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 like @usr0192’s answer fine, but I’d apply a more take-no-prisoners approach. You know that $K_1$ is ramified at $2$ and nowhere else; and that $K_2$ ramified at $5$ and nowhere else. Ramification indices multiply in towers, and when you take $K\supset k$ unramified and any extension $F$ of $k$, then $KF$ is unramified over $F$.
It follows that $e(L/\Bbb Q)$ is equal to $2$ above each of $2,5$, and $1$ above every other prime.
