Let $M,S$ be two right $A$-modules, the latter simple. Assume that for a certain $m\in M$ we have that $f(m)=0$ for every $f\in \text{Hom}(M,S)$. I want to show that then $m\in \text{rad}(M)$, where rad denotes the Jacobson radical.
Actually the backwards implication is also true, and I managed to prove it. To do it, I used that the Jacobson radical of a simple module is $\{0\}$ and that $f(\text{rad}(M))\subset \text{rad}(S)$. I feel that I should use the same tools for this implication, but I have been fiddling around with them for a while and got nowhere. Could someone help me?
This is not true. Consider the case $A=\Bbb Z$ and $M=\Bbb Z/4\Bbb Z$, $S=\Bbb Z/3\Bbb Z$.
What is correct however is that if for a fixed $m \in M$, we have for all $S$ simple and all $f \in \mathrm{Hom}(M,S)$ that $f(m)=0$, then $m \in \operatorname{rad}(M)$.
Edit: Let's prove the statement you actually wanted to ask. Let's proceed by contraposition. So suppose that $m \notin \mathrm{rad}(M)$, then $m\notin N$ for some maximal submodule $N$ of $M$. But now $M/N$ is simple and the projection $M \to M/N$ is an element of $\mathrm{Hom}(M,M/N)$ that doesn't map $m$ to zero.