An identity for Ext for rings

Question

Let $A$ be a two-sided noetherian ring (which we should assume to be Gorenstein first so that everything is well defined, otherwise it is only well defined up to a conjecture, which states that every non-zero module has finite grade). For simplicity we can also assume first that $A$ is a finite dimensional algebra and modules are finitely-generated (but non-finitely generated examples or non-Gorenstein examples are also welcome but Im mainly interested in finite dimensional algebras).

For a module $M$, define the grade of $M$ as: $g_M:= \inf \{ i \geq 0 | Ext_A^i(M,A) \neq 0 \}$. Define the Ext-dual of $M$ to be $U(M):=Ext_A^{g_M}(M,A)$ and the double Ext-dual of $M$ as $G(M):=U(U(M))=Ext_{A^{op}}^{g_{Ext_A^{g_M}(M,A)}}(Ext_A^{g_M}(M,A),A)$. Note that $G(M)$ is always non-zero.

Question: Do we have always that $G^l(M) \cong G^{l-1}(M)$ (at least in the stable category of $A$) for some $l \geq 1$ and indecomposable modules $M$, so that the sequence of the $G^l(M)$ becomse stationary?

This is true for $A$ selfinjective or hereditary (in those cases we have $G(M) \cong M$ for all $M$). In all examples this was even true for $l \leq 2$ so I wonder whether we have $G^2(M) \cong G(M)$ (at least in the stable category).

Answer 1

It seems that any module $M$ whose double $A$-dual $M^{**}$ is not reflexive gives a counterexample. In this case $G(M) = M^{**}$ is a summand of $G^2(M) = (M^{**})^{**}$ with non-trivial complement, and $G^2(M)$ can't be reflexive since this property is inherited by summands. Repeating this argument with $G(M)$ instead of $M$, we see that $G^l(M)$ is a summand of $G^{l+1}(M)$ with non-trivial complement for each $l \geq 0$.

Here's an example over a non-Gorenstein ring:

Take $S$ to be the unique simple module over $A = k[x,y]/(x,y)^2$. Then $g_S = 0$ and its $A$-dual is $S^* \cong S^{\oplus 2}$ with $g_{S^*} = 0$, so that $G(S) = S^{**} \cong S^{\oplus 4}$ and in general $G^{l}(S) = S^{\oplus 4l}$.