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Source of the following Problem:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition).

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 $d \leq n.$

I proved that $K$ is isomorphic to a subfield of the ring $M_n(F)$. I stuck in the next part, though from my previous question this it is clear that whenever $d | n$ it is true. But

Is it true if $d<n$ and $d$ does not divide $n.$

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    $\begingroup$ Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $d\times d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $d\times d$ matrices into upper left corners of $n\times n$ matrices. $\endgroup$ – Jyrki Lahtonen Dec 29 '18 at 10:47
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    $\begingroup$ On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $\ell$, then, when viewed as a vector space over $F$, it will have dimension $d\ell$. So $d\ell=n$ forcing $d$ to be a factor of $n$. $\endgroup$ – Jyrki Lahtonen Dec 29 '18 at 10:52
  • $\begingroup$ Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-) $\endgroup$ – Jyrki Lahtonen Dec 29 '18 at 10:54
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That's simply false, at least if one assumes the identity of $K$ maps to the identity of $M_n(F)$. For an extension $K/F$ of degree $d$, $K$ can only have an $F$-embedding in $M_n(F)$ if $d\mid n$.

Suppose $K$ embeds in $M_n(F)$. Then the space $C$ of column vectors of height $n$ is a left $M_n(F)$-module, and so by restriction of scalars becomes a vector space over $K$. If $\dim_K C=m$, then $n=\dim_K C=md$. Therefore $d\mid n$.

But if one admits non-unital ring homomorphisms, then $M_d(F)$ embeds in $M_n(F)$ for $d\le n$ by $$A\mapsto\pmatrix{A&0\\0&0}$$ so the answer becomes yes if you accept this.

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  • $\begingroup$ Amen :-) ${}{}$ $\endgroup$ – Jyrki Lahtonen Dec 29 '18 at 10:57
  • $\begingroup$ @Lord Shark the Unknown: Can you explain what $C$ is ? $\endgroup$ – user371231 Dec 29 '18 at 15:26
  • $\begingroup$ @user371231 The space of $1\times n$ column vectors over $F$. $\endgroup$ – Lord Shark the Unknown Dec 29 '18 at 15:28
  • $\begingroup$ @Lord Shark the Unknown: $n \times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $\text{dim}_{K}C=m.$ I guess it will be $n=\text{dim}_{F}C$ and considering the tower we get $n=md.$ $\endgroup$ – user371231 Dec 29 '18 at 15:47

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