Let us consider the field of rational functions in one variable with real coefficients, $\mathbb{R}(t)$. The algebraic closure is the field of algebraic functions with real coefficients. What is known of the absolute galois group here? Can we say anything, and if sp, what?
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What do you mean by "algebraic functions with real coefficients"? It doesn't sound like it includes $i$. – Chris Eagle Mar 02 '13 at 16:11
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2The function f(x)=i is algebraic, it satisfies the equation x^2+1. So all functions satisfying an algebraic equations with real polynomial coefficients are there. – Heidar Svan Mar 02 '13 at 16:18
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Replace polynomial coefficients with rational functions :) – Heidar Svan Mar 02 '13 at 16:20
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@ChrisEagle See the wiki article on it. – JSchlather Mar 02 '13 at 16:30
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Related: http://mathoverflow.net/questions/206362 – Watson Feb 03 '17 at 15:19
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@JSchlather Do you know of any source proving that the algebraic functions are algebraically closed? – Jose Brox Nov 02 '20 at 10:57
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Let $k$ be an algebraically closed field of cardinality of $\kappa$. Then the absolute Galois group of $k(t)$ is the free profinite group $\hat{F}_\kappa$ on $\kappa$ generators. See here.
So in your case we have that the absolute Galois group of $\mathbb R(t)$ is $\mathbb Z/ 2\mathbb Z \rtimes \hat{F}_\kappa$. Since the absolute Galois group will be generated by complex conjugation and the absolute Galois group of $\mathbb C(t)$.
JSchlather
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Dear @JSchlather, Let $E$ be an algebraic closure of $\mathbb{C}(t)$. Since $\mathbb{C}(t)$ is a Galois extension of $\mathbb{R}(t)$ of degree $2$, $G(E/\mathbb{C}(t))$ is a normal open subgroup of $G(E/\mathbb{R}(t))$ of index $2$. In particular, the absolute Galois group of $\mathbb{R}(t)$ is $\mathbb{Z}/2\ltimes \hat{F}_k$, no? – Doug Oct 27 '23 at 09:15