Two definitions of an associated prime of an $R$-module

Question

I have come across two definitions of an associated prime for an $R$-module $M$, one of which specifies that $R$ is Noetherian, however, I can't see the reason they would coincide.

First one: A prime ideal $P$ is an associated prime of $M$ if $P = Ann(m)$ for some $m \in M$, where $M$ is a finitely generated $R$-module for a commutative, Noetherian ring $R$.

Second one: An associated prime of an $R$-module $M$ for a commutative ring $R$ is the minimal prime over $Ann(m)$ for some non-zero $m \in M$

Why does the Noetherian property mean that $Ann(m)$ would be prime?

Answer 1

The Noetherian assumption does not imply $Ann(m)$ must be prime; rather, it implies that if $P$ is a minimal prime over $Ann(m)$, then you can find a different element $m'\in M$ such that $Ann(m')=P$. To prove this, note that we may replace $M$ with the submodule generated by $m$ and $R$ with $R/Ann(m)$ to suppose that $M=R$ and $P$ is a minimal prime. That is, we just need to prove that any minimal prime in a Noetherian ring is an associated prime (here and below "associated prime" will always mean "prime ideal that is the annihilator of an element", and all rings are assumed to be commutative).

This fact is a basic result in the theory of associated primes; here is one way to prove it. First, a Lemma:

Lemma: Let $R$ be a Noetherian ring and $M$ be a nonzero $R$-module. Then $M$ has an associated prime.

Proof: Let $S$ be the set of ideals of $R$ that are annihilators of nonzero elements of $M$. Since $R$ is Noetherian, $S$ has a maximal element $P$; we claim $P$ is prime. Since $P\in S$, there is some $m\in M$ such that $Ann(m)=P$. Suppose $fg\in P$ and $f\not\in P$, so $fgm=0$ but $fm\neq 0$. Then $Ann(fm)\in S$ and $P=Ann(m)\subseteq Ann(fm)$ and so $P=Ann(fm)$ by maximality of $P$. Since $fgm=0$, $g\in Ann(fm)=P$. Thus $P$ is prime, and is an associated prime of $M$.

Now let us return to our setting above: $R$ is a Noetherian ring and $P$ is a minimal prime of $R$, and we wish to show $P$ is an associated prime of $R$. By the Lemma, the localization $R_P$ has an associated prime $Q$ as an $R$-module. If $Q\neq P$, then $Q\not\subseteq P$ by minimality of $P$, so there is some $f\in Q\setminus P$. But then $f$ acts as a unit on $R_P$ and so cannot be in the annihilator of a nonzero element of $R_P$, contradicting that $Q$ is an associated prime of $R_P$. Thus we must have $Q=P$; that is, $P$ is an associated prime of $R_P$.

Now pick an element $\frac{r}{s}\in R_P$ ($r\in R,s\in R\setminus P$) whose annihilator is $P$. Then for each $f\in P$, $f\cdot\frac{r}{s}=0$, meaning that $fr$ is annihilated by some element of $R\setminus P$. Since $P$ is finitely generated, we can multiply together such elements of $R\setminus P$ where $f$ ranges over a finite set of generators of $P$ to get a single element $t\in R\setminus P$ such that $ftr=0$ for all $f\in P$.

I now claim that $Ann(tr)=P$, so that $P$ is an associated prime of $R$. By our choice of $t$, we have $P\subseteq Ann(tr)$. On the other hand, suppose $u\in R\setminus P$. Then $ut\not\in P$ and so $ut\cdot\frac{r}{s}\neq 0$. That implies $utr\neq 0$ so $u\not\in Ann(tr)$.