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The question is as follows:

A finite-dimensional central division $\mathbb K$-algebra $D$ is a $\mathbb K$-algebra isomorphic to a subalgebra of $M_r(\mathbb K)$ if and only if $\dim_{\mathbb K} D \mid r$.

In what comes, we already know that the other direction is true by a beautiful answer from @Algeboy.

Also, I understood that the comment by @Jyrki give a very nice and reasonable way for to show the statement. But I don't know how to use it.

For to prove this statement, I think for to show the statement, I think we need to use the notion of splitting field for the central division algebra $D$ of dimension say $d^2$ and to show that $[L:K]=rd$ by using $[D:K]=[L:K]^2$. I know how to prove these facts but I still do not know how to apply it for to prove our statement?

Also there is some hint for to prove it in the book Finite dimensional division algebras over fields (1996 Springer) by Nathan Jacobson and you can download it by clicking on this link, in the beginning of the section 2.12 is written that this is a direct sequence from the theorem 2.3.17 and some statements from its proof. But I still cannot see it!

Thanks!

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    $\begingroup$ Can't you use the fact that $D$ acts faithfully on itself by $\Bbb{K}$-linear transformation? So if $\dim_{\Bbb{K}}D=r$ this action can be viewed as embedding $D$ into $M_r(\Bbb{K})$. Use action on $D^\ell$ iff $\dim_{\Bbb{K}}D=r/\ell$. $\endgroup$ – Jyrki Lahtonen Apr 8 '18 at 19:35
  • $\begingroup$ @JyrkiLahtonen Many thanks for your comment! Can you please explain a bit more? Thanks! $\endgroup$ – Nikita Apr 8 '18 at 21:11
  • $\begingroup$ @JyrkiLahtonen Many thanks for your comment! But I think for to show the statement we need to use the notion of splitting field for the central division algebra $D$ of dimension say $d^2$ and to show that $[L:K]=rd$ by using $[D:K] =[L:K]^2$. I know how to prove these facts but I still do not know how to apply it for to prove the above mentioned statement? Can you please help me to prove it? Thanks! $\endgroup$ – Nikita Apr 9 '18 at 15:39
  • $\begingroup$ The question is senseless as stated. I took time to understand that the claim is that $D$ is isomorphic as a $K$-algebra to a subalgebra of $M_r(K)$. To claim that something is a subring of something consists in proving that it's stable under ring operations... $\endgroup$ – YCor Apr 10 '18 at 21:22
  • $\begingroup$ Btw by editing you made Algeboy's answer obsolete... you were asking for an iff statement ad he proved the only nontrivial implication. If I understand correctly (trying to make sense of the question), the remaining implication is trivial. $\endgroup$ – YCor Apr 10 '18 at 21:26
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Write $d=\dim_K(D)$. Consider the set $D^m$ for a natural number $m$. Then $D^m$ is a right $D$-module, and has dimension $md$ as a vector space over $D$. Choose a basis $B$ for $D^m$ over $K$. Each right multiplication map $\rho_a:x\to xa$ on $D^m$ is a $K$-vector endomorphism of $D^m$. Expressing it in terms of the basis $B$ gives a matrix $N_a$. Then $a\to N_a$ is a homomorphism from $D$ to $M_{md}(K)$. As $D$ is simple, its image is isomorphic to $D$ and is a subring of $M_{md}(K)$.

Conversely consider an embedding $\phi:D\to M_n(K)$. Then $K^n$ becomes a right $D$-module via $v\cdot a=v\phi(a)$. For a division ring, module theory follows the same lines as vector space theory over a field: every module is free. Thus $K^n\cong D^m$ as a $D$-module for some $m$. Then $n=md$ on counting dimensions over $K$.

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  • $\begingroup$ Uh very nice and knowledgeable answer! Thank you very much! I understood whole part of it! Many thanks! $\endgroup$ – Nikita Apr 11 '18 at 14:34
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If $D$ is a unital subring of $M_r(K)$ then $K^r$ is a unital $D$-module. But $D$ is a division ring so this means $K^r$ is a $D$-vector spaces, i.e. $K^r=D^m$ for $m=\dim_D K^r$. Evidently $r=m\dim D$.

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  • $\begingroup$ Thank you very much for your answer! But sorry I cannot understand some part of it. Why $K^r$ is an unital $D$-module? and why $K^r$ becomes a $D$-vector space? And how to show the sufficiency part, please? Thanks! $\endgroup$ – Nikita Apr 6 '18 at 22:05
  • $\begingroup$ Many thanks for your answer! Can you please explain it a bit more? And also can you please let me know how to do the other part? Thanks! $\endgroup$ – Nikita Apr 7 '18 at 22:01
  • $\begingroup$ Many thanks for your answer! I think for to show the statement we need to use the notion of splitting field for the central division algebra $D$ of dimension say $d^2$ and to show that $[L:K]=rd$ by using $[D:K] =[L:K]^2$. I know how to prove these facts but I still do not know how to apply it for to prove the above mentioned statement? Can you please help me to prove it? Thanks! $\endgroup$ – Nikita Apr 9 '18 at 15:40
  • $\begingroup$ Can you please also let me know how to show the other part? Thanks! $\endgroup$ – Nikita Apr 10 '18 at 18:34

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