Let $k$ be a field and $Q$ be the quiver with two vertices 1 and 2 and three arrows: $a$ from 2 to 1, $b$ from 2 to 1 and $c$ from 2 to 2. Let $I_1=\langle ab-c^2,ba\rangle$ and $I_2=\langle ab-c^2,c^4,ba-bca\rangle$ and $A_1:=kQ/I_1$ and $A_2:=kQ/I_2$ be the corresponding quiver algebras. It is known that $A_1$ and $A_2$ are isomorphic if and only if the characteristic of the field is not equal to two. Let $M_1$ be the direct sum of all indecomposable $A_1$ modules and $M_2$ be the direct sum of all indecomposable $A_2$ modules. Let $B_1=\operatorname{End}_{A_1}(M_1)$ and $B_2=\operatorname{End}_{A_2}(M_2)$ be the Auslander algebras of $A_1$ and $A_2$. Let $C_1=\underline{\operatorname{End}_{A_1}}(M_1)$ and $C_2=\underline{\operatorname{End}_{A_2}}(M_2)$ be the stable Auslander algebra of $A_1$ and $A_2$.

Question 1: Exercise 2 (e) in chapter VII. in the book on Artin algebras by Auslander, Reiten and Smalo asks for a prove that when the characteristic of the field $k$ is equal to two, $A_1$ and $A_2$ have the same Auslander–Reiten quiver (this follows because they are socle-equivalent), but $B_1$ and $B_2$ are not isomorphic. Is there an easy prove/argument for this using only the techniques developed in the book? A direct proof calculating quiver and relations of $B_1$ and $B_2$ would probably be very tedious (both algebras have 20 indecomposable modules) and longer than 10 pages. But maybe there is an elementary argument.

Question 2: Are $B_1$ and $B_2$ derived or stable equivalent in characteristic 2?

Question 3: Are $C_1$ and $C_2$ isomorphic?

  • $\begingroup$ Standard notation is $I_1=\langle ab-c^2,ba\rangle,$ not $I_1=< ab-c^2,ba>.\,\,\,{}$ I edited accordingly. $\endgroup$ – Michael Hardy Jun 12 '19 at 20:51

Question 1: Yes, there is an elementary argument that $B_1$ and $B_2$ are not isomorphic. One uses Theorem VI.5.7, which states that there is a bijection between Morita equivalence classes of nonsemisimple artin algebras of finite type and Morita equivalence classes of nonsemisimple Auslander algebras. If $B_1 \cong B_2$, then they would be Morita equivalent Auslander algebras, and hence $A_1$ and $A_2$ would be Morita equivalent. Since $A_1$ and $A_2$ are both basic, they must be isomorphic, a contradiction.

Question 3: It is known that the stable Auslander algebras $C_1$ and $C_2$ are not isomorphic (in characteristic $2$), but this is much harder to show. This is equivalent to the fact that $A_1$ and $A_2$ (as well as the other standard-nonstandard pairs of socle equivalent self-injective algebras of finite type) are not stably equivalent, which was shown by Asashiba (The derived equivalence classification of representation-finite selfinjective algebras. J. Algebra 214 (1999), no. 1, 182–221).

Question 2: I would guess that $B_1$ and $B_2$ are neither derived nor stably equivalent (in characteristic $2$), but I have not checked the details. For stably equivalent, I expect that one could apply Martinez-Villa's result that a stable equivalence induces a stable equivalence between certain associated self-injective algebras, which should turn out to be $A_1$ and $A_2$ (Properties that are left invariant under stable equivalence. Comm. Algebra 18 (1990), no. 12, 4141–4169). However, we know these are not stably equivalent (see above). For derived equivalence, one might be able to apply ideas of Hu and Xi on almost $\nu$-stable derived equivalences, as in [Derived equivalences and stable equivalences of Morita type, I. Nagoya Math. J. 200 (2010), 107–152] and subsequent papers.

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  • $\begingroup$ Thanks, for question 1 I forgot that the Auslander correspondence was done in the book. So it is indeed easy with that result (or more general the Morita-Tachikawa correspondence). To question 3: It seems plausible that for selfinjective algebras, the stable categories are equivalent if and only if the stable Auslander algebras are isomorphic. Is this true for general representation-finite algebras? $\endgroup$ – Mare Jun 11 '19 at 22:19
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    $\begingroup$ Yes, I believe the stable Auslander algebras are isomorphic iff the stable categories are equivalent, and it's almost a tautology. I just noticed your other question to this effect, but haven't had a chance to answer yet. $\endgroup$ – Alex Dugas Jun 11 '19 at 22:26
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    $\begingroup$ Ok, thanks. One direction is easy of course (stable equivalent implies isomorphic stable Auslander algebras), but I am not completely sure about the other direction as I am not very experienced with stable equivalences. $\endgroup$ – Mare Jun 11 '19 at 22:31

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