**Proof.**
We first prove the lemma for $S = R[x_1, \ldots , x_ n]$, and then we deduce the result in general.

Assume $S = R[x_1, \ldots , x_ n]$. Choose elements $m_1, \ldots , m_ t$ which generate $M$. This gives a short exact sequence

\[ 0 \to N \to S^{\oplus t} \xrightarrow {(m_1, \ldots , m_ t)} M \to 0. \]

Denote $K$ the fraction field of $R$. Denote $S_ K = K \otimes _ R S = K[x_1, \ldots , x_ n]$, and similarly $N_ K = K \otimes _ R N$, $M_ K = K \otimes _ R M$. As $R \to K$ is flat the sequence remains exact after tensoring with $K$. As $S_ K = K[x_1, \ldots , x_ n]$ is a Noetherian ring (see Lemma 10.31.1) we can find finitely many elements $n'_1, \ldots , n'_ s \in N_ K$ which generate it. Choose $n_1, \ldots , n_ r \in N$ such that $n'_ i = \sum a_{ij}n_ j$ for some $a_{ij} \in K$. Set

\[ M' = S^{\oplus t}/\sum \nolimits _{i = 1, \ldots , r} Sn_ i \]

By construction $M'$ is a finitely presented $S$-module, and there is a surjection $M' \to M$ which induces an isomorphism $M'_ K \cong M_ K$. We may apply Lemma 10.118.2 to $R \to S$ and $M'$ and we find an $f \in R$ such that $M'_ f$ is a free $R_ f$-module. Thus $M'_ f \to M_ f$ is a surjection of modules over the domain $R_ f$ where the source is a free module and which becomes an isomorphism upon tensoring with $K$. Thus it is injective as $M'_ f \subset M'_ K$ as it is free over the domain $R_ f$. Hence $M'_ f \to M_ f$ is an isomorphism and the result is proved.

For the general case, choose a surjection $R[x_1, \ldots , x_ n] \to S$. Think of both $S$ and $M$ as finite modules over $R[x_1, \ldots , x_ n]$. By the special case proved above there exists a nonzero $f \in R$ such that both $S_ f$ and $M_ f$ are free as $R_ f$-modules and finitely presented as $R_ f[x_1, \ldots , x_ n]$-modules. Clearly this implies that $S_ f$ is a finitely presented $R_ f$-algebra and that $M_ f$ is a finitely presented $S_ f$-module.
$\square$

## Comments (2)

Comment #3265 by Samir Canning on

Comment #3360 by Johan on