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Exercise 15.1.32 in Dummit & Foote, along with the included hint, is

Suppose that $M$ is a $R$-module and that $P$ is a maximal element in the collection of ideals of the form $\operatorname{Ann}(m)$, for $m\in M$. Prove that $P$ is a prime ideal. [If $P=\operatorname{Ann}(m)$ and $ab\in P$, show that $bm\neq 0$ implies $\operatorname{Ann}(m)\subseteq \operatorname{Ann}(bm)$ and use the maximality of $P$ to deduce that $a\in \operatorname{Ann}(bm)=P$.]

(Note that $R$ is commutative with $1\neq0$.)

It seems to me that this is false unless we replace "for $m\in M$" with "for $0\neq m\in M$"; otherwise, $R=\operatorname{Ann}(0)$ is the only such maximal element yet is never prime.

I also have an issue with the hint: $\operatorname{Ann}(m)\subseteq \operatorname{Ann}(bm)$ is true whether or not $bm=0$, since $rm=0$ implies $rbm=brm=b0=0$. Should I asusme the above correction, and replace this with "$\operatorname{Ann}(m)\subseteq \operatorname{Ann}(bm)\neq R$"?

Avi Steiner
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  • Yes. For the first part they surely just mean to consider the collection of proper ideals. I don't see the issue with the hint. It is just to get you started. As you point out you have to notice and provide an argument that $Ann(bm)=P$ by noting $bm\neq 0$ implies $Ann(bm)\neq R$ so that it is actually of the form under consideration. – Matt Jun 16 '13 at 02:17

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As Matt points out, the correct statement is "$P$ is a maximal ideal in the set of ideals of the form $Ann(m)$ for $0 \neq m \in M$. The hint is structured like that, because of the following: how would you prove that $Ann(m)$ is prime? You need to take a product $ab \in Ann(m)$, assume that $b \notin Ann(m)$ and prove that $a \in Ann(m)$. Since you assume that $b \notin Ann(m)$, this gives $b m \neq 0$ and so $Ann(bm) \neq R$ as we want. So, you are right that $Ann(m) \subset Ann(bm)$ irrespectively of whether $bm=0$, but as i explained, because of the way that one would try to prove that $Ann(m)$ is prime, we already have that $bm \neq 0$.

Manos
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