Showing that a simple ring is either the set containing 0 or one of its products must be 0

Question

I am asked:

I am focusing on (a).

My proof for it is as follows:

Assume $n=0$, then trivially $nR = 0$.

Assume $n\neq0$, then $nr = 0 \implies r=0$

I suspect, due to how short my answer is. That my proof is in fact wrong, or that I am missing some important steps in order to show it.

Answer 1

You are right to be suspicious (though I would say shortness of the proof is not a reason to be suspicious; rather, the fact that you have never used the assumption that $R$ is simple is). Your proof in the case $n\neq 0$ is incorrect: why would $nr=0$ imply $r=0$? This is true for, say, real numbers, but you can't just assume it is true in any arbitrary ring. For instance, notice that if your ring $R$ is the integers mod $4$, then $2\cdot 2=0$ even though $2\neq 0$ and in fact $2R\neq 0$.

So to get a correct proof you will need to somehow use the fact that the ring $R$ is simple. As a hint, you might first prove that $nR$ is an ideal in $R$.

Answer 2

Sorry, your proof is wrong. For instance, if $R=\mathbb{Z}/6\mathbb{Z}$, it holds that $6R=0$.

Hint for a correct proof of (a): given a nonnegative integer $n$, $nR$ and $\{r\in R:nr=0\}$ are ideals of $R$. Since $R$ is simple…

For (b): if $m$ divides $n$, then $nR\subseteq mR$.