1
$\begingroup$

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 center $K$ and $[L:K]<\infty$?

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.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Boyarsky
    Commented Jun 13, 2010 at 18:58
  • $\begingroup$ That's great, thanks. I'll take a look at the book. $\endgroup$
    – carlos
    Commented Jun 13, 2010 at 19:09
  • $\begingroup$ Another reference is Pierce's Associative Algebras. $\endgroup$
    – Robin Chapman
    Commented Jun 13, 2010 at 19:52

2 Answers 2

Reset to default
2
$\begingroup$

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?

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you very much. What I meant was that if there's another way other than applying Tsen's theorem to solve the problem. $\endgroup$
    – carlos
    Commented Jun 13, 2010 at 19:12
  • $\begingroup$ 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... $\endgroup$
    – Vladimir Dotsenko
    Commented Jun 13, 2010 at 20:24
0
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ Sorry, I meant L is a skew field. I fixed the mistake. $\endgroup$
    – carlos
    Commented Jun 13, 2010 at 18:52
  • $\begingroup$ 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. $\endgroup$
    – Charles Matthews
    Commented Jun 13, 2010 at 18:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .