Let $R$ is a commutative ring, and let $M$ be an $R$-module. According to Wikipedia, the following are equivalent:
- $P={\rm Ann}_R(N)$ is a prime ideal, where $N$ is a prime $R$-submodule of $M$.
- $P$ is a prime ideal of $R$ of the form ${\rm Ann}_R(m)$, for some nonzero $m\in M$.
- $R/P\cong N$ for some $R$-submodule $N$ of $M$.
Namely, (1) is the definition of an associated prime of a (not necessarily commutative) ring, and (2) and (3) are equivalent to (1) if the ring is commutative. I would like to prove this; proving (2)$\iff$(3) is easy enough, but I cannot make the connection to (1). Here's what I've got:
(1)$\implies$(2) For some prime $R$-submodule $N$ of $M$, we have ${P={\rm Ann}_R(N)}$. Let $x\in P$. Then $xn=0_M$ and $x\in{\rm Ann}_R(n)$ for any $n\in N$. Hence, $P\subseteq {\rm Ann}(m)$ for some $m\in M$.
I'm not sure how to obtain equality.
(3)$\implies$(1) Let $f\colon R/P\to N$ be an $R$-linear isomorphism. Then for any $n\in N$ there exists $x\in R$ such that $f(x+P)=n$. Hence \begin{align*} r(x+P)=P\iff f(rx+P)=f(P)\iff rn=0_N, \end{align*} so ${\rm Ann}_R(x+P)={\rm Ann}_R(n).$ Since these $x$ and $n$ are in one-to-one correspondence, there exists $X\subseteq R$ such that ${\rm Ann}_R(X+P)={\rm Ann}_R(N)$.
To be honest, I'm not sure if this is the right direction. Any help would be appreciated.