I'm trying to solve the following problem: Let

\begin{equation*} f(x) = x^4 - 2x^2 - 2 \in \mathbb{Q}[x] \end{equation*}

and $ E $ be its splitting field. What is the degree $ [E: \mathbb{Q}] $?

First of all, the roots of this polynomial are $ \pm \sqrt{1 + \sqrt{3}}, \pm \sqrt{1 - \sqrt{3}} $. The first two are real and the other two are complex. It looks as if $ E $ was equal to $ \mathbb{Q}(\sqrt{1 + \sqrt{3}}, \sqrt{1 - \sqrt{3}}) $. The degree of these elements over $ \mathbb{Q} $ is $ 4 $ since $ f(x) $ is irreducible by Eistenstein's criterion. However, I can't tell what the degree of $ \sqrt{1 - \sqrt{3}} $ over $ \mathbb{Q}(\sqrt{1 + \sqrt{3}}) $ is, since I don't know how to check whether $ f $ is irreducible over this field.

I'd appreciate any ideas on how to get to that


Hint: $1-\sqrt{3}\in\Bbb Q(\sqrt{1+\sqrt{3}})$.

  • $\begingroup$ Wow, that was easy. Thanks! $\endgroup$ – Jytug May 10 '15 at 19:13

Note your field is $\mathbb{Q}$($\alpha$, $\beta$) where $\alpha$ and $\beta$ are your explicit roots and note then that $\alpha^2$ + $\beta^2$ = 2. Apply now the multiplication of degrees knowing that each of them is of degree 4 over $\mathbb{Q}$ and the other is of degree 2 over the first extension, by the above equation of sum of squares.


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