Isomorphism for Ext spaces for finite dimensional algebras

Question

Let $A$ be an Artin algebra with enveloping algebra $A^e$. Then we have $Hom_{A^e}(X,A^e) \cong Hom_A(D(A) \otimes_A X,A)$ for a bimodule $X$. (see for example in the article "A theorem of Green on the dual of the transpose" by Auslander and Reiten in corollary 4.2.)

Question: When do we have the Ext analogue: $Ext_{A^e}^i(X,A^e) \cong Ext_A^i(D(A) \otimes_A X,A)$ for all $i \geq 1$?

This holds for example for $X=A$. Maybe there is a nice condition and a reference for such isomorphisms.

My guess is that it should be true for $X=A^{*}$ (the dual of the bimodule $A$) for $i=1,..,n-2$ in case $A$ is $n$-torsionfree for $n \geq 3$.

Question 2: Is this true?

This would prove the equivalence of conditions a) and b) for finite dimensional algebras in question 2 in On properties of an algebra as a bimodule

edit:

Here is an example found with the computer that shows that the formula might not hold in general:

Let $A$ be the Nakayama algebra with Kupisch series [ 2, 3, 2, 1 ] and $X=D(A)$.

Then QPA says that $Ext_{A^e}^1(D(A),A^e)$ has dimension 1 while $Ext_A^1(D(A) \otimes_A D(A) , A)$ should have dimension 0.

The code: A:=NakayamaAlgebra([2,3,2,1],GF(3));D:=DualOfAlgebraAsModuleOverEnvelopingAlgebra(A);B:=EnvelopingAlgebra(A);RegB:=DirectSumOfQPAModules(IndecProjectiveModules(B)); t:=Size(ExtOverAlgebra(D,RegB)[2]); CoRegA:=DirectSumOfQPAModules(IndecInjectiveModules(A));U:=NakayamaFunctorOfModule(CoRegA);RegA:=DirectSumOfQPAModules(IndecProjectiveModules(A)); tt:=Size(ExtOverAlgebra(U,RegA)[2]);

I used that $D(A) \otimes_A D(A)$ is isomorphic to the Nakayama functor applied to $D(A)$.

Another example is again $X=D(A)$ and A linear oriented of Dynkin type $A_n$.

Answer 1

The natural way for proving such an isomorphism would be to take a projective resolution of $X$ (over $A^e$) and using the Hom-isomorphism to translate to one-sided modules. But then, even if $\operatorname{Tor}^A_i(D(A),X) = (0)$ for $i>0$, then to get an isomorphism on Ext-groups one would need either that (i) $D(A)\otimes_A A^e$ is projective as a left $A$-module or (ii) that $\operatorname{Ext}^1_A(D(A)\otimes_A A^e,A) = 0$. Then (i) is equivalent to $A$ being selfinjective, while (ii) is equivalent to $\operatorname{Ext}^1_A(D(A), A) = 0$.

Answer 2

Question 2 has an easy answer in case I made no mistake: Let $A$ be $n$-torsionfree as an $A$-bimodule for $n \geq 2$, then also $X=A^{*}$ in $n$-torsionfree. Choose a minimal projective resolution $(P_i)$ of $X$, then we get an exact sequence $0 \rightarrow X^{*} \rightarrow P_0^{*} \rightarrow P_1^{*} \rightarrow ... \rightarrow P_{n-2}^{*}$. Now we can apply the Hom-isomorphisms to get the result for $i=1,..,n-2$.

I was probably just confused because I first thought it should hold for all $i \geq 1$.