · · · · vm and w1 · · · · · wn in S(V ) is just by concatenation, giving v1 · · · · · vm · w1 · · · · · wn . Moreover, since · is symmetric, we can reorder this as we like in S(V ) to see that v1 · · · · · vm · w1 · · · · · wn = w1 · · · · · wn · v1 · · · · · vn . Hence, since these “pure” elements (the images of the pure tensors which generate T (V )) span S(V ) as an F -vector space, we see that S(V ) is a commutative F -algebra. Note not all elements of S(V ) can be written as v1 · · · · · vm , just as not all tensors are pure tensors. The symmetric algebra S(V ), together with the inclusion map i : V → S(V ), is characterized by the following universal property (compare with the universal property of the tensor algebra): Universal property of the symmetric algebra. Given an F -linear map f : V → A, where A is a commutative F -algebra, there exists a unique F -linear homomorphism f¯ : S(V ) → A such that f = f¯ ◦ i. Proof. Since A is commutative, the map f¯1 : T (V ) → A given by the universal property of the tensor algebra annihilates all elements v ⊗ w − w ⊗ v ∈ T 2 (V ). These generate the ideal I by the preceeding lemma. Hence, f¯1 factors through the quotient S(V ) of T (V ). Now suppose that V is finite dimensional on basis e1 , . . . , em . Then, we’ve seen S(V ) before! Indeed, let F [x1 , . . . , xm ] be the polynomial ring over F in indeterminates x1 , . . . , xm . The map ei $→ xi extends to a unique F -linear map V → F [x1 , . . . , xm ], hence since the polynomial ring is commutative, the universal property of symmetric algebras gives us a unique F -linear homomorphism S(V ) → F [x1 , . . . , xm ]. Now, S(V ) is spanned by the images of pure tensors of the form ei1 ⊗ · · · ⊗ ein . Moreover, any such can be reordered in S(V ) using the symmetric property to assume that i1 ≤ · · · ≤ in . Hence, S(V ) is spanned by the ordered monomials of the form ei1 . . . ein for all i1 ≤ · · · ≤ in and all n ≥ 0. Clearly, such a monomial maps to xi1 . . . xin in the polynomial ring F [x1 , . . . , xm ]. But we know (by definition) that the ordered monomials give a basis for the F -vector space F [x1 , . . . , xm ]. Hence, they must in fact be linearly independent in S(V ) too, and we’ve constructed an isomorphism: Basis theorem for symmetric powers. Let V be an F -vector space of dimension m. Then, S(V ) is isomorphic to F [x1 , . . . , xm ], the isomorphism mapping a basis element ei of V to the indeterminate xi in the polynomial ring. In particular, S n (V ) has basis given by all ordered monomials in the basis of the form ei1 · · · · · ein with 1 ≤ i1 ≤ · · · ≤ in ≤ m. In this language, the universal property of symmetric algebras gives that the polynomial algebra F [x1 , . . . , xm ] is the free commutative F -algebra on the generators x1 , . . . , xm . Note in particular that if V is finite dimensional, say of dimenion m, the basis theorem implies * + m+n−1 n dim S (V ) = n (exercise!). 102 CHAPTER 4. ALGEBRAS The last important topic in multilinear algebra I want to cover is the exterior algebra. The construction goes in much the same way as the symmetric algebra, however unlike there we do not have a known object like the polynomial algebra to compare it with – so we have to work harder to get the analogous basis theorem to the basis theorem for symmetric powers. Start with V being any vector space. Define K to be the two-sided ideal of the tensor algebra T (V ) generated by the elements {x ⊗ x | x ∈ V }. Note that (v + w) ⊗ (v + w) = v ⊗ v + w ⊗ v + v ⊗ w + w ⊗ w. So K also contains all the , elements {v ⊗ w + w ⊗ v | v, w ∈ V } automatically. Define the exterior (or Grassmann) algebra (V ) to be the quotient algebra , (V ) = T (V )/K. , We’ll write v1 ∧ · · · ∧ vn for the image in (V ) of the pure tensor v1 ⊗ · · · ⊗ vn ∈ T (V ). So, now we have “anti-symmetric” properties like: v1 ∧ · · · ∧ vi ∧ vi ∧ . . . vn = 0 and v1 ∧ · · · ∧ vi ∧ vi+1 ∧ . . . vn = −v1 ∧ · · · ∧ vi+1 ∧ vi ∧ . . . vn . ) Since the ideal K is generated by homogeneous elements, K = n≥0 Kn where Kn = K ∩ T n (V ). It follows that ( ,n , (V ) = (V ) n≥0 ,n ∼ where (V ,n) = T (V )/Kn is the subspace spanned by the images of all pure tensors of degree n. We’ll call (V ) the nth ,nexterior power. The exterior power (V ), together with the distinguished map ,n i : V × ··· × V → (V ), (v1 , . . . , vn ) $→ v1 ∧ · · · ∧ vn , n is characterized by the following universal property. First, call a multlinear map f : V ×· · ·×V → W to a vector space W alternating if we have that f (v1 , . . . , vj , vj+1 , . . . , vn ) = 0 ,n whenever vj = vj+1 for some j. The map i : V × · · · × V → V is multilinear and alternating. Moreover, given any other multilinear, alternating map f : V × · · · × V → W , there exists a unique ,n linear map f¯ : V → W such that f = f¯ ◦ i. The proof is the same as for the symmetric powers, you should compare the two constructions! Using this all important universal property, we can now prove: Basis , theorem for exterior powers. Suppose that V is finite dimensional, with basis e1 , . . . , em . n Then, (V ) has basis given by all monomials ei1 ∧ · · · ∧ ein for all sequences 1 ≤ i1 < · · · < in ≤ m. In particular, * + ,n m dim (V ) = , n and is zero for n > m. 103 4.2. SOME MULTILINEAR ALGEBRA ,n ,n Proof. Since (V ) is a quotient of T n (V ), we certainly have that (V ) is spanned by all monomials ei1 ∧ · · · ∧ ein . Now using the antisymmetric properties of ∧, we get that it is spanned by the given strictly ordered monomials. The problem is to prove that the given elements are linearly ,1independent. For this, we proceed by induction on n, the case n = 1 being immediate since (V ) = V . Now consider n > 1. ∗ Let f1 , . . . , fn be the basis for the given basis e1 , . . . , en for V . For each i = 1, . . . , n, ,nV dual to ,n−1 we wish to define a map Fi : (V ) → (V ). Start with the multilinear, alternating map ,n−1 f¯i : V × · · · × V → (V ) defined by f¯i (v1 , . . . , vn ) = n ! j=1 (−1)j fi (vj )v1 ∧ · · · ∧ vj−1 ∧ vj+1 ∧ · · · ∧ vn . Its clearly multilinear, and to check its alternating, we just need to see that if vj = vj+1 then: f¯i (v1 , . . . , vj , vj+1 , . . . , vn ) = (−1)j fi (vj )v1 ∧ · · · ∧ vj+1 ∧ · · · ∧ vn + (−1)j+1 fi (vj+1 )v1 ∧ · · · ∧ vj ∧ · · · ∧ vn , gives that f¯i induces a unique linear map Fi : which is zero. Now the univeral property of ,n ,n−1 V → V. Now we show that the given elements are linearly independent. Take a linear relation ! ai1 ,...,in ei1 ∧ · · · ∧ ein = 0. i1 <···<in Apply the linear map F1 , which annihilates all ej except for e1 by definition. We get: ! a1,i2 ,...,in ei2 ∧ · · · ∧ ein = 0. 1<i2 <···<in By the induction hypothesis, all such monomials are linearly independent, hence all a1,i2 ,...,in are zero. Now apply F2 in the same way to get that all a2,i2 ,...,in are zero, etc... ,n Now let’s focus -on.a special case. Suppose that dim V = n and consider V . Its dimension, n by the theorem, is n = 1, and it has basis just the element e1 ∧ · · · ∧ en . Let f : V → V be a linear map. Define ,n f¯ : V × · · · × V → (V ) by f¯(v1 , . . . , vn ) = f (v1 ) ∧ · · · ∧ f (vn ). One easily checks that this is multilinear and alternating. Hence, it induces a unique linear map we’ll denote ,n ,n ∧n f : (V ) → (V ). ,n But (V ) is one dimensional, so such a linear map is just multiplication by a scalar. In other words, there is a unique scalar which we’ll call det f determined by the equation f (e1 ) ∧ · · · ∧ f (en ) = (det f )e1 ∧ · · · ∧ en . Of course, det f is exactly the determinant of the linear transformation f . Note our definition of determinant is basis free, always a good thing. You can see what we’re calling det f is the same as usual, as follows. Suppose the matrix of f in our fixed basis is A, defined from ! f (ej ) = Ai,j ei . i 104 CHAPTER 4. ALGEBRAS Then, f (e1 ) ∧ · · · ∧ f (en ) = ! i1 ,...,in Ai1 ,1 Ai2 ,2 . . . Ain ,n ei1 ∧ · · · ∧ ein . Now terms on the right hand side are zero unless all i1 , . . . , in are distinct, i.e. (i1 , . . . , in ) = (g1, g2, . . . , gn) for some permutation g ∈ Sn . So: f (e1 ) ∧ · · · ∧ f (en ) = ! g∈Sn Ag1,1 Ag2,2 . . . Agn,n eg1 ∧ · · · ∧ egn . Finally, eg1 ∧ · · · ∧ egn = sgn(g)e1 ∧ · · · ∧ en where sgn denotes the sign of a permutation. So: f (e1 ) ∧ · · · ∧ f (en ) = This shows that det f = ! sgn(g)Ag1,1 Ag2,2 . . . Agn,n e1 ∧ · · · ∧ en . ! sgn(g)Ag1,1 Ag2,2 . . . Agn,n g∈Sn g∈Sn which is exactly the usual “Laplace expansion” definition of determinant. Here’s a final result to illustrate how nicely the universal property definition of determinant gives its properties: Multiplicativity of determinant. For linear transformations f, g : V → V , we have that det(f ◦ g) = det f det g. Proof. By definition, ∧n [f ◦ g] is the map uniquely determined by (∧n [f ◦ g])(v1 ∧ · · · ∧ vn ) = f (g(v1 )) ∧ · · · ∧ f (g(vn )). But ∧n f ◦ ∧n g also satisfies this equation. So, ∧n (f ◦ g) = ∧n f ◦ ∧n g. The left hand side is scalar multiplication by det(f ◦ g), the right hand side is scalar multiplication by det f det g. 4.3 Chain conditions Let M be a (left or right) R-module. Then, M is called Noetherian if it satisfies the ascending chain condition (ACC) on submodules. This means that every ascending chain M1 ⊆ M2 ⊆ M3 ⊆ . . . of submodules of M eventually stabilizes, i.e. Mn = Mn+1 = Mn+2 = . . . for sufficiently large n. Similarly, M is called Artinian if it satisfies the descending chain condition (DCC) on submodules. So every descending chain M1 ⊇ M2 ⊇ M3 ⊇ . . . of submodules of M eventually stabilizes, i.e. Mn = Mn+1 = Mn+2 = . . . for sufficiently large n. A ring R is called left (resp. right) Noetherian if it is Noetherian viewed as a left (resp. right) R-module. Similarly, R is called left (resp. right) Artinian if it is Artinian viewed as a left (resp. right) R-module. In the case of commutative rings, we can omit the left or right here, but we cannot in general as the pathological examples on the homework show. In this chapter, we’ll mainly be concerned with the Artinian property, but it doesn’t make sense to introduce one chain condition without the other. We will discuss Noetherian rings in detail in the chapter on commutative algebra. By the way, you should think of the Noetherian property 105 4.3. CHAIN CONDITIONS as a quite weak finiteness property on a ring, whereas the Artinian property is rather strong (see Hopkin’s theorem below for the justification of this). We already know plenty of examples of Noetherian and Artinian rings. For instance, any PID is Noetherian (Lemma 2.4.1), but it need not be Artinian (e.g. Z is not Artinian: (p) ⊃ (p2 ) ⊃ . . . is an infinitely descending chain). The ring Zpn for p prime is both Noetherian and Artinian: indeed, here there are only finitely many ideals in total, the (pi ) for 0 ≤ i ≤ n. The main source of Artinian rings is as follows. Let F be a field and suppose that R is an F -algebra. Then I claim that every R-module M which is finite dimensional as an F -vector space is both Artinian and Noetherian. Well, R-submodules of M are in particular F -vector subspaces. And clearly in a finite dimensional vector space, you cannot have infinite chains of proper subspaces. So finite dimensionality of M does the job immediately! In particular, if the F -algebra R is finite dimensional as a vector space over F , then R is both left and right Artinian and Noetherian. Now you see why finite dimensional algebras over a field (e.g. Mn (F ), the group algebra F G for G a finite group, etc...) are particularly nice things! i π 4.3.1. Lemma. Let 0 −→ K −→ M −→ Q −→ 0 be a short exact sequence of R-modules. Then M is Noetherian (resp. Artinian) if and only if both K and Q are Noetherian (resp. Artinian). Proof. I just prove the result for the Noetherian property, the Artinian case being analogous. First, suppose M satisfies ACC. Then obviously K does as it is isomorphic to a submodule of M . Similarly Q does by the lattice isomorphism theorem for modules. Conversely, suppose K and Q both satisfy ACC. Let M1 ⊆ M2 ⊆ . . . be an ascending chain of R-submodules of M . Set Ki = i−1 (i(K) ∩ Mi ), Qi = π(Mi ). Then, K1 ⊆ K2 ⊆ . . . is an ascending chain of submodules of K, and Q1 ⊆ Q2 ⊆ . . . is an ascending chain of submodules of Q. So by assumption, there exists n ≥ 1 such that KN = Kn and QN = Qn for all N ≥ n. Now, we have evident short exact sequences 0 −→ Kn −→ Mn −→ Qn −→ 0 and 0 −→ KN −→ MN −→ QN −→ 0 for each N ≥ n. Letting α : Kn → KN , β : Mn → MN and γ : Qn → QN be the inclusions, one obtains a commutative diagram with α and γ being isomorphisms. Then the five lemma implies that β is surjective, hence MN = Mn for all N ≥ n as required. 4.3.2. Theorem. If R is left (resp. right) Noetherian, then every finitely generated left (resp. right) R-module is Noetherian. Similarly, if R is left (resp. right) Artinian, then every finitely generated left (resp. right) R-module is Artinian. Proof. Let’s do this for the Artinian case, the Noetherian case being similar. So assume that R is left Artinian and M is a finitely generated left R-module. Then, M is a quotient of a free R-module with a finite basis. So it suffices to show that R⊕n is an Artinian left R-module. We proceed by induction on n, the case n = 1 being given. For n > 1, the submodule of R⊕n spanned by the first (n − 1) basis elements is isomorphic to R⊕(n−1) , and the quotient by this submodule is isomorphic to R. By induction both R⊕(n−1) and R are Artinian. Hence, R⊕n is Artinian by Lemma 4.3.1. The next results are special ones about Noetherian rings (but see Hopkin’s theorem below!) 4.3.3. Lemma. Let R be a left Noetherian ring and M be a finitely generated left R-module. Then every R-submodule of M is also finitely generated. 106 CHAPTER 4. ALGEBRAS Proof. By Theorem 4.3.2, M is Noetherian. Let N be an arbitrary R-submodule of M and let A be the set of finitely generated R-submodules of M contained in N . Note A is non-empty as the zero module is certainly there! Now I claim that A contains a maximal element. Well, pick M1 ∈ A . If M1 is maximal in A , we are done. Else, we can find M2 ∈ A with M1 ⊃ M2 . Repeat. The process must terminate, else we construct an infinite ascending chain of submodules of N ⊆ M , contradicting the fact that M is Noetherian. (Warning: as with all such arguments we have secretly appealed to the axiom of choice here! So now let N # be a maximal element of A . Say N # is generated by m1 , . . . , mn . Now take any m ∈ N . Then, (m1 , . . . , mn , m) is a finitely generated submodule of M contained in N and containing N # . By maximality of N # , we therefore have that N # = (m1 , . . . , mn , m), i.e. m ∈ N # . This shows that N = N # , hence N is finitely generated. 4.3.4. Corollary. Let R be a ring. Then, The following are equivalent: (i) R is left Noetherian; (ii) every left ideal of R is finitely generated; (Similar statements hold on the right, of course). Proof. (i)⇒(ii). This is a special case of Lemma 4.3.3, taking M = R. / (ii)⇒(i). Let I1 ⊆ I2 ⊆ . . . be an ascending chain of left ideals of R. Then, I = n≥1 In is also a left ideal of R, hence finitely generated, by a1 , . . . , am say. Then for some sufficiently large n, all of a1 , . . . , am lie in In . Hence In = I and R is Noetherian. Now you see that – for commutative rings – Noetherian is an obvious generalization of a PID. Instead of insisting all ideals are generated by a single element, one has that every ideal is generated by finitely many elements. 4.4 Wedderburn structure theorems Now we have the basic language of algebras and of Artinian rings and modules, we can begin to discuss the structure of rings. The first step is to understand the structure of semisimple rings. Recall that an R-module M is simple or irreducible if it is non-zero and has no submodules other than M and (0). Schur’s lemma. Let M be a simple R-module. Then, EndR (M ) is a division ring. Proof. Let f : M → M be a non-zero R-endomorphism of M . Then, ker f and im f are both R-submodules of M , with ker f 4= M and im f 4= (0) since f is non-zero. Hence, since M is simple, ker f = (0) and im f = M . This shows that f is a bijection, hence invertible. Throughout the section, we will also develop parallel but stronger versions of the results dealing with finite dimensional algebras over algebraically closed fields (recall a field F is algebraically closed if every monic f (X) ∈ F [X] has a root in F ). Relative Schur’s lemma. Let F be an algebraically closed field and A be an F -algebra. Let M be a simple A-module which is finite dimensional as an F -vector space. Then, EndR (M ) = F . Proof. Let f : M → M be an A-endomorphism of M . Then in particular, f : M → M is an F -linear endomorphism of a finite dimensional vector space. Since F is algebraically closed, f has an eigenvector vλ ∈ M of eigenvalue λ ∈ F (because the characteristic polynomial of f has a root in F ). 107 4.4. WEDDERBURN STRUCTURE THEOREMS Now consider f −λ idM . It is also an A-endomorphism of M , and moreover its kernel is non-zero as vλ is annihilated by f − λ idM . Hence since M is irreducible, the kernel of f − λ idM is all of M , i.e. f = λ idM . This shows that the only R-endomorphisms of M are the scalars, as required. Now, let R be a non-zero ring. Call R left semisimple if R R is semisimple as a left R-module (recall section 3.4). Similarly, R is right semisimple if RR is semisimple as a right R-module. Finally, R is simple if it has no two-sided ideals other than R and (0). Here’s a lemma from the previous chapter that should be quite familiar by now. 4.4.1. Lemma. Let A be an R-algebra, M a left A-module and D = EndA (M ). Then, EndA (M ⊕n ) ∼ = Mn (D) as R-algebras. Proof. Let us write elements of M ⊕n as column vectors m1 .. . mn with mi ∈ M . Suppose we have f ∈ EndA (M ⊕n ). Let fi,j : M → M be the map sending m ∈ M to the ith coordinate of the vector 0 .. . 