Involution, Orbit and Galois Extenison...

The motivation of this blog is for this question. Let $E$$E$ be the splitting field of $f(x)=1+x+x^2+x^3+x^4$$f(x)=1+x+x^2+x^3+x^4$ over $\mathbb{Q}$$\mathbb Q$.

Then the extension $E/\mathbb{Q}$$E/\mathbb Q$ is a Galois extension with degree $4$$4$, hence $|\mathrm{Gal}(E/\mathbb{Q})|=4$$|\mathrm{Gal}(E/\mathbb Q)|=4$

Since $f(x)$$f(x)$ is irreducible, $\mathrm{Gal}(E/\mathbb{Q})$$\mathrm{Gal}(E/\mathbb Q)$ act transitive on

$$\begin{array}{}\text{(1)}& \{\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4\}\end{array}$$

Let us consider $\sigma (\zeta )={\zeta }^2$$\sigma(\zeta)=\zeta^2$, then ${\sigma }^2(\zeta )=\sigma ({\zeta }^2)=\sigma (\zeta {)}^2={\zeta }^4$$\sigma^2(\zeta)=\sigma(\zeta^2)=\sigma(\zeta)^2=\zeta^4$, ${\sigma }^3(\zeta )=\sigma (\zeta {)}^4={\zeta }^8={\zeta }^3$$\sigma^3(\zeta)=\sigma(\zeta)^4=\zeta^8=\zeta^3$

Hence $\mathrm{Gal}(E/\mathbb{Q})\cong \mathbb{Z}/4\mathbb{Z}$$\mathrm{Gal}(E/\mathbb Q)\cong\mathbb Z/4\mathbb Z$​. What is the fixed field respect to $⟨{\sigma }^2⟩$$\langle\sigma^2\rangle$?

Let us consider sth in general.

Universal Property of Orbit Space:

Let $f:G\to {\mathrm{Aut}}_{\mathrm{Set}}(X)$$f: G \to \mathrm{Aut}_{\mathrm{Set}}(X)$ be a group action, and $\pi :X\to X/G$$\pi: X \to X/G$ be the orbit space. Then for any $G$$G$-invariant map $f:X\to Y$$f: X \to Y$, that is, $\mathrm{\forall }g\in G,x\in X$$\forall g \in G, x \in X$, $f(g\cdot x)=f(x)$$f(g \cdot x) = f(x)$, there exists a unique map $\psi :X/G\to Y$$\psi: X/G \to Y$ such that $f=\psi \circ \pi$$f = \psi \circ \pi$.

Let $F:\mathbb{Z}/2\mathbb{Z}\to {\mathrm{Aut}}_{\mathrm{Set}}(X)$$F: \mathbb{Z}/2\mathbb{Z} \to \mathrm{Aut}_{\mathrm{Set}}(X)$ be a faithful representation, and denote $F(1)$$F(1)$ as $\ast$$*$, we call $\ast$$*$ an involution on $X$$X$.

Then the set of all orbits of $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$ will be $X/(\mathbb{Z}/2\mathbb{Z})=\coprod _{x\in X}\{x,x^{\ast }\}$$X/(\mathbb{Z}/2\mathbb{Z}) = \coprod_{x \in X}\{x, x^*\}$. If the orbit of $x$$x$ is equal to $\{x\}$$\{x\}$, then we say $x$$x$ is a self-adjoint element.

Let $G$$G$ be a group and consider $\ast :g⟼g^{-1}$$*: g \longmapsto g^{-1}$, then $\ast$$*$ is an involution.

The orbit of an element $g$$g$ is either $\{g,g^{-1}\}$$\{g, g^{-1}\}$ or $\{g\}$$\{g\}$. If $|G|\equiv 0\mathrm{mod}2$$|G| \equiv 0 \mod 2$, then there exists a subgroup of $G$$G$ that is isomorphic to $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$. Since we know that the orbit of $e$$e$ is $\{e\}$$\{e\}$, there exists at least one $g$$g$ such that ${\mathcal{O}}_g=\{g\}$$\mathcal{O}_g = \{g\}$.

And $g$$g$ is an involution as well!

Let $E/F$$E/F$ be a Galois extension and $[E:F]\equiv 0\mathrm{mod}2$$[E:F] \equiv 0 \mod 2$, then $|\mathrm{Gal}(E/F)|\equiv 0\mathrm{mod}2$$|\mathrm{Gal}(E/F)| \equiv 0 \mod 2$, hence $\mathrm{Gal}(E/F)$$\mathrm{Gal}(E/F)$ has a subgroup that is isomorphic to $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$, and therefore an intermediate field $F\subseteq K\subseteq E$$F \subseteq K \subseteq E$ fixed by $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$​.

Easy to see that

$$\begin{array}{}\text{(2)}& K=\{a+a^{\ast }:a\in E\}\end{array}$$

As an example of universal property of orbit space, we have

$$\begin{array}{}\text{(3)}& f(x)=x+x^{\ast },\psi :E/(\mathbb{Z}/2\mathbb{Z})\to E,\{x,x^{\ast }\}⟼x+x^{\ast },\mathrm{Im}(f)=K.\end{array}$$