## Galois Group (Example)

This post is about the Galois group of $K$ over $\mathbb{Q}$, where $K$ is the splitting field of $f(x)=x^p-2$, where $p$ is an odd prime.

First we show that the polynomial $f(x)=x^p-2$ is irreducible over $\mathbb{Q}$. This follows immediately by Eisenstein’s Criterion, since $2\mid (-2)$, $2\nmid 1$ and $2^2\nmid (-2)$.

Next, we show that the splitting field $K$ of $f(x)$ in $\mathbb{C}$ is $Q(\sqrt[p]{2},\omega)$, where $\omega=e^{2\pi i/p}$ is a primitive p-th root of unity. The roots of $f$ are $\sqrt[p]2, \sqrt[p]2\omega, \sqrt[p]2\omega^2, \dots, \sqrt[p]2\omega^{p-1}$.

The splitting field $K$ contains $\sqrt[p]2$ and $\omega=\frac{\sqrt[p]2\omega^2}{\sqrt[p]2\omega}$. Thus $\mathbb{Q}(\sqrt[p]2,\omega)\subseteq K$.

On the other hand, $\mathbb{Q}(\sqrt[p]2,\omega)$ contains all the roots of $f$, hence $f$ splits in $\mathbb{Q}(\sqrt[p]2, \omega)$. Thus $K\subseteq\mathbb{Q}(\sqrt[p]2,\omega)$, since $K$ is the smallest field that contains $\mathbb{Q}$ and all the roots of $f$. All in all, we have that the splitting field $K=\mathbb{Q}(\sqrt[p]2, \omega)$.

The next part involves determining the Galois group of $K$ over $\mathbb{Q}$. We have $|Gal(K/\mathbb{Q})|=[K:\mathbb{Q}]$. Since $[\mathbb{Q}(\sqrt[p]2):\mathbb{Q}]=p$ (minimal polynomial $x^p-2$), and $[\mathbb{Q}(\omega):\mathbb{Q}]=p-1$ (minimal polynomial the cyclotomic polynomial $1+x+x^2+\dots+x^{p-1}$), thus $|Gal(K/\mathbb{Q})|=p(p-1)$. Here we have used the lemma that suppose $[F(\alpha):F]=m$ and $[F(\beta):F]=n$ with $\gcd(m,n)=1$, then $[F(\alpha,\beta):F]=mn$.

What the Galois group does is it permutes the roots of $f$. Let $\sigma$ be an element of the Galois group. $\sigma(\sqrt[p]2)$ can possibly be $\sqrt[p]2, \sqrt[p]\omega, \dots, \sqrt[p]2\omega^{p-1}$, a total of $p$ choices. Similarly, $\sigma(\omega)=\omega, \omega^2, \dots,\omega^{p-1}$, a total of $p-1$ choices. All these total up to $p(p-1)$ elements, which is exactly the size of the Galois group.

The above Galois group $Gal(K/\mathbb{Q})$ is described by how its elements act on the generators. For a more concrete representation, we can actually prove that the Galois group above is isomorphic to the group of matrices $\begin{pmatrix}a&b\\0&1\end{pmatrix}$, where $a,b\in\mathbb{F}_p$, $a\neq 0$. We denote the group of matrices as $M$.

To show the isomorphism, we define a map $\phi: Gal(K/\mathbb{Q})\to M$, mapping $\sigma_{a,b}$ to $\begin{pmatrix}a&b\\0&1\end{pmatrix}$.

Notation: $\sigma_{a,b}$ is defined on the generators as follows, $\sigma_{a,b}(\sqrt[p]2)=\sqrt[p]2\omega^b$, $\sigma_{a,b}(\omega)=\omega^a$.

We can clearly see that the map $\phi$ is bijective. To see it is a homomorphism, we compute $\phi(\sigma_{a,b}\circ\sigma_{c,d})=\begin{pmatrix}ac&a+bd\\0&1\end{pmatrix}=\phi(\sigma_{a,b})\phi(\sigma_{c,d})$. 