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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