Assignment 10 { Part 1 { Math 611 (1) A torsion-free module over Z that is not free. Here is a good `counterexample' to keep in mind. In class, we saw that a finitely-generated torsion-free module over a PID is free. This `counterexample' shows that the finitely-generated condition is necessary. (a) Show that Q is a non-free torsion-free Z-module. (b) More generally, show that for any integral domain R, its field of fractions is torsion-free, but cannot be embedded into a free R-module. (c) Give an example that shows that the PID condition is also necessary by showing that a finitely-generated torsion-free module over a UFD need not be free. (2) Let F be a free R-module. Explain why every short exact sequence 0 ! N ! M ! F ! 0 of R-modules splits. (3) Invariant factors versus Elementary divisors. In class, we will prove the Elementary divisors version of the Structure theorem for finitely-generated modules over a PID. Show that the two versions are equivalent. I.e. show that having one decomposi- tion gives the other, and that one uniqueness statement implies the other. (Hint: both directions use the Chinese Remainder Theorem, in one direction to break up invariant factors, in the other to combine the elementary divisors.) (4) The order and index of a torsion R-module. Let R be a PID and let M be a finitely- ∼ L generated torsion R-module, so that M = R=(bi) with b1 j b2 j · · · j b` for some non-unit bi 2 R. Define the order of M (or the R-order of M) to be the ideal ordR(M) E R generated by the product b1b2 ··· b`. (a) Consider R = Z. Show that the Z-order of a finitely-generated torsion Z- module is the ideal generated by its cardinality. (b) It is in fact true that the order of a finitely-generated torsion R-module can be computed from any decomposition into a direct sum n ∼ M M = R=ci; i=1 1 i.e. ordR(M) = (c1c2 : : : cn). I won't make you show this in general, but do show that the order is the ideal generated by the elementary divisors of M. (c) It is also true that if 0 ! M 0 ! M ! M 00 ! 0 is a short exact sequence of 0 00 finitely-generated torsion R-modules, then ordR(M) = ordR(M ) ordR(M ). You may assume this from now on. Show that if 0 ! M1 ! M2 !···! Mk ! 0 is an exact sequence of fintiely-generated torsion R-modules, then k Y i (−1) ordR(Mi) = (1): i=1 (d) If M is a finitely-generated (but maybe not torsion) R-module and M 0 ≤ M is such that M=M 0 is torsion, then define the R-index of M 0 in M by 0 0 [M : M ]R := ordR(M=M ): Suppose M is a finitely-generated R-module and M 00 ≤ M 0 ≤ M with M=M 00 torsion. Show that 00 0 0 00 [M : M ]R = [M : M ]R · [M : M ]R: 2.