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For complex, dual and split-complex numbers there are matrix representations: $$a+b \cdot i \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\ \end{pmatrix} +b\begin{pmatrix}0 & -1\\1 & 0 \\ \end{pmatrix}, i^2=-1,$$ $$a+b \cdot \epsilon \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\ \end{pmatrix} +b\begin{pmatrix}0 & 1\\0 & 0 \\ \end{pmatrix}, \epsilon^2=0,$$ $$a+b \cdot j \equiv a\begin{pmatrix} 1 & 0\\0 & 1 \\ \end{pmatrix} +b\begin{pmatrix}0 & 1\\1 & 0 \\ \end{pmatrix}, j^2=1,$$ and I know (as a fact learned 20 years ago, not quite familiar with it) too, that these sets are isomorphic to $\Bbb R[X]/(X^2+1)$, $\Bbb R[X]/(X^2)$ and $\Bbb R[X]/(X^2-1)$. While the quotient ring contructions "looks quite complete", I wonder:
If one varies the "imaginary matrices", e.g. if one considers $\begin{pmatrix}0 & 0\\1 & 0 \\ \end{pmatrix}$ , are then the sets one gets in any way "special", "meaningful", well-known or named?

Sascha
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1 Answers1

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In general, any matrix $B$ that you choose as the imaginary unit will have a minimal polynomial $p(x) $ such that $p(B) =0$. Any first or second degree minimal polynomial is possible. The resulting ring is $\mathbb R[x] /(p(x)) $. You can get rings that are not isomorphic to any of the three you mention this way, for example if $p(x) =(x-1)(x-2)$ and $$B=\left(\begin{array} {cc} 1&0\\0&2\end{array}\right)$$

Matt Samuel
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