Is a submodule of a free module also free?

Question

Is a submodule of a free module also free ? For me it looks natural that yes, but in my course it's written that it's only true for a module over a PID and I don't really understand why. Any example ?

Answer 1

A counter-example

If $K$ is a field, consider the ring of polynomials in two indeterminates $R=K[X,Y]$. The ideal $I=(X,Y)$ is a submodule of the rank $1$ free $R$-module $R$. It is not free, since if it were, it would have a single generator. Can you find a polynomial $F(X,Y)$ which is a divisor of both $X$ and $Y$?

Answer 2

Small (smallest?) counterexample: $(2)\subseteq \Bbb Z_4$ as $\Bbb Z_4$-modules.