Lemma 38.14.3. 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 isomorphic as an $R$-module to a direct sum of finite $R$-modules. 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 $\mathfrak q + \mathfrak m S \not= S$.
Proof. Write $N = \bigoplus _{i \in I} M_ i$ with each $M_ i$ a finite $R$-module. Let $M$ be an $R$-module and let $\mathfrak q \subset S$ be an associated prime of $N \otimes _ R M$ such that $\mathfrak q + \mathfrak m S = S$. Let $z \in N \otimes _ R M$ be an element with $\mathfrak q = \text{Ann}_ S(z)$. After modifying the direct sum decomposition a little bit we may assume that $z \in M_1 \otimes _ R M$ for some element $1 \in I$. Write $1 = f + \sum x_ j g_ j$ for some $f \in \mathfrak q$, $x_ j \in \mathfrak m$, and $g_ j \in S$. For any $g \in S$ denote $g'$ the $R$-linear map
\[ M_1 \to N \xrightarrow {g} N \to M_1 \]
where the first arrow is the inclusion map, the second arrow is multiplication by $g$ and the third arrow is the projection map. Because each $x_ j \in R$ we obtain the equality
\[ f' + \sum x_ j g'_ j = \text{id}_{M_1} \in \text{End}_ R(M_1) \]
By Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $f'$ is surjective, hence by Algebra, Lemma 10.16.4 we see that $f'$ is an isomorphism. In particular the map
\[ M_1 \otimes _ R M \to N \otimes _ R M \xrightarrow {f} N \otimes _ R M \to M_1 \otimes _ R M \]
is an isomorphism. This contradicts the assumption that $fz = 0$. $\square$
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