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I initially thought they were all isomorphic to some subring of the $n \times n$ diagonal matrices $\mathcal{D} \cong \mathbb{R} \times \dots \times \mathbb{R}$, but this was wrong: Every commutative ring of matrices over $\mathbb{R}$ is isomorphic to the diagonals?. One counterexample is matrices of the form (using block matrix notation) $\begin{bmatrix} \alpha I_1 & A \\ 0 & \alpha I_{n-1} \\ \end{bmatrix}$ for some $1 \times (n-1)$ real matrix block $A$ and some $\alpha \in \mathbb{R}$, which forms a commutative ring $(\mathcal{U}, +, *)$.

Are there other counterexamples? Can we classify all such rings up to isomorphism?

I use "ring" to mean "unital ring," but a similar classification for rngs would also be interesting.

From Unital rings within matrices, it seems that matrices in $M_2(\mathbb{R})$ of the form

$\begin{bmatrix} a & b \\ -b & a-b \\ \end{bmatrix}$

will be another example?

EDIT: If we require the commutative subring of $M_n(\mathbb{R})$ be closed under scalar multiplication, then it is a commutative $\mathbb{R}$-algebra.

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    $\begingroup$ Do you require them to be closed under multiplication by scalars? If not then you can get many examples by restricting the coefficients e.g. to the rationals or integers. $\endgroup$
    – badjohn
    Commented Jun 20, 2020 at 6:44
  • $\begingroup$ Yes, closed under scalar multiplication. $\endgroup$
    – jskattt797
    Commented Jun 20, 2020 at 6:48
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    $\begingroup$ @jskattt797 Then you should change the question to commutative algebras, not commutative rings. $\endgroup$
    – Anonymous
    Commented Jun 20, 2020 at 8:14

1 Answer 1

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This would be equivalent to classifying all commutative $\mathbb R$ algebras of dimension $n$. It’s a basic fact that every $n$ dimensional $\mathbb R$ algebra is isomorphic to a subring of $M_n(\mathbb R)$.

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  • $\begingroup$ Does this revelation give us any new information? If I understood correctly, the goal was already to classify all commutative matrix $\mathbb{R}$-algebras. $\endgroup$
    – Anonymous
    Commented Jun 20, 2020 at 11:17
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    $\begingroup$ @Anonymous the revelation is that the class being classified is way too big. No distinguishing features to speak of, so hard to classify. $\endgroup$
    – rschwieb
    Commented Jun 20, 2020 at 11:42
  • $\begingroup$ We are trying to classify all commutative $n$-dimensional $\mathbb{R}$-algebras in $M_n(\mathbb{R})$ up to isomorphism. But all $n$-dimensional $\mathbb{R}$-algebras will be isomorphic to such an algebra, so there are at least as many commutative matrix $\mathbb{R}$-algebra isomorphism classes as there are commutative $\mathbb{R}$-algebra isomorphism classes in general (and at most as many). Is my understanding correct? $\endgroup$
    – jskattt797
    Commented Jun 20, 2020 at 18:30
  • $\begingroup$ @jskattt797 finite dimensional algebras, yes $\endgroup$
    – rschwieb
    Commented Jun 20, 2020 at 19:10

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