Show $\sqrt{2 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{2 + \sqrt{2}})$ I'm in the middle of a proof where I'd like to show that $\sqrt{2 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{2 + \sqrt{2}})$
The only way I can think of involves finding an explicit set representation for $\mathbb{Q}(\sqrt{2 + \sqrt{2}})$.
At first I tried showing $\mathbb{Q}(\sqrt{2 + \sqrt{2}}) = \{a + b\sqrt{2 + \sqrt{2}}: a,b\in\mathbb{Q}\}$
and then realised this is probably false, as it doesn't look like it contains $\sqrt{2}$.
I figured I could try $\mathbb{Q}(\sqrt{2 + \sqrt{2}}) = \{a + b\sqrt{2 + \sqrt{2}} + c\sqrt{2}: a,b,c\in\mathbb{Q}\}$
but this method seems really long-winded. I'm pretty sure there'll be plenty of shorter methods, but I don't know any method to show this.
Any pointers appreciated!
 A: These expressions show up when computing the sine and cosine of $\pi/8 = 22.5^\circ$. Since the corresponding extensions are cyclotomic, hence normal (and even abelian), any extension of the automorphism sending $\sqrt{2}$ to $-\sqrt{2}$ must fix the extension.
A: Actually, this statement is not true.
Using $(2+\sqrt{2})(2-\sqrt{2})=2$, taking square roots and dividing, we see that $$\sqrt{2-\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$$
So, since $\sqrt{2}\in\mathbb Q(\sqrt{2+\sqrt{2}})$, $\sqrt{2-\sqrt{2}}\in\mathbb Q(\sqrt{2+\sqrt{2}})$.
A: $\begin{eqnarray}{\rm\bf Hint}\ \ w-1 && =\ \  (w+1)\,(w-1)^2\ \ \ {\rm by}\ \ \ w^2-1 = 1\ \ \ {\rm for}\ \ \ w := \sqrt{2}  \\  
\Rightarrow\ \ \ \ 2-w && =\ \  (2+w)\, (w-1)^2\ \ \ {\rm by}\ \ \ w * {\rm prior,\ \ using}\,\ \ w^2 = 2  \\
\Rightarrow\ \sqrt{2-w} && =\, \sqrt{2+w}\,\ (w-1) \ \ \ \ \ {\rm by\ taking\ \sqrt{prior}} \\
\Rightarrow\ \sqrt{2-w} && \in\Bbb Q(\sqrt{2+w}) = {\rm R}\ \ \ \ \ \ {\rm by}\ \ \sqrt{2+w}^{\,2}\! =\, 2+w\in {\rm R}\:\Rightarrow\: w \in\rm R
\end{eqnarray} $
A: Let $\alpha = \sqrt{2 + \sqrt{2}}$ and $\beta = \sqrt{2 - \sqrt{2}}$. I want to find out if $\mathbb Q(\alpha)$ is equal to $\mathbb Q(\beta)$ or not. They are equal but I was wondering about proving it with Galois theory instead of direct computation.
Since $\alpha,-\alpha,\beta,-\beta$ are the roots of the polynomial $P(x) = x^4 - 4x^2 + 2$ we could try building the splitting field of $P$ (which has degree 4 since the polynomial is irreducible with degree 4) by a chain of extensions like this:



the bottom ones all have degree 2 and the full tower has degree 4, so I think that implies the greyed out ones at the very top have degree 1 so it's a trivial extension, proving $\mathbb Q(\alpha) = \mathbb Q(\beta)$.
