Lemma 10.99.4 (00MH)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.99: Criteria for flatness
Lemma 10.99.4 (cite)

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Lemma 10.99.4. Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $\mathfrak m$ be the maximal ideal of $R$ . Let $M$ be a finite $S$ -module. Suppose that (a) $M/\mathfrak mM$ is a free $S/\mathfrak mS$ -module, and (b) $M$ is flat over $R$ . Then $M$ is free and $S$ is flat over $R$ .

Proof. Let $\overline{x}_1, \ldots , \overline{x}_ n$ be a basis for the free module $M/\mathfrak mM$ . Choose $x_1, \ldots , x_ n \in M$ with $x_ i$ mapping to $\overline{x}_ i$ . Let $u : S^{\oplus n} \to M$ be the map which maps the $i$ th standard basis vector to $x_ i$ . By Lemma 10.99.1 we see that $u$ is injective. On the other hand, by Nakayama's Lemma 10.20.1 the map is surjective. The lemma follows. $\square$

Comments (5)

Comment #2473 by Matthieu Romagny on April 03, 2017 at 17:23

Replace "modules" by "module" in statement of Lemma.

Comment #2506 by Johan on April 14, 2017 at 00:07

Thanks, fixed here.

Comment #4523 by Ronnie on September 06, 2019 at 04:57

To show $u$ is injective, may be we can also argue as follows. Let $K$ denote the kernel of $u$ , it is a finite $S$ module. Since $M$ is flat over $R$ , applying $-\otimes_RR/\mathfrak{m}$ we get $K\otimes_RR/\mathfrak{m}=0$ . It follows that $K\otimes_SS/\mathfrak{m}S=0$ , which shows, by Nakayama's lemma, that $K=0$ .

Comment #4524 by Johan on September 06, 2019 at 06:40

Yes very good. The key is that here $u$ is surjective whereas in Lemma 10.99.1 it doesn't need to be.

Comment #9776 by Jonas on September 21, 2024 at 19:06

For flatness of the ring map one needs $M$ to be non-zero.

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