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

1. 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,

2. $M$ is a finite $S$-module,

3. $M$ flat over $R$, and

4. 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

$0 \to K \to S^{\oplus r} \to M \to 0$

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

$\xymatrix{ M'_ g \ar[r] \ar[d] & M_ g \ar[d] \\ \prod (M'_ g)_{\mathfrak p_ j} \ar[r] & \prod (M_ g)_{\mathfrak p_ j} }$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).