Let K be the splitting field of the polynomial $x^4-2x^2-2$. Find an automorphism $\sigma \in Gal(K/\mathbb{Q})$ of order 4. My attempt:
The roots of the polynomials are $x=\sqrt{1+\sqrt{3}}, -\sqrt{1+\sqrt{3}}, -\sqrt{1-\sqrt{3}}, -\sqrt{1-\sqrt{3}}$. and the automorphisms $\sigma \in Gal(K/\mathbb{Q})$ permutes the root of the polynomial. But none of such automorphisms are of order 4! I don't know how to proceed further.
Thanks for any help! 
 A: First observe that $K$ must be a degree 8 extension over $\Bbb Q$. Here $\Bbb Q(\sqrt{1+\sqrt{3}})$ is a degree 4 extension, but note that it cannot contain $\sqrt{1-\sqrt{3}}$ since $\Bbb Q(\sqrt{1+\sqrt{3}}) \subset \Bbb R$ whereas  $\sqrt{1-\sqrt{3}} \not \in \Bbb R$. Thus, we can only have all of the roots of $x^4-2x^2-2$ by a further quadratic extension. (Here $\sqrt{1-\sqrt{3}}$ is only degree 2 over $\Bbb Q(\sqrt{1+\sqrt{3}})$ since it satisfies $x^2 -1+\sqrt{3} = 0$ and the latter field contains $\sqrt{3}$.)
Now we know that automorphisms in the Galois group $\text{Gal}(F/\Bbb Q)$ are determined by how they permute the elements of $x^4-2x^2-2$, so the only possibility for a group of order 8 must be given by the Sylow 2-subgroups of $S_4$, which contain two 4-cycles apiece.  The only suitable candidates for these 4-cycles are
$$
\sqrt{1+\sqrt{3}} \to 
\sqrt{1-\sqrt{3}} \to
-\sqrt{1+\sqrt{3}} \to
-\sqrt{1-\sqrt{3}} \to
\sqrt{1+\sqrt{3}}
$$
or its inverse.  This follows since if $\sqrt{1+\sqrt{3}}$ is sent to $-\sqrt{1+\sqrt{3}}$ that will force a 2-cycle, and similarly for the pair $\pm\sqrt{1-\sqrt{3}}$.
If you don't want to appeal to abstract Galois theory and want a concrete justification that this permutation of elements corresponds to a well-defined field automorphism, this can be done directly by modeling $K$ as $\Bbb Q[x,y]/(x^4-2x^2-2, y^2+x^2-2)$, and checking that the self-map given by $x\mapsto y, y\mapsto -x$ is well-defined on the quotient. Here the justification for the $y^2+x^2-2$ is given by $(\sqrt{1+\sqrt{3}})^2 + (\sqrt{1-\sqrt{3}})^2 = 1+\sqrt{3}+1-\sqrt{3} = 2$.
