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Let $A$ be a finite dimensional algebra with finite global dimension $g$ and $e$ the idempotent such that $eA$ is the direct sum of all indecomposable projective-injective $A$-modules. Do we have $g=injdim(A/AeA)?$

I am mostly interested in the case when $eA$ is injective and faithful, which is also the only case where I really tested it. I tested it for Nakayama algebras and found no counterexamples. I found a proof for higher Auslander algebras. Note that this is also true in case there are no projective-injective modules since then $A/AeA=A$ and the assertion is then well-known. It is not true for algebras of infinite global dimension: The Nakayama algebra with Kupisch series [3,4] has infinite global dimension but $A/AeA$ has injective dimension equal to one.

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$AeA$ is a quotient of a direct sum of copies of $eA$, which is injective. Hence $AeA$ is a cosyzygy, and so $\text{Ext}^g(-,AeA)=0$.

The $\text{Ext}$ long exact sequence from the short exact sequence $$0\to AeA\to A\to A/AeA\to 0$$ therefore ends with $$0=\text{Ext}^g(-,AeA)\to\text{Ext}^g(-,A)\to\text{Ext}^g(-,A/AeA)\to0,$$ and so $\text{Ext}^g(-,A/AeA)\cong\text{Ext}^g(-,A)$, which is nonzero.

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  • $\begingroup$ Very nice, seeing it now, it looks very easy. By the way the same argument should more general also show that the Gorenstein dimension is equal to the Gorenstein injective dimension of A/AeA. $\endgroup$
    – Mare
    May 13 '18 at 22:30
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    $\begingroup$ By the way you seem to accidently confuse $eAe$ with $AeA$ in your answer in some places it seems. $\endgroup$
    – Mare
    May 13 '18 at 23:37

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