I'm trying to find the ring of integers $A_L$ of $\mathbb{Q}(i,\sqrt{5})$. I know that the ring of integers of $\mathbb{Q}(i)$ is $\mathbb{Z}[i]$ and that the one of $\mathbb{Q}(\sqrt{5})$ is $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$. I would like to say that $A_L=\mathbb{Z}\left[\frac{1+\sqrt{5}}{2},i\right]$.
From Integral basis of the ring of integers of an extension, given integral bases of the rings of integers of subfields I understood it is possible to say $A_L=\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]\mathbb{Z}\left[i\right]$ using the fact the discriminants of the two integer basis are coprime (Here we use a result that is possible to find in Marcus' book).
Is there a way to find $A_L$ in a more direct way without using that result?
Write $\omega = \frac{1 + \sqrt{5}}2$. Then all elements $\alpha = a + bi + c\omega + di\omega$ where $a, b, c, d$ are integers. If the ring of integers is larger, there must be algebraic integers of the form $\alpha/2$ or $\alpha/5$ since $2$ and $5$ are the only prime divisors of the discriminant of the subring generated by $i$ and $\omega$. Now all you have to do is show that if $\alpha/2$ is an algebraic integer, then $a$, $b$, $c$, $d$ are divisible by $2$, and then do the same with respect to the prime $5$.