Lemma 10.101.1 (051F)—The Stacks project

Lemma 10.101.1. Let $(R, \mathfrak m)$ be a local ring with nilpotent maximal ideal $\mathfrak m$. Let $M$ be a flat $R$-module. If $A$ is a set and $x_\alpha \in M$, $\alpha \in A$ is a collection of elements of $M$, then the following are equivalent:

$\{ \overline{x}_\alpha \} _{\alpha \in A}$ forms a basis for the vector space $M/\mathfrak mM$ over $R/\mathfrak m$, and
$\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M$ over $R$.

Proof. The implication (2) $\Rightarrow $ (1) is immediate. Assume (1). By Nakayama's Lemma 10.20.1 the elements $x_\alpha $ generate $M$. Then one gets a short exact sequence

\[ 0 \to K \to \bigoplus \nolimits _{\alpha \in A} R \to M \to 0 \]

Tensoring with $R/\mathfrak m$ and using Lemma 10.39.12 we obtain $K/\mathfrak mK = 0$. By Nakayama's Lemma 10.20.1 we conclude $K = 0$. $\square$

Comments (2)

Comment #3621 by JuanPablo on October 11, 2018 at 18:44

Alternative proof: $x_{\alpha}$ generate $M$ by Nakayama's Lemma 00DV. Then one gets a short exact sequence $0\to K\to \bigoplus_{\alpha\in A}R\to M\to 0$ . Then one tensors by $R/\mathfrak{m}$ and by Lemma 00HL one gets $K=\mathfrak{m}K$ . Then by Nakayama's Lemma 00DV one gets $K=0$ .

Comment #3724 by Johan on October 23, 2018 at 13:43

Much, much better. Thanks. See changes here.

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