Questions are about this paper: http://users.uoi.gr/abeligia/gorenstein.pdf and all algebras are finite dimensional

Question 1: In corollary 6.21 (2) there is a proof of the direction (e) implies (a): It is said that if an algebra has finite finitistic dimension d, then $\Omega^d(mod-A)=\Omega^{d+i}(mod-A)$ for all $i \geq 0$ (a paper of auslander reiten is quoted, but I cant find the exact statement there, there is however a similar statement with the additional assumption that $\Omega^i(mod-A)$ is extension closed for all i).

So is the assertion of this proof in the paper correct?

Here an (possible counter) example: Let A be the Nakayama algebra with Kupisch series [3,4]. This algebra has finitistic dimension 1 and I think the first simple module is in $\Omega^{1}(mod-A)$ but not in $\Omega^2(mod-A)$, is this correct?

Question 2: In the same place in the paper the author claims that (a) implies for example (b), so that any Gorenstein algebra should be k-Gorenstein for all k. But on page 12 on the top there are counterexamples http://ac.els-cdn.com/0022404994900442/1-s2.0-0022404994900442-main.pdf?_tid=8201a554-7aab-11e7-87c3-00000aab0f6c&acdnat=1502026449_b3cbd17ed58cec1c97d44c6e3d508e59 . Do I confuse something?

Question 3: Are there easy example of algebras $A$ such that $\Omega^i(mod-A) \subset \Omega^{i-1}(mod-A)$ strictly for every i? Note that such algebras have to be a little exotice since $\Omega^i(mod-A)$ cant be representation-finite for any i (this excludes for example representation-finite or monomial algebras). I have many candidates for such algebras (including algebras which might have infinite finitstic dimension :) ) but it seems to be not so easy to prove it.

To question 1: I used the following in my calculations to test wheter an indecomposable non-projective module M is in $\Omega^i(mod-A)$: $M$ is in $\Omega^i(mod-A)$ iff $M$ is a summand of $\Omega^{i}(\Omega^{-i}(M))$, using https://folk.ntnu.no/oyvinso/Papers/rel1.pdf proposition 3.2. Hope this is correct.

edit: Assume an algebras is nearly Gorenstein (meaning that Gorenstein projectives are equal to $^{\perp}A$ and the dual property). For example all representation-finite algebras are nearly Gorenstein. Then the condition for being Co-Gorenstein is $\Omega^{\infty} \subseteq ^{\perp}A=Gp(A)$, but always $Gp(A) \subseteq \Omega^{\infty}$, so the condition is $Gp(A) = \Omega^{\infty}$. Now if the algebra is $\Omega^{r}$-finite for some $r$ (as representation-finite algebras are), then $\Omega^{\infty}=\Omega^{s}$ for some s and thus the condition is $\Omega^{s}=Gp(A)$, but this is equivalent to $A$ being s-Gorenstein. Thus in this situation Co-Gorenstein implies Gorenstien. This applies for example to torsionless-finite algebras (which include all representation-finite algebras).

edit2: Here an example of a Gorenstein algebra (thus CoGorenstein algebra) with $\Omega^{i}$ not extension closed for all i: Take the Nakayama algebra with Kupisch series [2,3,3,2,1]. By the main result of the paper "Syzygy modules for noetherian rings" by Auslander and Reiten, $\Omega^i$ is extension closed for $1 \leq i \leq k$ iff the projective dimesnion of $I_i$ is bounded by $i+1$ for all $i < k$. Now for the nakayama algebra corresponding to [2,3,3,2,1] with minimal projective resolution $P_i$ of D(A), the module $P_1$ has injective dimension 3 and 3>2 thus the condition is not satisfied.

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    $\begingroup$ I checked the calculations for the Kupisch series $[3,4]$. It is true that it is of finitistic dimension $1$ and that one of the simples is in $\Omega^1$, but not in $\Omega^2$. In fact, $\Omega^2$ consists of just one module and coincides with all higher $\Omega^t$. Did you check whether the algebra is co-Gorenstein? $\endgroup$ – Julian Kuelshammer Aug 7 '17 at 11:04
  • $\begingroup$ @JulianKuelshammer The argument in the proof is wrong, expect if Co-Gorenstein implies that $\Omega^i$ is extension closed for all i. But this should be wrong, since it is not correct for Gorenstein algebras in general I think, or? How did you calculate the syzygies? Is there a better way than what I used? I edit something about Co-Gorenstein related to representation-finite algebras. $\endgroup$ – Mare Aug 7 '17 at 11:35
  • $\begingroup$ I just calculated all syzygies by hand. Since Nakayama algebras are representation-finite you can just write down all $7$ indecomposables and their respective projective resolutions. According to Corollary 3.1 of AR $\Omega^k$ is extension closed for all $k$ if the algebra is $k$-Gorenstein for all $k$. $\endgroup$ – Julian Kuelshammer Aug 7 '17 at 12:07
  • $\begingroup$ According to the article "Syzygy modules and for noetherian rings" by Auslander and Reiten $\Omega^i$ is extension closed for all $i$ (What is needed in the proof) iff the projective dimension of $I_i$ is $\leq i+1$ when $I_i$ is the minimal injective coresolution of $A$. So take an algebra of global dimension larger than or equal to 2 and dominant dimension zero. Then $I_0$ is not projetive and thus has projective dimension at least two. Thus the condition is not satisfied even for Gorenstein algebras. Thus also not for CoGorenstein algebras and the given proof is wrong in my opinion. $\endgroup$ – Mare Aug 7 '17 at 12:11
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    $\begingroup$ @Mare I spent ages confused about your "edit 2" example until I realized that (slightly confusingly) Auslander and Reiten generally deal with left modules, but state their condition in terms of an injective resolution of $\Lambda$ as a right module. This means that your example should be the algebra with Kupisch series $[2,3,3,2,1]$ and not its opposite algebra. Even without the Auslander-Reiten result it's easy to check directly that $\Omega^2$ is not closed under extensions (but it is for the opposite algebra, which is what was confusing me). $\endgroup$ – Jeremy Rickard Aug 9 '17 at 9:27

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