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. $\square$

Comment #2473 by Matthieu Romagny on

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

Comment #4523 by Ronnie on

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 on

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

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