Factoring primes in $\mathbb{Q}(\sqrt{-1},\sqrt{5})$ 
Show that $2,5$ are the only primes which ramify in $L:=\mathbb{Q}(\sqrt{-1},\sqrt{5})$, and that their ramification indices are both $2$.

Obviously $K_1:=\mathbb{Q}(\sqrt{-1})$ and $K_2:=\mathbb{Q}(\sqrt{5})$ are both Galois and it's easy to check that: 
$$K_1K_2=\mathbb{Q}(\sqrt{-1},\sqrt{5})$$
$$K_1\cap K_2=\mathbb{Q}$$
$$(d_{K_1},d_{K_2})=(-4,5)=1$$
Since $\mathcal{O}_{K_1}=\langle 1,i\rangle_\mathbb{Z}$ and $\mathcal{O}_{K_2}=\langle 1,\omega\rangle_\mathbb{Z}$, there's a famous proposition which says that $\mathcal{O}_L=\langle 1,i,\omega,i\omega\rangle_\mathbb{Z}$ and $d_L=d_{K_1}^2d_{K_2}^2=2^45^2$. 
This proves $2,5$ are the only primes which ramify in $L$. I would know how to calculate the ramification indices if $\mathcal{O}_L=\mathbb{Z}[\theta]$ for some $\theta\in L$ by factoring the minimal polynomial of $\theta$ modulo $2$ and $5$.
I can't find such $\theta$, so I'm kind of lost.

I'll make my comment and answer and add the case of $5$.
See what happens to $(2)$ in $_2$ and then from there see what happens to $K_1K_2$. From $\mathbb{Q}$ to $K_2$ we have $e=1$ and $f=2$ and $g=1$ (since $^2−x−1$ is irreducible mod 2). Then going from $_2$ to $_1_2$ a degree 2 extension we know $=1$ or $e=2$ but it must be 2 since we know the rational prime 2 ramifies from discriminant calculation. So from $\mathbb{Q}$ to $_1_2$ we have $=2$ and $=2$ and $g=1$ for the prime $2$.
For the prime $5$, use $K_1$.  Since $X^2 + 1 = (X+2)(x+3)$ mod 5, we have that $5$ splits in $K_1$ into $g=2$ primes, and $e=1$ and $f=1$. Then going from $K_1$ to $K_1K_2$ we know 5 will end up being ramified, so $e=2$. So from $\mathbb{Q}$ to $K_1K_2$ we have $g=2, e=2, f=1$.
