Lemma 38.14.1. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume
$N$ is projective as an $R$-module, and
$S/\mathfrak mS$ is Noetherian and $N/\mathfrak mN$ is a finite $S/\mathfrak mS$-module.
Then for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ where $\mathfrak p = R \cap \mathfrak q$ we have $\mathfrak q + \mathfrak m S \not= S$.
Proof. Note that the hypotheses of Lemmas 38.7.1 and 38.7.6 are satisfied. We will use the conclusions of these lemmas without further mention. Let $\Sigma \subset S$ be the multiplicative set of elements which are not zerodivisors on $N/\mathfrak mN$. The map $N \to \Sigma ^{-1}N$ is $R$-universally injective. Hence we see that any $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ is also an associated prime of $\Sigma ^{-1}N \otimes _ R \kappa (\mathfrak p)$. Clearly this implies that $\mathfrak q$ corresponds to a prime of $\Sigma ^{-1}S$. Thus $\mathfrak q \subset \mathfrak q'$ where $\mathfrak q'$ corresponds to an associated prime of $N/\mathfrak mN$ and we win. $\square$
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