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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


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.

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In your counterexample, why is pd$_{R/I}(M)=1$? –  Kevin Ventullo Jul 27 '11 at 20:18
@JSE: That helped too. –  David Hansen Jul 27 '11 at 22:04

2 Answers 2

up vote 9 down vote accepted

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.

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I am probably confused, but isn't $0\to \mathbf{C}[x] \overset{p(x) \mapsto x^2p(x)}{\to} \mathbf{C}[x] \to \mathbf{C}[x] / (x^2) \to 0$ a projective resolution, so $\mathrm{pd}_{R}(R/I)=1$? It seems to me you have shown that $pd_{R/I}(M)= \infty$. –  David Hansen Jul 27 '11 at 19:31
Sorry! I read your statement too sloppily. The essential explanation is the same, but I've corrected it accordingly. –  Moosbrugger Jul 27 '11 at 19:46

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 $(x,y,z,w)^3 + (x,z)$.

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Thanks for pointing this out! (For applications I have in mind, $R$ is regular, so this isn't a problem.) –  David Hansen Jul 27 '11 at 20:31

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