Finding roots of polynomial over $\mathbb{Q}$ satisfying isomorphism

Question

Let $f(x)=x^4-2x^2-2 \in \mathbb{Q}[x]$. Find the the roots $\alpha,\; \beta$ of $f(x)$ such that $\mathbb{Q}(\alpha) \cong \mathbb{Q}(\beta)$.What is the splitting field of $f(x)$.

Attemption :

I consider the theorem saying:

$$\text{If F is a field and } f(x) \in \text{F[x] is irreducible over F with } f(a)= 0, \text{a} \in \text{F(a). F(a) is some extension of F that contains the root of } f(x) \text{, then } \dfrac{F[x]}{\langle f\rangle} \cong F(a)$$

We observe that $f(x)$ has $4$ roots and we also observe that neither of those roots does not contained in $\mathbb{Q}$ since by Eisenstein Criterion with $p=2$ we see $f$ is irreducible over $\mathbb{Q}$. Now, let $x^2:=u$ using these I defined $g(u)=u^2-2u-2 \in \mathbb{Q}[u]$

$$(u-1)^2-3=0 \rightarrow u=1\pm\sqrt{3}$$ Using back substitution we have;

$$x_{1,2}=\pm\sqrt{1+\sqrt{3}}, \; x_{3,4}=\pm i\sqrt{\sqrt{3}-1}$$

Using the theorem above we have:

$$\dfrac{Q[x]}{\langle x^4-2x^2-2\rangle} \cong \mathbb{Q}\left(\sqrt{1+\sqrt{3}}\right)$$ and $$\dfrac{Q[x]}{\langle x^4-2x^2-2\rangle} \cong \mathbb{Q}\left(i\sqrt{\sqrt{3}-1}\right)$$

Means that $\mathbb{Q}\left(\sqrt{1+\sqrt{3}}\right) \cong \mathbb{Q}\left(i\sqrt{\sqrt{3}-1}\right)$

And finally the splitting field is $$\mathbb{Q}\left(\left(\sqrt{1+\sqrt{3}}\right),\left(i\sqrt{\sqrt{3}-1}\right)\right)$$

Do I make any mistake? Thanks in advance!

Answer 1

Your solution is fine. As ancient mathematician commented, you might spell out your choices of $\alpha$ and $\beta$ though arguably you did :-)

Just in case the exercise continues in a predictable way, you may want to modify the second generator. Instead of $\alpha=\sqrt{1+\sqrt3}$ and $\beta=i\sqrt{\sqrt3-1}$ you may consider using $\alpha$ and $\alpha\beta=i\sqrt2$. With those two generators on display it is, for example, obvious what the extension degree of the splitting field will be.

It is also possible that this splitting field will be revisited after studying Galois groups. At that point do check out this recipe. There are many other threads linked to it, and they contain a wealth of illuminating examples.