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 $n.$
Best Answer
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.