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Suppose $k$ is a field and $A$ is a $k$-algebra of dimension no larger than 3. If $A$ is semi-simple, then $A$ can be written as a direct sum of simple $k$-algebras. Further one can find $A$ is commutative by exhausting all the cases.

Without the semi-simple condition, what can we say about $A$? Is it still commutative?

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The upper triangular $2\times 2$ matrix ring over a field is three dimensional and noncommutative.

Here's what I mean, if you're not familiar with it:

$\{\begin{bmatrix}a&b\\0&c\end{bmatrix}\mid a,b,c\in\mathbb{F}\}$

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    In this case, there is an ideal of matrices with diagonal zeros. Thus the algebra is non commutative, right?2012-08-22
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    @Honghao I don't see how having such an ideal makes an algebra noncommutative, but yes, it's noncommutative. In any case $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ are noncommuting.2012-08-22
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    Sorry for the terrible phrasing. I would have been saying it is a precise example of non-commutative not semi-simple algebra. Thank you for the quick answer.2012-08-22
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    @Honghao Ah! Yes then you're exactly correct. The subset of matrices which are zero on the diagonal form a nonzero nilpotent ideal.2012-08-22