0 f m 0 . .. 0 where the m is in the jth row. Then, fi,j is an Moreover, f1,1 . . . f1,n m1 .. .. .. .. f . = . . . fn,1 . . . fn,n mn A-module homomorphism, so is an element of D. m1 .. . mn (matrix multiplication!), so that f is uniquely determined by the fi,j . Thus, we obtain an R-algebra isomorphism f $→ (fi,j ) between EndA (M ⊕n ) and Mn (D). Wedderburn’s first structure theorem. The following conditions on a non-zero ring R are equivalent: (i) R is simple and left Artinian; (i# ) R is simple and right Artinian; (ii) R is left semisimple and all simple left R-modules are isomorphic; (ii# ) R is right semisimple and all simple right R-modules are isomorphic; (iii) R is isomorphic to the matrix ring Mn (D) for some n ≥ 1 and D a division ring. Moreover, the integer n and the division ring D in (iii) are determined uniquely by R (up to isomorphism). Proof. First, note the condition (iii) is left-right symmetric. So let us just prove the equivalence of (i),(ii) and (iii). 108 CHAPTER 4. ALGEBRAS (i)⇒(ii). Let U be a minimal left ideal of R. This exists because R is left Artinian so we have DCC on left ideals! Then, U = Rx for any non-zero x ∈ U , and Rx is a simple left R-module. Since R is a simple ring, the non-zero two-sided ideal RxR must be all of R. Hence, ! Rxa. R = RxR = a∈R Note each Rxa is a homomorphic image of the simple left R-module Rx, so is either isomorphic to Rx or zero. Thus we have written R R as a sum of simple R-modules, so by Lemma 3.4.1, there exists a subset S ⊆ R such that ( R= Rxa a∈S with each Rxa for a ∈ S being isomorphic to U . Hence, R is left semisimple. Moreover, if M is any simple left R-module, then M ∼ = R/J for a maximal left ideal J of R. We can pick a ∈ S such that xa ∈ / J. Then, the quotient map R → R/J maps the simple submodule Rxa of R to a non-zero submodule of M . Hence using simplicity, M ∼ = Rxa. This shows all simple R-modules are isomorphic to U . (ii)⇒(iii). Let U be a simple left R-module. We have that ( R= Ra a∈S for some subset S ) of R, where each left R-module Ra is isomorphic to U . I claim that S is finite. Indeed, 1R lies in a∈S Ra hence in Ra1 ⊕ · · · ⊕ Ran for some finite subset {a1 , . . . , an } of S. But 1R generates R as a left R-module, so in fact {a1 , . . . , an } = S. Thus, R∼ = U ⊕n for some n, uniquely determined as the composition length of R R. Now, by Schur’s lemma, EndR (U ) ∼ = D, where D a Division ring uniquely determined up to isomorphism as the endomorphism ring of a simple left R-module. Hence applying Lemma 4.4.1, EndR (R R) ∼ = EndR (U ⊕n ) ∼ = Mn (D). Now finally, we observe that EndR (R R) ∼ = Rop , the isomorphism in the forward direction being determined by evaluation at 1R . To see that this is an isomorphism, one constructs the inverse map, namely, the map sending r ∈ R to fr :R R →R R with fr (s) = sr for each s ∈ R (“fr is right multiplication by r”). Now we have shown that Rop ∼ = Mn (D). Hence, noting that matrix transposition is an isomorphism between Mn (D)op and Mn (Dop ), we get that R∼ = Mn (Dop ) and Dop is also a division ring. (iii)⇒(i). We know that Mn (D) is simple as it is Morita equivalent to D, which is a division ring (or you can prove this directly!). It is left Artinian for instance because Mn (D) is finite dimensional over the division ring D. This gives (i). Corollary. Every simple left (or right) Artinian ring R is an algebra over a field. Proof. Observe Mn (D) is an algebra over Z(D), which is a field. The relative version for finite dimensional algebras over algebraically closed fields is as follows: 4.4. WEDDERBURN STRUCTURE THEOREMS 109 Relative first structure theorem. Let F be an algebraically closed field and A be a finite dimensional F -algebra (hence automatically left and right Artinian). (i) A is simple; (ii) A is left semisimple and all simple left R-modules are isomorphic; (ii# ) A is right semisimple and all simple right R-modules are isomorphic; (iii) A is isomorphic to the matrix ring Mn (F ) for some n (indeed, n2 = dim A). Proof. The proof is the same, except that one gets that the division ring D equals F since one can use the stronger relative Schur’s lemma. Wedderburn’s second structure theorem. Every left (or right) semisimple ring R is isomorphic to a finite product of matrix rings over division rings: R∼ = Mn1 (D1 ) × · · · × Mnr (Dr ) for uniquely determined ni ≥ 1 and division rings Di . Conversely, any such ring is both left and right semisimple. ) Proof. Since R is left semisimple, we can decompose R R as i∈I Ui where each Ui is simple. Note 1R already lies in a sum of finitely many of the Ui , hence the index set I is actually finite. Now gather together isomorphic Ui ’s to write RR where = H1 ⊕ · · · ⊕ Hm Hi ∼ = Ui⊕ni for irreducible R-modules Ui and integers ni ≥ 1, with Ui ∼ 4 Uj for i 4= j. By Schur’s lemma, = Di = EndR (Ui ) is a division ring. Moreover, there are no R-module homomorphisms between Hi and Hj for i 4= j, because none of their composition factors are isomorphic. Hence: ∼ EndR (H1 ⊕ · · · ⊕ Hm )op ∼ R∼ = EndR (R R)op = = EndR (H1 )op × · · · × EndR (Hm )op op op ∼ ). = Mn1 (D1 )op × · · · × Mnm (Dm )op ∼ = Mn1 (D1 ) × · · · × Mnm (Dm The integers ni are uniquely determined as the multiplicities of the irreducible left R-modules in a composition series of R R, while the division rings Di are uniquely determined up to isomorphism as the endomorphism rings of the simple left R-modules. For the converse, we need to show that a product of matrix algebras over division rings is left and right semisimple. This follows because a single matrix algebra Mn (D) over a division ring is both left and right semisimple according to the first Wedderburn structure theorem. The theorem shows: Corollary. A ring R is left semisimple if and only if it is right semisimple. So we can now just call R semisimple if it is either left or right semisimple in the old sense. Also: Corollary. Any semisimple ring is both left and right Artinian. Proof. A product of finitely many matrix rings over division algebras is left and right Artinian. Finally, we need to give the relative version: 110 CHAPTER 4. ALGEBRAS Relative second structure theorem. Let F be an algebraically closed field and A be a finite dimensional F -algebra that is semisimple. Then, A∼ = Mn1 (F ) × · · · × Mnr (F ) for uniquely determined integers ni ≥ 1. Proof. Just use the relative Schur’s lemma in the same proof. This theorem lays bare the structure of finite dimensional semisimple algebras over algebraically closed fields: A is uniquely determined up to isomorphism by the integers ni , which are precisely the dimensions of the simple A-modules. 4.5 The Jacobson radical I want to explain in this section how understanding of the case of semisimple rings gives information about more general rings. The key notion is that of the Jacobson radical, which will be defined using the following equivalent properties: Jacobson radical theorem. Let R be a ring, a ∈ R. The following are equivalent: (i) a annihilates every simple left R-module; (i)# a annihilates every simple right R-module; (ii) a lies in every maximal left ideal of R; (ii)# a lies in every maximal right ideal of R; (iii) 1 − xa has a left inverse for every x ∈ R; (iii)# 1 − ay has a right inverse for every y ∈ R; (iv) 1 − xay is a unit for every x, y ∈ R. Proof. Since (iv) is left-right symmetric, I will only prove equivalence of (i)–(iv). (i)⇒(ii). If I is a maximal left ideal of R, then R/I is a simple left R-module. So, a(R/I) = 0, i.e. a ∈ I. (ii)⇒(iii). Assume (ii) holds but 1 − xa does not have a left inverse for some x ∈ R. Then, R(1 − xa) 4= R. So, there exists a maximal left ideal I with R(1 − xa) ≤ I < R. But then, 1 − xa ∈ I, and a ∈ I by assumption. Hence, 1 ∈ I so I = R, a contradiction. (iii)⇒(i). Let M be a simple left R-module. Take u ∈ M . If au 4= 0, then Rau = M as M is simple, so u = rau for some r ∈ R. But this implies that (1 − ra)u = 0, hence since (1 − ra) has a left inverse by assumption, we get that u = 0. This contradiction shows that in fact au = 0 for all u ∈ M , i.e. aM = 0. (iv)⇒(iii). This is trivial (take y = 1). (i),(iii)⇒(iv). For any y ∈ R and any simple left R-module M , ayM ⊆ aM = 0. So ay also satisfies the condition in (i), hence (iii). So for every x ∈ R, 1 − xay has a left inverse, 1 − b