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Let $R$ be a commutative ring with unity and $M$ be a simple $R$ module, i.e. it has no proper non-zero submodules.

Then $M\cong R/P$ for some maximal ideal $P$.

D. Eisenbud claims on p. 72 of his book "Commutative Algebra" that $P$ is in fact the annihilator of $M$.

Is this true? I don't think that $\operatorname{Ann} M$ is even maximal. I think that the statement becomes true when replacing this with $\operatorname{Ann} m$ for any element $0\neq m\in M$:

Let $0\neq m\in M$ and $\varphi_m\colon R\to M$ be the map $a\mapsto am$. Then $\varphi_m$ is surjective (as $M$ is simple, i.e. the image of $\varphi_m$ can only be $0$ or $M$, where the former is impossible due to $m\neq 0$). So $M\cong R/\ker\varphi_m$ (where $\ker\varphi_m=\operatorname{Ann}(m)$). Now let $I$ be an ideal strictly containing $\operatorname{Ann} m$ (which is possible as $\operatorname{Ann}m\neq R$ due to $m\neq 0$), and choose $b\in I\setminus \operatorname{Ann} m$. Then $bm\neq 0$, and due to surjectivity of $\varphi_{bm}$, we get some $b'\in R$ with $b'bm=\varphi_{bm}(b')=m$. Then $(b'b-1)m=0$, which yields $b'b-1\in\operatorname{Ann}m\subseteq I$. Now $b'b\in I$ implies $1\in I$, i.e. $I=R$, and $\operatorname{Ann}m$ is maximal.

Is this correct? Is the initial claim concerning $\operatorname{Ann} M$ correct/false? Thanks!

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You are right and so is the claim concerning $Ann(M)$.

As you noted $\varphi_m$ is an epimorphism for every $0 \neq m \in M$. Now if $n \in M$, we have $n = am$ for some $a \in R$. Now use that $R$ is commutative to note that this implies $Ann(m) \subseteq Ann(n)$. But since $0 \neq m \in M$ was arbitrary, this implies $Ann(m) = Ann(n)$ for all $m,n \in M$ and thus $Ann(M) = Ann(m)$ for all $m \in M$.

By the way, your idea of looking at $\varphi_m$ and noting that $R/\ker(\varphi_m) \cong M$ (if $M$ is simple; by the way, you wrote im$(\varphi_m)$, but I guess this was just a typo) and concluding that $\ker(\varphi_m) = Ann(m)$ is maximal, works for any ring, commutative or not. The only thing that you get extra in the commutative case is that $Ann(m) = Ann(M)$ if $M$ is simple.

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