If $P \in \operatorname{Ass}M$, then $R/P \subset M$.

Question

Let $R$ be a commutative ring with unity. $M$ an $R$-module. Then $P \in \operatorname{Ass}M$ if and only if there is a submodule $N\subset M$ such that $R/P \cong N$.

$$\operatorname{Ass}M:=\{P\text{ prime $R$-ideal}\mid P\text{ is the annihilator of an element of }M\}$$

I think all I need is a hint. Could somebody please give me a hint?

Answer 1

I assume that, for you, an associated prime $P$ of $R$ is a prime ideal that is the annihilator of some nonzero $m\in M$, and that you want to show that this implies that there is some submodule of $M$ isomorphic to $R/P$.

Here is my hint: As part of the condition, we are given a nonzero $m\in M$. Using this, construct a homomorphism of modules $R\to M$, and compute its kernel.

Answer 2

let $p=Ann\ x$ for some $x\in M$. then $$R/P\cong R/Ann\ x\cong Rx\subset M$$