Q:

Computing a Galois group and well-definedness of isomorphism

Thank you for watching this question.  On the way of studying Galois theory I had one question about well-definedness of isomorphism. 
For example, let $\alpha $ be $\sqrt{3}+\sqrt{5}$ and $F$ be $\mathbb{Q}(\alpha)$ .  I can understand the minimal polynomial of $\alpha$ is $f(x)=x^4-16x^2+4$ and $|\mathrm{Gal}(F/\mathbb{Q})|=4$. Roots of $f(x)=0$ are $\alpha_1=\alpha,\alpha_2=-\sqrt{3}+\sqrt{5},\alpha_3=\sqrt{3}-\sqrt{5}$ and $\alpha_4=-\sqrt{3}-\sqrt{5}$, so there is a isomorphism $\phi$ over $\mathbb{Q}$ such that $\phi(\alpha_1)=\alpha_2,\phi(\alpha_3)=\alpha_4$.
My question:Why is there no isomorphism $\alpha_1\rightarrow \alpha_2,\alpha_3\rightarrow \alpha_1$?  Of corse $\phi(\alpha_3)$ is automatically decided because of $\alpha_1\alpha_3=-2$. But this method is "special" for this question.  I want to know methods which can be used on any question.  Is there anyone who give me some advice? 

A:

As mentioned in the comments in this particular case an automorphism of the field $\Bbb{Q}[\alpha]$ is determined by its action on $\sqrt{3}$ and $\sqrt{5}$.  To answer your question as to what is going on more generally one can look at the discriminant $\Delta = \prod_{i<j}(\alpha_i-\alpha_j)^2$.  $\Delta$ is fixed by the action of the Galois group, but if we consider the action of the Galois group on $\sqrt{\Delta}$, we find for $\sigma$ in the Galois group, $\sigma(\sqrt{\Delta}) = \text{sign}(\sigma)\sqrt{\Delta}$.  It's obvious that $\Delta$ lies in the ground field (since it's fixed by the Galois group), but if $\Delta$ is a square, then $\sqrt{\Delta}$ also lies in the ground field from which we can conclude that all the elements of the Galois group have even signature.   Finally getting back to your question about $\Bbb{Q}[\alpha]$.  Since the discriminant of $f(x)$ equals $2^{14}\,3^2\,5^2$, we can conclude the discriminant of $\Bbb{Q}[\alpha]$ is a square (in fact, $2^4\,3^2\,5^2$).   The reason there can be no isomorphism $\alpha_1\rightarrow \alpha_2,\alpha_3\rightarrow \alpha_1$ is that such an isomorphism would necessarily have signature $-1$ contradicting $\sqrt{\Delta}$ is fixed by the Galois group.