I think I have a complete list for all the commutative ones, maybe with possible repeats (I did try my best to make sure none are same up to isomorphism):

  1. $\mathbb{R}^4 \simeq \begin{bmatrix}a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&d\end{bmatrix}$

  2. $\mathbb{R}^2 \times \mathbb{C} \simeq \begin{bmatrix}a&0&0&0\\0&b&0&0\\0&0&c&-d\\0&0&d&c\end{bmatrix}$

  3. $\mathbb{R}^2 \times \mathbb{R}[x]/(x^2) \simeq \begin{bmatrix}a&0&0&0\\0&b&0&0\\0&0&c&d\\0&0&0&c\end{bmatrix}$

  4. $\mathbb{R} \times \mathbb{R}[x]/(x^3) \simeq \begin{bmatrix}a&0&0&0\\0&b&c&d\\0&0&b&c\\0&0&0&b\end{bmatrix}$

  5. $\mathbb{C}^2 \simeq \begin{bmatrix}a&-b&0&0\\b&a&0&0\\0&0&c&-d\\0&0&d&c\end{bmatrix}$

  6. $\mathbb{C} \times \mathbb{R}[x]/(x^2) \simeq \begin{bmatrix}a&-b&0&0\\b&a&0&0\\0&0&c&d\\0&0&0&c\end{bmatrix}$

  7. $\mathbb{R}[x]/(x^4) \simeq \begin{bmatrix}a&b&c&d\\0&a&b&c\\0&0&a&b\\0&0&0&d\end{bmatrix}$

  8. $\mathbb{R}[x,y]/(x^2,y^2+1) \simeq \mathbb{C}[x]/(x^2) \simeq \begin{bmatrix}a&-b&c&-d\\b&a&d&c\\0&0&a&-b\\0&0&b&a\end{bmatrix}$

and here are what I think are noncommutative ones:

  1. $\mathbb{H} \simeq \mathbb{R}\langle x,y\rangle/(x^2+1,y^2+1,xy+yx)$

  2. $\mathbb{R}\langle x,y\rangle/(x^2+1,y^2-1,xy+yx) \simeq \mathbb{R}\langle x,y\rangle/(x^2-1,y^2-1,xy+yx)$

  3. $\mathbb{R}\langle x,y\rangle/(x^2-1,y^2,xy+yx)$

  4. $\mathbb{R}\langle x,y\rangle/(x^2+1,y^2,xy+yx)$

  5. $M_2(\mathbb{R})$

I am wondering is this is the complete list, I did try to make sure there are no repeats up to isomorphism, but there is no guarantee. And maybe the list isn't complete, maybe there is a subalgebra of $M_3(\mathbb{R})$ thats not on this list?

  • 2
    $\begingroup$ In the commutative list you're missing at least $\mathbb{R}[x]/x^2 \times \mathbb{R}[y]/y^2$ as well as $\mathbb{R}[x, y, z]/(x^2, xy, y^2, yz, z^2, zx)$. Generally speaking the difficulty with this sort of thing is getting a handle on what nilpotents can do. $\endgroup$ Sep 10, 2023 at 21:03
  • 2
    $\begingroup$ Oh, and $\mathbb{R}[x, y]/(x^2, y^2)$. Nilpotents can do a lot of stuff! $\endgroup$ Sep 10, 2023 at 21:10
  • $\begingroup$ I forgot about $\mathbb{R}[x]/(x^2)$ but wow I didn't think about how $i^2=j^2=k^2=ij=jk=ki = 0$ could be a solution as well! I wonder if I should repost this in mathoverflow? $\endgroup$
    – Leon Kim
    Sep 11, 2023 at 3:11


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