# Simple module isomorphic to quotient of ring by annihilator?

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!

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.