# Determine $Gal(K/F)$ and find all the intermediate subfields of $K/F$.

Let $F = \mathbb{Q}$ and $K$ be the splitting field of $x^4 - 7$ over $F$. Determine $Gal(K/F)$ and find all the intermediate subfields of $K/F$.

We have $x^4 - 7 = (x-\sqrt{7})(x + \sqrt{7})(x - i\sqrt{7})(x + i\sqrt{7})$ and so, $K = \mathbb{Q}(\sqrt{7},i)$. Now, $\mathbb{Q}(\sqrt{7})$ is a intermediate subfield of $K/F$ and is not Galois, then the correspondent groups given by Fundamental Theorem of Galois Theory is not abelian so, $Gal(K/F) \simeq D_{8}$.

By definition, $D_{8} = \lbrace \alpha, \beta \mid \alpha^{4} = \beta^{2} = e; \; \alpha^{3} = \beta^{-1}\alpha\beta \rbrace$. I'm having trouble to determine the automorphisms and so finding the generators of $Gal(K/F)$. Can someone help me?

• You have "determine[d] $\mathrm{Gal}(K/F)$" as required by the problem statement. Why are you looking for automorphisms to "find all intermediate subfields of $K/F$"? Jun 26, 2018 at 17:59

We have the following automorphisms in the Galois group:

$$Id : \begin{array}{lr} \sqrt{7} \to \sqrt{7}\\ i \to i \end{array}\quad \quad \tau: \begin{array}{lr} \sqrt{7} \to i\sqrt{7}\\ i \to i \end{array} \quad \quad \sigma: \begin{array}{lr} \sqrt{7} \to \sqrt{7}\\ i \to -i \end{array}$$

It's not hard to notice that $\tau$ and $\sigma$ generate the Galois group. In fact we have that $\text{Gal}(K/F) = \{Id,\tau,\tau^2,\tau^3,\sigma,\tau\sigma,\tau^2\sigma,\tau^3\sigma\}$. Now you can apply the automorphisms to a general element of the extension to see what kind of elements remain fixed. After some lengthy calculation you should be able to derive the following correspondences between subgroups of $\text{Gal}(K/F)$ and intermediate fields of $K/F$

$$G \longleftrightarrow \mathbb{Q}$$ $$\{Id,\tau,\tau^2,\tau^3\} \longleftrightarrow \mathbb{Q}(i)$$ $$\{Id,\sigma, \tau^2, \tau^2\sigma\} \longleftrightarrow \mathbb{Q}(\sqrt{7})$$ $$\{Id,\tau\sigma, \tau^2, \tau^3\sigma\} \longleftrightarrow \mathbb{Q}(i\sqrt{7})$$ $$\{Id,\tau^2\} \longleftrightarrow \mathbb{Q}(\sqrt{7},i)$$ $$\{Id,\sigma \} \longleftrightarrow \mathbb{Q}(\sqrt{7})$$ $$\{Id,\tau\sigma\} \longleftrightarrow \mathbb{Q}((1+i)\sqrt{7})$$ $$\{Id,\tau^2\sigma\} \longleftrightarrow \mathbb{Q}(i\sqrt{7})$$ $$\{Id,\tau^3\sigma\} \longleftrightarrow \mathbb{Q}((1-i)\sqrt{7})$$ $$\{Id\} \longleftrightarrow \mathbb{Q}(\sqrt{7},i)$$

There's an element of order $4$ (shall we call it $\alpha$?) fixing $i$ and taking $\sqrt7$ to $i\sqrt7$ (and so cycling the $i^k\sqrt7$).

There's also complex conjugation (shall we call that $\beta$?) which fixes each of $\pm\sqrt7$ and swaps $\pm i\sqrt7$.

Let $G:=\operatorname{Gal}(K/F)$. Since $K/\Bbb Q(i)$ is Galois, there exists $\sigma\in \operatorname{Gal}(K/\Bbb Q(i))\leq G$ with $\sigma(\sqrt{7}) = i\sqrt{7}$. This has order $4$. Similarly, there exists $\tau\in \operatorname{Gal}(K/\Bbb Q(\sqrt{7}))\leq G$ with $\tau(i\sqrt{7})) = -i\sqrt{7}$. This has order $2$.

Check that $\tau\sigma\tau = \sigma^{-1}$.