say. The equation (1 − b)(1 − xay) = 1 implies b = (b − 1)xay. Hence bM = 0 for all simple left R-modules M , so b satisfies (i) hence (iii). So, 1 − b has a left inverse, 1 − c, say. Then, 1 − c = (1 − c)(1 − b)(1 − xay) = 1 − xay, hence c = xay. Now we have that 1 − b is a left inverse to 1 − xay, and 1 − xay is a left inverse to 1 − b. Hence, 1 − xay is a unit with inverse 1 − b. 111 4.5. THE JACOBSON RADICAL Now define the Jacobson radical J(R) to be J(R) = {a ∈ R | a satisfies the equivalent conditions in the theorem}. Thus for instance using (i), J(R) is the intersection of the annihilators in R of all the simple left R-modules, or using (ii) 8 J(R) = I I where I runs over all maximal left ideals of R. This implies that J(R) is a left ideal of R, but equally well using (ii)# , 8 I J(R) = I where I runs over all maximal right ideals of R so that J is a right ideal of R. Hence, J(R) is a two-sided ideal of R. The following result is a basic trick in ring theory and maybe gives a first clue as to why the Jacobson radical is so important. Nakayama’s lemma. Let R be a ring and M be a finitely generated left R-module. If J(R)M = M then M = 0. Proof. Suppose that M is non-zero and set J = J(R) for short. Let X = {m1 , . . . , mn } be a minimal set of generators of M , so m1 4= 0. Since JM = M , we can write m1 = j1 m1 + · · · + jn mn for some ji ∈ J. So, (1R − j1 )m1 = j2 m2 + · · · + jn mn . But 1R − j1 is a unit by the definition of J(R). So we get that m1 = (1R − j1 )−1 j2 m2 + · · · + (1R − j1 )−1 jn mn . This contradicts the minimality of the initial set of generators chosen. We will apply Nakayama’s lemma on the next chapter, but actually never in this chapter. The remainder of the section is concerned with the Jacobson radical in an Artinian ring!!! All these results are false if the ring is not Artinian... First characterization of the Jacobson radical for Artinian rings. Suppose that R is left (or right) Artinian. Then, J(R) is the unique smallest two-sided ideal of R such that R/J(R) is a semisimple algebra. Proof. Pick a maximal left ideal I1 of R. Then (if possible) pick a maximal left ideal I2 such that I2 4⊇ I1 (hence I1 ∩ I2 is strictly smaller than I1 ). Then (if possible) pick a maximal left ideal I3 such that I3 4⊇ I1 ∩ I2 (hence I1 ∩ I2 ∩ I3 is strictly smaller than I1 ∩ I2 ). Keep going! The process must terminate after finitely many steps, else you construct an infinite descending chain R ⊃ I1 ⊃ I1 ∩ I2 ⊃ I1 ∩ I2 ∩ I3 ⊃ . . . of left ideals of R, contradicting the fact that R is left Artinian. We thus obtain finitely many maximal left ideals I1 , I2 , . . . , Ir of R such that I1 ∩ · · · ∩ Ir is contained in every maximal left ideal of R. In other words, J(R) = I1 ∩ · · · ∩ Ir , so the Jacobson radical of an Artinian ring is the intersection of finitely many maximal left ideals. 112 CHAPTER 4. ALGEBRAS Consider the map a $→ (a + I1 , . . . , a + Ir ). R → R/I1 ⊕ · · · ⊕ R/Ir , It is an R-module map with kernel I1 ∩ · · · ∩ Ir = J(R). So it induces an embedding R/J '→ R/I1 ⊕ · · · ⊕ R/Ir . The right hand side is a semisimple R-module, as it is a direct sum of simples, hence R/J is also a semisimple R-module. This shows that the quotient R/J is a semisimple ring. Now let K be another two-sided ideal of R such that R/K is a semisimple ring. By the lattice isomorphism theorem, we can write ( Bi /K R/K = where9 Bi is a left ideal of R containing K, Ci = j'=i Bj for short. Then, 9 i∈I i∈I 9 Bi = R and Bi ∩ ( j'=i Bj ) = K for each i. Set R/Ci = (Ci + Bi )/Ci ∼ = Bi /(Bi ∩ Ci ) = Bi /K. This is simple, so Ci is a maximal left ideal of R. Hence, each Ci ⊇ J(R). So K = contains J(R). Thus J(R) is the unique smallest such ideal. : Ci also Thus you see the first step to understanding the structure of an Arinian ring: understand the semisimple ring R/J(R) in the sense of Wedderburn’s structure theorem. 4.5.1. Corollary. Let R be a left Artinian ring and M be a left R-module. Then, M is semisimple if and only if J(R)M = 0. Proof. If M is semisimple, it is a direct sum of simples. Now, J(R) annihilates all simple left R-modules by definition, hence J(R) annihilates M . Conversely, if J(R) annihilates M , then we can view M as an R/J(R)-module by defining (a + J(R))m = am for all a ∈ R, m ∈ M (one needs J(R) to annihilate M for this to be well-defined!). Since R/J(R) is semisimple by the previous theorem, M is a semisimple R/J(R)-module. But the R-module structure on M is just obtained by lifting the R/J(R)-module structure, so this means that M is semisimple as an R-module too. Second characterization of the Jacobson radical for Artinian rings. Let R be a left (or right) Artinian ring. Then, J(R) is a nilpotent two-sided ideal of R (i.e. J(R)n = 0 for some n > 0) and is equal to the sum of all nilpotent left ideals of R. Proof. Suppose x ∈ R is nilpotent, say xn = 0. Then, (1 − x) is a unit, indeed, (1 − x)(1 + x + x2 + · · · + xn−1 ) = 1. So if I is a nilpotent left ideal of R, then every x ∈ I satisfies condition (iii) of the Jacobson radical theorem. This shows every nilpotent ideal of R is contained in J(R). It therefore just remains to prove that J(R) is itself a nilpotent ideal. Set J = J(R). Consider the chain J ⊇ J2 ⊇ J3 ⊇ . . . of two-sided ideals of R. Since R is left Artinian, the chain stabilizes, so J k = J k+1 = . . . for some k. Set I = J k , so I 2 = I. We need to prove that I = 0. Well, suppose for a contradiction that I 4= 0. Choose a left ideal K of R minimal such that IK 4= 0 (use the fact that R is left Artinian). Take any a ∈ K with Ia 4= 0. Then, I 2 a = Ia 4= 0, so the left ideal Ia of R coincides with K by the minimality of K. Hence, a ∈ K lies in Ia, so we can write a = xa for some x ∈ I. So, (1 − x)a = 0. But x ∈ J, so 1 − x is a unit, hence a = 0, which is a contradiction. 4.6. CHARACTER THEORY OF FINITE GROUPS 113 4.5.2. Corollary. In particular, a left (or right) Artinian ring R is semisimple if and only if it has no non-zero nilpotent ideals. We end with an important application: Hopkin’s theorem. Let R be a left Artinian ring and M be a left R-module. The following are equivalent: (i) M is Artinian; (ii) M is Noetherian; (iii) M has a composition series; (iv) M is finitely generated. Proof. (i)⇒(iii) and (ii)⇒(iii). Let J = J(R). Then, J is nilpotent by the second characterization of the Jacobson radical. So, J n = 0 for some n. Consider M ⊇ JM ⊇ J 2 M ⊇ · · · ⊇ J n M = 0. It is a descending chain of R-submodules of M . Set Fi = J i M/J i+1 M . Then, Fi is annihilated by J, hence is a semisimple left R-module. Now, if M is Artinian (resp. Noetherian), so is each Fi , so each Fi is in fact a direct sum of finitely many simple R-modules. Thus, each Fi obviously has a composition series, so M does too by the lattice isomorphism theorem. (iii)⇒(iv). Let M = M1 ⊃ · · · ⊃ Mn = 0 be a composition series of M . Pick mi ∈ Mi − Mi+1 for each i = 1, . . . , n − 1. Then, the image of mi in Mi /Mi+1 generates Mi /Mi+1 as it is a simple R-module. It follows that the mi generate M . Hence, M is finitely generated. (iv)⇒(i) is Theorem 4.3.2. (iii)⇒(ii) and (iii)⇒(i). These both follow immediately from the Jordan-Hölder theorem (actually, the Schreier refinement lemma using that any refinement of a composition series is trivial). Now we can show that “Artinian implies Noetherian”: 4.5.3. Corollary. If R is left (resp. right) Artinian, then R is left (resp. right) Noetherian. Proof. Apply Hopkin’s theorem to the Artinian R-module R R. 4.6 Character theory of finite groups Wedderburn’s theorem has a very important application to the study of finite groups, indeed, it is the starting point for the study of character theory of finite groups. In this section, I’ll follow Rotman section 8.5 fairly closely. Maschke’s theorem. Let G be a finite group and F be a field either of characteristic 0 or of characteristic 0 < p ! |G|. Then, the group algebra F G is a semisimple algebra. Proof. It is certainly enough to show that every left F G-module M is semisimple. We just need to show that every short exact sequence π 0 −→ K −→ M −→ Q −→ 0 of F G-modules splits. Well, pick any F -linear map θ : Q → M such that π ◦ θ = idQ . The problem is that θ need not be an F G-module map. Well, for g ∈ G, define a new F -linear map g θ : Q → M, x $→ gθ(g −1 x). 114 Then, CHAPTER 4. ALGEBRAS (π ◦ g θ)(x) = π(gθ(g −1 x)) = g(π ◦ θ)(g −1 x) = gg −1 x = x so each g θ also satisfies π ◦ g θ = idQ . Now define Avθ = 1 !g θ |G| g∈G (note |G| is invertible in F by assumption on characteristic). This is another F -linear map such that π ◦ Avθ = idQ . Moreover, Avθ is even an F G-module map: hAvθ(x) = 1 ! 1 ! hgθ(g −1 x) = gθ(g −1 hx) = Avθ(hx). |G| |G| g∈G g∈G Hence, Avθ defines the required splitting. 