I'm currently battling with the following question:
Over an $R$-module $M$, for a given non-zero $m \in M$, is the map $F : R \rightarrow M$ given by $F(r) = rm$ necessarily non-zero? i.e. is there some element $r \in R$ such that $rm \neq 0$? ($R$ here is assumed to be a non-trivial commutative ring with unity).
Also, given instead the map $F_s : M \rightarrow M$ for some non-zero $s \in R$ given by $F_s (m) = sm$ for all $m \in M$, is this map also non-zero?
This has led me to investigate the following:
In an $R$-module $M$ does $rm = 0$ with ($m \in R$ non-zero) necessarily imply $r = 0$?
I'm pretty sure this is false, as otherwise in particular torsion elements wouldn't exist. I'm also fairly certain that the first statement can hold when we strengthen the assumptions a bit - e.g. with the added assumptions that $M$ is torsion-free and $R$ has at least one regular element. Can these assumptions be loosened? In particular, in question I'm looking at has $M$ is a simple module, so is there a way to relate this property to either of these statements?
Any time I've seen people use maps similar to $F$ as above, they always seem to say "clearly $F$ is non-zero" or something to that effect. Am I missing something?
The answer to your first question is yes, take $r=1$. The answer to your second question is yes, take the $\mathbb{Z}$ module $\mathbb{Z}_2$. Then $2 \cdot \overline{1}=\overline{0}.$