# Automorphism Group of $\Q(\sqrt[3]{2})$ Over $\Q$.

## Problem 97

Determine the automorphism group of $\Q(\sqrt[3]{2})$ over $\Q$.

## Proof.

Let $\sigma \in \Aut(\Q(\sqrt[3]{2}/\Q)$ be an automorphism of $\Q(\sqrt[3]{2})$ over $\Q$.

Then $\sigma$ is determined by the value $\sigma(\sqrt[3]{2})$ since any element $\alpha$ of $\Q(\sqrt[3]{2})$ can be written as $\alpha=a+b\sqrt[3]{2}+c\sqrt[3]{2}^2$ for some $a,b,c \in \Q$ and
\begin{align*}
\sigma(\alpha) &=\sigma(a+b\sqrt[3]{2}+c\sqrt[3]{2}^2)\\
&=\sigma(a)+\sigma(b)\sigma(\sqrt[3]{2})+\sigma(c)\sigma(\sqrt[3]{2})^2\\
&=a+b\sigma(\sqrt[3]{2})+c\sigma(\sqrt[3]{2})^2
\end{align*}
since $\sigma$ fixes the elements in $\Q$.

Note that $\sigma(\sqrt[3]{2})$ is a root of the minimal polynomial $x^3-2$ of $\sqrt[3]{2}$.
But the roots of $x^3-2$ are not real except $\sqrt[3]{2}$, hence not in $\Q(\sqrt[3]{2})$.

Thus we must have $\sigma(\sqrt[3]{2})=\sqrt[3]{2}$. Hence $\sigma$ is trivial.
In conclusion, we have
$\Aut(\Q(\sqrt[3]{2})/Q)=\{1\}.$