Algebra (Sept 2015) - University of Manitoba
... (a) Show that one can define a k[x]-module M by taking the underlying abelian group of V together with a k[x]-action where x acts on v ∈ V by x · v = T (v). (b) Show that there is a one-to-one correspondence between subspaces U ⊂ V satisfying T (U ) = U and submodules of M . ...
... (a) Show that one can define a k[x]-module M by taking the underlying abelian group of V together with a k[x]-action where x acts on v ∈ V by x · v = T (v). (b) Show that there is a one-to-one correspondence between subspaces U ⊂ V satisfying T (U ) = U and submodules of M . ...
Chapter A.1. Basic Algebra
... associativity, and distributivity. The typical examples are the field of rational numbers (usually denoted Q), the field of real numbers R, and the field of complex numbers C; however as just mentioned not all examples are so familiar. The integers do not constitute a field because in general it is ...
... associativity, and distributivity. The typical examples are the field of rational numbers (usually denoted Q), the field of real numbers R, and the field of complex numbers C; however as just mentioned not all examples are so familiar. The integers do not constitute a field because in general it is ...
2008-09
... Let I, J be ideals of a ring R such that I + J = R. Prove that I J = IJ and if IJ = 0, then R ; R R . ...
... Let I, J be ideals of a ring R such that I + J = R. Prove that I J = IJ and if IJ = 0, then R ; R R . ...
Math 3333: Fields, Ordering, Completeness and the Real Numbers
... is clearly bounded since e.g. −3/2 ≤ x ≤ 3/2 for every x ∈ S . Moreover, it can be shown that if r = sup S then this number must satisfy r2 = 2; see exercise 6 below. In class however we showed there is no rational number whose square is two. This exemplifies the fundamental flaw of the ordered fiel ...
... is clearly bounded since e.g. −3/2 ≤ x ≤ 3/2 for every x ∈ S . Moreover, it can be shown that if r = sup S then this number must satisfy r2 = 2; see exercise 6 below. In class however we showed there is no rational number whose square is two. This exemplifies the fundamental flaw of the ordered fiel ...
2009-04-02 - Stony Brook Mathematics
... associative, meaning that we have a set N and an operation (+) so that the following is true for any a, b, c € N: a + b € N (closure); a + b = b + a (commutativity); and a + (b + c) = (a + b) + c (associativity). Note that N is both an additive semigroup and a multiplicative one. A monoid is a set M ...
... associative, meaning that we have a set N and an operation (+) so that the following is true for any a, b, c € N: a + b € N (closure); a + b = b + a (commutativity); and a + (b + c) = (a + b) + c (associativity). Note that N is both an additive semigroup and a multiplicative one. A monoid is a set M ...
Algebra I Section 1-1 - MrsHonomichlsMathCorner
... Verbal Phrases more than, sum, plus, increased by, added to less than, subtracted from, difference, decreased by, minus ...
... Verbal Phrases more than, sum, plus, increased by, added to less than, subtracted from, difference, decreased by, minus ...
Math 711, Fall 2007 Problem Set #5 Solutions 1. (a) The extension
... 1. (a) The extension is module-finite since the equation is monic in Z, and generically étale since adjoining a cube root gives a separable field extension in characteristic 6= 3. The matrix has as entries the traces of the elements z i−1+j−1 , 1 ≤ i, j ≤ 3. The trace of 1 is 3, and the trace of mu ...
... 1. (a) The extension is module-finite since the equation is monic in Z, and generically étale since adjoining a cube root gives a separable field extension in characteristic 6= 3. The matrix has as entries the traces of the elements z i−1+j−1 , 1 ≤ i, j ≤ 3. The trace of 1 is 3, and the trace of mu ...
On the Reducibility of Cyclotomic Polynomials over Finite Fields
... The following well-known lemmata will be used in the development of the main theorem. Lemma 3.1. For a ∈ Z, the equation x2 = a has a solution in Z if and only if x2 ≡ a (mod n) has a solution for all n ∈ Z. Lemma 3.2. Let Fp denote the finite field of p elements, Z/pZ. For all field extensions ...
... The following well-known lemmata will be used in the development of the main theorem. Lemma 3.1. For a ∈ Z, the equation x2 = a has a solution in Z if and only if x2 ≡ a (mod n) has a solution for all n ∈ Z. Lemma 3.2. Let Fp denote the finite field of p elements, Z/pZ. For all field extensions ...
Take-Home Final
... M = A[x]/(p(x)); this is a free A-module of rank n. (a) Let ϕx : M −→ M be the A-module homomorphism which is “multiplication by x”, i.e., by the image of x under the quotient map A[x] −→ M. Find the characteristic polynomial of ϕx . Now suppose that A is a domain, with quotient field K, and L any e ...
... M = A[x]/(p(x)); this is a free A-module of rank n. (a) Let ϕx : M −→ M be the A-module homomorphism which is “multiplication by x”, i.e., by the image of x under the quotient map A[x] −→ M. Find the characteristic polynomial of ϕx . Now suppose that A is a domain, with quotient field K, and L any e ...
WITT`S PROOF THAT EVERY FINITE DIVISION RING IS A FIELD
... with identity, there is a unique ring homomorphism—call it χR —from the ring Z of integers to R that respects the identity (i.e., such that χR (1) = 1R ). The image of χR is a subring of R and lies in the centre of R. The kernel of χR is an ideal in Z and so equals jZ for some unique integer j ≥ 0. ...
... with identity, there is a unique ring homomorphism—call it χR —from the ring Z of integers to R that respects the identity (i.e., such that χR (1) = 1R ). The image of χR is a subring of R and lies in the centre of R. The kernel of χR is an ideal in Z and so equals jZ for some unique integer j ≥ 0. ...
LOCAL CLASS GROUPS All rings considered here are commutative
... finitely generated. The class of noetherian rings include fields (there is only one ideal, namely (0)!) and PIDs (ideals are generated by one element). If R is noetherian, then so are R[x], any localization S −1 R, and any quotient ring R/I. In particular, if X ⊂ Cn is an algebraic variety, then the ...
... finitely generated. The class of noetherian rings include fields (there is only one ideal, namely (0)!) and PIDs (ideals are generated by one element). If R is noetherian, then so are R[x], any localization S −1 R, and any quotient ring R/I. In particular, if X ⊂ Cn is an algebraic variety, then the ...
Section 6.5 Rings and Fields
... these rings have zero divisors since ab ≠ 0 for nonzero a, b . Note: Although a general ring is not the integers with usual addition and multilplication, thinking of everything you know about the integers when working with general rings often helps bridge the gap to the abstract world. Example 2 (Po ...
... these rings have zero divisors since ab ≠ 0 for nonzero a, b . Note: Although a general ring is not the integers with usual addition and multilplication, thinking of everything you know about the integers when working with general rings often helps bridge the gap to the abstract world. Example 2 (Po ...