Given the algebra $A=K[x]/(x^n)$ for some field $K$ and natural number $n \geq 2$ with enveloping algebra $A^e=A \otimes\_K A$.
It is easy to see that the 1. Hochschild cohomology of $A$ is nonzero since $Ext\_{A^e}^{1}(A,A)=\underline{Hom\_{A^e}}(A,\Omega^{2}(A))=\underline{Hom\_{A^e}}(A,A) \neq 0)$, where the 2. equality is by the Auslander-Reiten formula and the third equality uses that $\Omega^{2}(A)=A$ as $A^e$-modules.


Question:
Can one give an explicit non-split short exact sequence of $A^e$-modules:
$0 \rightarrow A \rightarrow K \rightarrow A \rightarrow 0$ and say what $K$ is decomposed into indecomposables?



First Hochschild cohomology of $A=K[x]/(x^n)$
Consider $M=A\oplus A$ as a left $A$-module. Denote an $A$-basis by $e\_1,e\_2$. In order to define the action of $A$ from the right, we just need to give an $A$-linear map $M\to M$ whose $n$-th power is zero. So define $e\_1x = xe\_1$ and $e\_2x = xe\_2 + x^{n-1}e\_1$. This module is not isomorphic with the direct sum $A\oplus A$, because in the direct sum the map $r\mapsto rx - xr$ is zero. 


