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Tag 00MH

Chapter 10: Commutative Algebra > Section 10.98: Criteria for flatness

Lemma 10.98.4. 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 10.98.1 we see that $u$ is injective. On the other hand, by Nakayama's Lemma 10.19.1 the map is surjective. The lemma follows. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 22355–22363 (see updates for more information).

    \begin{lemma}
    \label{lemma-free-fibre-flat-free}
    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 $R$.
    \end{lemma}
    
    \begin{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 \ref{lemma-mod-injective}
    we see that $u$ is injective. On the other hand, by
    Nakayama's Lemma \ref{lemma-NAK} the map is surjective. The
    lemma follows.
    \end{proof}

    Comments (2)

    Comment #2473 by Matthieu Romagny on April 3, 2017 a 5:23 pm UTC

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

    Comment #2506 by Johan (site) on April 14, 2017 a 12:07 am UTC

    Thanks, fixed here.

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