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$\DeclareMathOperator{\ass}{Ass}$

Let $R$ be a Noetherian ring and $S$ be a multiplicative set in $R$. Then show that $\ass_R(S^{-1}M)=\ass_R(M)\cap \{P\in \text{Spec}(R):P\cap S=\emptyset\}$

Attempt: Let $P\in \ass_R(S^{-1}M)$. We argue that $S\cap P=\emptyset$. Say $P=\text{Ann}(m/s)$ for some $m/s\neq 0$ in $S^{-1}M$. If $r\in P\cap S$, then from $r\cdot (m/s)=0$ we get $(r/1)\cdot (m/s)=0$. But $r/1$ is invertible in $S^{-1}R$ and thus we have $m/s=0$, a contradiction. So we must have $S\cap P=\emptyset$.

The main difficulty I am having is in showing that $P\in \ass_R(M)$. I suspect that $P=\text{Ann}(m)$. Clearly, if $r\cdot m=0$, then $r\cdot (m/1)=0$ and therefore $r\in P$. So $\text{Ann}(m)\subseteq P$. On the other hand, if $r\in P$, then $(r\cdot m)/s=0$, giving an element $u\in S$ such that $(ur)\cdot m=0$. But I am getting nowhere with this.

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$\mathfrak p\in\operatorname{Ass}_{R}(S^{-1}M)\implies\exists x\in M,\ \mathfrak p=\operatorname{Ann}_{R}(x/1)$.
This shows that $\mathfrak p\supseteq\operatorname{Ann}_R(x)$. (If you think at this point that $\mathfrak p=\operatorname{Ann}_R(x)$, forget it!)
Moreover, $\mathfrak p$ is minimal over $\operatorname{Ann}_R(x)$: if $\mathfrak p\supseteq\mathfrak p'\supseteq\operatorname{Ann}_R(x)$, then let $a\in\mathfrak p$, and from $\frac a1\cdot\frac x1=\frac01$ deduce that there is $s\in S$ such that $sa\in\operatorname{Ann}_R(x)$, so $sa\in\mathfrak p'$ hence $a\in\mathfrak p'$.
Now one uses that $R$ is Noetherian and conclude that $\mathfrak p\in\operatorname{Ass}_{R}(M)$.

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  • $\begingroup$ Is it a general fact that for a Noetherian ring $R$, and an element $m$ of a finitely generated module $M$, the minimal primes containing $Ann_R(m)$ are associated? This makes geometric sense, since $V(Ann_R(m)$ in $Spec A$ is going to be the support of $m$, so the generic points of its irreducible components are among the associated primes of $M$. But I'm not sure how to see this from the definition: $P$ is associated iff there is an injection $R / P \to M$. If it is already known that localization is compatible with associated primes I can turn the geometry into a proof, but otherwise... $\endgroup$ – Lorenzo Najt Dec 27 '15 at 23:29
  • $\begingroup$ Yes, it is, and don't even need $M$ finitely generated. $\endgroup$ – user26857 Dec 27 '15 at 23:48

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