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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?

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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.

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  • $\begingroup$ Thanks, did you guess this or can one generalise (or give a method) to a more general situation (like commtuative finite dimensional local algebra)? $\endgroup$
    – Mare
    Commented Jul 25, 2017 at 15:34
  • $\begingroup$ This one I guessed. The same argument will work quite in general though, but if the description by generators and relations is more complicated then it becomes more difficult indeed. Another option is to really write down a resolution and get an extension out of it. $\endgroup$
    – Ehud Meir
    Commented Jul 25, 2017 at 15:45
  • $\begingroup$ The thing is that I can prove $Ext^{1}_{A^e}(A,A) \neq 0$ for local commutative (nonsemisimple) algebras, but I cant really write down a nice explicit short exact sequence. $\endgroup$
    – Mare
    Commented Jul 25, 2017 at 15:48
  • $\begingroup$ you can do the following thing: You do have the following short exact sequence $$0\to K\to A\otimes A\to A\to 0$$ where the second arrow is given by multiplication, and $K$ is simply the kernel, generated by $a\otimes 1 - 1\otimes a$. Now you just need to find a homomorphism of $A$-bimodules from $K$ to $A$, which cannot be extended to $A\otimes A$, and take the pushout. $\endgroup$
    – Ehud Meir
    Commented Jul 25, 2017 at 15:50

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