4.6.1. Remark. Maschke’s theorem is actually if and only if... F G is a semisimple algebra if and only if char F = 0 or char F ! |G|. Let me just give one example. Take F of characteristic p > 0 and G = Cp = *x,. The group algebra F G is commutative and artinian. Now recall a commutative artinian ring is semisimple if and only if it has no non-zero nilpotent elements. But The corollary can be applied in particular to commutative rings. In that case, if x ∈ R is a nilpotent element, the ideal (x) it generates is a nilpotent ideal. So, a commutative Artinian ring is semisimple if and only if it has no non-zero nilpotent elements. Now let G = Cp , the cyclic group of order p. Maschke’s theorem shows that if F is any field of characteristic different from p, then F G is a semisimple. Conversely, suppose F is a field of characterisitic p. The group algebra F G is finite dimensional, hence is a commutative Artinian ring. Consider the element x − 1 ∈ FG where x is a generator of G. We have that (x − 1)p = xp + (−1)p = 1 + (−1)p = 0. Hence, it is a non-zero nilpotent element. Thus, F G is not semisimple in this case. We are going to be interested from now on in the following situation: • G is a finite group (so that F G is artinian); • the field F is of characteristic 0 (so that F G is semisimple); • the field F is algebraically closed (so that the souped-up version of Wedderburn’s theorem holds). Actually I’ll just assume F = C from now on. Then, we have shown that (1) CG ∼ = Mn1 (C) × · · · × Mnr (C) where r is the number of isomorphism classes of irreducible CG-modules and n1 , . . . , nr are their dimensions. These give some interesting invariants of the group G defined using modules... Note first (considering dimension of each side as a C-vector space) that: |G| = (n1 )2 + · · · + (nr )2 . The number r has a simply group theoretic meaning too: 4.6.2. Lemma. The number r in (1) is equal to the number of conjugacy classes in the group G. 4.6. CHARACTER THEORY OF FINITE GROUPS 115 Proof. Let us compute dim Z(CG) in two different ways. First, since the center of Mn (C) is one dimensional spanned by the identity matrix, dim Z(CG) = r (one for each matrix algebra in the Wedderburn decomposition). On the other hand if ! ag g ∈ CG g∈G is central then conjugating by h ∈ G you see that ag = ahgh−1 for all h ∈ G. Hence the coefficients ag are constant on conjugacy classes. Hence if C1 , . . . , Cs are 9 the conjugacy classes of G, the elements zi = g∈Ci g form a basis {z1 , . . . , zs } for Z(CG). Hence r = dim Z(CG) = s. 4.6.3. Example. Say G = S3 . There are three conjugacy classes. Hence r = 3, i.e. there are three isomorphism classes of irreducible CS3 -modules. Moreover n21 + n22 + n23 = 6 so the dimensions can only be 1, 1 and 2. 4.6.4. Example. Say G is abelian. Then there are r = |G| conjugacy classes, and n21 +· · ·+n2r = r hence each ni = 1. This proves that there are n isomorphism classes of irreducible CG-module, and all of these modules are one dimensional. Before going any further I want to introduce a little language. A (finite dimensional complex) representation of G means a pair (V, ρ) where V is a finite dimensional complex vector space and ρ : G → GL(V ) is a group homomorphism. The representation is called faithful if this map ρ is injective. Note having a representation (V, ρ) of G is exactly the same as having a finite dimensional CG-module V : given a representation (V, ρ) we get a CG-module structure on V by defining gv := ρ(g)(v); conversely given a CG-module structure on V we get a representation ρ : G → GL(V ) by defining ρ(g) to be the linear map v $→ gv. This is all one big tautology. But I will often switch between calling things CG-modules and calling them representations. Sorry. Say ρ : G → GL(V ) is a representation of G. Pick a basis for V , v1 , . . . , vn . Then each element of GL(V ) becomes an invertible n × n matrix, and in this way you can regard ρ instead as a group homomorphism ρ : G → GLn (C). This is a matrix representation: ρ maps the group G to the group of n × n invertible matrices. You should compare the notion of matrix representation with that of permutation representation from before: that was a map from G to the symmetric group Sn for some n. It corresponded to having a G-set X of size n... Its all very analogous to what we’re doing now, but sets have been replaced by vector spaces. Before we called a permutation representation faithful if the map G → Sn was injective (then it embedded G as a subgroup of a symmetric group); now we are calling a (linear) representation faithful if the map G → GLn (C) is injective (then it embeds G as a subgroup of the matrix group)... 4.6.5. Example. Recall that GL(C) ∼ = GL1 (C) is just the multiplicative group C× . For any group G, there is always the trivial representation, namely, the homomorphism mapping every g ∈ G to 1 ∈ GL1 (C). This corresponds to the trivial CG-module equal to C as a vector space with every g ∈ G acting as 1. 4.6.6. Example. For the symmetric group Sn , there is a group homomorphism sgn : Sn → {±1}. The latter group sits inside C× , so you can view this as another 1-dimensional representation, the sign representation. The corresponding module is not isomorphic to the trivial module (providing n > 1). Now we’ve constructed both of the one dimensonal CS3 -modules: one is trivial, one is sign. What about the two dimensional CS3 -module? There is a simple way to construct (linear) representations of G out of permutation representations. Suppose X = {x1 , . . . , xn } is a finite G-set. Let CX be the C-vector space on basis X. Each 116 CHAPTER 4. ALGEBRAS g ∈ G defines an automorphism of the set X. Extending linearly, this means each g ∈ G can be viewed as a vector space automorphism of CX, i.e. an element of GL(CX). Thus we have defined a map ρ : g → GL(CX), i.e. we have turned the set X into a linear representation CX of G. In module language, CX is a CG-module. Identify GL(CX) with GLn (C) via the choice x1 , . . . , xn of basis. Then ρ(g) becomes an n × n matrix for each g ∈ G. Note the ij-entry of this matrix is 1 if gxj = xi and it is zero otherwise. Since X was a G-set this means that the matrix ρ(g) is a permutation matrix: all its entries are zeros and ones, and there is just one non-zero entry in every row and column. So amongst all matrix representations of G, the ones coming from permutation representations are very simple... 4.6.7. Example. Let’s just go back to S3 again. It IS a permutation group, so it has a natural permutation representation on the set X = {1, 2, 3}. The corresponding matrix representation S3 → GL3 (C) is one you all know perfectly well. For instance, the image of the three cycle (1 2 3) with respect to the standard basis v1 , v2 , v3 of CX labelled by the elements of the set X (I’d better not just call them 1, 2, 3!) is the matrix 0 0 1 1 0 0 . 0 1 0 Now clearly the vector v1 + v2 + v3 is fixed by all of G so it spans a one dimensional submodule (isomorphic to the trivial module). By Maschke’s theorem that had better have a complement. For instance {v1 − v2 , v2 − v3 } is a complement (all a1 v1 + a2 v2 + a3 v3 with a1 + a2 + a3 = 0). Let’s write down matrices with respect to the new basis v1 + v2 + v3 , v1 − v2 , v2 − v3 instead: we get the following matrices 1 0 0 ρ(1) = 0 1 0 , 0 0 1 1 0 0 ρ((1 2)) = 0 −1 1 , 0 0 1 1 0 0 ρ((2 3)) = 0 1 0 , 0 1 −1 1 0 0 −1 , ρ((1 3)) = 0 0 0 −1 0 1 0 0 ρ((1 2 3)) = 0 0 −1 , 0 1 −1 1 0 0 ρ((1 3 2)) = 0 −1 1 . 