Relation Between Free Quotients and Modules

Question

Here is my question:

Let $M$ and $M'$ be $R$-modules, where $R$ is a commutative ring, and $N \subseteq M$ and $N' \subseteq M'$ submodules. Suppose that $N \cong N'$ and $M/N \cong M'/N'$. Determine (by proof or cournterxample) which of the following statements holds:

(1) If $N$ is free, then $M \cong M'$;

(2) If $M/N$ is free, then $M \cong M'$.

As always, a hint as to how to get the the proof or a family to consider for a counterexample would be appreciated, not just a straight out answer. :)

(The previous part of the question was the following: Show that $M$ is free if there exists a submodule $N \subseteq M$ such that both $N$ and $M/N$ are free.)

I have considered the standard $\Bbb Z_4 \ncong \Bbb Z_2 \oplus \Bbb Z_2 $, but quotienting by $\Bbb Z_2$ gives the same in both cases, but these don't have a basis so aren't free. I've also tried the contrapositive... still no where!

Also, I have looked to see if this is a duplicate question (I thought that it would have been asked before), but I can't find one. If it is, please just let me know and I'll remove it - I've tried to find one so please no negative reputation!!) :P

Answer 1

$\newcommand{\Z}{\mathbb{Z}}$This is a problem of computing extension groups $\mathrm{Ext}_R^1(M/N, N)$.

1. $N$ being free doesn't imply $M \cong M'$. Consider $M = \Z \oplus \Z/2\Z$, $N = \Z \oplus 0 \subset M$ and $M' = \Z$, $N' = 2 \Z \subset M'$. Then both $N \cong N' \cong \Z$ are free, $M/N \cong M'/N' \cong \Z/2\Z$, but $M \not \cong M'$.
2. If $M/N$ is free the result is true. This is because in this case the exact sequence $0 \to N \to M \to M/N \to 0$ splits, and therefore $M \cong N \oplus M/N$ (both well-known results). Apply the same thing to $M'$ to get $M' \cong N' \oplus M'/N' \cong N \oplus M/N \cong M$.