Given a finite dimensional non-Gorenstein algebra $A$, do we have $Ext^i(D(A),A) \neq 0$ for infinitely many $i$? (We can assume A is local or commutative if that helps). All I can show is that such algebras would be pretty exotic when this holds only for finitely many i.
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My modules are right modules.
Let $M$ be a non-projective module for a self-injective algebra $B$ such that $\text{Ext}^i_B(M,M)=0$ for $i>t$ (i.e., what you've called a "strange" module in some recent questions). Let $$A=\pmatrix{k&M\\0&B}.$$ Then $A$ is not Gorenstein, as it has a one-dimensional injective module with infinite projective dimension, and according to my calculations, $\text{Ext}^i_A(DA,A)=0$ for $i>t+1$.
An $A$-module is a "row vector" $\pmatrix{V&X}$, where $V$ is a vector space and $X$ a $B$-module, together with a $B$-module map $V\otimes_kM\to X$.
$DA$ is the direct sum of two injectives $I_1=\pmatrix{k&0}$ and $I_2=\pmatrix{DM&DB}$ corresponding to the columns of $A$.
If $$\dots P_1\to P_0\to M\to0$$ is a $B$-projective resolution of $M$, then $$\dots\pmatrix{0&P_1}\to\pmatrix{0&P_0}\to\pmatrix{k&M}\to\pmatrix{k&0}\to0$$ is an $A$-projective resolution of $I_1$, and so $$\text{Ext}^{i+1}_A(I_1,A)=\text{Ext}^i_B(M,M\oplus B)$$ for $i>1$, which is zero for $i>t$ by the assumptions on $B$ and $M$.
There is a short exact sequence $$0\to\pmatrix{0&DB}\to I_2\to\pmatrix{DM&0}\to0.$$ The last term is a direct sum of copies of $I_1$, and the first term is a projective $A$-module, so the $\text{Ext}^*_A(-,A)$ long exact sequence gives $\text{Ext}^i_A(I_2,A)=0$ for $i>t+1$.
answered Sep 11, 2017 at 22:43
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$\begingroup$ Good idea to use strange modules! Is there a trick to obtain $Ext^i(D(A),A)$ in term of $M$ and $B$? At the moment I do not see how to calculate that in a good way. But I will try it at a concrete example. $\endgroup$– MareCommented Sep 12, 2017 at 7:14
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$\begingroup$ I choose $A=K<x,y>/(x^2,y^2,x*y-qy*x)$ and $M=(x+y)A$ (I interchange A and B here, since this selfinjective algebra was always an A for me). This should be a strange module (if you only need selfinjective instead of symmetric).But my calculations say that $Ext^i(D(A),A) \neq 0$ for all i>0. $\endgroup$– MareCommented Sep 12, 2017 at 7:59
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$\begingroup$ Here quiver and relations I obtained for the algebra which you call A: Q:=Quiver(2,[[1,2,"u"],[2,2,"s"],[2,2,"t"]]); q:=2;rel:=[ut-qus,s^2,t^2,st-qts]; And sorry I think you are right, we really have $Ext^i(D(A),A)=0$ for $i >2$, I made a mistake. $\endgroup$– MareCommented Sep 12, 2017 at 8:01
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$\begingroup$ The modules $\Omega^i(D(A))$ for $i \geq 2$ for this algebra are all 6-dimensional and maximal Cohen-Macaulay modules that are not Gorenstein projective. So you also found a new systematic construction of such exotic modules using strange modules. Of course, I have to ask whether those algebras satisfy the finitistic dimension conjecture! $\endgroup$– MareCommented Sep 12, 2017 at 8:16
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1$\begingroup$ @Mare They satisfy the FDC, since for any module $M$, $\pmatrix{k&0\\0&0}$ annihilates $\Omega M$, and so calculating a projective resolution of $M$ reduces to calculating a projective resolution of a module for the self-injective algebra $B$. Regarding your first comment, I'll add an outline of the calculation when I have some time. $\endgroup$ Commented Sep 12, 2017 at 8:37