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?

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.
