About weakly associated primes

Question

Let $A$ be a commutative ring, and $M$ an $A$-module. A prime ideal $\mathfrak{p}\subset A$ is said to be weakly associated to $M$ if it is minimal over some $\operatorname{ann}m$, where $m\in M$. I came across a statement that says, $$\operatorname{WeakAss}_{A}S^{-1}M=\operatorname{WeakAss}_{S^{-1}A}S^{-1}M.$$ (Source: Lemma 10.63.14 in http://stacks.math.columbia.edu/tag/0546)

It is a well known result for ordinary associated primes ($\operatorname{Ass}_{A}S^{-1}M=\operatorname{Ass}_{S^{-1}A}S^{-1}M$), but for weakly associated primes, I suspect it should be $$\operatorname{WeakAss}_{A}S^{-1}M\cap\operatorname{Spec}S^{-1}A=\operatorname{WeakAss}_{S^{-1}A}S^{-1}M.$$ Did I discover a typo, or am I missing something? (Surely I'm more likely to make a mistake than Columbia University?) Is every weakly associated prime of $S^{-1}M$ as an $A$-module necessarily disjoint from $S$ for some reason?

Answer 1

$\DeclareMathOperator{\p}{\mathfrak{p}}$$\DeclareMathOperator{\ann}{\operatorname{ann}}$Let $\p \in \operatorname{WeakAss}_A S^{-1}M$ be minimal over $\ann_A(\frac{m}{1})$. Since annihilators localize (Lemma), $\ann_A(\frac{m}{1}) = (\ann_A(m))^{ec}$, where extension/contraction are with respect to $A \to S^{-1}A$. Now suppose $s \in \p \cap S$. Then by assumption, $s$ is nilpotent in $(A/\ann_A(m)^{ec})_{\p}$, so $s^nx \in \ann_A(m)^{ec}$ for some $x \in A \setminus \p$, so $\frac{s^nx}{1} \in \ann_A(m)^{ece} = \ann_A(m)^e$. Thus $x \in (\ann_{A}(m))^{ec} \subseteq \p$, contradiction.

Lemma: $\ann_{S^{-1}A}(\frac{m}{1}) = S^{-1}(\ann_A(m)) \; (= \ann_A(m)^e)$ for any $m \in M$.

Answer 2

I think zcn's answer is great, but here's a minor modification that avoids using extensions/contractions. (It doesn't really avoid using them, just hides that we're using them but it is in any case slightly more direct and easier for me to understand.) Moreover, both zcn's answer and mine only prove one containment, but the other is not so hard so I'll leave it out. The argument can be found in my answer here, if you're willing to wade through some geometric content.

Suppose $\mathfrak{p} \in \operatorname{WeakAss}_{A}(S^{-1}M)$, say $\mathfrak{p}$ is minimal over $\operatorname{Ann}_A(\frac{m}{s})$. Now suppose $t \in \mathfrak{p}\cap S$. Then since $\mathfrak{p}_\mathfrak{p}$ is the unique minimal prime lying over $\operatorname{Ann}_A(\frac{m}{s})_{\mathfrak{p}}$ in $A_{\mathfrak{p}}$, $t$ is nilpotent in $A_{\mathfrak{p}}/\operatorname{Ann}_A(\frac{m}{s})_{\mathfrak{p}}\cong (A/\operatorname{Ann}_A(\frac{m}{s}))_{\mathfrak{p}}$, that is to say there exists some $x \in A\setminus \mathfrak{p}$ with $t^nx \in \operatorname{Ann}_A(\frac{m}{s})$. But then $x \in \operatorname{Ann}_A(\frac{m}{s}) \subset \mathfrak{p}$, a contradiction.

There are a few ways to see that $x$ annihilates $\frac{m}{s}$ if $t^nx$ does, if you don't see it immediately (I didn't!). One is to actually write down from definitions what it means for an element of $A$ to annihilate an element of $S^{-1}M$ and observe that for $a \in A, s \in S$, $a$ annihilates an element iff $sa$ does, or you can just write $\frac{m}{s} = \frac{t^nm}{t^ns}$ and then use that $x(\frac{m}{s}) = x\frac{t^nm}{t^ns}= \frac{xt^n}{s}\frac{1}{t^n} = 0$. The first method is perhaps more enlightening, and can be thought of as being a consequence of the fact that $s$ "acts as a unit on $M$".