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 $4$.

Galois Group of the Polynomial $x^2-2$
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 […]

Galois Group of the Polynomial $x^p-2$.
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 $\zeta$ […]

The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity
Prove that the polynomial $x^p-2$ for a prime number $p$ is irreducible over the field $\Q(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.
Hint.
Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of […]

Application of Field Extension to Linear Combination
Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.
Let $\alpha$ be any real root of $f(x)$.
Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.
Proof.
We first prove that the polynomial […]

$x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$
Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.
Hint.
Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.
Proof.
Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is […]

Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$
Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.
Proof.
Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein's criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the […]

Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.
(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?
(b) […]