If $E/F$ is Galois of degree 4 with $\sqrt{-1}\in F$ then $E=F(\sqrt[4]{a})$ or $E=F(\sqrt{a}, \sqrt{b})$

Question

Let $F$ be a field, $Char(F)\neq 2$ and $i=\sqrt{-1}\in F$.
Let $E/F$ be a Galois extension of degree 4.
Prove that $E=F(\sqrt[4]{a})$ or $E=F(\sqrt{a}, \sqrt{b})$, for $a,b\in F$.

Hand-waving for the first part:

$|Gal(E/F)|=4 \Rightarrow Gal(E/F)\cong\mathbb Z/4\mathbb Z$ or $Gal(E/F)\cong\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$.
If $Gal(E/F)\cong\mathbb Z/4\mathbb Z$ then it is cyclic with generator $\sigma\in Gal(E/F)$, and by a theorem, because $i\in F$, there's an $\alpha\in E$ such that $\sigma(\alpha)=i\alpha \Rightarrow \sigma^j(\alpha)=i^{j(mod 4)}\alpha \Rightarrow$ the set of F-conjugates of $\alpha$ is $\{\alpha, i\alpha, -\alpha, -i\alpha\}$ which is the set of solutions of $x^4-a=0$ $\Rightarrow \alpha=\sqrt[4]{a} \Rightarrow E=F(\sqrt[4]{a})$.

Help solving the second part would be much appreciated..

Answer 1

Let $E/F$ be Galois with degree 4. Since $i \in F$, $i$ will be unaffected by automorphisms of the Galois group. Complex conjugation is not an automorphism.

In both cases we will have a quadratic subfield since both groups have subgroups of index 2.

Let this quadratic subfield be $F(\sqrt{B})$. Our field $E$ is a quadratic extension of this field so we know that $E = F(\sqrt{A + \sqrt{B}})$. It is possible that $A=0$. Put $\alpha = \sqrt{A + \sqrt{B}}$.

The radical can be denested into the form $\sqrt{X} + \sqrt{Y}$ iff $A^2 − B$ is a square. In this case we will have the $C_2 \times C_2$ Galois group.

If the radical cannot be denested we must have a different Galois group than $C_2 \times C_2$ (because the two quadratic subfields would lead to a denesting). The only other group is $C_4$.

We will now use the fact that $i \in F$ to show that $A = 0$. We have an automorphism $\sigma$ of order 4. But we also have the automorphisms $\tau_{\pm} \alpha = \pm i \alpha$ which has order 4. Because the Galois group is $C_4$ we know that these automorphisms must coincide.

The minimal polynomial of $\alpha$ is $X^4 - 2 A X^2 + (A^2 - B) = 0$ the following conjugates must lie in the field $F$:

• $\alpha_1 = \sqrt{A + \sqrt{B}}$
• $\alpha_2 = \sqrt{A - \sqrt{B}}$
• $\alpha_3 = -\sqrt{A + \sqrt{B}}$
• $\alpha_4 = -\sqrt{A - \sqrt{B}}$

We can consider the possible cases:

• $\sigma \alpha_1 = \alpha_2$. This would imply that $\sigma \alpha_3 = \alpha_4$. So to have an order 4 automorphism we would need $\sigma \alpha_2 = \alpha_3$. Now $\sigma \alpha_1' = i \alpha_2 = \sqrt{-A + \sqrt{B}}$
• $\sigma \alpha_1 = \alpha_3$. This would have order 2, impossible.
• $\sigma \alpha_1 = \alpha_4$. This would be the same as earlier but $-i \alpha_1$. WLOG we will assume we are not in this case since it is symmetrical with the $\alpha_2$ case.

furthermore (WLOG we will consider $\sigma = \tau_+$)

• $\tau_+ \alpha_1 = i \sqrt{A + \sqrt{B}} = \sqrt{-A - \sqrt{B}}$

equating these proves $A = 0$, thus $E = F(\sqrt{\sqrt{B}})$.