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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?

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    $\begingroup$ 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? $\endgroup$
    – user14972
    Commented Jan 17, 2013 at 5:42
  • $\begingroup$ Right, ring here means unital ring. $\endgroup$
    – user15464
    Commented Jan 17, 2013 at 13:52
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    $\begingroup$ 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. $\endgroup$
    – lhf
    Commented Dec 20, 2017 at 23:43
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    $\begingroup$ 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. $\endgroup$ Commented Dec 22, 2017 at 11:47

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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.

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  • $\begingroup$ 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 $m$. $\endgroup$
    – reuns
    Commented Dec 22, 2017 at 11:22

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