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Diagonal matrices are an abelian group under addition, and with multiplication they become a commutative ring $(\mathcal{D},+, *)$.

More generally, the set of $n \times n$ matrices in $M_n(\mathbb{R})=\mathbb{R}^{n \times n}$ that are simultaneously diagonalized by a given eigenbasis (see Prove that simultaneously diagonalizable matrices commute) will also yield a commutative ring. I believe any such set will be isomorphic to the diagonals, since all elements are of the form $SDS^{-1}$ for fixed $S$ and any diagonal $D$.

My hypothesis: if $\mathcal{R} \subseteq M_n(\mathbb{R})$ forms a commutative ring $(\mathcal{R},+,*)$, then $\mathcal{R} \cong \mathcal{D}$. True or false?

EDIT: False, as scalar matrices $Z(M_n(\mathbb{R}))=kI_n \not \cong \mathcal{D}$. So the hypothesis should be $\mathcal{R} \cong \mathcal{D}$ OR some subring of $\mathcal{D}$. I am considering "rings" to be unital, although rng counterexamples are still interesting.

User JCAA provided an excellent counterexample. For $\alpha, a_i \in \mathbb{R}$, consider upper triangular matrices of the form

$$ \begin{bmatrix} \alpha & a_{2} & \dots & a_{n} \\ 0 & \alpha & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \alpha \\ \end{bmatrix} = \begin{bmatrix} \alpha I_1 & A_{1 \times (n-1)} \\ 0_{(n-1) \times 1} & \alpha I_{n-1} \\ \end{bmatrix} $$

(The right side is block matrix notation.) For distinct matrices in this set $\mathcal{U}$, we have

$$ \begin{bmatrix} \alpha I & A \\ 0 & \alpha I \\ \end{bmatrix} \begin{bmatrix} \beta I & A \\ 0 & \beta I \\ \end{bmatrix} = \begin{bmatrix} \alpha \beta I & \alpha A + \beta A \\ 0 & \alpha \beta I \\ \end{bmatrix} = \begin{bmatrix} \alpha \beta & (\alpha + \beta) a_{2} & \dots & (\alpha + \beta) a_{n} \\ 0 & \alpha \beta & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \alpha \beta \\ \end{bmatrix} $$

So multiplication is closed and commutative (and associative, distributive since "multiplication" is just composition of linear transformations); moreover, $I_n \in \mathcal{U}$ so this becomes a unital ring. Unlike $\mathcal{D} \cong \mathbb{R} \times \dots \times \mathbb{R}$, some $u \in \mathcal{U}$ satisfy $u^2 = 0$.

jskattt797
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The answer is "no". Consider the ring of matrices with first row $(0, x,y,...,z)$ and all other entries 0. This is a ring with zero product, so commutative. It is not isomorphic to a subring of $D$.

markvs
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    This example is not unital. Is there a unital example? – Geoffrey Trang Jun 20 '20 at 00:50
  • Add to this example all scalar matrices. You will obtain a unital example. – markvs Jun 20 '20 at 01:13
  • But

    $$ \begin{bmatrix} 1 & 0 \ 0 & 1 \ \end{bmatrix}

    \begin{bmatrix} 0 & 33 \ 0 & 0 \ \end{bmatrix} = \begin{bmatrix} 1 & 33 \ 0 & 1 \ \end{bmatrix} $$

    So it won't be closed under addition?

     – jskattt797 Jun 20 '20 at 01:57
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    It will be a ring. Let $S$ be the ring of scalar matrices and $R$ be the ring I suggested. Then $S+R={s+r | s\in S,r\in R}$ is a commutative ring of matrices with $I$ which is not isomorphic to any ring of diagonal matrices. That is because any ring of diagonal matrices satisfies $\forall x (x^2=0\implies x=0)$ and $S+T$ does not satisfy this formula. – markvs Jun 20 '20 at 02:14
  • @GeoffreyTrang There are several unital counterexamples mentioned in the comments, right? – rschwieb Jun 20 '20 at 10:30
  • @rscwieb: There are four comments. Which one do you have in mind? – markvs Jun 20 '20 at 15:26
  • @JCAA Thomas Andrews’s first two – rschwieb Jun 21 '20 at 00:25
  • @rschwieb: These comments address the question whether a commutative ring of matrices is isomorphic to $D$. But the real question asks if it embeds into $D$. – markvs Jun 21 '20 at 00:29
  • @JCAA the first one still takes care of that: $\mathbb C$ does not embed in $\mathbb R \times \mathbb R$ – rschwieb Jun 21 '20 at 01:00
  • To prove that $D$ does not contain $\mathbb C$ the first comment is not enough. One needs to show that every subring of $D$ without zero divisers is a subring of the reals. That is true but requires some argument which is not in that comment. – markvs Jun 21 '20 at 01:53
  • @JCAA I don’t see the need for any of that. It is obvious that $x^2+1$ has no solution in $\mathbb R\times \mathbb R$, and this is what most people would think of first, I believe. – rschwieb Jun 21 '20 at 19:39
  • @rscwieb: In any case just the comment is not enough. You need to say something else.In the firsr comment the key is zero divisors. It is an irrelevant observation even for your argument. – markvs Jun 21 '20 at 19:45
  • @JCAA Om I don’t know what your deal is, so ima stop talking. Why would the existence of a simple counterexample in the comments lead to this reaction anyhow... – rschwieb Jun 21 '20 at 22:45
  • @rschwieb: What is "om" and what is "your deal". Why don't you write as a human being? If English is too hard use some other language. – markvs Jun 21 '20 at 22:56