# $M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $d \leq n.$

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 $$d and $$d$$ does not divide $$n.$$

• 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. – Jyrki Lahtonen Dec 29 '18 at 10:47
• 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 $n$. – Jyrki Lahtonen Dec 29 '18 at 10:52
• 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 :-) – Jyrki Lahtonen Dec 29 '18 at 10:54

## 1 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.

• Amen :-) ${}{}$ – Jyrki Lahtonen Dec 29 '18 at 10:57
• @Lord Shark the Unknown: Can you explain what $C$ is ? – user371231 Dec 29 '18 at 15:26
• @user371231 The space of $1\times n$ column vectors over $F$. – Lord Shark the Unknown Dec 29 '18 at 15:28
• @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 $n=md.$ – user371231 Dec 29 '18 at 15:47