Lemma 10.106.6. Let $R$ be a regular local ring. Any maximal Cohen-Macaulay module over $R$ is free.

Proof. Let $M$ be a maximal Cohen-Macaulay module over $R$. Let $x \in \mathfrak m$ be part of a regular sequence generating $\mathfrak m$. Then $x$ is a nonzerodivisor on $M$ by Proposition 10.103.4, and $M/xM$ is a maximal Cohen-Macaulay module over $R/xR$. By induction on $\dim (R)$ we see that $M/xM$ is free. We win by Lemma 10.106.5. $\square$

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