0 −1 0 Note all these matrices are block diagonal matrices. The top 1×1 block is the trivial representation of G on V1 = C(v1 + v2 + v3 ), the bottom 2 × 2 block is a two dimensional representation of G on V2 = C(v1 − v2 ) ⊕ C(v2 − v3 ). Of course V2 is irreducible. The decomposition V = V1 ⊕ V2 of V into irreducibles corresponds in matrix language to choosing a basis so that each ρ(g) is block diagonal... and since the blocks are irreducible representations you cannot do any better. In other words, if V and W are two finite dimensional CG-modules, so is V ⊕ W . Pick basis, to view V as a matrix representation ρ : G → GLn (C) and W as a matrix representation σ : G → GLm (C). With respect to the basis for V ⊕ W obtained by concatenating the two basis, 4.6. CHARACTER THEORY OF FINITE GROUPS 117 the matrix representation ρ ⊕ σ : G → GLm+n (C) corresponding to the module V ⊕ W has all g mapping to block diagonal matrices diag(ρ(g), σ(g))... This is how you should think of direct sums of CG-modules in terms of matrices. Let V be a finite dimensional CG-module. Equivalently, let (V, ρ) be a representation of G. Define its character χV : G → C to be the function with χV (g) equal to the trace of the endomorphism ρ(g). (Recall that tr(AB) = tr(BA), hence if A is invertible tr(ABA−1 ) = tr(A−1 AB) = tr(B), so the trace of a matrix depends only on its similarity class, so it makes sense to define trace of an endomorphism as the trace of the matrix of the endomorphism in any basis...) Note χV extends linearly to a map also denoted χV : CG → C, sometimes we’ll apply χV to elements of the group algebra, but usually I won’t think of χV in this way. 4.6.8. Example. The character χ1 of the trivial CG-module is χ1 (g) = 1 for all g ∈ G. The character ψ of the regular CG-module CG is ψ(g) = |G| if g = 1 and ψ(g) = 0 for all other 1 4= g ∈ G. A class function is a function f from G to C that is constant on conjugacy classes, i.e. f (hgh−1 ) = f (g). For example, the character χV of any finite dimensional CG-module is a class function: (2) tr(ρ(hgh−1 )) = tr(ρ(h)ρ(g)ρ(h)−1 ) = tr(ρ(h)−1 ρ(h)ρ(g)) = tr(ρ(g)) hence χV (hgh−1 = chiV (g). Let C(G) denote the vector space of all class functions on G. There is an obvious basis for C(G): let C1 , . . . , Cr be the conjugacy classes of G. Let δi : G → C be the function with δi (g) = 1 if g ∈ Ci , 0 otherwise. Then C(G) has basis δ1 , . . . , δr . There is a less obvious basis for C(G): let L1 , . . . , Lr be a complete set of pairwise non-isomorphic irreducible CG-modules. (Recall the number of them is the number of conjugacy classes in G too...) Let χi = χLi be the corresponding characters. 4.6.9. Theorem. χ1 , . . . , χr is a basis for C(G). Proof. Consider Wedderburn: CG = Mn1 (C) × · · · × Mnr (C). In the product on the right hand side, let ei = (0, · · · , 0, Ini , 0, · · · , 0), i.e. the identity matrix of the ith matrix algebra. The irreducible representations on the right hand side are the modules L1 , . . . , Lr where Li is the module of column vectors for Mni (C) on which ei acts as 1 and all other ej act as zero. So χi (x) is just the trace of the ith matrix in the Wedderburn decomposition of an element of x ∈ CG... But χi (ej ) = δi,j nj . This proves that χ1 , . . . , χr are linearly independent. Hence they form a basis by dimension considerations. 4.6.10. Corollary. Two finite dimensional CG-modules V and W are isomorphic if and only if χV = χW , i.e. they have the same characters. )r )r i i Proof. Note V is a direct sum of irreducibles, say V ∼ . Similarly W ∼ . = i=1 L⊕a = 9i=1 L⊕b i i r ∼ Now V = W if and only if a = b for all i (say by Jordan-Holder theorem). But χ = a χ i i V i i i=1 9r and χW = i=1 bi χi . So χV = χW if and only if ai = bi for all i by the theorem... In the proof of the theorem, we exploited the idempotents ei ∈ CG coming from the identity matrices in the Wedderburn decomposition. Let us repeat this, because these are important: there are mutually orthogonal central idempotents e1 , . . . , er ∈ CG summing to the identity with the property that ei acts on the jth irreducible module Lj (of dimension nj ) as δi,j . Sorry there is a lot of fixed notation: r is the number of conjugacy classes, C1 , . . . , Cr are the conjugacy classes, L1 , . . . , Lr are representatives of the isomorphism classes of irreducible modules, χi is the character 118 CHAPTER 4. ALGEBRAS of Li , ni is the dimension of Li or χi (1), ei is the central idempotent corresponding to Li ... To make things worse I will always assume from now on that C1 = {1} and L1 is the trivial module. Also let ci = |Ci | be the size of the ith conjugacy class. Also I’ll pick once and for all a representative gi in the conjugacy class Ci . The character table of G is the r × r matrix with rows labelled by χ1 , . . . , χr and columns labelled by C1 , . . . , Cr with ij-entry equal to χi (gj ). Of course this is independent of the particular representative gj of Cj chosen because χi is a class function. (There are two choices in all of these that are important: the indexing of the conjugacy classes by 1, . . . , r determines the ordering of the columns of the character table and the indexing of the isoclasses of irreducible representation by 1, . . . , r determines the ordering of the rows of the character table). The character table of G is a crucial invariant of the group G... 4.6.11. Example. Here is the character table of S3 : χ1 χ2 χ3 1 (1 2) (1 2 3) 1 1 1 1 −1 1 2 0 −1 Here χ2 is the sign representation, χ3 is the two dimensional one sitting in the natural permutation representation. Note for example that given any finite dimensional CG-module V you can decompose it into9irreducibles given just χ and its character table: just work out how to )r its character r i ... In class I will apply this to the natural permutation write χ = i=1 ai χi then V ∼ = i=1 L⊕a i representation and the regular representation. Before we forget all the notation, let me prove a slightly technical lemma. 9 (i) 4.6.12. Lemma. If ei = g∈G ag g then a(i) g = ni χi (g −1 ) . |G| Proof. Let ψ be the character of the regular CG-module and g ∈ G. We’ll compute ψ(ei g −1 ) in two different ways. On the one hand, ! (i) ei g −1 = ah hg −1 . h So since ψ(1) = |G| and ψ is zero on all other group elements, On the other hand, ψ = 9r j=1 ψ(ei g −1 ) = a(i) g |G|. nj χj . So ψ(ei g −1 ) = ! nj χj (ei g −1 ). j But ei acts as zero on all Lj for j 4= i, and it acts as 1 on Li . So we get that ψ(ei g −1 ) = ni χi (g −1 ). Comparing these formulas proves the lemma. Okay, now let’s do something clever. Introduce a Hermitian form on the complex vector space C(G) by defining the pairing of two class functions χ and ψ from 1 ! (χ, ψ) = χ(g)ψ(g) |G| g∈G 9 1 2 where − denotes complex conjugation. Note that (χ, χ) = |G| g∈G |χ(g)| which is > 0 if and only if χ 4= 0. So it is a positive definite Hermitian form. So we can call it an inner product... 119 4.6. CHARACTER THEORY OF FINITE GROUPS 4.6.13. Theorem (Orthogonality relations). (i) With respect to the inner product just defined, χ1 , . . . , χr are orthonormal. Hence for any character χ, χ= r ! (χ, χi )χi . i=1 (ii) (row orthogonality relation) With our usual notation for the character table, we have for any i, j = 1, . . . , r that ; r ! 0 if i 4= j, ck χi (gk )χj (gk ) = |G| if i = j. k=1 (iii) (column orthonality relation) For any j, k = 1, . . . , r we have that ; r ! 0 if j 4= k, χi (gj )χi (gk ) = |G|/cj if j = k. i=1 Proof. (i) We know from the preceeding lemma that χi (ej ) = δi,j nj . Also ej = Hence, 1 ! δi,j nj = nj χj (g −1 )χi (g). |G| g 1 |G| 9 g nj χj (g −1 )g. You’ll show on the exercises that χj (g −1 ) is χj (g) = χj (g)−1 . Hence the right hand side is nj |G| (χi , χj ). (ii) This is just a restatement of (i) in the language of the character table. (iii) Let A be the character table, i.e. the matrix with ij-entry χi (gj ). Let B be the matrix with ij-entry ci χj (gi )/|G|. Let us compute AB: its ij-entry is r 1 ! ck χi (gk )χj (gk ) = δi,j . |G| k=1 Therefore AB = I. Therefore BA = I. Let’s compute its ij-entry: 1 ! ci χk (gi )χk (gj ) = δi,j . |G| k I think that’s it. This gives some rather strong arithmetic properties that the character table must satisfy. Using it you can compute character tables using only a relatively small amount of information. Here are some examples. 4.6.14. Example. Let us first compute the character table of the cyclic group C4 of order 4. Its abelian, so r = 4 and n1 = n2 = n3 = n4 = 1. Let x be a cyclic generator. A 1 dimensional representation must map x → ω with ω 4 = 1 in C. Let ω be a primitive 4th root of one. Then the four choices are x $→ ω, x $→ ω 2 = −1, x $→ ω 3 and x $→ ω 4 = 1. So the character table is: Ci ci χ1 χ2 χ3 χ4 1 x x2 x3 1 1 1 1 1 1 1 1 1 ω −1 ω 3 1 −1 1 −1 1 ω 3 −1 ω Note χ2 can be thought of as the composite of the map C4 ! C2 mapping x2 to 1, then the non-trivial character of C2 ... 120 CHAPTER 4. ALGEBRAS 4.6.15. Example. Next, let’s look at V4 = {1, x, y, xy}. There are four obvious one dimensional representations. For example you can map x to −1, y to 1 then xy has to map to −1. So you get Ci ci χ1 χ2 χ3 χ4 1 x y xy 1 1 1 1 1 1 1 1 1 −1 1 −1 1 1 −1 −1 1 −1 −1 1 You should check the orthogonality relations hold in both cases... These two examples give groups G, H with CG ∼ = CH but with different character tables: character tables are a finer invariant than just the information in the Wedderburn decomposition. 