Let $\mathbb{F}$ be a field and $d,n$ be positive integers with $d <n.$ 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 $\leq n.$

  • $\begingroup$ Could you say what the Dummit and Foote exercise is? $\endgroup$
    – Slade
    Dec 29, 2018 at 8:01
  • 1
    $\begingroup$ And such a map does exist when $d\mid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form. $\endgroup$
    – Slade
    Dec 29, 2018 at 8:04
  • $\begingroup$ Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote. $\endgroup$
    – user371231
    Dec 29, 2018 at 9:15

1 Answer 1


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.

  • $\begingroup$ Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this. $\endgroup$
    – user371231
    Dec 29, 2018 at 9:20
  • $\begingroup$ If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question. $\endgroup$
    – user371231
    Dec 29, 2018 at 9:22
  • 1
    $\begingroup$ 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. $\endgroup$
    – Carmeister
    Dec 29, 2018 at 9:32
  • $\begingroup$ Ok then I'll ask a separate question, thanks for the nice counter example. $\endgroup$
    – user371231
    Dec 29, 2018 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.