# Groups, inverse Galois problem and transcendence degrees

This is a curiosity of mine. I suspect there might be a trivial answer, but if there is none, this problem will probably haunt me for a long time...

The question is as follows : Given a group $G$, does there exists two field extensions $K_1/F_1$ and $K_2/F_2$ such that $\mathrm{Aut}(K_1/F_1) \simeq G \simeq \mathrm{Aut}(K_2/F_2)$ and the extensions $K_i/F_i$ have different transcendence degrees, or does $G$ determine the transcendence degree of the extensions of which it is the group of automorphisms?

Perhaps allowing $F_1 \neq F_2$ is probably too strong, but I would still be interested in the case $F_1 = F_2 = \mathbb Q$ or $\mathbb F_p$. I leave it open just in case ; maybe letting $F_i$ vary in some set of fields might not change the problem much.

Motivation : Finite extensions must have finite automorphism groups since they are algebraic, so you can write $K/F$ as $K = F(\alpha_1,\cdots,\alpha_n)$ and using the minimal polynomials of the generators we deduce the finiteness of the group $G$, so in this case the transcendence degree is always zero if $G$ is finite. If we have an extension $K/F$ of transcendence degree $1$, we can write $K/L/F$ where $L/F$ is purely transcendental and $K/L$ is algebraic, so I expect (huge faith here) the morphism $\mathrm{Aut}(K/F) \to \mathrm{Aut}(L/F)$ given by restriction to not change much about the "size" (the notion of size I want the transcendence degree to define), and Lüroth's theorem tells me that essentially all extensions over $F$ have the same "size" as $L/F$.

If you don't like my motivation, just dismiss it and try to answer the question... I'm just throwing my feelings out there!

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The fields $\mathbb{R}$ and $\mathbb{Q}_p$ are both 'rigid'. So, $\text{Gal}(\mathbb{R}/\mathbb{Q})=\text{Gal}(\mathbb{Q}_p/\mathbb{Q})=1$. That said, neither are algebraic over $\mathbb{Q}$ since they are uncountable.
If $K/\mathbb C$ is finitely generated and has transcendence degree $1$, then $K$ is the function field of a compact Riemann surface $X$. Moreover, $\text{Aut}(K/\mathbf C) \cong \text{Aut}(X)$. It can be shown that if $X$ is sufficiently generic and has genus $>2$, then $\text{Aut}(X)=1$. Hence $\text{Aut}(K/\mathbf C) \cong \text{Aut}(\mathbf C/\mathbf C)$, so the answer to your question is negative.
How is that possible? If it has transcendence degree $1$, then $K/L/\mathbb C$ where $L/\mathbb C$ is purely transcendental, and I thought automorphisms of $L$ could be lifted to $K$? I mean, I'm disturbed because I don't "see" your example. – Patrick Da Silva Aug 16 '14 at 3:19
Second question : I feel like there's some automorphisms missing... a bit like in the splitting field of $x^3-2$ over $\mathbb Q$. Maybe my question only make sense for "Galois" groups (I don't know what "Galois" would mean for non-algebraic extensions...) – Patrick Da Silva Aug 16 '14 at 3:22
+1 For @PatrickDaSilva this is known as Hurwitz's formula. It says that for a Riemann surface $X$ with $g(X)>1$, you have that $\text{Aut}_{hol}(X)$ has cardinality at most $84(g-1)$. And, since we're on a curve, the automorphism group of the curve is just the automorphism group of the function field. – Alex Youcis Aug 16 '14 at 3:37
Dear @Patrick: Lifting automorphisms is a lot more subtle for transcendental extensions. If you want to describe $K$ explicitly as a finite extension of $\mathbb C(z)$, this is the same as presenting $X$ as a finite ramified covering $X \to \mathbb P^1$. The question of whether an automorphism of $\mathbb P^1$ lifts to $X$ is quite subtle; for a start, there are possible obstructions coming from the side of algebraic topology. – Bruno Joyal Aug 16 '14 at 3:43