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Let K=F(x), where x is transcendental over F and F is an algebraically closed field. Does there exist a non-commutative division algebra L with the center K and [L : K] < infinity?

I think, but I'm not sure, that an old result due to Tsen implies that the answer is no? I'd like to know if there's another way, other than applying Tsen's theorem, to prove this? Thanks.

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Serre's Galois cohomology book gives an elegant purely algebraic treatment of behavior of cohomological dimension with respect to the formation of rational function fields, and the relation between vanishing of Brauer groups and low cohomological dimension. In particular, it gives a purely algebraic proof of Tsen's theorem, so not sure what more you could want in the direction of a "more algebraic" proof. –  Boyarsky Jun 13 '10 at 18:58
    
That's great, thanks. I'll take a look at the book. –  carlos Jun 13 '10 at 19:09
    
Another reference is Pierce's Associative Algebras. –  Robin Chapman Jun 13 '10 at 19:52

2 Answers 2

Yes, you are right to assume that there are no such division algebras. And your plan of proving that is correct: the norm function (a polynomial of degree $n$ in $n^2>n$ variables) will vanish at some point because of Tsen's theorem. And Tsen's theorem can be seen as quite algebraic, so maybe you can elaborate on what you would consider a less geometric proof?

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Thank you very much. What I meant was that if there's another way other than applying Tsen's theorem to solve the problem. –  carlos Jun 13 '10 at 19:12
    
Then I'd strongly recommend you to rewrite the title of the question. Maybe "How to prove that Br(F(t))=0 without Tsen's theorem?", or something like that. This way you stand a better chance to attract the attention of someone who actually knows the answer... –  Vladimir Dotsenko Jun 13 '10 at 20:24

The answer is yes, though. K is not algebraically closed. You may mean something else.

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Sorry, I meant L is a skew field. I fixed the mistake. –  carlos Jun 13 '10 at 18:52
    
en.wikipedia.org/wiki/Tsen%27s_theorem and references. The Brauer group is indeed trivial. I suppose this is proved in geometric analogues of class field theory. –  Charles Matthews Jun 13 '10 at 18:58

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