TITLE: Show that $\mathbb{Q}(\sqrt{2 +\sqrt{2}})$ is a cyclic quartic field i.e. is a galois extension of degree 4 with cyclic galois group
QUESTION [9 upvotes]: Show that $\mathbb{Q}(\sqrt{2 +\sqrt{2}})$ is a cyclic quartic field i.e. is a galois extension of degree 4 with cyclic galois group
with some elementary algebra,
$x - \sqrt{2 +\sqrt{2}} = 0 \implies x^2 = 2+\sqrt{2} \implies (x^2 - 2)^2 = 2 \implies x^4-4x^2+2=0$ is the minimal polynomial and is Eisenstein at 2 so irreducible which gives seperable.
and the roots of $x^4-4x^2+2=0$ are $\pm \sqrt{2 \pm \sqrt{2}}$ again found with elementary algebra. So the splitting field is clearly a degree 4 extension.
Now I've confused myself a bit trying to work out all the automorphisms
since the polynomial is seperable I know $\mid AUT(\mathbb{Q}(\sqrt{2 +\sqrt{2}})/\mathbb{Q}\mid = [\mathbb{Q}(\sqrt{2 +\sqrt{2}}) : \mathbb{Q}]$ so there are 4. So it needs to be ismorphic to $\mathbb{Z}_4$
I think the only automorphisms are id, $\sigma : \sqrt{2 +\sqrt{2}} \mapsto -\sqrt{2 +\sqrt{2}}, \tau : \sqrt{2 -\sqrt{2}} \mapsto -\sqrt{2 +\sqrt{2}} $

REPLY [7 votes]: Here’s a much more advanced argument than you wanted, but one that may prove instructive.
If it’s true that the field $K=\Bbb Q\bigl(\sqrt{2+\sqrt2}\,\bigr)$ is cyclic over $\Bbb Q$, then there’s really only one candidate for it. It’s quartic, supposed cyclic, and pretty clearly ramified only above $2$. But the abelian extensions of $\Bbb Q$ are just those contained in the cyclotomic fields $\mathscr C_n$, and ramification only above $2$ implies that $K$ would be contained in $\mathscr C_{2^m}=\Bbb Q(\zeta_{2^m})$ for some $m$.
You use one more fact about your field, that it is totally real, since all conjugates of your irrational quantity are real. So $K$ should be the real subfield of some $2^m$-power cyclotomic field. Since $[\mathscr C_n:\Bbb Q]=\deg(\Phi_n)=\phi(n)$, where $\Phi_n$ is the $n$-th cyclotomic polynomial, and $\phi$ is the Euler function, you guess that $K$ is the real subfield of $\mathscr C_{16}$.
Almost done now, $\Phi_{16}(X)=X^8+1$, and if $\zeta=\zeta_{16}$ is a root, we put $\xi=\zeta+\overline\zeta=\zeta+\zeta^{-1}$. We have:
\begin{align}
0&=\zeta^4+\zeta^{-4}\\
\xi^4&=\zeta^4+4\zeta^2+6+4\zeta^{-2}+\zeta^{-4}\\
&=4\zeta^2+6+4\zeta^{-2}\\
\xi^4-4\xi^2&=-2\,,
\end{align}
and there you are with your minimal polynomial for your irrational quantity, which necessarily is equal to $\zeta_{16}+\zeta_{16}^{-1}$. The Galois group is the quotient of $\text{Gal}(\mathscr C_{16}/\Bbb Q)\cong(\Bbb Z/16\Bbb Z)^*\cong C_4\oplus C_2$, where $C_n$ is cyclic group of order $n$. It’s quick that you’re taking the quotient by that second term of the direct sum, so you’re left with just the $C_4$ for the Galois group.