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 $n.$

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.