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Given a positive integer $n$, how to classify $n$-dimensional basic $K$-algebras?, where $K$ is algebraically closed.

For $n=3$, Let $A=\left[ \begin{array}{ccc} K &0& 0\\ 0& K& 0\\ 0 &0& K \\ \end{array} \right] ,B=\left[ \begin{array}{cc} K & 0 \\ K & K\\ \end{array} \right]$ and $ C=\left[ \begin{array}{ccc} K &0& 0\\ K&0& 0\\ K &0& 0\\ \end{array} \right] $. Then we have $\operatorname{rad}A=\left[ \begin{array}{ccc} 0 &0& 0\\ 0& 0& 0\\ 0 &0& 0 \\ \end{array} \right] ,\operatorname{rad}B=\left[ \begin{array}{cc} 0 & 0 \\ K & 0\\ \end{array} \right]$ and $ \operatorname{rad}C=\left[ \begin{array}{ccc} 0 &0& 0\\ K&0& 0\\ K &0& 0\\ \end{array} \right] $, and hence $A/\operatorname{rad}A\cong K\times K\times K, B/\operatorname{rad}B\cong K\times K, C/\operatorname{rad}C\cong K$, this implies that $A,B$ and $C$ are basic three-dimensional algebras. Let $ Q$ is the quiver

$\circlearrowright^{\beta} $ and $\mathcal{I}$ is the ideal of$ KQ$ generated by one zero relation $ \beta^3$. Then $D=KQ/I$ is basic, Are there other three dimensional basic algebra which are not isormorphic to the above? It is known that every bound quver algebra is basic, conversely, Is every $n$-dimensional basic $K$-algebra isomorphic to a bound quiver algebra?

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    @AiminXu Oh, sorry I meant $C/\operatorname{rad} C\cong K$, $\operatorname{rad} C/\operatorname{rad}^2 C\cong K$ and $\operatorname{rad}^2 C\cong K$, $\operatorname{rad}^3 C=0$.2012-12-06

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Is every n-dimensional basic K-algebra isomorphic to a bound quiver algebra?

Yes, over algebraically closed field every basic algebra isomorphic to a quiver algebra with relations. You can find proof of this fact in the book Auslander, Reiten, Smalo Representation theory of Artin algebras p.65.

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    Is every finite dimensional basic $K$-algebra connected?2012-12-09