Additivity of projective dimensions, or, help me lower my blood pressure - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:04:55Z http://mathoverflow.net/feeds/question/71425 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71425/additivity-of-projective-dimensions-or-help-me-lower-my-blood-pressure Additivity of projective dimensions, or, help me lower my blood pressure David Hansen 2011-07-27T18:38:38Z 2011-07-28T00:11:06Z <p>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. </p> <p>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. </p> <p><strong>Claim</strong>: Under the above hypotheses, we have <code>$\mathrm{pd}_R(R/I)+\mathrm{pd}_{R/I}(M)=\mathrm{pd}_R(M)$</code>. </p> <p><strong>Proof of claim</strong>: 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 <em>Commutative Ring Theory</em>). Hence Auslander-Buchsbaum, applied to $R$ and $R/I$, gives the equality</p> <p><code>$\mathrm{pd}_R(M)-\mathrm{pd}_{R/I}(M)=\mathrm{depth}_R(R)-\mathrm{depth}_{R/I}(R/I)$</code>.</p> <p>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 <code>$\mathrm{depth}_R(R)-\mathrm{depth}_{R}(R/I)$</code>, which is equal to $\mathrm{pd}_{R}(R/I)$ by another (!) application of Auslander-Buchsbaum. $\square$</p> <p><strong>Counterexample to claim</strong>: 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.</p> http://mathoverflow.net/questions/71425/additivity-of-projective-dimensions-or-help-me-lower-my-blood-pressure/71427#71427 Answer by Moosbrugger for Additivity of projective dimensions, or, help me lower my blood pressure Moosbrugger 2011-07-27T18:49:47Z 2011-07-27T19:43:22Z <p>Your proof only works if the projective dimensions of $M$ as an $R$-module <em>and</em> 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.</p> <p>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 </p> <p>$$\operatorname{Ext}^{i}_{\mathbb{C}[x]/x^2}(\mathbb{C},\mathbb{C})=\mathbb{C}$$ for all $i\geq{0}$. In particular, the projective dimension is infinite.</p> <p>In this case, your module is $\mathbb{C}\oplus\mathbb{C}[x]/x^2$, which by the above has infinite projective dimension.</p> http://mathoverflow.net/questions/71425/additivity-of-projective-dimensions-or-help-me-lower-my-blood-pressure/71430#71430 Answer by Graham Leuschke for Additivity of projective dimensions, or, help me lower my blood pressure Graham Leuschke 2011-07-27T19:46:33Z 2011-07-27T20:39:32Z <p>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.</p> <p>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.</p> <p><strong>Later:</strong> The famous <a href="http://www.springerlink.com/content/x6537160gu0v442v/" rel="nofollow">Dutta-Hochster-McLaughlin example</a> does the trick. It is a module $M$ of finite projective dimension over $R = k[x,y,z,w]/(xy-zw)$, with annihilator $(x,y,z,w)^3 + (x,z)$.</p>