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I know that it should be because is finite and separable ($\rm{char}(\mathbb Q)=0$). However I'm having some trouble in finding the primitive element.

First of all, given the irreducible polynomial $f(x)=x^4-10x^2+20$, I find the roots of $f$: $$\alpha_1 = \sqrt{5+\sqrt{5}},\,\alpha_2=-\sqrt{5+\sqrt{5}},\, \alpha_3 = \sqrt{5-\sqrt{5}},\, \alpha_4=-\sqrt{5-\sqrt{5}}$$

It is easy to see that $\sqrt{5}\in \mathbb Q(\alpha_1):$ $$\alpha_1^2 - 5 = \sqrt{5}$$

Then, $\alpha_1\,\alpha_3 = \sqrt{25-5}=2\sqrt{5}\iff \alpha_3=2\sqrt{5}\,\alpha_1^{-1}$. Since is a product of elements of $\mathbb Q(\sqrt{5})$ and it is a field, $\alpha_3\in\mathbb Q(\alpha_1)$.

Is this argument correct?

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If I remember right, the proof of the primitive element theorem is very constructive. Have you read through it? In particular I think your $\alpha_1+\alpha_3$ will do the job. – Derek Allums Jan 31 '13 at 15:58
I think you mean $\alpha_1$ instead of $\sqrt{5}$ in your last sentence. – marlu Jan 31 '13 at 15:59
@marlu: you're right – synack Jan 31 '13 at 16:04
up vote 1 down vote accepted

Your argument is mostly fine and you seem to have already found a primitive element: $\alpha_1$. The only lingering detail I see is why $x^4-10x^2+20$ is irreducible, this follows easily from Eistenstein just don't pick $2$. Of course you don't need to do the first part of your argumet. It's enough to observe that $\alpha_3 \in \mathbb Q(\alpha_1)$.

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