Let $A$ be a finite dimensional algebra and assume all modules are also finite dimensional. A module $M$ is said to have dominant dimension at least $n$ in case the term $I_i$ for $i=0,1,...,n-1$ are projective when $(I_i)$ denotes a minimal injective coresolution of $M$. The dominant dimension of the algebra is the dominant dimension of the regular module $A$. A module $M$ is said to be $r$-torsionfree in case $Ext_A^i(D(A),\tau(M))=0$ for $i=1,2,...,r$. Note that this generalises the classical concept of reflexive modules, which are exactly the 2-torsionfree modules.

I have the following conjecture:

Conjecture: Assume $A$ has dominant dimension at least $n-1$ and that $X$ is $n$-torsionfree. Then $Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X))) \cong Ext_A^1(D(A),\Omega^{-(n-1)}(\tau(X))$.

edit: This is trivially false as Alex Dugas noted in his answer. I made an error when testing this conjecture. I actually tested the conjecture with algebras of dominant dimension at least $n$ instead of $n-1$, where it is useless for the conjecture in the other thread.

Background information:

Note that this would imply Question on $n$-torsionless modules as a special case (I decided to make a new thread since this conjecture is more compact).

It is not too good tested except for some Nakayama algebras.

Here some facts that I think should lead to a prove but I have not found one yet:

  1. A module $M$ for an algebra $A$ of dominant dimension at least $n-1$ has dominant dimension at least $r$ iff it is $r$-torsionless iff it is an $r$-th syzygy for $r \leq n-1$.

2.By a result of Auslander-Reiten for $r$-th syzygy modules $M$ we have that $M \cong \Omega^r(Tr(\Omega^r(Tr(M)))$ in $\underline{mod}-A$. Note $M \cong \Omega^r(Tr(\Omega^r(Tr(M))) \cong \Omega^r(\tau^{-1} \Omega^{-(r-1)} \tau (M)))$ and this looks like exactly what we need here but sadly we are usually not allowed to use $\Omega^{-r} \Omega^r =id$, which is wrong in general.


1 Answer 1


The conjecture does not appear to be true as stated. However, this might just mean that it needs to be clarified or refined.

To see a counterexample, assume $A$ has dominant dimension exactly $n-1$ and take $X=A$. By your definition, $A$ is $n$-torsionfree since $\tau(A) = 0$. Then the conjectured isomorphism becomes $$\mathrm{Ext}^1(DA, \tau \Omega^{-(n-1)}(A)) \cong 0,$$ which would imply that $\Omega^{-(n-1)}(A)$ is 1-torsionfree, or equivalently has dominant dimension at least $1$. But then looking at the minimal injective resolution of $A$ would show that $A$ has dominant dimension at least $n$, a contradiction.

So maybe you want to assume that $X$ has no projective summands, which still might suffice for the application in your linked question.

  • $\begingroup$ Thanks, probably one should say that X has only non-projective direct summands as you say. But strangely enough the computer says that it is true also for projective modules. I have to check the computer program carefully now what is going on. $\endgroup$
    – Mare
    Aug 7, 2018 at 19:16

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