# Galois Group of $x^4-5$ over $\mathbb{Q}(\sqrt{5})$

Goal is to compute the Galois group of the polynomial $f(x)=x^4-5$ over $\mathbb{Q}(\sqrt{5})$. My difficulty is computing the degree of this particular Galois extension.

My attempt:

So over $\mathbb{Q}(\sqrt{5})$ we can factor $f$ as: $$x^4-5 = (x^2+\sqrt{5})(x^2-\sqrt{5})$$ Where $x^2-\sqrt{5}$ is the minimal polynomial of $\sqrt[4]{5}$ over $\mathbb{Q}(\sqrt{5})$. Hence irreducible. Then with the roots being $R(f)=\{\sqrt[4]{5},i\sqrt[4]{5}\}$ our splitting field would be $E:=\mathbb{Q}(\sqrt{5})[\sqrt[4]{5},i]$. Finally, since $f$ is separable and splits in $E$, we have that $E/\mathbb{Q}(\sqrt{5})$ is Galois.

Now this is where I have run into problems. I now want to compute the degree of the Galois extension: $$|Gal(E/\mathbb{Q}(\sqrt{5})|=[E:\mathbb{Q}(\sqrt{5})]$$ but in this case how can I do that?

• Use the basis $1, i, a, ia$ with $a=\sqrt[4]5$. – dan_fulea Aug 5 '18 at 17:07
• For the Galois group, see this question. – Dietrich Burde Aug 5 '18 at 20:40
• @DietrichBurde thanks, that link was very helpful! – math189925 Aug 6 '18 at 15:02

Clearly, $\big[E:\mathbb{Q}(\sqrt[4]{5})\big]>1$, as $\text{i}\in E$ is not a real number, whence it does not lie in $\mathbb{Q}(\sqrt[4]{5})=K(\sqrt[4]{5})$, where $K:=\mathbb{Q}(\sqrt{5})$. Because $x^2+1\in\big(\mathbb{Q}(\sqrt[4]{5})\big)[x]$ is a polynomial with root $\text{i}$, we conclude that it is irreducible over $\mathbb{Q}(\sqrt[4]{5})$, whence $$\big[E:\mathbb{Q}(\sqrt[4]{5})\big]=2\,,\text{ noting that }E=\big(\mathbb{Q}(\sqrt[4]{5})\big)(\text{i})\,.$$
Now, show that $\big[\mathbb{Q}(\sqrt[4]{5}):K\big]=2$. It should be easy to verify that $\sqrt[4]{5}\notin K$, and that there exists a polynomial $p(x)\in K[x]$ of degree $2$ with root $\sqrt[4]{5}$. You can also see that $\text{Gal}(E/K)$ is the Klein four-group.
$E=K(i)$ where $K=\Bbb Q(\sqrt[4]5)$. As $K\subseteq\Bbb R$ but $K\subseteq\Bbb Q$ then $|E:K|>1$. But of course $|E:\Bbb Q(\sqrt5)|=|E:K||K:\Bbb Q(\sqrt5)|=\cdots.$