If $N$ and $M/N$ are free modules of finite rank, so is $M$

Question

Definiton: A $R$-module $M$ is said to be free of finite rank if $M\cong R^k = R\times R\times\cdots \times R$ (k times).

Let $R$ be a ring with $1$ and $N$ be a submodule of an $R$-module $M$.

Prove that $M$ is free of finite rank if $N$ and $M/N$ are free of finite rank.

I think this is trivial. $N$ and $M/N$ are free of finite rank, so $N\cong R^k$ for some $k$ and $M/N\cong R^l$ for some $l$. Then $M = N\times M/N \cong R^{k+l}$. Thus, $M$ is free of finite rank.

Does my argument work?

Answer 1

It turns out to be right that $M\cong N\times M/N$, but not without justification. In fact it is somewhat specific to this case (where everything is free of finite rank); e.g. $M=\mathbb Z$ has a submodule isomorphic to $N=\mathbb Z$ such that the quotient $M/N\cong\mathbb Z/n\mathbb Z$ for any $n$, and of course $\mathbb Z\not\cong\mathbb Z\times\mathbb Z/n\mathbb Z$.

Answer 2

Any free $R$-module $P$ is a projective: any short exact sequence with $P$ on the right must split. There is an exact sequence $$ 0 \rightarrow N \rightarrow M \rightarrow M/N \rightarrow 0$$ with $M/N$ on the right. Since $M/N$ is free, this sequence splits, so there is an isomorphism of $R$-modules $M \cong N \oplus M/N$. Hence $M$ is free as a direct sum of free modules.