# Automorphism Group of $\Q(\sqrt{2})$ Over $\Q$. ## Problem 97

Determine the automorphism group of $\Q(\sqrt{2})$ over $\Q$. Add to solve later

## Proof.

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

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

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

Thus we must have $\sigma(\sqrt{2})=\sqrt{2}$. Hence $\sigma$ is trivial.
In conclusion, we have
$\Aut(\Q(\sqrt{2})/Q)=\{1\}.$ Add to solve later

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