# Injective dimension of $A/AeA$

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.

$AeA$ 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.
• By the way you seem to accidently confuse $eAe$ with $AeA$ in your answer in some places it seems. – Mare May 13 '18 at 23:37