Lemma 15.80.8. Let $R \to A$ be a finite type ring map. Let $M$ be an $A$-module. Let $f_1, \ldots , f_ r \in A$ generate the unit ideal. The following are equivalent

1. each $M_{f_ i}$ is finitely presented relative to $R$, and

2. $M$ is finitely presented relative to $R$.

Proof. The implication (2) $\Rightarrow$ (1) is in Lemma 15.80.4. Assume (1). Write $1 = \sum f_ ig_ i$ in $A$. Choose a surjection $R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r] \to A$. such that $y_ i$ maps to $f_ i$ and $z_ i$ maps to $g_ i$. Then we see that there exists a surjection

$P = R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r]/(\sum y_ iz_ i - 1) \longrightarrow A.$

By Lemma 15.80.1 we see that $M_{f_ i}$ is a finitely presented $A_{f_ i}$-module, hence by Algebra, Lemma 10.23.2 we see that $M$ is a finitely presented $A$-module. Hence $M$ is a finite $P$-module (with $P$ as above). Choose a surjection $P^{\oplus t} \to M$. We have to show that the kernel $K$ of this map is a finite $P$-module. Since $P_{y_ i}$ surjects onto $A_{f_ i}$ we see by Lemma 15.80.1 and Algebra, Lemma 10.5.3 that the localization $K_{y_ i}$ is a finitely generated $P_{y_ i}$-module. Choose elements $k_{i, j} \in K$, $i = 1, \ldots , r$, $j = 1, \ldots , s_ i$ such that the images of $k_{i, j}$ in $K_{y_ i}$ generate. Set $K' \subset K$ equal to the $P$-module generated by the elements $k_{i, j}$. Then $K/K'$ is a module whose localization at $y_ i$ is zero for all $i$. Since $(y_1, \ldots , y_ r) = P$ we see that $K/K' = 0$ as desired. $\square$

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