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

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}} $

Answer 1

Letting $\alpha=\sqrt{2+\sqrt{2}}$ and $\beta=\sqrt{2-\sqrt{2}}$, we can use the following fact: $$\alpha\beta=\alpha^2-2=\sqrt{2}$$ and `follow' an element around under one of these automorphisms. More precisely, if there is an automorphism sending $\alpha$ to $\beta$… calling it $\tau'$, what is $\tau'(\tau'(\alpha))=\tau'(\beta)$? In other words, lets examine the order of a non-identity element of the Automorphism group.

$$\tau'(\tau'(\alpha))=\tau'(\beta)=\tau'\left(\frac{\sqrt{2}}{\alpha}\right)=\frac{\tau'(\sqrt{2})}{\tau'(\alpha)}=\frac{-\sqrt{2}}{\beta}=-\alpha$$ What does this tell you about the order of $\tau'$? And what do we know about the order of an element of a group in relation to the order of the group?

Answer 2

using from what I have about about the minimal polynomial being $x^4-4x^2+2$ which gives $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ as the splitting field since this polynomial is irreducible and also separable further,

$\sqrt{2} = (\sqrt{2+\sqrt{2}})^2-2$ so it follows that

$\displaystyle\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} = \sqrt{2-\sqrt{2}}$ and now all roots sit inside this field so clearly a degree 4 extension.

Let $\alpha = \sqrt{2+\sqrt{2}}), \beta = \sqrt{2-\sqrt{2}})$

Now the automorphisms are identity,

$\sigma : \alpha \mapsto -\alpha, -\alpha \mapsto \alpha, \beta \mapsto -\beta, -\beta \mapsto \beta$

$\tau : \alpha \mapsto \beta, \beta \mapsto -\alpha, -\alpha \mapsto -\beta, -\beta \mapsto \alpha$

$\eta : \alpha \mapsto -\beta, -\alpha \mapsto \beta, \beta \mapsto \alpha, -\beta \mapsto -\alpha$

and clearly $\sigma\sigma = id, \tau\tau=\sigma, \sigma\tau = \eta, \sigma\eta = \tau, \eta\eta = \sigma, \tau\eta = id $

hence we have $C_4 \simeq \mathbb{Z}_4$

Answer 3

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.