Showing that a function is an automorphism of $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ I'm trying to prove that the function $\phi$ given $\phi(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ and $\phi(q)=q$ for all $q\in \mathbb{Q}$ is an automorphism of  the extension field $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})}))/\mathbb{Q}$ but I can't prove the surjectivity.
More precisely, I cannot find an element $x$ in the extension field such that $\phi(x)=\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$
I try this: Assuming $\phi$ is a homomorphism (which I already proved)
Let $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ and $\beta=\sqrt{(2-\sqrt{2})(3+\sqrt{3})})$
then
$\alpha=\frac{\alpha \beta}{\beta}=\frac{\sqrt{2}(3+\sqrt{3})}{\phi(\alpha)}$ and i would like that ${\sqrt{2}(3+\sqrt{3})}=\phi(y)$ some $y\in\mathbb{Q}(\alpha)$ but I get circular reasoning ...
 A: Mainlines for an answer
Denote
$$\begin{cases}
\pi &= \sqrt{2+ \sqrt 2}\\
\rho &= \sqrt{3+ \sqrt 3}
\end{cases}$$
$\pi, \rho$ are respectively roots of the rational polynomials
$$\begin{cases}
p(x) &= x^4 - 4 x^2 +2\\
r(x) &= x^4 - 9 x^2 + 3
\end{cases}$$
$p,r$ are irreducible according to Eisenstein's criterion. The set of roots

*

*of $p$ is $\{ \pi_1=\sqrt{2+ \sqrt 2}, \pi_2=\sqrt{2- \sqrt 2}, \pi_3=-\sqrt{2+ \sqrt 2},\pi_4=-\sqrt{2-\sqrt 2}\}$

*the one of $r$ is $\{ \rho_1=\sqrt{3+ \sqrt 3}, \rho_2=\sqrt{3- \sqrt 3}, \rho_3=-\sqrt{3+ \sqrt 3},\rho_4=-\sqrt{3-\sqrt 3}\}$.

$\alpha, \beta$ belong to the finite field extension $\mathbb Q(\pi, \rho)/\mathbb Q$ and the field homomorphism $\phi$ satisfies $\phi(\alpha) = \beta$.
You suppose that $\beta$ belongs to $\mathbb Q(\alpha) / \mathbb Q $ which remains to be proved.
To do so, notice that $\alpha \notin \mathbb Q(\pi, \sqrt 3)$. If that would be the case, $\rho$ would also belong to $\mathbb Q(\pi, \sqrt 3)$. Which in turn implies the contradiction that $\{1, \sqrt 2, \sqrt 3, \sqrt 6\}$ would be linearly dependent over $\mathbb Q$. Similarly $\beta$ doesn't belong to $\mathbb Q(\pi, \sqrt 3)$.
Therefore $\mathbb Q(\alpha) = \mathbb Q(\beta) = \mathbb Q(\pi, \rho)$ and if $\phi(\alpha) = \beta$, $\phi$ can be extended into an homeomorphism of $\mathbb Q(\alpha)$ which is also an automorphism.
