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Dec 14, 2019 at 18:58 comment added Fernando Muro Mare, for me the quiver algebra is actually the path algebra. Now that you added a condition on the ideal $I$ my previous answer doesn't work.
Dec 14, 2019 at 18:55 history edited Mare CC BY-SA 4.0
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Dec 14, 2019 at 18:50 comment added Mare @FernandoMuro For me in the definition of quiver algebra, I is an admissible ideal.
Dec 14, 2019 at 18:21 comment added Fernando Muro If you take $Q$ to be a loop and $K=\mathbb{Q}$ then $KQ=\mathbb{Q}[x]$. Among its quotients you find all finite field extensions of $\mathbb{Q}$. There are infinitely many of those and all of them have finite representation type and global dimension $0$.
Dec 14, 2019 at 11:47 history asked Mare CC BY-SA 4.0