Lemma 10.106.5. Let $R$ be a Noetherian local ring. Let $x \in \mathfrak m$. Let $M$ be a finite $R$-module such that $x$ is a nonzerodivisor on $M$ and $M/xM$ is free over $R/xR$. Then $M$ is free over $R$.

Proof. Let $m_1, \ldots , m_ r$ be elements of $M$ which map to a $R/xR$-basis of $M/xM$. By Nakayama's Lemma 10.20.1 $m_1, \ldots , m_ r$ generate $M$. If $\sum a_ i m_ i = 0$ is a relation, then $a_ i \in xR$ for all $i$. Hence $a_ i = b_ i x$ for some $b_ i \in R$. Hence the kernel $K$ of $R^ r \to M$ satisfies $xK = K$ and hence is zero by Nakayama's lemma. $\square$

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