The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.110 Auslander-Buchsbaum

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

Proposition 10.110.1. 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

\[ \text{depth}(R) = \text{pd}_ R(M) + \text{depth}(M) \]

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

\[ 0 \to R^{n_ e} \to R^{n_{e-1}} \to \ldots \to R^{n_0} \to M \to 0 \]

be a minimal finite free resolution, so $e = \text{pd}_ R(M)$. By Lemma 10.101.2 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 10.101.9 we see that $\text{depth}(R) \geq e$. On the other hand, breaking the long exact sequence into short exact sequences

\begin{align*} 0 \to R^{n_ e} \to R^{n_{e - 1}} \to K_{e - 2} \to 0,\\ 0 \to K_{e - 2} \to R^{n_{e - 2}} \to K_{e - 3} \to 0,\\ \ldots ,\\ 0 \to K_0 \to R^{n_0} \to M \to 0 \end{align*}

we see, using Lemma 10.71.6, that

\begin{align*} \text{depth}(K_{e - 2}) \geq \text{depth}(R) - 1,\\ \text{depth}(K_{e - 3}) \geq \text{depth}(R) - 2,\\ \ldots ,\\ \text{depth}(K_0) \geq \text{depth}(R) - (e - 1),\\ \text{depth}(M) \geq \text{depth}(R) - e \end{align*}

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 10.101.9 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 10.71.6 (for example) and we can find an $x \in \mathfrak m$ which is a nonzerodivisor on both $R$ and $M$. Let

\[ 0 \to R^{n_ e} \to R^{n_{e-1}} \to \ldots \to R^{n_0} \to M \to 0 \]

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

\[ 0 \to (R/xR)^{n_ e} \to (R/xR)^{n_{e-1}} \to \ldots \to (R/xR)^{n_0} \to M/xM \to 0 \]

is a minimal resolution too. Thus $\text{pd}_ R(M) = \text{pd}_{R/xR}(M/xM)$. By Lemma 10.71.7 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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