abstract algebra - Prove that $[mathbf{Q}(sqrt{1+i},sqrt{2}):mathbf{Q}]=8$.

I am trying to calculate the Galois group of the polynomial $f=X^4-2X^2+2$. $f$ is Eisenstein with $p=2$, so irreducible over $\mathbf{Q}$. I calculated the zeros to be
$\alpha_1=\sqrt{1+i},\alpha_2=\sqrt{1-i},\alpha_3=-\alpha_1$ and $\alpha_4=-\alpha_2$. Let $\Omega_f=\mathbf{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=\mathbf{Q}(\alpha_1,\alpha_2)$ be a splitting field of $f$ over $\mathbf{Q}$. Since $\alpha_1\alpha_2=\sqrt{1+i}\sqrt{1-i}=\sqrt{2}$, we have $\Omega_f=\mathbf{Q}(\sqrt{1+i},\sqrt{2})$.

So if we can prove that $[\Omega_f:\mathbf{Q}]=8$, then we have $\#\operatorname{Gal} (f)=8$ and for $\operatorname{Gal}(f)\subset S_4$, we must have that it is isomorphic to the dihedral group $D_4$.

How do I go about proving $[\mathbf{Q}(\sqrt{1+i},\sqrt{2})]=8$?

Answer

You have shown that your splitting field is $K=\mathbf Q(\sqrt {1+i}, \sqrt {1-i})$. The two fields $\mathbf Q(\sqrt {1\pm i})$ are obviously quadratic extensions of $\mathbf Q(i)$, and these are equal iff $(1+i)(1-i)=2$ is a square in $\mathbf Q(i)$, iff $\sqrt 2\in \mathbf Q(i)$: impossible. Hence $K$ is a biquadratic extension of $\mathbf Q(i)$, and $[K:\mathbf Q]=8$.