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I got this question from a problem book of Ring Theory, the question is:

Let $\mathbb{F}$ be a field. If $m>n$ are positive integers, prove that there exists no ring Homomorphism from $M_m(\mathbb{F})$ to $M_n(\mathbb{F})$.

My approach: Firsly let's take $\phi:M_m(\mathbb{F}) \to M_n(\mathbb{F})$ be a Ring Homomorphism, then for Ker$\phi$, there're only two possibility.Either {$0$} or $M_m(\mathbb{F})$ since there're only two ideals. But it cannot be $M_m(\mathbb{F})$ since we define Ring Homomorphism which sends unity to unity.

So $\phi$ must be injective.

Now to get the contradiction, I know we've to take a nilpotent matrix of nilpotency $m$ and have to look at the image of it (which also will be a nilpotent). But how to construct such matrix and what is the image precisely?

Any help will be appreciated.

ZetaF_2004
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2 Answers2

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Since $n < m$, there is a nilpotent $m \times m$-matrix $A$ with $A^{m-1} \neq 0$ and hence $A^n \neq 0$. For example take the matrix corresponding to the linear map $K^m \to K^m$ which sends $e_i \mapsto e_{i-1}$ for $i > 1$ resp. $\mapsto 0$ otherwise. (You can also write this down as a matrix, but computation of the powers is considerably more easy when working with linear maps.) Since $\varphi$ is an injective ring homomorphism, the image $B := \varphi(A)$ is a nilpotent $n \times n$-matrix with $B^n \neq 0$. But any nilpotent $n \times n$-matrix satisfies $B^n=0$.

Martin Brandenburg
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More generally, if a ring homomorphism from $M_m(\mathbb{F})$ to $M_n(\mathbb{F})$ which maps identity to identity exists, then $m$ necessarily divides $n$. In particular, $m\leq n$. For a proof of this result, see here. Briefly, using the linearity condition, we have $$I_n=f(I_m)=f\left(\sum_{i=1}^m E_{ii}\right)=\sum_{i=1}^m f(E_{ii})$$ and so $\mathbb{F}^n=\sum_{i=1}^m R(f(E_{ii}))$, where $R$ means the range. With a bit more work using the ring homomorphism condition and the fact that $f(E_{ii})$ is an idempotent, one can show that $R(f(E_{ii}))\cap R(f(E_{jj}))=0$ for $i\neq j$ (so that we actually have a direct sum decomposition $\displaystyle\mathbb{F}^n=\bigoplus_{i=1}^m R(f(E_{ii}))$) and that $\text{rank }f(E_{ii})$ are all equal for all $1\leq i\leq m$. Then we have $$n=m\cdot\text{rank }f(E_{11})$$ and it follows that $m$ divides $n$.

Alex Fok
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