TITLE: Maximal numbers of summands in middle terms of short exact sequences
QUESTION [8 upvotes]: Let $A$ be a finite dimensional algebra and $M$ and $N$ indecomposable $A$-modules. Denote by $\xi(M,N)$ the maximal number of indecomposable summands of a modules $X$ such that there is a non-split short exact sequence $0 \rightarrow N \rightarrow X \rightarrow M \rightarrow 0$.

Question 1: Is there an easy example of $A$ such that $\sup \{ \xi(M,N) | M, N $ indecomposable $\}$ is infinite?
Question 2: Is there an easy example of $A$ such that $\sup \{ \xi(M,\tau(M)) | M $ indecomposable $\}$ is infinite?

(here $\tau$ denotes the Auslander-Reiten translate. One might also ask for an algebra where the number of middle term indecomposable summands of Auslander-Reiten sequences can be arbitrarily large)

Question 3: Do we have $\sup \{ \xi(M,N) | M, N $ indecomposable $\}=\sup \{ \xi(M,\tau(M)) | M $ indecomposable $\}$?

Bonusquestion: Can $\xi(M,N)$ be calculated with the GAP-package QPA?

REPLY [5 votes]: For question 1:
For a self-injective algebra, if $0\to X\to Y\to Z\to0$ is a short exact sequence with $Y$ and $Z$ indecomposable, then there is a short exact sequence $0\to\Omega Z\to X\oplus P\to Y\to0$, for some projective module $P$.
It's easy to come up with examples for, say, $k[x,y]/(x^3,y^3)$ where $X=\text{soc }Y$ has arbitrarily many summands (and $Y/X$ is indecomposable).