Lemma 10.101.6. Let $R$ be an Artinian local ring. Let $M$ be an $R$-module. Let $I \subset R$ be a proper ideal. The following are equivalent

$M$ is flat over $R$, and

$M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$.

Lemma 10.101.6. Let $R$ be an Artinian local ring. Let $M$ be an $R$-module. Let $I \subset R$ be a proper ideal. The following are equivalent

$M$ is flat over $R$, and

$M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$.

**Proof.**
The implication (1) $\Rightarrow $ (2) follows immediately from the definitions. Assume $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$. By Lemma 10.101.2 this implies that $M/IM$ is free over $R/I$. Pick a set $A$ and elements $x_\alpha \in M$ such that the images in $M/IM$ form a basis. By Lemma 10.101.3 we conclude that $M$ is free and in particular flat.
$\square$

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