Question about the roots of an irreducible polynomial How can we deduce that each field is isomorphic?
$g(x) := x^4 - 10x^2 + 1$ is an irreducible polynomial with $\alpha_{1} := \sqrt{2}+ \sqrt{3}$ as a zero. Show that $\alpha_{2} := \sqrt{2}- \sqrt{3},\alpha_{3} :=- \sqrt{2}+ \sqrt{3}, \alpha_{4} := - \sqrt{2}- \sqrt{3}$ are also zeros of $g(x)$ in $\mathbb{C}$ and deduce that the four fields $\mathbb{Q}(\alpha _i), \quad i = 1,...,4$ are isomorphic to each other.$
Thanks
 A: You have
$$
x^4 - 10x^2 + 1 = (x^2 - 5)^2 - 24.
$$
Note that $\pm \sqrt 2 \pm \sqrt 3 = \pm 5 \pm 2 \sqrt 6$, so that 
$$
(\pm \sqrt 2 \pm \sqrt 3)^2 = (\sqrt 2 \pm \sqrt 3)^2 = 2 \pm 2 \sqrt 6 + 3 = 5 \pm 2 \sqrt 6.
$$
Now if you remove $5$ and square again you get $24$. So essentially the other roots are the same as the first root up to two signs because you square twice in the process and these "up to a sign" things are destroyed when squaring. 
Now the automorphisms of both $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ lift to automorphisms of $\mathbb Q(\sqrt 2, \sqrt 3)$, and since each of the automorphisms of $\mathbb Q(\sqrt 2,\sqrt 3)$ must restrict to a pair of automorphisms for both $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$, you know that the four automorphisms are just sign changes, hence giving you these four roots. 
Hope that helps,
A: No, these are the roots indeed (check it!).
Now, to see those fields are isomorphic, not that when $\alpha$ is a root of $g(x) \in F[x]$ irreducible (here $F$ is a field), then $$ F(\alpha) \simeq \frac{F[x]}{\left(g(x)\right)} $$
(prove this)
If you also want to check it is indeed irreducible, check that ${\mathbb Q}(\alpha_1) = {\mathbb Q}(\sqrt{2},\sqrt{3})$ (hint: $1/\alpha_1 = \sqrt{3} - \sqrt{2}$), and then it is easier to see that $[{\mathbb Q}(\sqrt{2},\sqrt{3}) : {\mathbb Q}] = 4$.
