Degree of extension fields

Question

I've been working my way through some problems to get this idea solidified but now I've run into something odd. How do I find the degree of $\mathbb{Q}(\sqrt{1+ \sqrt{2}})$ over $\mathbb{Q}$? Would it be degree 4 because we have the polynomial $X^{4}-2X^{2}-1 = (X^{2}-1)^{2}-2$?

Answer 1

Yes.

Or, you can also argue that $$ \Bbb Q\subset\Bbb Q(\sqrt{2})\subset\Bbb Q(\sqrt{1+\sqrt{2}}) $$ and each intermediate extension has degree $2$.

Answer 2

Yes. Here's a hint to show your polynomial is irreducible (ie. And therefore a minimum polynomial):

Let $X = Y+1$, then use Eisenstein's Criterion.

Answer 3

You can observe that this biquadratic polynomial has two real roots and two complex conjugate roots, so it splits, over the reals, as $$ (X-\sqrt{1+\sqrt{2}})(X+\sqrt{1+\sqrt{2}})(X^2+aX+b) $$ where $a^2-4b<0$. By uniqueness of decomposition over the reals, the polynomial is not reducible over the rationals: just consider the possible factors.

You can use the irrationality of $r=\sqrt{1+\sqrt{2}}$, which follows from $\sqrt{2}=r^2-1$.