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Construct the splitting field for the polynomial $(x^4 - x^2 - 2)$ over $\mathbb Q$ (rationals) . What is the degree of the extension? Why?

How would one go about tackling this question? I'm a bit lost, the notes I have a aren't great and there's no clear worked example. I'm working within finite fields and haven't started Galois theory yet.

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    $\begingroup$ The first thing to do is to look at $g(t)=t^2-t-2$, and find its splitting field (easy, right?). Then maybe take the square roots of your two $t$’s? $\endgroup$ – Lubin Mar 20 '14 at 20:58
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    $\begingroup$ $x^4-x^2-2 = (x^2-2)(x^2+1) = (x-\sqrt{2})(x+\sqrt{2})(x+i)(x-i)$ where $i = \sqrt{-1}$. $\endgroup$ – Dilip Sarwate Mar 20 '14 at 20:58
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    $\begingroup$ Tha last time I checked $\Bbb{Q}$ was an infinite set, so I removed a tag :-) $\endgroup$ – Jyrki Lahtonen Mar 21 '14 at 5:54
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Notice that $x^4-x^2-2$ is quadratic in $x^2$ so we can factor easily: $(x^2-2)(x^2+1)$. Thus by adjoining $\sqrt{2}$ and $i=\sqrt{-1}$ to the rationals, we'll have all of the necessary roots to make this split: $(x-\sqrt{2})(x+\sqrt{2})(x-i)(x+i)$.

The splitting field is $\mathbb{Q}(\sqrt{2},i)$.

The degree of this extension is $4$ since $\{1,\sqrt{2},i,i\sqrt{2}\}$ forms a basis (as a vector space) for $\mathbb{Q}(\sqrt{2},i)$ over $\mathbb{Q}$. OR... notice that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2})]=2$ so $[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}]=2 \cdot 2 = 4$.

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