Mathematics 893 Problem Set 9

Required reading: Section 7.6 (Chinese Remainder Theorem). Section 12.1. Suggested reading: sections 13.1 to 13.4.

Solutions due Monday December 5 in my office.

(a) Let k be a field and R = k[x_{1}, …, x_{n}]. Let σ ∈ S_{n} and define a kalgebra automorphism σ̂ of R by σ̂(x_{i}) = x_{σ(i)}. Let Δ = ∏_{i<j}(x_{j}− x_{i}). Prove that if σ is a transposition then σ̂(Δ) = −Δ. From this conclude that every element of S_{n} is either the product of an even number of transpositions or an odd number of transpositions.

(b) Let L be the matrix [ \begin{matrix}2 & 2 & −4 & −4 & 8 & 8 \\ −4 & 8 & 2 & 8 & 2 & −4 \\ 8 & −4 & 8 & 2 & −4 & 2 \end{matrix} ] and let A

be the subgroup of ℤ^{6} generated by the rows of L. Find bases for ℤ^{6} and A as described by Theorem 4 of page 460. Write ℤ^{6}/A (up to isomorphism) as the direct sum of cyclic groups as in Theorem 5, page 462, and as in Theorem 6, page 464.

(c) Let L be the matrix of the previous question. Find (ℤ-module) bases for ℤ^{6} ∩ V where V is the subspace of ℚ^{6} spanned by the rows of A and for ker L (where L is regarded as a linear map from ℤ^{6} → ℤ^{3} by left multiplication). (Both of these are subgroups of ℤ^{6}, hence must be free, of rank less than or equal to 6.)

(d) Page 529 #2,3

(e) Page 530 #8,9,10 Notes

1. A given permutation σ ∈ S_{n} can be written in many ways as the product of transpositions. Problem (a) shows that no matter how this is done, the number of the transpositions is either always even or always odd. The subset of S_{n} consisting of all permutations that are the product of an even number of transpositions is obviously closed under multiplication, hence is a subgroup A_{n} of S_{n}, called the alternating group. Our text spends several pages on this (pp 107-110). However the proof of the existence of the alternating group should only be a few lines if one makes more explicit use of group actions.

2. Our proof of Theorem 4, page 460, for a general PID was not very constructive. However over a Euclidian ring (such as ℤ) one can work in a much more constructive manner using matrices. This idea is outlined in Exercises 17-19 of pages 470-471. These ideas can be used in questions (b) and (c).

3. Suppose that M is a finitely generated module over a PID R, with generators g_{1}, …, g_{n}. Then there is a surjective homomorphism of R-modules ϕ : R^{n} → M defined by sending e_{i} to g_{i}, 1 ≤ i ≤ n. Let N = ker ϕ, so that M ≅ R^{n}/N. If we apply Theorem 4, page 460, to the pair (R^{n}, N) (with m =rank N and a_{i} as in the statement of the Theorem) then we conclude that M ≅ R^{n−m} ⨁ ⊕_{i=1}^{m}R/a_{i}R, i.e. that every finitely generated R-module is the direct sum of cyclic R-modules. The cyclic Rmodules R/a_{i}R and R occuring in the decomposition are unique (up to order) if either a_{1}|a_{2} … |a_{m} (as in the statement of Theorem 4), or if they are of the form R or R/p^{i}R where p is a prime element of R. (Of course one chooses one prime from each class of associates). One can convert one representation into the other by using the Chinese Remainder Theorem. These uniqueness results are stated more precisely and proved in pages 462-468. The idea is as follows (simplifying somewhat): if N = R/p^{a}R, a ≥ 1 then p^{i}N/p^{i+1}N is isomorphic to R/pR (i.e. is a one dimensional vector space over the field k = R/pR) if 0 ≤ i ≤ a − 1 and is equal to 0 if i ≥ a. If N = R then p^{i}N/p^{i+1}N is isomorphic to R/pR for all i ≥ 0. The dimensions of the finite dimensional k-vector spaces p^{i}M/p^{i+1}M thus determine the number of invariant factors R or R/p^{a}R so the invariant factor decomposition in Theorem 6, page 464, is thus uniquely determined. From the invariant factor decomposition the a_{i} in the decomposition of Theorem 4, page 460, are uniquely determined.