# Galois Group of the Polynomial $x^p-2$.

## Problem 110

Let $p \in \Z$ be a prime number.

Then describe the elements of the Galois group of the polynomial $x^p-2$.

## Solution.

The roots of the polynomial $x^p-2$ are
$\sqrt[p]{2}\zeta^k, k=0,1, \dots, p-1$ where $\sqrt[p]{2}$ is a real $p$-th root of $2$ and $\zeta$ is a primitive $p$-th root of unity.
(Explicitly, you may take $\zeta=e^{2\pi i/p}$.)

Thus $x^p-2$ is a separable polynomial over $\Q$. The Galois group of $x^p-2$ is the Galois group of the splitting field of $x^p-2$.
The splitting field of $x^p-2$ is $K:=\Q(\sqrt[p]{2}, \zeta)$ of extension degree $p(p-1)$.
(Check this.)

Let $G=\Gal(K/\Q)$ be the Galois group of $x^p-2$. The order of the Galois group $G$ is $p(p-1)$. Let $\sigma \in G$ be an automorphism.
Then $\sigma$ sends an element of $G$ to its conjugate (a root of the minimal polynomial of the element.)
The minimal polynomial of $\sqrt[p]{2}$ is $x^p-2$ since it is irreducible by Eisenstein’s criteria.
The minimal polynomial of $\zeta$ is the cyclotomic polynomial
$\Phi(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1.$

Therefore $\sigma$ maps
\begin{align*}
\sqrt[p]{2} &\mapsto \sqrt[p]{2}\zeta^a \\
\zeta & \mapsto \zeta^b
\end{align*}
for some $a=0, 1, \dots, p-1$ and $b=1, 2, \dots, p-1$.

Thus there are $p(p-1)$ possible maps for $\sigma$.
Since the order of $G$ is $p(p-1)$, these are exactly the elements of the Galois group $G$ of the polynomial $x^p-2$.

##### Two Quadratic Fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are Not Isomorphic
Prove that the quadratic fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are not isomorphic.