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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.

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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.

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  • $\begingroup$ Franz, how do you relate that first observation to the ring of integers? Of course, knowing the discriminant you find which are the primes that ramify, but I don't see how you can relate that to explicitly find the ring of integers. $\endgroup$
    – GSF
    Nov 12, 2016 at 12:53
  • $\begingroup$ I do not understand your question - the discriminant of the ring of integers is the field discriminant. If $D$ is the discriminant of an order, then the field discriminant divides $D$. $\endgroup$
    – franz lemmermeyer
    Nov 12, 2016 at 14:49
  • $\begingroup$ Yes, but how can you write the ring of integers explicitly? e.g, the ring of integers of $K=\mathbb{Q}(\sqrt{-3})$ is $\mathcal{O}_{K}=\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$. $\endgroup$
    – GSF
    Nov 12, 2016 at 14:51
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    $\begingroup$ You've written it down in your question; it is generated by $\frac{1+\sqrt{-3}}2$, $\frac{1+\sqrt{5}}2$ and $\frac{1+\sqrt{-15}}2$. $\endgroup$
    – franz lemmermeyer
    Nov 12, 2016 at 14:53
  • $\begingroup$ Sorry, that's true. Now I see it! Thank you. $\endgroup$
    – GSF
    Nov 12, 2016 at 14:56

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