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Is there a field K such that $\operatorname{Gal}(\overline{K}/K)$ is the profinite free group with two generators?

For one generator I know that for all the $\mathbb{F}_p$ we have $\operatorname{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)\simeq\widehat{\mathbb{Z}}$.

Gabriel Soranzo
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1 Answers1

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Yes, there is. In fact, for every projective profinite group $G$, there is a perfect and pseudo-algebraically closed field whose absolute Galois group is $G$.

You may want to check corollary 23.1.2 of Field Arithmetic by Fried and Jarden for a proof.

Captain Lama
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  • Thanks. It does not seem to be simple. Does this result have a classical name? – Gabriel Soranzo Mar 29 '16 at 19:44
  • I don't know about a name. As for difficulty, it's probably a non-trivial statement in any case, but the fact is that in the book they take some pain to construct a PAC field that has the property, so maybe if you just want any field there is a simpler construction. I just happen to have this book so I noticed that your question was one treated there, maybe someone has another source with a more elementary solution. :) – Captain Lama Mar 29 '16 at 19:50
  • Yes, I should change my title: Is there a simple construction for a field $K$ such that... ;-) – Gabriel Soranzo Mar 29 '16 at 20:02