Explanation behind $\text{Gal}(K/K^H)=H$.

Question

I am studying Galois correspondence theorem i.e. the fundamental theorem of Galois theory.The first part of the theorem says that given a finite Galois extension $K/F$ there exists an inclusion reversing bijection between the subgroups of $\text{Gal}(K/F)$ and the intermediate fields of $K/F$.The maps in the two directions are given by $H\mapsto K^H$ and $L\mapsto \text{Gal}(K/L)$.Now I am doing the proof of the fact that these two maps are inverses of each other.For that I need to show that $\text{Gal}(K/K^H)=H$ for a subgroup $H$ of $\text{Gal}(K/F)$.But I cannot find a proper reasoning for this.Can someone help me with this because this is the only part of the proof where I am stuck.

Answer 1

The fundamental theorem of Galois theory claims that the maps $H\mapsto\mathbb{K}^H$ (also denoted $\operatorname{Fix}_\mathbb{K}(H)$) and $\mathbb{L}\mapsto\operatorname{Gal}(\mathbb{K}/\mathbb{L})$ form an antitone Galois connection.

1. Let $\mathbb{L}_1<\mathbb{L}_2$, then we have $\operatorname{Gal}(\mathbb{K}/\mathbb{L}_2)<\operatorname{Gal}(\mathbb{K}/\mathbb{L}_1)$ because: $$\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L}_2) \Leftrightarrow \forall x\in\mathbb{L}_2\colon \varphi(x)=x \Rightarrow \forall x\in\mathbb{L}_1\colon \varphi(x)=x \Leftrightarrow \varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L}_1).$$
2. Let $H_1<H_2$, then we have $\mathbb{K}^{H_2}<\mathbb{K}^{H_1}$ because: $$x\in\mathbb{K}^{H_2} \Leftrightarrow \forall\varphi\in H_2\colon \varphi(x)=x \Rightarrow \forall\varphi\in H_1\colon \varphi(x)=x \Leftrightarrow x\in\mathbb{K}^{H_1}.$$
3. We have $\mathbb{L}<\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}$ since: $$x\in\mathbb{L} \Rightarrow \forall\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L})\colon \varphi(x)=x \Rightarrow x\in\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}.$$
4. We have $H<\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H)$ since: $$\varphi\in H \Rightarrow \forall x\in\mathbb{K}^H\colon\varphi(x)=x \Rightarrow \varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H).$$

That is all that is necessary to establish a Galois connection, but when we take a Galois extension like $\mathbb{K}/\mathbb{F}$, where all intermediate extensions $\mathbb{K}/\mathbb{L}$ for $\mathbb{F}<\mathbb{L}<\mathbb{K}$ are also galois, we even get equalities for the last two results.

3. Because of Artin's theorem and since $\mathbb{K}/\mathbb{L}$ is a Galois extension, we have: $$[\mathbb{K}:\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}] =|\operatorname{Gal}(\mathbb{K}/\mathbb{L})| =[\mathbb{K}:\mathbb{L}],$$ which implies $\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}=\mathbb{L}$ together with the upper inclusion.
4. Since $\mathbb{K}/\mathbb{K}^H$ is a Galois extension and because of Artin's theorem, we have: $$|\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H)| =[\mathbb{K}:\mathbb{K}^H] =|H|,$$ which implies $\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H)=H$ together with the upper inclusion.

You can also find this calcuation in a little different way here.