Degree of $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$ over $\mathbb{Q}$

Question

I looked at some exercises from Dummit and Foote, which cover degree of $\mathbb{Q}(\sqrt{a + \sqrt{b}})$, when $a^{2} - b$ is a square in the ground field, but here, $a^{2} - b = -1$, so these results don't apply. However, I think I may have found the minimal polynomial just by playing with it: $x^{4} - 2x^{2} - 1$. How do I show this is irreducible though? Or is it not, in fact?

Answer 1

Use $x=y+1$ and Eisenstein's criterion.

Answer 2

A standard way to prove that a general polynomial of degree 4 is irreducible is to show that it has no roots in the base field, then suppose it decomposes into polynomials of degree 2 so compute:

$$ x^4-2x^2-1 = (x^2+ax+b)(x^2+cx+d)$$

and show that there are no solutions $a,b,c,d$ in the base field.