# Additivity of projective dimensions, or, help me lower my blood pressure

Sorry for the shameless title. I'm rather stuck on a lemma in commutative algebra - namely, I have both a proof and a counterexample! I have tried rather strenuously and frustratingly to find the error here, without success; any help from the community in debugging this would be greatly appreciated.

Suppose $$R$$ is a Noetherian local ring and $$M$$ is a finite $$R$$-module of finite projective dimension ($$\mathrm{pd}$$ for short); write $$I=\mathrm{Ann}_R(M)$$ in all that follows.

Claim: Under the above hypotheses, we have $$\mathrm{pd}_R(R/I)+\mathrm{pd}_{R/I}(M)=\mathrm{pd}_R(M)$$.

Proof of claim: Recall the Auslander-Buchsbaum formula, namely $$\mathrm{pd}_R(M)+\mathrm{depth}_R(M)=\mathrm{depth}_R(R)$$. Since a sequence $$r_1,\dots,r_n \in R$$ is $$M$$-regular if and only if $$\overline{r}_1,\dots,\overline{r}_n \in R/I$$ is $$M$$-regular, the $$R$$-depth and $$R/I$$-depth of $$M$$ agree. (This is well-known, see e.g. pp. 130-131 of Matsumura's Commutative Ring Theory). Hence Auslander-Buchsbaum, applied to $$R$$ and $$R/I$$, gives the equality

$$\mathrm{pd}_R(M)-\mathrm{pd}_{R/I}(M)=\mathrm{depth}_R(R)-\mathrm{depth}_{R/I}(R/I)$$.

By the same reasoning as previously, the $$R$$-depth and the $$R/I$$-depth of $$R/I$$ are equal, so the right-hand side of this formula can be rewritten as $$\mathrm{depth}_R(R)-\mathrm{depth}_{R}(R/I)$$, which is equal to $$\mathrm{pd}_{R}(R/I)$$ by another (!) application of Auslander-Buchsbaum. $$\square$$

Counterexample to claim: Take $$R$$ local Noetherian, $$a\in R$$ a nonunit, $$M=R/(a) \oplus R/(a^2)$$, so $$I=(a^2)$$. If I have done this right, each of the projective dimensions in my claim is exactly $$1$$, and I believe $$1+1\neq 1$$ was known in antiquity.

• In your counterexample, why is pd$_{R/I}(M)=1$? Commented Jul 27, 2011 at 20:18
• @JSE: That helped too. Commented Jul 27, 2011 at 22:04

Your proof only works if the projective dimensions of $M$ as an $R$-module and as an $R/I$-module are finite. Indeed, finite projective dimension is a hypothesis for the Auslander-Buchsbaum formula, and you used the AB-formula for $M$ as an $R/I$-module in your argument.

In the case of your counter-example, the projective dimension of $M$ is $\infty$. E.g., if $R=\mathbb{C}[x]$ (or its localization at $0$ if you like), $a=x$, then you have the resolution: $$\ldots\to\mathbb{C}[x]/x^2\overset{\cdot x}\to\mathbb{C}[x]/x^2\overset{\cdot x}{\to}\mathbb{C}[x]/x^2\to 0\to \ldots$$ of the module $\mathbb{C}$. Using this to compute $\operatorname{Ext}^{\cdot}_{\mathbb{C}[x]/x^2}(\mathbb{C},\mathbb{C})$, one sees that

$$\operatorname{Ext}^{i}_{\mathbb{C}[x]/x^2}(\mathbb{C},\mathbb{C})=\mathbb{C}$$ for all $i\geq{0}$. In particular, the projective dimension is infinite.

In this case, your module is $\mathbb{C}\oplus\mathbb{C}[x]/x^2$, which by the above has infinite projective dimension.

One problem in the proof is that $$\operatorname{pd}_R(R/I)$$ might be infinite, so you can't apply Auslander-Buchsbaum the final time.

I can't seem to come up with an example where $$M$$ has finite projective dimension while $$\operatorname{ann} M$$ has infinite projective dimension, but they must exist.

Later: The famous Dutta-Hochster-McLaughlin example does the trick. It is a module $$M$$ of finite projective dimension over $$R = k[x,y,z,w]/(xy-zw)$$, with annihilator $$(x,y,z,w)^3 + (x,z)$$.

• Thanks for pointing this out! (For applications I have in mind, $R$ is regular, so this isn't a problem.) Commented Jul 27, 2011 at 20:31