Vanishing of $Ext^i_R(N, R)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:29:22Z http://mathoverflow.net/feeds/question/104595 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104595/vanishing-of-exti-rn-r Vanishing of $Ext^i_R(N, R)$ Pham Hung Quy 2012-08-13T06:23:44Z 2012-09-11T20:21:29Z <p>This question is related to <a href="http://mathoverflow.net/questions/32310/are-the-supports-of-extim-n-eventually-periodic" rel="nofollow">http://mathoverflow.net/questions/32310/are-the-supports-of-extim-n-eventually-periodic</a>. </p> <p>Let $(R, \mathfrak{m}, k)$ be a Noetherian local ring of dimension $d$. It well known that</p> <p><strong>Theorem</strong> $R$ is Gorenstein iff $Ext^i_R(k, R) = 0$ for some $i > d$, and iff $Ext^i_R(k, R) = 0$ for all $i > d$.</p> <p>By above Theorem we have: $Ext^i_R(k, R) = 0$ for some $i > d$ iff $Ext^i_R(k, R) = 0$ for all $i > d$.</p> <p>The following question consider in a generalization of this situation for modules of finite length.</p> <p><strong>Question:</strong> Let $(R, \mathfrak{m}, k)$ be a Noetherian local ring of dimension $d$, $N$ an $R$-module of finite length. Is it true that: $Ext^i_R(N, R) = 0$ for some $i > d$ iff $Ext^i_R(N, R) = 0$ for all $i > d$?</p> http://mathoverflow.net/questions/104595/vanishing-of-exti-rn-r/106954#106954 Answer by Graham Leuschke for Vanishing of $Ext^i_R(N, R)$ Graham Leuschke 2012-09-11T20:21:29Z 2012-09-11T20:21:29Z <p>No.</p> <p>Let's work with an Artinian local ring $R$. Let $X$ be a module satisfying $\mathrm{Ext}_R^i(X,R)\neq 0$ for all $i$. (These are plentiful; for example, assume $R$ is non-Gorenstein and let $X$ be the residue field.) Define a module $Q$ by "Serre's trick": take generators $\chi_1, \dots, \chi_r$ for $\mathrm{Ext}_R^1(X,R)$ and consider the short exact sequence corresponding to $(\chi_1, \dots, \chi_r) \in \mathrm{Ext}_R^1(X,R^r)$: $$0 \to R^r \to Q \to X \to 0$$ Then $\mathrm{Ext}_R^1(Q,R) =0$, since the long exact sequence of $\mathrm{Ext}$ looks like $$\cdots \to \mathrm{Hom}_R(R^r,R) \to \mathrm{Ext}_R^1(X,R) \to \mathrm{Ext}_R^1(Q,R) \to \mathrm{Ext}_R^1(R^r,R)=0$$ and the map $\mathrm{Hom}_R(R^r,R) \to \mathrm{Ext}_R^1(X,R)$ is cooked up precisely to be surjective. On the other hand, $\mathrm{Ext}_R^i(Q,R) = \mathrm{Ext}_R^{i}(X,R) \neq 0$ for all $i\geq 2$. </p> <p>One can jazz this up a bit, using a result of <a href="http://www.ams.org/mathscinet-getitem?mr=2238367" rel="nofollow">Jorgensen-Sega</a> (<a href="http://www.springerlink.com/content/k6u4336j24653761/?MUD=MP" rel="nofollow">journal link</a>): There exists a local Artinian ring $R$ and a family <code>$\{M_s\}_{s\geq 1}$</code> of reflexive $R$-modules such that (among other things) $\mathrm{Ext}_R^i(M_s,R)\neq 0$ if and only if $1 \leq i \leq s-1$. They give a completely explicit construction of $R$ and the modules $M_s$. Taking one of the $M_s$ in place of $X$ above, one obtains modules $Q_{a,b}$ for which <code>$\mathrm{Ext}_R^i(Q_{a,b},R)$</code> vanishes up to $i=a-1$, is nonzero for $a \leq i \leq b$, and vanishes again for $i \geq b+1$. Then taking direct sums gives essentially arbitrary behavior of vanishing and non-vanishing.</p>