## Fun with the Auslander-Buchsbaum formula

Let ${R}$ be a Noetherian local ring, and let ${M}$ be a finite ${R}$-module. When ${M}$ has finite projective dimension over ${R}$, we have the remarkable equality

$\displaystyle \mathrm{projdim}_{R}(M)+\mathrm{depth}_{R}(M)=\mathrm{depth}_{R}(R).$

This is the famous Auslander-Buchsbaum formula, and is one of my favorite tools in commutative algebra. In this post I want to record the following innocent application of Auslander-Buchsbaum, which I found in September 2011 but didn’t have any use for until recently.

Proposition. Let ${R}$ be a Noetherian local ring, ${\mathfrak{a}}$ an ideal of ${R}$, and ${M}$ a finite ${R/\mathfrak{a}}$-module. Then

$\displaystyle \mathrm{projdim}_{R}(M)=\mathrm{projdim}_{R}(R/\mathfrak{a})+\mathrm{projdim}_{R/\mathfrak{a}}(M)$

whenever both number on the right-hand side are finite.

Proof. Supposing ${\mathrm{projdim}_{R}(R/\mathfrak{a})+\mathrm{projdim}_{R/\mathfrak{a}}(M)<\infty}$, a direct construction with complexes yields a finite resolution of ${M}$ by finite free ${R}$-modules, so ${\mathrm{projdim}_{R}(M)<\infty}$ as well.

Now, a sequence ${r_{1},\dots,r_{n}\in R}$ is ${M}$-regular if and only if ${ }$ is ${M}$-regular, so the ${R}$-depth and the ${R/\mathfrak{a}}$-depth of ${M}$ are equal. Equating the Auslander-Buchsbaum formulas for${M}$ over ${R}$ and ${R/\mathfrak{a}}$ then yields the equality

$\displaystyle \mathrm{depth}_{R}(R)-\mathrm{projdim}_{R}(M)=\mathrm{depth}_{R/\mathfrak{a}}(R/\mathfrak{a})-\mathrm{projdim}_{R/\mathfrak{a}}(M).$

Rearranging, we compute

$\displaystyle \begin{array}{rcl} \mathrm{projdim}_{R}(M)-\mathrm{projdim}_{R/\mathfrak{a}}(M) & = & \mathrm{depth}_{R}(R)-\mathrm{depth}_{R/\mathfrak{a}}(R/\mathfrak{a})\\ & = & \mathrm{depth}_{R}(R)-\mathrm{depth}_{R}(R/\mathfrak{a})\\ & = & \mathrm{projdim}_{R}(R/\mathfrak{a}),\end{array}$

where the second line follows from the equality ${ }$, and the third line follows from another application of the Auslander-Buchsbaum formula, which is allowed by our assumption that ${\mathrm{projdim}_{R}(R/\mathfrak{a})<\infty}$. ${\square}$

Theorem. Let ${(R,\mathfrak{m})}$ be a regular local ring, and let ${M}$ be a finite ${R}$-module of projective dimension ${d\geq0}$. Then the following are equivalent:
i. ${\mathrm{dim}_{R/\mathfrak{m}}\left((\mathrm{Ann}_{R}(M)+\mathfrak{m}^{2})/\mathfrak{m}^{2}\right)\geq d}$,
ii. ${\mathrm{dim}_{R/\mathfrak{m}}\left((\mathrm{Ann}_{R}(M)+\mathfrak{m}^{2})/\mathfrak{m}^{2}\right)=d}$,
iii. ${R/\mathrm{Ann}_{R}(M)}$ is a regular local ring of dimension ${\mathrm{dim}R-d}$, and ${M}$ is free over ${R/\mathrm{Ann}_{R}(M)}$.

Proof. It’s trivial to check that iii. implies ii. implies i. We prove that i. implies iii. By hypothesis, we may choose elements ${r_{1},\dots,r_{d}}$ of ${\mathrm{Ann}_{R}(M)}$ whose reductions modulo ${\mathfrak{m}^{2}}$ are linearly independent over ${R/\mathfrak{m}}$; set ${\mathfrak{a}=(r_{1},\dots,r_{d})}$. By e.g. Theorem 14.2 of Matsumura, the elements ${r_{1},\dots,r_{d}}$ form a subset of a regular system of parameters for ${R}$, and so the ring ${R/\mathfrak{a}}$ is regular of dimension ${\mathrm{dim}R-d}$. Furthermore, we have ${\mathrm{projdim}_{R}(R/\mathfrak{a})=d}$. To see this, note that since regular local rings are Cohen-Macaulay, Theorem 17.4 of Matsumura implies that ${r_{1},\dots,r_{d}}$ is an ${R}$-regular sequence, so the Koszul complex yields a minimal projective resolution of length ${d}$. Now, the ${R}$-module structure of ${M}$ naturally factors through the map ${R\rightarrow R/\mathfrak{a}}$, and applying the previous Proposition, we find

$\displaystyle \begin{array}{rcl} \mathrm{projdim}_{R/\mathfrak{a}}(M) & = & \mathrm{projdim}_{R}(M)-\mathrm{projdim}_{R}(R/\mathfrak{a})\\ & = & d-d=0,\end{array}$

so ${M}$ is projective and hence free over ${R/\mathfrak{a}}$. Since ${M}$ is also a faithful module over ${R/\mathrm{Ann}_{R}(M)}$, the natural surjection ${R/\mathfrak{a}\twoheadrightarrow R/\mathrm{Ann}_{R}(M)}$ is necessarily an isomorphism. ${\square}$