# Does there exist a $1-1$ ring homomorphism from $M_d(\mathbb{F})$ to $M_n(\mathbb{F})$ for $d <n$?

Let $$\mathbb{F}$$ be a field and $$d,n$$ be positive integers with $$d Then does there exist a injective ring homomorphism from $$M_d(\mathbb{F})$$ to $$M_n(\mathbb{F})$$ ? (BTW A ring map sends $$1$$ to $$1$$)

I failed to produce a $$1-1$$ ring map. This appears in the process of solving a field theory exercise from Dummit and Foote. For your ref. it is Problem number $$19.(b)$$ in sec. $$13$$ (Second Edition). Any help will be appreciated. Thanks.

Edited Later:Prob. $$19.(b).,$$ Section $$13.2,$$ from Abstract Algebra by Dummit and Foote(Second Edition) is stated below.

Let $$K$$ be an extension of $$F$$ of degree $$n.$$ Prove that $$K$$ is isomorphic to a subfield of the ring $$M_n(F),$$ so $$M_n(F)$$ contains an isomorphic copy of every extension of $$F$$ of degree $$\leq n.$$

• Could you say what the Dummit and Foote exercise is? Dec 29, 2018 at 8:01
• And such a map does exist when $d\mid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form. Dec 29, 2018 at 8:04
• Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote. Dec 29, 2018 at 9:15

As Slade notes in a comment, this is true when $$d\mid n$$, and the homomorphism from $$M_d(\mathbb F)$$ to $$M_n(\mathbb F)$$ can be defined by sending a $$d\times d$$ matrix $$A$$ to a block diagonal matrix with $$n/d$$ copies of $$A$$.

However, it is not true for general $$d$$ and $$n$$. We can see this, for example, with $$\mathbb F=\mathbb C$$, $$d=2$$, $$n=3$$. If we have such an injection, let $$A$$ be the image of $$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and $$B$$ be the image of $$\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$ Then $$A^2=0$$, $$B^2=0$$, and $$(A+B)^2=I$$. We can see (e.g. using the Jordan form) that the only $$3\times 3$$ matrix $$A$$ satisfying $$A\ne0$$ but $$A^2=0$$ is, up to similarity, $$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix},$$ so $$A$$ must have rank $$1$$. Similarly $$B$$ also has rank $$1$$, so their sum $$A+B$$ has rank at most $$2$$. But this contradicts $$(A+B)^2=I$$, which would require $$A+B$$ to be invertible and so have full rank.

This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $$d\not\mid n$$, although I expect that it is so.

• Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this. Dec 29, 2018 at 9:20
• If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question. Dec 29, 2018 at 9:22
• Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $\le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that. Dec 29, 2018 at 9:32
• Ok then I'll ask a separate question, thanks for the nice counter example. Dec 29, 2018 at 9:38