Suppose one has a (commutative) ring $R$ and a free $R$-module $M$ of rank $n$. Let $N \subset M$ be a free $R$-submodule of $M$. Of course, in general, it is not true that $N$ is the set of solutions of a system of linear equations (take for instance $M$ to be the $\mathbb Z$-module $M=\mathbb Z$, and $N = 2\mathbb Z$).
I was wondering the following: If $rk(N) = n-k$, what are conditions on $N$, such that $N$ is the set of solutions of $k$ linear equations inside $M$?
The example above shows that this is not true unconditionally. The problem with the example seems to be that $M/N$ is not torsion free as a $\mathbb Z$-module.
So I think that a necessary condition is that $M/N$ is a torsion free $R$-module. Is it also sufficient?