Proposition 10.111.1. Let $R$ be a Noetherian local ring. Let $M$ be a nonzero finite $R$-module which has finite projective dimension $\text{pd}_ R(M)$. Then we have

## 10.111 Auslander-Buchsbaum

The following result can be found in [Auslander-Buchsbaum].

**Proof.**
We prove this by induction on $\text{depth}(M)$. The most interesting case is the case $\text{depth}(M) = 0$. In this case, let

be a minimal finite free resolution, so $e = \text{pd}_ R(M)$. By Lemma 10.102.2 we may assume all matrix coefficients of the maps in the complex are contained in the maximal ideal of $R$. Then on the one hand, by Proposition 10.102.9 we see that $\text{depth}(R) \geq e$. On the other hand, breaking the long exact sequence into short exact sequences

we see, using Lemma 10.72.6, that

and since $\text{depth}(M) = 0$ we conclude $\text{depth}(R) \leq e$. This finishes the proof of the case $\text{depth}(M) = 0$.

Induction step. If $\text{depth}(M) > 0$, then we pick $x \in \mathfrak m$ which is a nonzerodivisor on both $M$ and $R$. This is possible, because either $\text{pd}_ R(M) > 0$ and $\text{depth}(R) > 0$ by the aforementioned Proposition 10.102.9 or $\text{pd}_ R(M) = 0$ in which case $M$ is finite free hence also $\text{depth}(R) = \text{depth}(M) > 0$. Thus $\text{depth}(R \oplus M) > 0$ by Lemma 10.72.6 (for example) and we can find an $x \in \mathfrak m$ which is a nonzerodivisor on both $R$ and $M$. Let

be a minimal resolution as above. An application of the snake lemma shows that

is a minimal resolution too. Thus $\text{pd}_ R(M) = \text{pd}_{R/xR}(M/xM)$. By Lemma 10.72.7 we have $\text{depth}(R/xR) = \text{depth}(R) - 1$ and $\text{depth}(M/xM) = \text{depth}(M) - 1$. Till now depths have all been depths as $R$ modules, but we observe that $\text{depth}_ R(M/xM) = \text{depth}_{R/xR}(M/xM)$ and similarly for $R/xR$. By induction hypothesis we see that the Auslander-Buchsbaum formula holds for $M/xM$ over $R/xR$. Since the depths of both $R/xR$ and $M/xM$ have decreased by one and the projective dimension has not changed we conclude. $\square$

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