4.6.16. Example. Next lets look at D4 = {1, x, x2 , x3 , y, xy, x2 y, x3 y}. Recall that x2 is central. So there is a surjection D4 ! V4 under which x2 maps to 1. Using this the four irreducible characters of V4 lift to give four irreducible characters of D4 . There are just 5 conjugacy classes, and you can work out the last irreducible character just using the orthogonality relations. Ci ci χ1 χ2 χ3 χ4 χ5 1 x2 x y xy 1 1 2 2 2 1 1 1 1 1 1 1 −1 1 −1 1 1 1 −1 −1 1 1 −1 −1 1 2 −2 0 0 0 4.6.17. Example. Finally lets look at Q3 = {1, 1̄, i, ī, j, j̄, k, k̄}. The conjugacy classes are {1}, {1̄}, {i, ī}, {j, j̄}, {k, k̄}. There’s a map Q3 ! V4 , 1̄ $→ 1}. This gives four irreducible characters, the last comes for free. Ci ci χ1 χ2 χ3 χ4 χ5 1 1̄ i j k 1 1 2 2 2 1 1 1 1 1 1 1 −1 1 −1 1 1 1 −1 −1 1 1 −1 −1 1 2 −2 0 0 0 Note this is the same as the character table in the preceeding example. So we’ve found groups G 4∼ = H with the same character tables, just in case you were wondering if character tables separated groups... The purpose of the next theorem is to convince you that the character table of a finite group does actually say quite a bit about the structure of the group. In fact character theory is absolutely crucial in many parts of finite group theory, e.g. the classification of finite simple groups would never have been possible without it. 4.6.18. Theorem. Let ρ : G → GL(V ) be a representation of G, and χ = χV be its character. Let ker χ = {g ∈ G | χ(g) = χ(1)}. Then, (i) |χ(g)| ≤ χ(1) = dim V ; (ii) ker ρ = ker χ, hence ker χ is a normal subgroup of G; 121 4.6. CHARACTER THEORY OF FINITE GROUPS (iii) if χ = 9t s=1 as χis with as > 0 then ker χ = : i ker χis ; (iv) if N is :tany normal subgroup of G, then there are irreducible characters χi1 , . . . , χis such that N = s=1 ker χi . Proof. (i) Note that g |G| = 1 for any g ∈ G. Hence the minimal polynomial of ρ(g) divides x|G| −1. In particular, it has distinct linear factors, so ρ(g) is diagonalizable, and all its eigenvalues are |G|th roots of unity. Hence, χ(g) is a sum of χ(1) roots of unity. Now the triangle inequality implies that |χ(g)| ≤ χ(1). (ii) If g ∈ ker ρ, then |χ(g)| = χ(1), hence g ∈ ker χ. Conversely, if g ∈ ker χ then χ(g) = χ(1) so equality holds in the triangle inequality, so all of the eigenvalues of ρ(g) are equal. Say ρ(g) = ωI, a |G|th root of 1. So χ(g) = χ(1)ω. But χ(g) = χ(1) by hypothesis, so in fact ω = 1 and g ∈ ker ρ as ρ(g) = I. 9t 9 9 (iii) Say χ =<< 9 s=1 as χis <<with9as > 0. Take g ∈ ker χ. Then χ(g) = s as χis (g). Hence χ(1) = and all s as χis (g) = s as |χis (g)|. So equality holds in the triangle inequality, s as χis (1) = 9 χ (g) are equal to some complex number z (say) times χ (1). But then χ(g) = a χ (1)z = i i s i s s s 9s : a χ (1), hence z = 1. Therefore χ (1) = χ (g) for all s, so g ∈ ker χ . The reverse s i i i i s s s s s s inclusion is clear. (iv) Let N be a normal subgroup. Lift the regular character of G/N to G. This gives a character ψ of G with ψ(g) = [G : N ] if g ∈ N , ψ(g) = 0 if g ∈ / N . For this, ker ψ = N . Let us now compute the character table of S4 . C |C| χ1 χ2 χ3 χ4 χ5 1 1 1 1 2 3 3 (1 2) 6 1 −1 0 1 −1 (1 2 3) 8 1 1 −1 0 0 (1 2 3 4) 6 1 −1 0 −1 1 (1 2)(3 4) 3 1 1 2 −1 −1 Here’s how you get it. Start with the trivial and sign character. (These are lifts of the two irreducible characters of the cyclic group C2 via the surjection sgn : S4 ! C2 ). Next look at the surjection S4 ! S4 /V4 ∼ = S3 . Lifting the two dimensional irreducible character of S3 through this gives χ3 . Next look at the natural permutation representation χ. Its values on these classes are the numbers of fixed points: 4, 2, 1, 0, 0. Of course it contains χ1 as a constituent. Subtracting χ − χ1 gives χ4 (check its irreducible by showing (χ4 , χ4 ) = 1). Finally χ5 is completed by using the orthogonality relations, or noting χ5 = χ2 χ4 (pointwise multiplication). I want to end the section by discussing the various tricks around to compute character tables. By a character of G I mean the character of some CG-module. So characters are special class functions: the ones which are Z≥0 -linear combinations of irreducible characters. These techniques all give ways to find some genuine characters. They need not be irreducible (a character is irreducible if and only if (χ, χ) = 1). But they are a starting point... You can then reduce them bit by bit to eventually pin down the irreducible characters. • Characters of finite abelian groups are easy. • For any group, there’s always the trivial character χ1 , and the regular character ψ with 9r ψ(1) = |G|, ψ(g) = 0 for g 4= 1. Note ψ = i=1 ni χi (by Wedderburn’s theorem), so ψ contains everything else ... but its hard to decompose it without more data. • Lifts. If N is a normal subgroup of G, then you can lift irreducible characters of G/N to G through G ! G/N . The character table of G/N gives some but not all rows of the character table of G... 122 CHAPTER 4. ALGEBRAS • Permutation characters. If G acts on a set X, the corresponding CG-module CX gives rise to a character χ of G with χ(g) equal to the number of fixed points of g on X. This is an easy source of characters... Its always the case that (χ, χ1 ) (where χ1 is the trivial character of G) is the number of orbits of G on X. (This is the orbit counting lemma: the number of orbits of G on X is the average number of fixed points ... that is the definition of the inner product on C(G)). • Tensor products. There is an obvious pointwise multiplication on the vector space C(G) making it into a commutative C-algebra: (χ · ψ)(g) = χ(g)ψ(g). In fact the product of two characters is always a character. Indeed, let V and W be two CG-modules with characters χV and χW . Make the vector space V ⊗ W (tensor product over C) into a G-module so that g(v ⊗ w) = (gv) ⊗ (gw) for g ∈ G, v ∈ V, w ∈ W . Then, I claim that χV χW = χV ⊗W . To see this, use the following fact: suppose α : V → V, β : W → W are diagonalizable endomorphisms. Then tr(α ⊗ β) = tr(alpha) tr(β). I’ll leave the proof as an exercise: pick bases in which everything is diagonal then its not too bad. Note this statement is actually true even if things are not diagonalizable, but its a little harder to prove. But of course we know for any finite dimensional representation ρ : G → GL(V ) that all ρ(g) ARE diagonalizable matrices, because their minimal polynomials have distinct linear factors. • *Induction/restriction*. Let H be a subgroup of G. Then CH is a subalgebra of CG. So any CG-module is a CH-module by restriction. At the level of characters, any character χ of G restricts to a character χ ↓H of H... its just the restriction of the function on G to a function on H. Conversely, any CH-module V gives rise to a CG-module CG ⊗CH V . This is called the “induced module”. At the level of characters, there is a corresponding notion of “induced character”, denoted χ ↑G if χ was a character of H. In class, I’ll prove the following formula for χ ↑G : if t1 , . . . , tk is a system of G/H-coset representatives then χ ↑ (g) = G k ! χ̇(t−1 i gti ) i=1 where χ̇(g) is χ(g) if g ∈ H, χ̇(g) = 0 otherwise. For example, if G acts transitively on a set X and you let H be the stabilizer of a point x ∈ X, hence X is identified with the coset space G/H via the orbit map, then this formula gives that 1H ↑G (g) is the number of fixed points of g on X. Here 1H denotes the trivial character of H. So permutation characters are a special case of induced characters: they are precisely the characters induced from trivial representations of subgroups... There is one important fact about induced characters which is a good excuse to remind you some more about tensor products. Let us write (., .)G for the inner product on C(G) and (., .)H for the inner product on C(H) to avoid confusing the two. The following formula is known as Frobenius reciprocity: (χ ↑G , ψ)G = (χ, ψ ↓H )H for any χ ∈ C(G), ψ ∈ C(H). Modern proof. First suppose that V and W are two finite dimensional CG-modules. I want to observe that the inner product (χV , χW )G computes the dimension of the vector ) ⊕ai ) b space HomCG (V, and W = j Ljj . So χV = 9 9W ). To prove this, decompose 9 V = i Li ai χi , χW = bj χj . Hence (χV , χW )G = i ai bi . This equals dim HomCG (V, W ), since dim HomCG (Li , Lj ) = δi,j . Now to prove the Frobenius reciprocity formula, it suffices to consider the case that χ = χV for some CH-module V and ψ = χW for some CG-module W . In terms of homomorphisms, the formula claims that HomCG (CG ⊗CH V, W ) ∼ = HomCH (V, W ↓H ). 