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Given a Dynkin quiver $Q$ and a field $K$.

Question 1: For which such $Q$ are there only finitely many indecomposable representations over the dual numbers $K[x]/(x^2)$?

Note that those representations are exactly those of $KQ \otimes_K K[x]/(x^2)$. This is for example true for $Q$ being of type $A_1, A_2$ or $A_3$.

Question 2: For which combinations of $Q$ and $n$ is $KQ \otimes_K K[x]/(x^n)$ representation-finite?

I think question two has an easy answer for $n \geq 3$, namely only when $Q$ is of type $A_1$ for arbitrary $n$ or $A_2$ for $n=3$, but is there an elementary argument?

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    $\begingroup$ What is the relationship between $k$ and $K$? $\endgroup$ – LSpice Jul 10 at 16:51
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    $\begingroup$ @LSpice Thanks, they are the same. I corrected it. $\endgroup$ – Mare Jul 10 at 16:52
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You can find this result in [Geiss, Leclerc, Schröer: Quivers with relations for symmetrizable Cartan matrices I: Foundations] as Proposition 13.1. They consider a more general class of algebras. The class of algebras you are considering is obtained by setting their parameters $(c_1,\dots,c_m)$ to all the same number, which you call $n$.

Translating to your setup, their result reads:

The algebra $KQ\otimes_K K[x]/(x^n)$ for $Q$ a Dynkin quiver is representation-finite if and only if you are in one of the following cases:

  1. $n=1$ and $Q$ arbitrary, in which case you just recover $KQ$.
  2. $Q=A_1$, the quiver with only one vertex, in this case $n$ is arbitrary.
  3. $Q=A_2$ and $n=2$ or $n=3$.
  4. $Q=A_3$ and $n=2$.

The proof they give (unfortunately) uses the Bongartz-Gabriel and the Happel-Vossieck list. The question for a more elementary argument thus remains open. For this one should try to construct an infinite family of modules.

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