Small fix

Thanks to  Elías Guisado
https://stacks.math.columbia.edu/tag/05BZ#comment-9034

algebra.tex

@@ -15336,11 +15336,15 @@ \section{Associated primes}
 \end{lemma}
 
 \begin{proof}
-The first equality follows, since if $m \in S^{-1}M$, then the annihilator
-of $m$ in $R$ is the intersection of the annihilator of $m$ in $S^{-1}R$
-with $R$.
-The displayed inclusion and equality in the Noetherian case follows from
-Lemma \ref{lemma-associated-primes-localize}
+For $m \in S^{-1}M$, let $I \subset R$ and $J \subset S^{-1}R$ be the
+annihilators of $m$. Then $I$ is the inverse image of $J$ by the map
+$R \to S^{-1}R$ and $J = S^{-1}I$. The equality in (1) follows by the
+description of the map $\Spec(S^{-1}R) \to \Spec(R)$ in
+Lemma \ref{lemma-spec-localization}.
+For $m \in M$, let $I \subset R$ be the annihilator of $m$ in $R$
+and let $J \subset S^{-1}R$ be the annihilator of $m/1 \in S^{-1}M$.
+We have $J = S^{-1}I$ which implies (2). The equality in the
+Noetherian case follows from Lemma \ref{lemma-associated-primes-localize}
 since for $\mathfrak p \in R$, $S \cap \mathfrak p = \emptyset$ we have
 $M_{\mathfrak p} = (S^{-1}M)_{S^{-1}\mathfrak p}$.
 \end{proof}