I have recently learned the following theorem.

Let $E$ be a field and let $H$ be a finite group of automorphisms of $E.$ Then $(E:E^H)=\operatorname{card}(H),$ where $E^H$ is the subfield of elements which are fixed by all automorphisms in $H.$

The proof depends on the finiteness of $H$ because it uses linear equations in $n$ (and also $n+1$) variables, where $n=\operatorname{card}(H).$ Is the theorem still true if we drop this assumption? If it is, is it possible to give a proof of the general case without distinguishing the cases of $H$ finite and $H$ infinite?

Re Pierre-Yves Gaillard's answer

$(1)$ Why is the Galois group $G=\operatorname{Aut}(K/F_p)$ uncountable?

Clearly, $\operatorname{card}(G)\leq \operatorname{card}(K)^{\operatorname{card}(K)}=\aleph_0^{\aleph_0}=\mathfrak c.$ I'm having trouble with the other inequality. Here's a slightly "more accurate" way of bounding $\operatorname{card}(G)$ from above.

We have $$K=\bigcup_{n=1}^\infty F_{p^n}.$$ Therefore, in particular $K$ is countable as a union of countably many finite sets. Let $\alpha\in G$ and $x\in K.$ Then there is $k\in \mathbb N$ such that $x^{p^k}-x=0$ and so $(\alpha(x))^{p^k}-\alpha(x)=0.$ So every element $x\in F_{p^k}$ is mapped to one of the $p^k$ elements of $F_{p^k}.$ So to define an automorphism we have to make countably many finite choices, which again means that $\operatorname{card}(G)\leq\mathfrak c.$ But it doesn't prove that there is equality because certainly not all such choices are OK. How can this be improved?

$(2)$ I understand that you're using the following formula. $$K^G=F_p.$$ For this I would have to show that if $\xi\in K\setminus F_p$ then there exists $\alpha\in G$ such that $\alpha(x)\neq x.$ This looks very true. Let $f$ be the minimal polynomial of $\xi$ over $F_p.$ Its degree is greater than $1.$ Suppose it has no roots in $K$ other than $\xi.$ Then $\xi$ is a multiple root of $f$ and so $f'(\xi)=0.$ But then $f'=0$ because $\deg(f')<\deg(f).$ So $f$ must be of the form $$f(x)=a_nx^{pn}+a_{n-1}x^{p(n-1)}+\ldots+a_1x^p+a_0.$$ There is $k\in\mathbb N$ such that $\xi\in F_{p^k}.$ But $\phi(x)=x^p$ is an automorphism of $F_{p^k}$ and therefore so is $\phi^{-1}$. So since $$a_n(\xi^p)^n+a_{n-1}(\xi^p)^{n-1}+\ldots+a_1\xi^p+a_0=0,$$ we also have $$g(\xi):=a_n\xi^n+a_{n-1}\xi^{n-1}+\ldots+a_1\xi+a_0=0.$$ Now $\deg f=pn$ and $\deg g=n$ and both have $\xi$ as their zero. Therefore $n$ must be zero, because otherwise $pn>n.$ But then $\deg f=pn=0,$ a contradiction.

So $f$ has another root $\eta\neq\xi.$ We should be able to construct an automorphism of $K$ sending $\xi$ to $\eta.$ I think there is such an automorphism of $F_{p^n}$ by a theorem about splitting fields. Why can I extend it to $K?$

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    $\begingroup$ You should pick up a book that addresses infinite Galois theory, which settles both of your follow-up questions. One choice is Szamuely's Galois Groups and Fundamental Groups. In particular, 1) follows from the fact that the Galois group in question is the profinite completion of $\mathbb{Z}$, which follows from standard facts in infinite Galois theory, and 2) follows from the fact that automorphisms of an algebraic extension always extend to a larger algebraic extension (by Zorn's lemma). $\endgroup$ – Qiaochu Yuan Mar 26 '12 at 21:37
  • $\begingroup$ @QiaochuYuan Thanks! I'll see if I can access that book. $\endgroup$ – user23211 Mar 26 '12 at 21:58
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    $\begingroup$ A simpler way to (2) in this specific case: $\mathbb{F}_p$ is the fixed field of the single automorphism $x \mapsto x^p$. $\endgroup$ – Chris Eagle Mar 26 '12 at 21:59

The answer is no:

If $K$ is an algebraic closure of $\mathbb F_p$, then the Galois group has the power of the continuum, but the degree of $K/\mathbb F_p$ is countable.

Edit. The question in the edit has been answered by comments of Qiaochu Yuan and Chris Eagle, but here are a few more sentences.

Let $K$ be an algebraic closure of $K_1:=\mathbb F_2$.

View $K_n:=\mathbb F_{2^n}$ as a subfield of $K$, note that $K_d\subset K_n$ if and only if $d$ divides $n$, define the automorphism $f$ of $K$ by $fx:=x^2$, let $E$ be the union of the $K_n$ where $n$ is square-free. Then $E$ is a countable subfield of $K$, and, for each prime $p$, the Galois group of $K_p/K_1$ is cyclic of order $p$ generated by $f$.

Moreover, if $$ g=(g_p)\in G:=\prod_{p\text{ prime}}\text{Aut}(K_p/K_1)\simeq\prod_{p\text{ prime}}\mathbb Z/(p), $$ then there is a (unique) automorphism of $E$ whose restriction to $K_p$ is $g_p$ for all $p$.

As $G$ is uncountable, so is $\text{Aut}(E)$, and $K_1$ is the fixed field of $\text{Aut}(E)$.

  • $\begingroup$ Thank you for the answer. Could you please see my questions about it in the edit? $\endgroup$ – user23211 Mar 26 '12 at 21:28
  • $\begingroup$ Dear @ymar: I edited the answer. $\endgroup$ – Pierre-Yves Gaillard Mar 27 '12 at 1:59

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