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}

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