Finding degree of $\mathbb{Q}(i, \sqrt{-1 + \sqrt{-3}}) : \mathbb{Q}$ Letting $\alpha = \sqrt{-1 + \sqrt{-3}}$, I have already that $|\mathbb{Q}(\alpha):\mathbb{Q}| = 4$
So one could do:
$|\mathbb{Q}(i, \alpha):\mathbb{Q}| = |\mathbb{Q}(i, \alpha):\mathbb{Q}(\alpha)| \cdot |\mathbb{Q}(\alpha):\mathbb{Q}|$
It's clear that  $|\mathbb{Q}(i, \alpha):\mathbb{Q}(\alpha)|$ is at most 2 (as $i$ is a root of polynomial $x^2 + 1$ in $\mathbb{Q}(\alpha)[x]$).
To show the degree of $|\mathbb{Q}(i, \alpha):\mathbb{Q}(\alpha)|$  is not 1 could be done by showing $i \notin \mathbb{Q}(\alpha)$, which would involve showing $i = a + b\alpha + c\alpha^2 + d\alpha^3$ for $a, b, c, d \in \mathbb{Q}$ leads to a contradiction...
However this gets messy... so I was wondering if anyone could see an easier way of doing this?
 A: $\alpha$ generates a biquadratic extension: $\alpha=\dfrac{\sqrt2+\sqrt{-6}}{2}$
A: I find your solution really not too bad. We have $\alpha^4+2\alpha^2+4=0$. 
If $i=a+b\alpha+c\alpha^2+d\alpha^3$ then
$i^2+1=0$ says that
$ad + bc - 2cd = 0$ 
$2ac + b^2 - 4bd - 2c^2 = 0$ 
$ab - 4cd = 0$ 
$a^2 - 8bd - 4c^2 + 8d^2 + 1 = 0$
These equations obviously have no solution so that $i\not\in \mathbb{Q}(\alpha)$ and 
$\mathbb{Q}(i,\alpha):\mathbb{Q})=8$.
Checking that $a^4+2a^2+4$ is irreducible over $\mathbb{Q}(i)$ also involves some calculation to see that we cannot write the polynomial as a product of two quadratic ones.
A: There are many ways to do this. Easiest, I think, is to show that the whole field is $\mathbb Q(\sqrt2,\sqrt3,i)$, a field which I will denote $K$, and  which is obviously of degree eight over $\mathbb Q$, being an imaginary quadratic extension of the real biquadratic field $\mathbb Q(\sqrt2,\sqrt3)$. Let’s see why our $\mathbb Q(\alpha, i)$ has the above-stated form.
First notice that $\alpha^2=-1+\sqrt{-3}=2\omega$, that is it’s twice a primitive cube root of unity, so that $(-1+\sqrt{-3})^3=8$, as you may also calculate directly. Consequently, $\mathbb Q(\alpha)$ certainly contains $\sqrt2$, since it contains a cube root of unity and a cube root of $8$. And $\mathbb Q(\alpha)$ certainly contains $\sqrt{-3}$, so that $\mathbb Q(\alpha,i)$ also contains $\sqrt3$. Thus $\mathbb Q(\alpha,i)\supset K$, which is enough for our purposes, though the opposite inclusion is easy to show.
EDIT: A quicker, slicker, but more advanced argument takes $\alpha/(1+i)$ and squares it to get $(-1+\sqrt{-3}\,)/(2i)=\omega/i=\cos30^\circ+i\sin30^\circ$, a primitive twelfth root of unity. Thus $\alpha/(1+i)$ is a primitive $24$-th root of unity, a root of the (well-known to be) irreducible $24$-th cyclotomic polynomial $\Phi_{24}(X)=X^8-X^4+1$.
