# Alternative proof of Wedderburn's little theorem

I have this exercise where I'm proving: "Every finite division ring is a field". I need only a part (c) of it:

(a) show that a subalgebra of a finite dimensional central division algebra is a finite dimensional division algebra. (DONE)

(b) show that if $D$ is a finite dimensional central division algebra and $K\neq Z(D)$ is any subfield, then $D$ is generated as an $Z(D)$-algebra by $\bigcup_{d\in D^*}d^{-1}Kd$. (DONE)

(c) Conclude, without using the Noether-Skolem theorem, that a finite division ring is a field. (NEEDED...)

Thanks, G.

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Write using LateX: it's very hard to guess what you meant by "d^-1Kd", for example – DonAntonio Jan 13 '13 at 22:40
have'nt had the chance to learn how to. i meant was Union over all invertible elements "d" in D, of "the inverse of d" * "K" * "d" – cruvadom Jan 13 '13 at 22:42
Go to the FAQ section for directions on LaTeX. – DonAntonio Jan 13 '13 at 22:46
Excuse me, but what does it mean to add $G$ in the end? Regards. – awllower Feb 27 '13 at 16:18
@awllower: See here – Prism Jul 26 '13 at 21:56

To provide an alternate, maybe somewhat too over-loaded proof of this fact: every finite division ring is commutative.
It amounts to the same thing as showing that the Brauer group of any finite field is trivial, for then the finite division rings are all matrix rings. Since they are division rings, this implies that they are fields. Now, by a theorem in the theory of central simple algebras, $\mathbb {Br}(K/\kappa)$, for a finite Galois extension of a finite field $\kappa$, is isomorphic with $H^2(Gal(K/\kappa),K^\times)$. But finite extensions of finite fields are cyclic, so this is a cohomology group of a finite cyclic group. Since such cohomology is periodic of period $2$, we find that it is just the norm residue group $\kappa^\times/N_{\mathbb K\mid \kappa}\mathbb K^\times$. Since it is well-known that norms of finite field extensions are surjective, this tells us that the relative Brauer groups are trivial. As Brauer groups are colimits of relative ones, this finishes the proof.
P.S. I learned this proof from Mariano Suarez-Alvarez. As this proof is quite remarkable, I make it CW, to keep it as a reference. Thanks.

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It is enteresting to pose this problem for division rings which multiplication unilateral distributs the law of addition (a+b)c=ac+bc, i. e. the multiplication by an element f(x)=ax is not an additive. May be such finite division rings also will be commutative?

Gintaras

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You can ask this as a question. – HDE 226868 Dec 31 '14 at 19:11
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. – Grigory M Dec 31 '14 at 19:29
@Gintaras: Your friendly community moderator here. The commenters nailed it. You should ask this is a separate question. Try to fully follow the instructions. Remember to make the question as self-contained as possible, and add your own thoughts. Do link to this question when doing that, please. That gives helpful context. – Jyrki Lahtonen Dec 31 '14 at 19:36
And it is probably best that you delete this post. I like to think that as a question your post will be well received. – Jyrki Lahtonen Dec 31 '14 at 19:37