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

$\endgroup$
3
  • 2
    $\begingroup$ You should visit Osamu Iyama and get all these questions resolved! $\endgroup$ Jan 1, 2018 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, 2018 at 0:19
  • 2
    $\begingroup$ I see, so my Mare number is 2! $\endgroup$ Jan 1, 2018 at 0:23

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.