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Finite generation descends along faithfully flat ring maps

Statement

Assume the Axiom of Choice.

Let $R \to S$ be a faithfully flat homomorphism of commutative rings and let $M$ be an $R$ -module. If $M \otimes_{R} S$ is finitely generated as an $S$ -module, then $M$ is finitely generated as an $R$ -module.

Facts & Assumptions

Given: The Axiom of Choice, a faithfully flat ring map $R \to S$ , and an $R$ -module $M$ such that $M \otimes_{R} S$ is finitely generated over $S$ .

[L1]

A faithfully flat module detects nonzero quotients (For a flat module, faithful flatness is equivalent to detecting nonzero modules and residue fields).

[L2]

The given map is faithfully flat as an $R$ -module map (A flat ring map is faithfully flat exactly when it detects proper ideals and is surjective on spectra).

Proof

technique · direct

1.1givenchoose

Choose generators $u_{1}, \dots, u_{r}$ of $M \otimes_{R} S$ . Write each $u_{i}$ as a finite sum of simple tensors and collect the finitely many elements of $M$ occurring there, say $m_{1}, \dots, m_{n}$ . Let $N \subseteq M$ be the submodule they generate.

1.2L1L2

By construction the images of the $m_{j} \otimes 1$ generate $M \otimes_{R} S$ , so $(M / N) \otimes_{R} S = 0.$ If $M / N \neq 0$ , then [L1] applied to the faithfully flat $R$ -module $S$ from [L2] would force $(M / N) \otimes_{R} S \neq 0$ , contradiction. Therefore $M / N = 0$ .

2.1algebra∎

Thus $M = N$ is finitely generated.

Depends on

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, Proposition 11.10 (standard reference, not scraped)