# Every commutative ring of matrices over $\mathbb{R}$ is isomorphic to the diagonals?

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 $$u^2 = 0$$.

• There is a sub ring isomorphic to $\mathbb C,$ and $D$ has zero divisors. Commented Jun 19, 2020 at 23:28
• Also, there are sub-rings of $D$ that are not isomorphic to $D.$ Commented Jun 19, 2020 at 23:30
• Assuming $D$ includes complex values on the diagonal, then I think every commutative sub-ring is isomorphic to a sub-ring of $D.$ Commented Jun 20, 2020 at 0:04
• There are nonzero nilpotent matrices but there aren't any nonzero nilpotent diagonal matrices. Commented Jun 20, 2020 at 0:14

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 $$D$$.
• 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? Commented Jun 20, 2020 at 1:57
• 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. Commented Jun 20, 2020 at 2:14