Groups, inverse Galois problem and transcendence degrees

Question

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!

Answer 1

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.

Answer 2

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.