<s>[INST] Determining the minimal polynomial over $\Bbb{Q}$ I was working on a homework assignment from Hungerford:

Find the minimal polynomial of the element $\sqrt{1+\sqrt{5}}$ over $\Bbb{Q}$.

Naturally the solution would be the polynomial with roots 
$$ \pm \sqrt{1 \pm \sqrt{5}} $$ 
Which is found as 
$$ x = \pm \sqrt{1 \pm \sqrt{5}}  \rightarrow x^2 -1 = \pm \sqrt{5} \rightarrow (x^2-1)^2 = 5 \rightarrow $$
$$\text{Minimal Polynomial = } (x^2-1)^2-5$$
Problem is, I quite frankly don't know how to prove this. Hypothetically what if, in the same vein that 
$$ \sqrt{5 + 2\sqrt{6}} = \sqrt{2} + \sqrt{3}$$ 
There exists some subtle factorization for $\sqrt{1 + \sqrt{5}}$ with the additionaly property that it is the root of a cubic or quadratic. How do I definitely rule out any of those cases?
 [/INST] It is enough to show that $(x^2-1)^2-5$, that is, $x^4-2x^2-4$ is irreducible over the rationals. There are no rational roots, so we only need to rule out factorization as a product of quadratics with integer coefficients. 
Since there is no $x^3$ term. we can confine attention to factorizations $(x^2+ax+b)(x^2-ax+c)$. If $a=0$, we get $b+c=-2$, $bc=-4$, impossible in integers. If $a\ne 0$, then because there is no $x$ term we need $b=c$, but then $bc$ cannot be $-4$.   
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