The Stacks project

Proof. One of the inclusions has already been proved, see Remark 10.66.12. To prove the other assume $\mathfrak q \in \text{WeakAss}_ S(M)$ and let $\mathfrak p$ be the corresponding prime of $R$. Let $m \in M$ be an element such that $\mathfrak q$ is a minimal prime over $J = \{ g \in S \mid gm = 0\}$. Thus the radical of $JS_{\mathfrak q}$ is $\mathfrak qS_{\mathfrak q}$. As $R \to S$ is finite there are finitely many primes $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ l$ over $\mathfrak p$, see Lemma 10.36.21. Pick $x \in \mathfrak q$ with $x \not\in \mathfrak q_ i$ for $i > 1$, see Lemma 10.15.2. By the above there exists an element $y \in S$, $y \not\in \mathfrak q$ and an integer $t > 0$ such that $y x^ t m = 0$. Thus the element $ym \in M$ is annihilated by $x^ t$, hence $ym$ maps to zero in $M_{\mathfrak q_ i}$, $i = 2, \ldots , l$. To be sure, $ym$ does not map to zero in $S_{\mathfrak q}$.

The ring $S_{\mathfrak p}$ is semi-local with maximal ideals $\mathfrak q_ i S_{\mathfrak p}$ by going up for finite ring maps, see Lemma 10.36.22. If $f \in \mathfrak pR_{\mathfrak p}$ then some power of $f$ ends up in $JS_{\mathfrak q}$ hence for some $t > 0$ we see that $f^ t ym$ maps to zero in $M_{\mathfrak q}$. As $ym$ vanishes at the other maximal ideals of $S_{\mathfrak p}$ we conclude that $f^ t ym$ is zero in $M_{\mathfrak p}$, see Lemma 10.23.1. In this way we see that $\mathfrak p$ is a minimal prime over the annihilator of $ym$ in $R$ and we win. $\square$