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.

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 $4$.

The post Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic Group first appeared on Problems in Mathematics.

Let $\Q$ be the field of rational numbers.

(a) Is the polynomial $f(x)=x^2-2$ separable over $\Q$?

(b) Find the Galois group of $f(x)$ over $\Q$.

Solution.

(a) The polynomial $f(x)=x^2-2$ is separable over $\Q$

The roots of the polynomial $f(x)$ are $\pm \sqrt{2}$. Since all the roots of $f(x)$ are distinct, $f(x)=x^2-2$ is separable.

(b) The Galois group of $f(x)$ over $\Q$

The Galois group of the separable polynomial $f(x)=x^2-2$ is the Galois group of the splitting field of $f(x)$ over $\Q$.
Since the roots of $f(x)$ are $\pm \sqrt{2}$, the splitting field of $f(x)$ is $\Q(\sqrt{2})$. Thus, we want to determine the Galois group
\[\Gal(\Q(\sqrt{2})/\Q).\] Let $\sigma \in \Gal(\Q(\sqrt{2})/\Q)$. Then the automorphism $\sigma$ permutes the roots of the irreducible polynomial $f(x)=x^2-2$.

Thus $\sigma(\sqrt{2})$ is either $\sqrt{2}$ or $-\sqrt{2}$. Since $\sigma$ fixes the elements of $\Q$, this determines $\sigma$ completely as we have
\[\sigma(a+b\sqrt{2})=a+b\sigma(\sqrt{2})=a\pm \sqrt{2}.\]

The map $\sqrt{2} \mapsto \sqrt{2}$ is the identity automorphism $1$ of $\Q \sqrt{2}$.
The other map $\sqrt{2} \mapsto -\sqrt{2}$ gives non identity automorphism $\tau$. Therefore, the Galois group $\Gal(\Q(\sqrt{2})/\Q)=\{1, \tau\}$ is a cyclic group of order $2$.

In summary, the Galois group of the polynomial $f(x)=x^2-2$ is isomorphic to a cyclic group of order $2$.

The post Galois Group of the Polynomial $x^2-2$ first appeared on Problems in Mathematics.