An lower bound for the injective dimension of a module

Question

Let $A$ be a finite dimensional algebra and $M$ a non-injective $A$-module.

Question: Do we have idim $M>$domdim $\tau^{-1}(M)$?

Here idim $N$ denotes the injective dimension of a module $N$ and the dominant dimension domdim $N$ of a module N with minimal injective coresolution $$0 \rightarrow N \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots $$ is defined as the smallest $n$ such that $I^n$ is not projective or infinite if no such $n$ exists. $\tau^{-1}$ denotes the inverse Auslander-Reiten translate.

I can prove the above inequality for idim $M=1$. I think such an elementary inequality should have an easy proof or it must be wrong. But I was not able to find a proof and on the other hand many computer experiments found no counterexample.

Answer 1

Let $M$ be indecomposable and non-injective. Let $\operatorname{domdim} \tau^{-1}(M) = t$. Then $$\tau^{-1}(M) = \Omega^t_A(\Omega^{-t}_A(\tau^{-1}(M)))$$ Since $M$ is non-injective, \begin{align} 0\neq \operatorname{Ext}^1_A(\tau^{-1}(M),M) & = \operatorname{Ext}^1_A(\Omega^t_A(\Omega^{-t}_A(\tau^{-1}(M))), M)\notag\\ & \simeq \operatorname{Ext}^{t+1}_A(\Omega^{-t}_A(\tau^{-1}(M)),M)\notag \end{align} Hence, $\operatorname{idim} M \geq t+1 > t = \operatorname{domdim} \tau^{-1}(M)$.