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Given a finite dimensional connected algebra $A$ with $Ext^{1}(M,M) \neq 0$ for any non-projective and non-injective indecomposable module $M$, with the condition that at least one such module exists.

  1. Is $A$ selfinjective?

  2. Is $A$ local?

(answer is no,see the answer by Jeremy Rickard)

Two other questions:

  1. Is it local when it is selfinjective?

  2. Is it selfinjective when it is local?

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  • $\begingroup$ Same counterexample, just because for the path algebra of $A_2$ there are no non-projective non-injective modules? $\endgroup$
    – Julian Kuelshammer
    Jul 24, 2017 at 8:45
  • $\begingroup$ Ok I added that at least one such module exists. So no hereditary algebras should work. $\endgroup$
    – Mare
    Jul 24, 2017 at 8:49
  • $\begingroup$ Is the title grammatically correct? $\endgroup$
    – Qfwfq
    Jul 24, 2017 at 9:27

1 Answer 1

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Let $A=kQ/\text{rad}(kQ)^2$, where $Q$ has two vertices, a loop at vertex $1$ and an arrow from vertex $1$ to vertex $2$. I think it has five indecomposable modules, all of which are either projective or injective apart from the simple $S_1$ at vertex $1$, and $\text{Ext}^1_A(S_1,S_1)\neq0$.

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  • $\begingroup$ QPA confirms it. I leave the question open for 3. and 4. $\endgroup$
    – Mare
    Jul 24, 2017 at 11:13

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