Annihalator which is maximal among all other proper annihalators is prime.

Question

The following is an exercise in Vakils notes on algebraic geometry and I came up with a really weird result and wanted to verify that it was correct.

If $R$ is a commutative ring and $M$ a nonzero $R-$module let $\text{Ann}(m)$ where $m \in M$ be maximal among all proper ideals which are annihalators of elements of $M$. Then $\text{Ann}(m)$ is a prime ideal.

$\text{Ann}(x)$ where $x \in M$ is defined as $\{r \in R: rx = 0\}$

Proof:

If $rs \in \text{Ann}(m)$ then $r \in \text{Ann}(sm)$. We can assume that $sm \neq 0$ because if $sm = 0,$ $s \in \text{Ann}(m)$ and we're done. This implies that $1 \notin \text{Ann}(sm)$ so $\text{Ann}(sm)$ is proper.

Clearly $\text{Ann}(m) \subset \text{Ann}(sm)$ so $\text{Ann}(m) = \text{Ann}(sm)$ because $\text{Ann}(m)$ is maximal. Thus $r \in \text{Ann}(m)$. But this argument is symmetric in $s$ and $r$ so $s \in \text{Ann}(m)$ aswell? This seems really weird... but the proof looks ok.

Thanks in advance for helping me out!

Answer 1

Edit : this was wrong You write "Clearly $\text{Ann}(m) \subset \text{Ann}(sm)$", but this doesn't seem clear at all to me. I think this statement is wrong, and I'm trying to come up with a counterexample.

Nvm, we have $\text{Ann}(m) \subset \text{Ann}(sm)$ indeed, since (as shown by Alex Wertheim and Noel Lundström in the comments below) if $rm = 0$ then $rsm = srm = 0$.

What you actualy prove : Assume $rs \in \operatorname{Ann}(m)$. If $s \notin \operatorname{Ann}(m)$ then $r \in \operatorname{Ann}(m)$.

This is logically equivalent to $$rs \in \operatorname{Ann}(m) \Longrightarrow r \in \operatorname{Ann}(m) \textrm{ or } s \in \operatorname{Ann}(m)$$ Which is the definition of $\operatorname{Ann}(m)$ being prime.