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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 $\leq n.$

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  • $\begingroup$ Could you say what the Dummit and Foote exercise is? $\endgroup$
    – Slade
    Dec 29, 2018 at 8:01
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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