Lemma 38.14.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. If $N$ is $I$-adically complete, then for any $R$-module $M$ and for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R M$ we have $\mathfrak q + I S \not= S$.
Proof. Let $S^\wedge $ denote the $I$-adic completion of $S$. Note that $N$ is an $S^\wedge $-module, hence also $N \otimes _ R M$ is an $S^\wedge $-module. Let $z \in N \otimes _ R M$ be an element such that $\mathfrak q = \text{Ann}_ S(z)$. Since $z \not= 0$ we see that $\text{Ann}_{S^\wedge }(z) \not= S^\wedge $. Hence $\mathfrak q S^\wedge \not= S^\wedge $. Hence there exists a maximal ideal $\mathfrak m \subset S^\wedge $ with $\mathfrak q S^\wedge \subset \mathfrak m$. Since $IS^\wedge \subset \mathfrak m$ by Algebra, Lemma 10.96.6 we win. $\square$
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