By Proposition 10.9.10 this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$, so we conclude by Lemma 10.23.1.

For each $i$ take a finite generating set $X_ i$ of $M_{f_ i}$. Without loss of generality, we may assume that the elements of $X_ i$ are in the image of the localization map $M \rightarrow M_{f_ i}$, so we take a finite set $Y_ i$ of preimages of the elements of $X_ i$ in $M$. Let $Y$ be the union of these sets. This is still a finite set. Consider the obvious $R$-linear map $R^ Y \rightarrow M$ sending the basis element $e_ y$ to $y$. By assumption this map is surjective after localizing at an arbitrary prime ideal $\mathfrak p$ of $R$, so it is surjective by Lemma 10.23.1 and $M$ is finitely generated.

By (2) we have a short exact sequence

\[ 0 \rightarrow K \rightarrow R^ n \rightarrow M \rightarrow 0 \]

Since localization is an exact functor and $M_{f_ i}$ is finitely presented we see that $K_{f_ i}$ is finitely generated for all $1 \leq i \leq n$ by Lemma 10.5.3. By (2) this implies that $K$ is a finite $R$-module and therefore $M$ is finitely presented.

By Proposition 10.9.10 the assumption implies that the induced morphism on localizations at all prime ideals is an isomorphism, so we conclude by Lemma 10.23.1.

By Proposition 10.9.10 the assumption implies that the induced sequence of localizations at all prime ideals is short exact, so we conclude by Lemma 10.23.1.

We will show that every ideal of $R$ has a finite generating set: For this, let $I \subset R$ be an arbitrary ideal. By Proposition 10.9.12 each $I_{f_ i} \subset R_{f_ i}$ is an ideal. These are all finitely generated by assumption, so we conclude by (2).

For each $i$ take a finite generating set $X_ i$ of $S_{f_ i}$. Without loss of generality, we may assume that the elements of $X_ i$ are in the image of the localization map $S \rightarrow S_{f_ i}$, so we take a finite set $Y_ i$ of preimages of the elements of $X_ i$ in $S$. Let $Y$ be the union of these sets. This is still a finite set. Consider the algebra homomorphism $R[X_ y]_{y \in Y} \rightarrow S$ induced by $Y$. Since it is an algebra homomorphism, the image $T$ is an $R$-submodule of the $R$-module $S$, so we can consider the quotient module $S/T$. By assumption, this is zero if we localize at the $f_ i$, so it is zero by (1) and therefore $S$ is an $R$-algebra of finite type.

By the previous item, there exists a surjective $R$-algebra homomorphism $R[X_1, \ldots , X_ n] \rightarrow S$. Let $K$ be the kernel of this map. This is an ideal in $R[X_1, \ldots , X_ n]$, finitely generated in each localization at $f_ i$. Since the $f_ i$ generate the unit ideal in $R$, they also generate the unit ideal in $R[X_1, \ldots , X_ n]$, so an application of (2) finishes the proof.

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