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 $[\mathbf{Q}(\sqrt{1+i},\sqrt{2})]=8$?
 A: 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 $[K:\mathbf Q]=8$.
A: Break it into towers.
Look at the two extensions $\mathbb Q ( \sqrt {1+i} )|_{Q(i)} $ and $\mathbb Q ( \sqrt {1-i} )|_{Q(i)} $. 
Each has degree $2$ ( since $1+i$ and $1-i$ are primes in $\mathbb Z[i] $ which is a UFD ).
Their compositum is the field you are interested in i.e. $ \mathbb Q (\sqrt {1+i} , \sqrt {1-i} )$ 
At this point you can have a look at the question Finding degree of a finite field extension
I hope you have seen the similarity. 
You have a UFD $\mathbb Z[i]$ , it's field of fractions $\mathbb Q(i)$ and you have adjoined square roots of two distinct primes $1+i , 1-i$. 
By a similar argument as in the question above we argue the Galois group is of the form $\mathbb Z_2 ^k$ with $k\leq 2$. 
But then you got $3$ distinct degree $2$ subextensions $\mathbb Q(\sqrt {1+i}), \mathbb Q(\sqrt {1-i}) ,  \mathbb Q(\sqrt {2}) $ 
This gives you $k\geq 2$
So the upshot is $$\mathbb Q(\sqrt {1+i}, \sqrt 2): \mathbb Q(i)=4$$ 
And conclusion follows.
