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 | b2 | · · · | b` for some

non-unit bi ∈ 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.

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