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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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    http://en.wikipedia.org/wiki/Algebra_over_a_field#Structure_coefficients2012-11-29
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    In principle you could write down all quivers with number of vertices + number of arrows $\leq n$. Then try to write down all relations making it an $n$-dimensional algebra. This could be done with some effort, as long as $n$ is not too big. But the reallly hard task is to decide whether two such algebras are isomorphic. I doubt there is an easy receipe unless $n$ is really small, maybe $\leq 10$.2012-11-29
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    What is a “basic” algebra?2012-12-03
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    Assume that $A$ is a $K$-algebra with a complete set ${e_1,\cdots , e_n}$ of primitive orthogonal idempotents. The algebra $A$ is called basic if $e_iA \ncong e_jA$, for all $i\neq j$.2012-12-03
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    @AiminXu There is one more $3$-dimensional algebra: There are two algebras with $C/\operatorname{rad} C\cong K$, one has $\operatorname{rad}^2 C=0$, another one has $\operatorname{rad}^2 C\cong \operatorname{rad} C\cong K$.2012-12-04
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    If a finite dimensional algebra $C$ with $C/radC\cong K$ and $rad^2C\cong radC\cong K$, then $C$ is two-dimensional.2012-12-05
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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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