
Lemma 10.100.1. 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:

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

2. $\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M$ over $R$.

Proof. The implication (2) $\Rightarrow$ (1) is immediate. Assume (1). By Nakayama's Lemma 10.19.1 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 10.38.12 we obtain $K/\mathfrak mK = 0$. By Nakayama's Lemma 10.19.1 we conclude $K = 0$. $\square$

Comments (2)

Comment #3621 by JuanPablo on

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 $K=0$.

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• 2 comment(s) on Section 10.100: Flatness criteria over Artinian rings

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