# Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic Group ## Problem 231

Show that $\Q(\sqrt{2+\sqrt{2}})$ is a cyclic quartic field, that is, it is a Galois extension of degree $4$ with cyclic Galois group. Add to solve later

## Proof.

Put $\alpha=\sqrt{2+\sqrt{2}}$. Then we have $\alpha^2=2+\sqrt{2}$. Taking square of $\alpha^2-2=\sqrt{2}$, we obtain $\alpha^4-4\alpha^2+4=2$. Hence $\alpha$ is a root of the polynomial
$f(x)=x^4-4x+2.$ By the Eisenstein’s criteria, $f(x)$ is an irreducible polynomial over $\Q$.

There are four roots of $f(x)$:
$\pm \sqrt{2 \pm \sqrt{2}}.$ Note that we have a relation
$(\sqrt{2+\sqrt{2}})(\sqrt{2-\sqrt{2}})=\sqrt{2}.$ Thus we have
$\sqrt{2-\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} \in \Q(\sqrt{2+\sqrt{2}}).$

Hence all the roots of $f(x)$ are in the field $\Q(\sqrt{2+\sqrt{2}})$, hence $\Q(\sqrt{2+\sqrt{2}})$ is the splitting field of the separable polynomial $f(x)=x^4-4x+2$.
Thus the field $\Q(\sqrt{2+\sqrt{2}})$ is Galois over $\Q$ of degree $4$.

Let $\sigma \in \Gal(\Q(\sqrt{2+\sqrt{2}})/ \Q)$ be the automorphism sending
$\sqrt{2+\sqrt{2}} \mapsto \sqrt{2-\sqrt{2}}.$ Then we have
\begin{align*}
2+\sigma(\sqrt{2})&=\sigma(2+\sqrt{2})\\
&=\sigma\left((\sqrt{2+\sqrt{2}}) ^2 \right)\\
&=\sigma \left(\sqrt{2+\sqrt{2}} \right) ^2\\
&= \left(\sqrt{2-\sqrt{2}} \right)^2=2-\sqrt{2}.
\end{align*}
Thus we obtain $\sigma(\sqrt{2})=-\sqrt{2}$.

Using this, we have
\begin{align*}
\sigma^2(\sqrt{2+\sqrt{2}})&=\sigma(\sqrt{2-\sqrt{2}})\\
&=\sigma \left(\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} \right)\\
&=\frac{\sigma(\sqrt{2})}{\sigma(\sqrt{2+\sqrt{2}})} \\
&=\frac{-\sqrt{2}}{\sqrt{2-\sqrt{2}}} \\
&=-\sqrt{2-\sqrt{2}}.
\end{align*}
Therefore $\sigma^2$ is not the identity automorphism. This implies the Galois group $\Gal(\Q(\sqrt{2+\sqrt{2}})/ \Q)$ is generated by $\sigma$, that is, the Galois group is a cyclic group of order $4$. Add to solve later

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