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$\DeclareMathOperator\Ext{Ext}$This question is related to Are the supports of $Ext^i(M,N)$ eventually periodic? .

Let $(R, \mathfrak{m}, k)$ be a Noetherian local ring of dimension $d$. It well known that

Theorem $R$ is Gorenstein iff $\Ext^i_R(k, R) = 0$ for some $i > d$, and iff $\Ext^i_R(k, R) = 0$ for all $i > d$.

By above Theorem we have: $\Ext^i_R(k, R) = 0$ for some $i > d$ iff $\Ext^i_R(k, R) = 0$ for all $i > d$.

The following question consider in a generalization of this situation for modules of finite length.

Question: 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 $i > d$?

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  • $\begingroup$ The theorem that you mentioned implies: if $\mathrm{inj\;dim}\;R=\infty$ then $\mathrm{Ext}^i_R(k,R)\neq0$ for all $i>d$. Perhaps a simpler question would be: if $\mathrm{inj\;dim}\;R=\infty$, can there be an $R$-module of finite length $N$ such that $\mathrm{Ext}^i_R(N,R)=0$ for some $i>d$? $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Aug 14, 2012 at 5:18
  • $\begingroup$ @ Mahdi Majidi-Zolbanin: if $pd_R(N)$ ($N$ has finite projective dimension) then $Ext^i_R(N, R) = 0$ for all $i > pd_R(N)$ but $pd_R(N) = d$ in this case. $\endgroup$
    – Pham Hung Quy
    Commented Aug 14, 2012 at 8:43
  • $\begingroup$ $pd_R(N) < \infty$ $\endgroup$
    – Pham Hung Quy
    Commented Aug 14, 2012 at 8:44
  • $\begingroup$ @ Pham Hung Quy: Good point. If that happens then $R$ is Cohen-Macaulay. What if $R$ is not Cohen-Macaulay? Do you know an example of an $R$-module of finite length $N$ such that $\mathrm{Ext}_i^R(N,R)=0$ for some $i>d$? $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Aug 14, 2012 at 14:33
  • $\begingroup$ @ Mahdi Majidi-Zolbanin: In general we have $pd(N) = depth(R) - depth(N)$ if $pd(N) < \infty$. In our case $pd(N) = depth(R) = t$. By passing a regular sequence $x_1,...,x_t$ contained in $Ann(N)$ we may assume that $depth(R) = 0$, so $N$ is projective and so free. Thus $\dim R = 0$ since $\ell(N) < \infty$. Hence if $R$ is not Cohen-Macaulay, then $Ext^i_R(N, R) \neq 0$ for all $i > d$. $\endgroup$
    – Pham Hung Quy
    Commented Aug 15, 2012 at 1:22

1 Answer 1

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No.

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 $i\geq 2$.

One can jazz this up a bit, using a result of Jorgensen–Şega (journal link): There exists a local Artinian ring $R$ and a family $\{M_s\}_{s\geq 1}$ 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 $\mathrm{Ext}_R^i(Q_{a,b},R)$ 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.

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