123 4.7. BURNSIDE’S P A QB THEOREM Now notice that W ↓H is isomorphic as a CH-module to HomCG (CG, W ), where CG is being viewed as a (CG, CH)-bimodule by right multiplication for H. Now the formula is exactly adjointness of tensor and hom! • *Geometry*. Well we won’t go there. But there are lots of beautiful ways to get characters from groups acting on topological spaces and their cohomology.... Let’s end this introduction to character theory by computing the characters of the symmetric group S5 . Remark: the symmetric groups are remarkable in that the characters of all Sn are integers. Usually character values really are complex numbers, but for a character of degree n we do at least know that all χ(g)’s are sums of n |G|th roots of unity... In particular all character values are algebraic integers. C |C| χ1 χ2 χ3 χ4 χ5 χ6 χ7 1 (12) 1 10 1 1 1 −1 4 2 4 −2 5 1 5 −1 6 0 (123) (1234) 20 30 1 1 1 −1 1 0 1 0 −1 −1 −1 1 0 0 (12345) 24 1 1 −1 −1 0 0 1 (12)(34) 15 1 1 0 0 1 1 −2 (123)(45) 20 1 −1 −1 1 1 −1 0 Working: Work out the conjugacy classes and their orders. Then we have the trivial and sign representations. Next look at the natural permutation character which has values 5, 3, 2, 1, 0, 1, 0 on the conjugacy classes. It contains the trivial character, so subtract it... Then what is left is irreducible since its inner product with itself is 1. This gives a degree 4 irreducible character, and tensoring that with sign gives another one. The remaining three characters either have degrees (5, 5, 6) or (1, 6, 7). But the degree of a character divides the order of the group (see next section) so the second thing couldn’t happen. Now we need some input. Consider the permutation action on the set of pairs {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}. Computing fixed points gives the character values 10, 4, 1, 0, 0, 2, 1 on the conjugacy classes in the order listed. Subtract the trivial character. We’ve got 9, 3, 0, −1, −1, 1, 0. Now take the inner product with χ3 . You get 1. So subtract χ3 . Then check that is indeed irreducible. Now tensor with sign to get χ6 . Finally work out χ7 using orthogonality relations. 4.7 Burnside’s pa q b theorem You know that simple arithmetic plays an important role in finite group theory. The character table is the way that “higher arithmetic” comes into play. By “higher arithmetic” I mean algebraic number theory... Don’t forget: if χ is a character of degree n, then each character value χ(g) is a sum of n |G|th roots of unity. We need to understand such numbers to make further progress... In class we’ll digress at this point to cover sections 5.1 and 6.1 from these notes, in particular, we’ll meet the definition of an algebraic integer: an element of C that is a root of a monic polynomial in Z[x]. We’ll prove in §6.1 that the sum or the product of two algebraic integers is again an algebraic integer. Hence all character values χ(g) are algebraic integers. Here are a couple of basic lemmas. 124 CHAPTER 4. ALGEBRAS 4.7.1. Lemma. Let α be an algebraic number and mα (x) ∈ Q[x] be its minimal polynomial (the monic polynomial with coefficients in Q of smallest possible degree having α as a root). Then α is an algebraic integer if and only if mα (x) ∈ Z[x]. Proof. Let α be an algebraic integer. Let p(x) be a monic polynomial of minmal degree in Z[x] having α as a root. Note p(x) is irreducible in Z[x] (if it factorizes then by looking at the leading coefficients both factors are monic and one of them already has α as a root). Hence by Gauss’ lemma it is irreducible in Q[x]. Clearly mα (x)|p(x). Together these imply that mα (x) = p(x). Note we will prove the next lemma again in §6.1 – in fancy language it says that Z is integrally closed in Q. 4.7.2. Lemma. A rational algebraic integer (i.e. an algebraic integer lying in Q) is actually an integer (i.e. an element of Z). Proof. Let α be an algebraic integer that can be written as a rational number p/q in lowest terms. Say f (x) = xn + a1 xn−1 + · · · + an is a monic polynomial in Z[x] with α as a root. Then, q n−1 f (α) = 0 = It follows that pn q pn + a1 pn−1 + · · · + an q n−1 . q is already an integer, i.e. q|pn . This contradicts that p/q was in its lowest terms. Finally we need the following lemma about algebraic integers, which I’ll prove again in §6.1 without using the noetherianness of Z (it will make more sense then when it is put into the right context – right now it looks like magic). 4.7.3. Lemma. Let α ∈ C. Suppose there exists a faithful Z[α]-submodule U of C that is finitely generated as an abelian group. (Here, faithful means that annZ[α] (U ) = {0}). Then α is an algebraic integer. Proof. The main task is to prove that Z[α] is a noetherian Z-module. Once we’ve done that, the lemma is easy: look at 1, α, α2 , . . . . By ACC there must be some k such that αk ∈ Z + Zα + · · · + Zαk−1 . Hence α is a root of a monic polynomial in Z[x] of degree k. Hence α is an algebraic integer. To prove that Z[α] is a noetherian Z-module, note that since U is a finitely generated Zmodule, so is EndZ (U ). To prove this statement, let F be a free Z-module of finite rank having U as a quotient. Then EndZ (U ) is a Z-submodule of HomZ (F, U ) (it’s the homomorphism that factor through the quotient U of F ...). The latter is a direct sum of finitely many copies of HomZ (Z, U ) ∼ = U , which is finitely generated. But Z is noetherian so submodules of finitely generated Z-module are finitely generated. Hence the submodule EndZ (U ) is finitely generated too. But now the assumption that U is faithful means that the representation Z[α] → EndZ (U ) is injective. So Z[α] is a Z-submodule of a noetherian Z-module, hence its noetherian as Z is a noetherian ring. Now I’m going to prove an important theorem which is really just a warm up to proving Burnside’s pa q b theorem. It is our first application of algebraic integers to character theory. 4.7.4. Theorem. With the notation of the previous section, the degrees ni of the irreducible characters of a finite group G all divide |G|. 9 Proof. Recall that ei = g∈G α1 χi (g −1 )g where α = |G|/ni . Since ei is an idempotent, e2i = ei , so ! αei = χi (g −1 )gei . g∈G 125 4.7. BURNSIDE’S P A QB THEOREM Now for any g, h ∈ G, define ah,g ∈ C from the equation ! gei = ah,g h. h∈G The above equation shows that ! ! ! αah,g h = αgei = χi (k −1 )(gk)ei = χi (k −1 )ah,gk h. h∈G k∈G h,k∈G 9 Equating coefficients shows that αah,g = k∈G χi (k −1 )ah,gk . Now let M be the Z-submodule of C generated by all {za ∈ G, z |G| = 1}. Clearly M is finitely generated as a Z-module. 9 h,g | h, g −1 Moreover, αzah,g = k∈G χi (k )zah,gk . The coefficient χi (k −1 )z is a sum of |G|th roots of 1, because character values are. This shows that αzah,g ∈ M . Hence M is a Z[α]-submodule of C, clearly faithful as C is a field, that is finitely generated as a Z-module. Hence by the previous lemma, α is an algebraic integer. But α ∈ Q, hence by Lemma 4.7.2 α is an integer. Hence ni ||G|. Okay, now we’re going to do a similar trick to prove Burnside’s pa q b theorem. We need a little more preparation, which needs a little Galois theory. Let me recall: if f (x) ∈ Q[x] is a polynomial, its splitting field E is the subfield of C obtained by adjoining to Q all the roots of the polynomial f (x). The Galois group of f (x) over Q is the Galois group Gal(E/Q) of the field extension Q ⊆ E. That is, it is the group of all field automorphisms of E leaving Q invariant (that last condition is guaranteed in this case that the small field is Q since Q doesn’t have any non-trivial automorphisms...). Note the fixed subfield of the Galois group {α ∈ E | g(α) = α for all g ∈ Gal(E/Q)} is precisely the field Q again. This follows from the fundamental theorem of Galois theory. Note the Galois group of f (x) permutes the set of roots of the polynomial f (x), hence Gal(E/Q) embeds into the symmetric group on the set of roots of f (x). This is where group theory came from originally. You probably met the following result when you first studied Galois theory. 4.7.5. Lemma. If f (x) ∈ Q[x] is monic and irreducible, then its Galois group acts transitively on the set of roots of f (x). # Proof. Note = f (x) has distinct roots (else (f (x), f (x)) 4= 1). So if R is the set of roots of f (x), f (x) == α∈R (x − α). Suppose the Galois group has two orbits on R, R = R1 8 R2 . Then g(x) = α∈R1 (x − α) divides f (x). But the Galois group fixes g(x), hence all its coefficients lie in the fixed subfield, namely, Q. Hence f (x) is reducible. Now here is the main difficult thing that we’ll need about algebraic numbers. 4.7.6. Lemma. Suppose ε1 , . . . , εn are N th roots of unity and α = integer with |α| < 1. Then α = 0. 1 n (ε1 + · · · + εn ) is an algebraic Proof. Let mα (x) ∈ Z[x] be the minimal polynomial of α. Its irreducible over Q[x]. Let β be any root of mα (x). Let G be its Galois group, i.e. G = Gal(E/Q) where E is the splitting field of mα (x) over Q. By the previous lemma, G acts transitively on the roots of mα (x). Hence there exists g ∈ G such that g(α) = β. Let F be the splitting field of xN − 1 over E. Since g maps xN − 1 to itself, the automorphism g of E extends to an automorphism also denoted g of F (e.g. by Existence and uniqueness of splitting fields in 5.3). Hence β= 1 (gε1 + · · · gεn ) n and each gεi is also an N th root of unity, hence has |gεi | ≤ 1. It follows from this and the triangle inequality that |β| ≤ 1.