Show that if $M$ is a R-module free, for all $r \in R$ and $m \in M$ such that $r \neq 0$ e $m \neq 0$ we have $rm \neq 0$.

Question

is it true that

if $M$ is a $R$-module free, for all $r \in R$ and $m \in M$ such that $r \neq 0$ e $m \neq 0$ we have $rm \neq 0$.

Comments: I tried to do the following:

Suppose that $rm = 0$. As $M$ is free, let $\{ x_i\}_{i \in I}$ basis of M. Thus, given $m \in M$, $m \neq 0$ then $m = \sum_{i \in I} \lambda_i x_i$, with $( \lambda _i)_{i \in I} \in R^{(I)}$. Next $rm = r \sum_{i \in I} \lambda_i x_i = \sum_{i \in I} r \lambda_i x_i = 0 \Rightarrow r\lambda_i=0$ because $\{ x_i\}_{i \in I}$ is basis.

I want to show that $r=0$, but I can not.

I thank you for your help.

Answer 1

This is not true:
If $R=\mathbb Z/(6)$ and $M=R$, we have $r\cdot m=0\in M$ for $r=2\neq 0\in R$ and $m=3\neq 0\in M$.

However, the result is true if $R$ is an integral domain, i.e. a ring without zero divisors.