Let $A$ be a local selfinjective algebra with indecomposable module $M$. Let $N=A \oplus M$.

When there is an indecomposable module $U$ not in $add(N)$, having finite $add(N)$-resolution for some choice of $A$ and $M$?

This is not possible in general due to the following examples:

  • $A=K[x]/(x^n)$ for arbitary $M$

  • $A$ arbitrary and $M$ simple.


1 Answer 1


I think that this is possible if and only if $\operatorname{Ext}^1_A(M,M)=0$, and so the question reduces to another question Ext^1 for a local finite dimensional selfinjective algebra that you have asked (and which I don't know the answer to).

If $\operatorname{Ext}^1(M,M)=0$ then take $U=\Omega^{-1}M$.

Conversely, if there is such a $U$, then by truncating the $\text{add}(N)$-resolution (and changing $U$) we can assume the resolution $0\to N_1\to N_0\to U\to 0$ has length $2$ and is minimal.

By minimality, $N_1$ is a direct sum of copies of $M$. By the long exact sequence of $\text{Hom}(M,-)$, the map $\text{Ext}^1(M,N_1)\to\text{Ext}^1(M,N_0)$ is injective.

So, removing the free summands from $N_0$, we have a map $\alpha:N_1\to N'_0$ between direct sums of copies of $M$, not a split injection on any summand of $N_1$, that becomes injective upon applying $\text{Ext}^1(M,-)$. But since all components of $\alpha$ are in the radical of $\text{End}(M)$, this is not possible unless $\text{Ext}^1(M,N_1)=0$.

More generally, if $\alpha: M^m\to M^n$ ($m>0$) is a map between direct sums of copies of $M$ all of whose components are in the radical of $\text{End}(M)$, and $F$ is an additive functor such that $F(\alpha)$ is injective, then $F(M)=0$.

To see this, let $E=\text{End}(M)$, and suppose $\beta: M\to M$ is in the radical of $E$. If $F(M)\neq0$ then choose $x\neq0$ in $\text{rad}^s(E)F(M)$ where $s$ is maximal subject to $\text{rad}^s(E)F(M)\neq0$. Then $F(\beta)(x)=\beta x\in\text{rad}^{s+1}(E)F(M)=0$. Therefore the element of $F(M^n)=F(M)^n$ with all components equal to $x$ is a nonzero element of $\ker(\alpha)$.

[This is the sillier proof of the last claim that I originally posted:

Without loss of generality, $n$ is a multiple of $m$, as otherwise we could add extra summands to $M^n$ until it is. Say $n=dm$. Splitting $M^{dm}$ into $d$ summands, and mapping each to $M^{dm}$ by $\alpha$, we get a map $M^{dm}\to M^{d^2m}$. Continuing in this way, we get a sequence of maps $$M^m\to M^{dm}\to M^{d^2m}\to\dots\to M^{d^km},$$ all of which become injective upon applying $F$. But for $k$ greater than the Loewy length of $\text{End}(M)$ the composition of this sequence of maps is zero, so $F(M^m)=0$.]

  • $\begingroup$ thanks, can you explain the last step a little more? What is the conenction between Ext^1 and the radical there? $\endgroup$
    – Mare
    Dec 22, 2016 at 19:41
  • $\begingroup$ @Mare That part is nothing specific to $\text{Ext}^1$. I'll add more explanation. $\endgroup$ Dec 23, 2016 at 10:16
  • $\begingroup$ Thanks, thats some weird looking argument. That generalises a result I have found. Ill write you an email about that. $\endgroup$
    – Mare
    Dec 23, 2016 at 10:38
  • $\begingroup$ @Mare Probably the proof I gave makes it look weirder than it really is: I was replying to your comment in a hurry, so was content with any proof that worked. The picture in my head, though, is that $F(M)$ is a module for $E=\text{End}(M)$ and that $F(\alpha)$ maps into the radical (as an $E$-module) of $F(M^n)$. If you fill in the gaps in that argument, I think you'll get something less weird. $\endgroup$ Dec 23, 2016 at 16:28
  • 1
    $\begingroup$ @Mare I've included a more sensible proof in my answer. $\endgroup$ Dec 24, 2016 at 9:04

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