Classification of 4 dimensional real associative unital algebra

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?

• 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. Sep 10, 2023 at 21:03
• Oh, and $\mathbb{R}[x, y]/(x^2, y^2)$. Nilpotents can do a lot of stuff! Sep 10, 2023 at 21:10
• 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? Sep 11, 2023 at 3:11