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



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> Question 1: For which such $Q$ are there only finitely many indecomposable representations over the dual numbers $K[x]/(x^2)$?
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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 $A\_3$.



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



Representation-finite quivers over dual numbers
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: 
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> 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 $n=2$.
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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. 


