Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.