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Let $\mathbb{R}(t)$ be the field of rational functions over $\mathbb{R}$ (the fraction field of $\mathbb{R}[x]$).

I am looking for elements in the Brauer group of the field, and the current idea I have to follow on is to find infinitely many cyclic field extensions, and use those to create cyclic division algebras.

My Galois theory experience is not very rich with transcendental extensions of $\mathbb{R}$ and I'm a bit lost. Am I even on the right path towards the Brauer group? Any ideas on how to prove there are many cyclic extensions?

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There are many, many quadratic extensions... –  Mariano Suárez-Alvarez Feb 13 '11 at 0:45

1 Answer 1

up vote 6 down vote accepted

By a famous theorem of Tsen and Lang, the Brauer group of $\mathbb{C}(t)$ is zero. Thus if you take any class in the Brauer group of $\mathbb{R}(t)$ and restrict it to the quadratic extension $\mathbb{C}(t)$, it becomes zero.

From this it follows that every element of $\operatorname{Br}(\mathbb{R}(t))$ can be represented by a quaternion algebra, so you're on the right track by considering cyclic (quadratic) extensions, of which there are infinitely many. You need to think about when $\mathbb{R}(t)(\sqrt{f(t)}) = \mathbb{R}(t)(\sqrt{g(t)})$ when $f$ and $g$ are rational functions. Hint: it is enough to consider the case of squarefree polynomials, and then the above equality implies that $f$ and $g$ have the same roots (in $\mathbb{C}$). It is very tempting for me to speak geometrically in terms of ramification but I don't know if you would be comfortable with that.

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By the way: free to ask more questions about this, if you have them. –  Pete L. Clark Feb 13 '11 at 1:03
Many thanks. It took a few cups of coffee, but all the work I've put into this before, and after your answer, paid off. I hope. –  Asaf Karagila Feb 13 '11 at 5:40

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