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Slight generalization of [Lemme 2(a), Beauville-Laszlo].

Lemma 15.90.5. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. An $R$-module $M$ is finitely generated if and only if the ($R' \oplus R_ f$)-module $M \otimes _ R (R' \oplus R_ f)$ is finitely generated. For example, if $M \otimes _ R (R^\wedge \oplus R_ f)$ is finitely generated as a module over $R^\wedge \oplus R_ f$, then $M$ is a finitely generated $R$-module.

Proof. The ‘only if' is clear, so we assume that $M \otimes _ R (R' \oplus R_ f)$ is finitely generated. In this case, by writing each generator as a sum of simple tensors, $M \otimes _ R (R' \oplus R_ f)$ admits a finite generating set consisting of elements of $M$. That is, there exists a morphism from a finite free $R$-module to $M$ whose cokernel is killed by tensoring with $R' \oplus R_ f$; we may thus deduce $M$ is finite generated by applying Lemma 15.90.3 to this cokernel. The last statement is a special case of the first statement by Lemma 15.90.1. $\square$

Comments (2)

Comment #3823 by Kestutis Cesnavicius on

The statement can evidently be strengthened to `if and only if'.

Comment #3926 by on

This was already fixed thanks to your own edits! Ha!

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  • 4 comment(s) on Section 15.90: The Beauville-Laszlo theorem

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