Let $A$ be a Nakayama algebra, that is an Artin algebra such that any indecomposable module has a unique composition series. The easiest examples of such algebras are $K[x]/(x^n)$.



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> Question 1: In case $Ext^1(X,Y) \neq 0$ for two indecomposable $A$-modules $X,Y$, can one write down an explicit nice non-split short exact sequence $0 \rightarrow Y \rightarrow Z \rightarrow X \rightarrow 0$, where $Z$ is constructed in some canonical way out of $X$ and $Y$?
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> Question 2: Can such an $Z$ as above have at most 2 indecomposble summands?
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Ext in Nakayama algebras
I'll answer Question 2 first. The answer is yes.


If $Z$ has more than two indecomposable summands, then $\text{soc}(Z)$ has at least three summands, but $X$ has simple socle since it's uniserial, so $Y$, the kernel of $Z\to X$, has at least two summands. But $Y$ is also uniserial, and so has simple socle.


For Question 1:


If $Z$ is indecomposable (uniserial), then it's easy to see what the exact sequence must be, and when it exists, so suppose $Z=Z\_1\oplus Z\_2$ is a direct sum of two indecomposables.


An indecomposable is determined by the list of its composition factors from head to socle, which can be represented by a word $w$ in a set indexing the isomorphism classes of simples. Denote the module corresponding to a word $w$ by $[w]$, if there is such a module.


Since $Y$ has simple socle, its map to one of the $Z\_i$, say $Z\_1$, must be injective, but not surjective since the exact sequence doesn't split. But then its map to $Z\_2$ must be surjective, or else $X$, the cokernel of $X\to Z$, has nonsimple head.


So if $X=[w]$ and $Y=[w']$ then the extensions all look like the following, with the obvious maps.


If $[ww']$ is a module, then we can take $Z=[ww']$


If $w=uv$ and $w'=vt$, where $[uvt]$ is a module, then $Z$ could be $[uvt]\oplus[v]$.


There could be more than one choice that works. 


