
Lemma 10.100.2. Let $R$ be a local ring with nilpotent maximal ideal. Let $M$ be an $R$-module. The following are equivalent

1. $M$ is flat over $R$,

2. $M$ is a free $R$-module, and

3. $M$ is a projective $R$-module.

Proof. Since any projective module is flat (as a direct summand of a free module) and every free module is projective, it suffices to prove that a flat module is free. Let $M$ be a flat module. Let $A$ be a set and let $x_\alpha \in M$, $\alpha \in A$ be elements such that $\overline{x_\alpha } \in M/\mathfrak m M$ forms a basis over the residue field of $R$. By Lemma 10.100.1 the $x_\alpha$ are a basis for $M$ over $R$ and we win. $\square$

There are also:

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).