TITLE: Ring of integers of $\mathbb{Q}(\sqrt{-3},\sqrt{5})|\mathbb{Q}$ and group of units
QUESTION [6 upvotes]: I'm having some problems finding the ring of integers of $\mathbb{Q}(\sqrt{-3},\sqrt{5})|\mathbb{Q}$. How can I find it?  
Also, I'd like to prove that $\alpha:=\frac{1+\sqrt{-3}+\sqrt{5}+\sqrt{-15}}{4}$ generates a subgroup of finite index of the group of units of $\mathbb{Q}(\sqrt{-3},\sqrt{5})|\mathbb{Q}$, but I don't know how to apply the Dirichlet theorem to prove it (that's the only point I've got to so far).
I already know how to compute its discriminant and therefore I know which primes ramify, the problem is that I really don't know how to find its ring of integers, even though I've felt the temptation of writting $\mathcal{O}_{\mathbb{Q}(\sqrt{-3},\sqrt{5})}=\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2},\frac{1+\sqrt{5}}{2}\right]$
Thanks in advance.

REPLY [5 votes]: If $L/K$ is a quadratic extension, then the discriminant of $L$ is divisible by the square of the discriminant of $K$ (Hilbert's report). Since $L = {\mathbb Q}(\sqrt{-3},\sqrt{5})$ is a quadratic extension of  ${\mathbb Q}(\sqrt{-15})$,
you know that $15^2$ divides the discriminant of $L$. The discriminant of the order you have written down is $15^2$, so you're done.
Since the field $L$ has unit rank $1$, any unit that is not a root of unity generates  a subgroup of the unit group with finite index. Your unit is, up to a root of unity, just the fundamental unit of ${\mathbb Q}(\sqrt{5})$, and this is easily shown to be the fundamental unit of $L$ using techniques that are explained in articles e.g. by Wada on Kuroda's class number formula.