Show $Gal(E/\mathbb{Q})\cong \mathbb{\mathbb{Z_2\times Z_2}}$

Solution 1:

I want to give a thourough explanation of what is going on. Our goal ist to understand $Gal(E/\mathbb{Q})$ where $E$ is the splitting field of $X^4-4\in \mathbb{Q}[X]$. Clearly in $\mathbb{C}[X]$ this polynomial factors $(X^2-2)(X^2+2)=(X-\sqrt{2})(X+\sqrt{2})(X+i\sqrt{2})(X-i\sqrt{2})$ and the splitting field is thus genereated by the roots of this polynomial, hence $E=\mathbb{Q}(\sqrt{2},i\sqrt{2})$. We know that $\#Gal(E/\mathbb{Q})=[E:\mathbb{Q}]=4$, because we have $[E:\mathbb{Q}]=[E:\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2\cdot 2=4$.

Consider the canonical embedding $\sigma:\mathbb{Q}\to \mathbb{Q}(\sqrt{2})$ by a standard theorem we know that for every root of $X^2-2$ in $\mathbb{Q}(\sqrt{2})$ we get a continuation of $\sigma$ that send $\sqrt{2}$ to the root. Let $\sigma':\mathbb{Q}(\sqrt{2})\to \mathbb{Q}(\sqrt{2})\hookrightarrow E$ be the canonical continuation of such a continuation to $E$ by inclusion. Apllying the same argument again we obtain the four elements of the galois group: $$ \sigma_1=\operatorname{id}_E; \ \sigma_2:\sqrt{2}\to -\sqrt{2},i\sqrt{2}\to i\sqrt{2};\ \sigma_3:\sqrt{2}\to \sqrt{2},i\sqrt{2}\to -i\sqrt{2}; \ \sigma_4:\sqrt{2}\to -\sqrt{2},i\sqrt{2}\to -i\sqrt{2}$$ In particular we know that $Gal(E/\mathbb{Q})$ is $\cong \mathbb{Z}/4\mathbb{Z}$ or $\cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. In order to find out which of those is the case we look at the order of the elements in $Gal(E/\mathbb{Q})$, but one easily sees that no element has order $4$. Hence, $Gal(E/\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$.