Condition for module to not be free

Question

Let $M$ be a module over a ring $R$ with $1$. Suppose that $rm=0$ for some nonzero $r \in R$ and $m \in M$. Can we conclude $M$ is not free?

Here is my attempt. Suppose $M$ is free with basis $m_1,\cdots m_N$. Write $m=c_1m_1+\cdots+c_Nm_N$. Then $rc_1m_1+\cdots rc_Nm_N=0$. Now, if some $rc_i \neq 0$, then we are done since we've shown that $\{m_i\}$ is in fact not a basis.

How do we proceed?

Answer 1

As you have learned the condition $\operatorname{Ann}_R(m)\ne0$ for all $m\in M$ is not enough to conclude that $M$ is not a free module. (Simple counterexamples are given by $M=R$, and $R$ a ring with zero divisors, e.g. $R=\mathbb Z/6\mathbb Z$.)

If you assume $\operatorname{Ann}_R(M)\ne0$, then the module can't be free (unless it is $0$).

In the particular example you mentioned in the comments, $R=R_1\times R_2$, and $M=R_1\times\{0\}$ we have $\operatorname{Ann}_R(M)=\{0\}\times R_2\ne0$.