# Automorphism group gives the Galois group over the fixed field.

Question. Let $$E$$ be a field and let $$G=\operatorname{Aut}(E)$$ be the group of ring automorphisms. Let $$F=\operatorname{Fix}(G)$$ be the fixed field of $$G$$. Is $$E$$ Galois over $$F$$? Is $$G=\operatorname{Gal}(E/F)$$?

Let $$p(x)\in F[x]$$ be irreducible over $$F$$ and let $$p(x)=q_1(x)...q_k(x)$$be the factorization of $$p(x)$$ in $$E$$ into irreducible factors $$q_1(x),...,q_k(x)$$. Since any automorphism $$\sigma: E\to E$$ fixes the coefficient of $$p(x)$$, $$q_1(x)...q_k(x)=p(x)=(\sigma\cdot p)(x)=(\sigma\cdot q_1)(x)...(\sigma\cdot q_k)(x)$$where $$\sigma$$ acts on the polynomial by evaluating its coefficient. As $$F[x]$$ is a UFD, $$\sigma$$ permutes the irreducible factors $$q_1,...,q_k$$. Thus once I showed $$\operatorname{Aut}(E)$$ acts transitively on $$q_1,...,q_k$$, then $$q_1,...,q_k$$ must have the same degree.

In particular when $$p(x)$$ has a root $$\alpha\in E$$, i.e., when $$p(x)$$ is the minimal polynomial of some element in $$E$$, then $$p(x)$$ must split over $$E$$. This would prove that $$E/F$$ is normal.(But I still haven't shown the transitivity)

The equality $$G=\operatorname{Gal}(E/F)$$ is clear. In general I do not trust $$E/F$$ to be separable, but I am struggling to come up with a counterexample.

• This is how E. Artin introduces Galois extensions in his lovely book on Galois Theory. Take a field F and a finite group G of automorphisms F. If E is the fixed field, then F/E is Galois with Galois group G. Later he shows that this is equivalent to the finite, normal, separable characterisation Commented May 19 at 15:10
• @AndrewHubery That's a classical result in Galois theory. But I guess OP doesn't assume $G$ is finite.
– Mark
Commented May 19 at 15:20
• Let $E=K(T)$, where $T$ is an indeterminate over $K.$ Assume that $K$ is infinite. Then $G=Aut(E)$ contains $H=Aut(E/K)$. But if $K$ is infinite, then it is known that $E^H=K$ (I do not have any reference at the moment). Hence $E/E^H$ is transcendental, and this , so is $E/E^G$. In particular,$E/E^G$ is not Galois. Commented May 19 at 17:10
• For infinite groups there is a lemma by Rosenlicht in ‘Automorphisms of function fields’ TAMS 79 (1955), p4. It says that if F is a field, G any group of automorphisms, and E the fixed field, then F/E is separable generated. Commented May 19 at 17:52