I got this question while I am studying Galois correspondence.
Let $K/F$ be an infinite extension and $G = \mathrm{Aut}(K/F)$.
Let $H$ be a subgroup of $G$ with finite index and $K^H$
be the fixed field of $H$. Is it true that $[K^H:F]= (G:H)$?
For finite extension, I verified this
is true. Is it true in infinite case also?
Thanks
Best Answer
KCd's comment answers the question, but let's be a little more explicit. Let $F = \mathbb{Q}(x_1, x_2, ...)$ and let $K = \mathbb{Q}(\sqrt{x_1}, \sqrt{x_2}, ...)$. The Galois group $G = \text{Aut}(K/F)$ is $\prod_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ with the product topology. The subgroup $\bigoplus_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ is countable, so is a proper subgroup of $G$. Conditional on the axiom of choice (but actually we only need the ultrafilter lemma here) this subgroup is contained in a maximal subgroup $H$. The quotient $G/H$ is a simple group in which every element has order $2$, so must be $\mathbb{Z}/2\mathbb{Z}$.
So $H$ has index $2$. But since it contains the topological generators of $G$ we conclude that $K^H = F$.