4
$\begingroup$

Let $A$ be a finite dimensional algebra. Recall that a module $M$ is called $n$-torsionfree in case $Ext_A^i(D(A),\tau(M))=0$ for all $i=1,...,n$ when $\tau$ denotes the Auslander-Reiten translate. For example for $n=1$ this is equivalent to $M$ being torsionless and for $n=2$ this is equivalent to $M$ being reflexive. Recall that a module $M$ with minimal injective coresolution $(I_i)$ is said to have dominant dimension at least $n$ in case $I_0,I_1,...,I_{n-1}$ are projective. Let $P$ be an indecomposable projective non-injective module. Let $TF_n$ denote the full subcategory of $n$-torsionfree modules and $Dom_n$ the full subcategory of modules with dominant dimension at least $n$. Let $0 \rightarrow P \rightarrow X \rightarrow \tau^{-1}(P) \rightarrow 0$ be an almost split sequence and assume that $X$ and $\tau^{-1}(P)$ are $n$-torsionfree for some $n \geq 1$ and each indecomposable projective non-injective module $P$.

Conjecture: Then the regular module $A$ has dominant dimension at least $n$.

The case $n=2$ follows from a result of Tachikawa and a proof of the conjecture would improve the main result of Tachikawa (see his paper "Reflexive Auslander-Reiten sequences"). The computer gives good evidence that the conjecture is true.

Here my attempt to prove it by induction for $n \geq 2$ (The case $n=1$ (which is the induction start) is true and can be proved analogous to what Tachikawa did):

Assume the conjecture holds for $n-1$ and assume that each such almost split sequence as above has $X$ and $\tau^{-1}(P)$ being $n$-torsionfree (of course $X$ depends on $P$, so probably should write $X_P$ better). We prove it for $n$ then. By induction we have that $A$ has dominant dimension at least $n-1$. Showing that $A$ has dominant dimension at least $n$ is equivalent to showing that each indecomposable projective non-injective module $P$ has dominant dimension at least $n$. Now by a result of Auslander and Reiten, $A$ having dominant dimension at least $n-1$ implies that $TF_{l}=Dom_{l}$ for all $l \leq n-1$. Thus $X$ and $N:= \tau^{-1}(P)$ have dominant dimension at least $n-1$. In case we can show that $X$ has dominant dimension at least $n$ we are finished since this would imply that $P$ has dominant dimension at least $n$. Now $X$ has dominant dimension at least $n$ iff $\Omega^{-(n-1)}(X)$ has domiant dimension at least one, which is equivalent to $\Omega^{-(n-1)}(X)$ being torsionless. That the module $\Omega^{-(n-1)}(X)$ is torsionless is equivalent to $Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X)))=0$, which is what we need to show. The assumption that $X$ is $n$-torsionfree this gives us that $0=Ext_A^n(D(A),\tau(X)) \cong Ext_A^1(D(A) , \Omega^{-(n-1)}(\tau(X))$.

Now the result would follow in case we have: $0=Ext_A^1(D(A) , \Omega^{-(n-1)}(\tau(X))=Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X)))$, but I see at the moment no reason why we can "exchange" $\Omega^{-(n-1)}$ with $\tau$ here.

So my question is:

Do we have: $Ext_A^1(D(A) , \Omega^{-(n-1)}(\tau(X))=Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X)))=0$?

$\endgroup$
  • 2
    $\begingroup$ You should visit Osamu Iyama and get all these questions resolved! $\endgroup$ – Hailong Dao Jan 1 '18 at 0:17
  • 1
    $\begingroup$ @HailongDao I actually have a paper with him and Aaron Chan and he liked my questions when I met him. But this one might be too easy and I just miss something easy I fear. $\endgroup$ – Mare Jan 1 '18 at 0:19
  • 2
    $\begingroup$ I see, so my Mare number is 2! $\endgroup$ – Hailong Dao Jan 1 '18 at 0:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.