not free modules

Question

I'm not sure if this result is true or not. Let $R$ be a commutative ring and M a left $R$-module, if P is not free as a sub-module of M, can we say that M is not free either? thank you for your time

Answer 1

You are asking if a submodule of a free module is necessarily free. This is not true in general. It is, however, true if $R$ is a PID.

A counter example is given by $R=\mathbb{Z}/\mathbb4{Z}$ as a module over itself. The submodule $2\mathbb{Z}/4\mathbb{Z}$ is not free.

Answer 2

For an explicit counterexample, consider $R = k[X,Y]$ for any field $k$, which is free over itself. Then the ideal $\mathfrak m = RX+RY$ is not free.