In the article http://www.sciencedirect.com/science/article/pii/S0021869301991306?via%3Dihub (see also the MO thread Why does there exist a non-split sequence with the condition that $\mathrm{pd} M=\infty$? ) the authors proved the following:

Theorem: In case $A$ (every algebra is an Artin algebra in the following) has infinite global dimension, there exists an indecomposable module $M$ with infinite projective and infinite injective dimension.

Now I looked at the following, which I formulate as a conjecture in the following:

Conjecture: In case $A$ has infinite Gorenstein dimension, there exists an indecomposable module $M$ with infinite Gorenstein projective and infinite Gorenstein injective dimension.

There are two motivations to conjecture this (appart from being the natural Gorenstein homological version of the first result):

1. My computer found no counterexample.

2. A proof of the conjecture would give a new proof of the theorem and thus a generalisation of the theorem.

Proof of the theorem using the conjecture: Assume $A$ has infinite global dimension. We have 2 cases:

Case a) A is Gorenstein. In this case there is an indecomposable non-projective Gorenstein projective module. But such modules have infinite projective and infinite injective dimensions.

Case b) A is not Gorenstein. If the conjecture is true, there exists indecomposable M of infinite Gorenstein projective and infinite Gorenstein injective dimension. But then also the projective and injective dimension of $M$ are infinite. $\blacksquare$

Now I try to give a proof of the conjecture, which imitates the proof in the paper (expect that I take $D(A)$ instead of the simple module $S$ as in the paper). One just hast to find a short exact sequence of the form:

$$0 \rightarrow T \rightarrow E \rightarrow M \rightarrow 0 $$

as in the paper but with projective and injective dimension replaced by Gorenstein projective dimension (Gpd) and Gorenstein injective dimension (Gid) everywhere.

Since $A$ is not Gorenstein, the Gorenstein projective dimension of $D(A)$ is infinite (why???). Let $n:= \sup \{ Y | Y $ simple and $Gid(Y) < \infty \}$. There is a summand $M'$ of $\Omega^n(D(A))$ of infinite Gorenstein projective dimension. Since $M'$ also has infinite projective dimension, there exists a simple module T with $Ext^l(M',T) \neq 0$ for infinitely many $l$. Let $r$ be such that $Ext^r(M',T) \neq 0$ and $M$ an indecomposable summand of $\Omega^{r-1}(M')$ (which is a summand of $\Omega^{n+r-1}(D(A))$) of infinite Gorenstein projective dimension such that $Ext^1(M,T) \neq 0$. Then $M$ has infinite Gorenstein projective dimension and $Ext^1(M,T) \neq 0$. $Ext^l(M',T) \neq 0$ for infinitely many $l$ implies that $Ext^q(D(A),T) \neq 0$ for infinitely many $q$ and thus $T$ has infinite Gorenstein injective dimension. (here I use that the Gorenstein injective dimension of a module $X$ is larger than or equal to $\sup \{ t \geq 1 | Ext^t(D(A),X) \neq 0 \}$, see for example the dual of proposition 3.2.2. in http://home.ustc.edu.cn/~xwchen/Personal%20Papers/postdoc-Xiao-Wu%20Chen%202010.pdf .)

This gives the desired short exact sequence and one can finish the proof now exactly as in the rest of the paper, in case I made no mistake.

Questions:

1. I had to assume that for a non-Gorenstein algebra $A$ one has that the Gorenstein projective dimension of $D(A)$ is infinite. Is that correct? I see no proof at the moment (but it is late...). So in short: Does $pd(D(A))= \infty$ imply that $Gpd(D(A))= \infty $?

2. Is there maybe another proof of the conjecture (in case 1. fails)?