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Let $\operatorname{Mat}(n,\mathbb{K})$ be the ring of $n\times n$ matrices with entries in a field $\mathbb{K}$. Is $\operatorname{Mat}(2,\mathbb{R})$ isomorphic to $\operatorname{Mat}(3,\mathbb{R})$?

I want to know whether there is ring isomorphism between the two. I cannot seem to find any bijection that will do. Thanks.

An Idea.

Let $f:\text{Mat}(2,\mathbb{R})\to\text{Mat}(3,\mathbb{R})$ be a ring homomorphism. Then, $\ker f$ is a two-sided ideal of $\text{Mat}(2,\mathbb{R})$. It is known that $\text{Mat}(2,\mathbb{R})$ is a simple ring. Therefore, either $\ker f=0$ or $\ker f=\text{Mat}(2,\mathbb{R})$. In the case $\ker f=\text{Mat}(2,\mathbb{R})$, it follows that $f=0$, so it is not an isomorphism. If $\ker f=0$, then $f$ is an embedding. This is where the argument stops. There are many ways to embed $\text{Mat}(2,\mathbb{R})$ into $\text{Mat}(3,\mathbb{R})$. But is there an embedding that is a surjection, i.e., it is an isomorphism?

Batominovski
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    Does e.g. $Mat(2,\mathbb{R})$ mean 2 by 2 matrices with real coefficients? Also by isomorphic do you mean as rings? as vector spaces? – coffeemath May 02 '16 at 22:29
  • "I cannot seem to find any bijection that will do" suggests you believe the answer is "the two are isomorphic". Does that seem reasonable intuitively...? – Andrew D. Hwang May 03 '16 at 01:38

1 Answers1

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There are no nilpotent elements of order $3$ in $\text{Mat}_{2\times2}(\mathbb{R})$, while there are many of them in $\text{Mat}_{3\times3}(\mathbb{R})$. Hence, the two cannot be isomorphic as rings.

In general, for two positive integers $m$ and $n$, and for any unital (associative) ring $R$, the rings of matrices $\text{Mat}_{m\times m}(R)$ and $\text{Mat}_{n\times n}(R)$ are isomorphic as rings if and only if $m=n$. If the assumption that the ring $R$ is unital is dropped, then it is possible that $\text{Mat}_{m\times m}(R)$ and $\text{Mat}_{n\times n}(R)$ are isomorphic even if $m\neq n$. For example, by taking $R=\bigoplus\limits_{i=1}^\infty\,\mathbb{Z}$ with trivial multiplication (i.e., $x\cdot y=0$ for all $x,y\in R$), we have $$\text{Mat}_{m\times m}(R)\cong R\cong\text{Mat}_{n\times n}(R)$$ for all positive integers $m$ and $n$.

On the other hand, if there are two unital rings $R$ and $S$, then it is possible that $\text{Mat}_{m\times m}(R)$ and $\text{Mat}_{n\times n}(S)$ are isomorphic. An easy example, is when $S=\text{Mat}_{k\times k}(R)$. Then, $$\text{Mat}_{(nk)\times (nk)}(R)\cong \text{Mat}_{n\times n}(S)\,.$$

Batominovski
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