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Given a representation-finite (finite dimensional over a field) quiver algebra of finite global dimension. Is $eAe$ isomorphic to the field for at least one primitive idempotent $e$? This is true for Nakayama algebras (and trivially for acyclic algebras).

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Edit: When writing this, I missed the condition that the algebra should be representation-finite. The algebra below is not.

Not necessarily. Consider the path algebra of the quiver with two vertices and three arrows $$ 1 \substack{\xrightarrow{a} \\ \xleftarrow[b,c]{\xleftarrow{}} } 2 $$ modulo the relations $ab=0$ and $ca=0$ (composing arrows from left to right). This algebra $A$ has dimension $8$ over the base field, and its global dimension $3$, since the projective resolutions of the simple modules have the form $$ 0\to P_1\to P_2\to P_1\to S_1 \to 0 $$ and $$ 0\to P_1 \to P_2 \to P_1\oplus P_1 \to P_2 \to S_2 \to 0, $$ where $P_i$ is the projective cover of the simple module $S_i$.

However, we see that $e_1Ae_1 \cong k[x]/(x^2) \cong e_2Ae_2$.

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  • $\begingroup$ Thanks, Ill check that with the computer now. However in the question I asked for representation-finite algebra. But your example is interesting so it would be good to leave it. $\endgroup$ – Mare Sep 1 '17 at 8:25
  • $\begingroup$ Oh, sorry, I missed the representation-finite condition. I'll add a comment. $\endgroup$ – Pierre-Guy Plamondon Sep 1 '17 at 8:25

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