Consider $p(x)=x^4 -2x^2 -2 \in \mathbb{Q}[x]$. I have to find the roots $\alpha, \beta$ of $p(x)$ such that $\mathbb{Q}(\alpha)\cong\mathbb{Q}(\beta)$.
I already know that all the roots are $\sqrt{1+\sqrt{3}} ,  -\sqrt{1+\sqrt{3}}, \sqrt{1-\sqrt{3}},  -\sqrt{1-\sqrt{3}} $, so I think I have to take $\alpha=\sqrt{1+ \sqrt{3}}$ and $\beta=\sqrt{1-\sqrt{3}}$, and also I need to find the decomposition field of $p(x)$ so I think is enough to take $\mathbb{Q}(\alpha,\beta)$ because, let be $K$ the decomposition fiel of $p(x)$ then by definition it is the smallest in the one $p(x)$ factorizes completely, and clearly it factorizes in $\mathbb{Q}(\alpha,\beta)$ so, $K$$\subseteq$$\mathbb{Q}(\alpha,\beta)$, and in the other hand , notice that $\mathbb{Q}$$\subseteq$$K$, and $\alpha, \beta$ $\in$$K$, then $K$$\subseteq$$\mathbb{Q}(\alpha,\beta)$ that proves $K$=$\mathbb{Q}(\alpha,\beta)$. doesn't it?
Could someone give me some hint about it, be more accurate with the isomorphism?, thanks.
 A: As mentioned in the comment, $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ are isomorphic for any two roots of the irreducible polynomial $p$. I will just give the details. Define a ring homomorphism $\phi$ from $\mathbb{Q}[x]$ to $\mathbb{Q}(\alpha)$ where $\alpha$ is a root of the irreducible polynomial $p$. $\phi$ is defined by $\phi (h)=h(\alpha)$ for $h\in \mathbb{Q}[x]$ and one can verify that this is indeed a ring homomorphism.  Now the kernel of $\phi$ consists of those polynomials $h$ which vanish at $\alpha$. Since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is a principal ideal domain. Since the kernel contains the irreducible polynomial $p$, the kernel is nothing but the principal ideal $(p)$. Now, by the usual isomorphism theorem for rings, $\mathbb{Q}[x]/(p)$ is isomorphic to the image of $\phi$. Since, $p$ is irreducible and $\mathbb{Q}[x]$ is an Euclidean ring, the ideal $(p)$ is maximal. Hence, $\mathbb{Q}[x]/(p)$ is a field and so the image of $\phi$ is also a field. But $Im(\phi)$ contains $\alpha$ since $\phi(x)=\alpha$. It also contains all $q\in \mathbb{Q}$ as $\phi(q)=q$. Hence, by the definition of $\mathbb{Q}(\alpha)$, $Im(\phi)=\mathbb{Q}(\alpha)$. Hence, we have that $\mathbb{Q}[x]/(p)\equiv \mathbb{Q}(\alpha)$. Since $\alpha$ was any root, and the object $\mathbb{Q}[x]/(p)$ is common across all roots, we have $\mathbb{Q}(\alpha)\equiv \mathbb{Q}(\beta)$.
