# Subrings of matrix rings over a field

Let $M_n(F)$ be the matrix ring over a field $F$. Suppose there is a subring $R$ of $M_n(F)$ which is isomorphic to $M_m(F)$ for some $m$. I am trying to show that $m$ divides $n$.

I tried looking at various actions of $R$ on vector spaces, and using simplicity of matrix rings, but didn't really get anywhere. Does anyone know how to prove this?

• Just to check: the definition of "ring" you're using includes a multiplicative unit $1$, so that subrings must have the same multiplicative unit as the enclosing ring?
– user14972
Commented Jan 17, 2013 at 5:42
• Right, ring here means unital ring. Commented Jan 17, 2013 at 13:52
• I think this has to do with the fact that $M_n(F)$ contains copies all field extensions $E/F$ of degree $n$. But I can't see the details right now.
– lhf
Commented Dec 20, 2017 at 23:43
• See chapter 4 of Jacobson's Basic Algebra II. Consider the $M_n(F)$-module $F^n$ of column vectors. $M_m(F)$ is a simple ring so $F^n$ decomposes into a direct sum of simple $M_m(F)$ modules. But those are all isomorphic to $F^m$. The claim follows. Commented Dec 22, 2017 at 11:47

Here is a solution for $F=\mathbb Q$.
Let $f$ be an irreducible polynomial of degree $m$. For instance, $f(x)=x^m-2$.
Let $A$ be the companion matrix of $f$. Then $f$ is the minimal polynomial of $A$ in $M_m(F)$ and so also its image $\tilde{A}$ in $R \subset M_n(F)$.
The characteristic polynomial of $\tilde{A}$ in $M_n(F)$ has the same irreducible factors as the minimal polynomial of $\tilde{A}$ and $A$. Since the minimal polynomial $f$ is irreducible in our case, the characteristic polynomial must be a power of $f$. Now compare degrees.
• I think your answer is hard to read. Let the ring isomorphism $\phi : M_m(\mathbb{Q})\to R \subset M_n(\mathbb{Q})$. For $A \in M_m(\mathbb{Q})$ such that $\text{charpoly}(A)=\text{minpoly}(A)$ is irreducible of degree $m$, then $\text{minpoly}(A) = \text{minpoly}(\phi(A))$. The latter divides and has the same roots as $\text{charpoly}(\phi(A))$, thus $\text{charpoly}(\phi(A))= \text{minpoly}(\phi(A))^d$ where $d= n/m$. Thanks to the companion matrix, the same works for any field such that $F[x]$ has some irreducible polynomials of degree $m$. Commented Dec 22, 2017 at 11:22