Lemma 15.25.1. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]$ be a polynomial ring over $R$. Let $M$ be an $S$-module. Assume
there exist finitely many primes $\mathfrak p_1, \ldots , \mathfrak p_ m$ of $R$ such that the map $R \to \prod R_{\mathfrak p_ j}$ is injective,
$M$ is a finite $S$-module,
$M$ flat over $R$, and
for every prime $\mathfrak p$ of $R$ the module $M_{\mathfrak p}$ is of finite presentation over $S_{\mathfrak p}$.
Then $M$ is of finite presentation over $S$.
Proof. Choose a presentation
of $M$ as an $S$-module. Let $\mathfrak q$ be a prime ideal of $S$ lying over a prime $\mathfrak p$ of $R$. By assumption there exist finitely many elements $k_1, \ldots , k_ t \in K$ such that if we set $K' = \sum Sk_ j \subset K$ then $K'_{\mathfrak p} = K_{\mathfrak p}$ and $K'_{\mathfrak p_ j} = K_{\mathfrak p_ j}$ for $j = 1, \ldots , m$. Setting $M' = S^{\oplus r}/K'$ we deduce that in particular $M'_{\mathfrak q} = M_{\mathfrak q}$. By openness of flatness, see Algebra, Theorem 10.129.4 we conclude that there exists a $g \in S$, $g \not\in \mathfrak q$ such that $M'_ g$ is flat over $R$. Thus $M'_ g \to M_ g$ is a surjective map of flat $R$-modules. Consider the commutative diagram
The bottom arrow is an isomorphism by choice of $k_1, \ldots , k_ t$. The left vertical arrow is an injective map as $R \to \prod R_{\mathfrak p_ j}$ is injective and $M'_ g$ is flat over $R$. Hence the top horizontal arrow is injective, hence an isomorphism. This proves that $M_ g$ is of finite presentation over $S_ g$. We conclude by applying Algebra, Lemma 10.23.2. $\square$
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