# Understanding a proof in JS Milne's Fields and Galois Theory (Prop 7.10)

The following is proposition 7.10 in Milne's Fields and Galois Theory:

Let $$G$$ be a group of automorphisms of a field $$E$$, and let $$F=E^G$$ (ie. $$F$$ is fixed field of $$G$$). If $$G$$ is compact and the stabilizer of each element of $$E$$ is open in $$G$$ (wrt Krull Topology), then $$E$$ is a Galois extension of $$F$$ with Galois group $$G$$.

The author starts by considering a finite subset of $$E$$ say $$\{x_1,x_2, \cdots x_n\}$$. Then using compactness, he argues that orbit of each $$x_i$$ is finite. Let $$N$$ be the intersection of all the conjugates of $$H_i$$'s where $$H_i$$ is stabilizer of $$x_i$$. Then by previous observations, its straightforward to see that $$N$$ is a normal open subgroup of $$G$$. Then the following claim is made:

The fixed field of $$N$$ is the subfield of $$E$$ that is generated over $$F$$ by the elements of orbit of $$x_i$$'s.

I am able to see that field generated over $$F$$ by the elements of orbit of $$x_i$$'s is a subset of the fixed field of $$M$$ but I can't show the converse. I tried many times. One approach which seemed promising was for show $$E$$ was algebraic extension of $$F$$, then take an arbitrary element $$\alpha \in M$$ ($$M$$ is fixed field of $$N$$) and consider its minimal polynomial $$m_{\alpha}$$ over $$F$$ and using similar arguments from finite Galois Theory, I showed $$m_{\alpha}$$ splits in $$M$$ into distinct linear factors. Thus $$M$$ is Galois over $$F$$. However this goes nowhere (and moreover I found a flaw in my reasoning where I showed $$E$$ is algebraic over $$F$$).

I would really appreciate a hint.

PS: I'm referring to version 5.10 of JS Milne's Fields and Galois Theory. The proposition is in pg. 97.