Integral basis of the ring of integers of an extension, given integral bases of the rings of integers of subfields

Question

Let $E/ F_2 / \mathbb{Q}$ and $E/ F_3 / \mathbb{Q}$ be towers of number fields, where $F_2$ is a quadratic extension of $\mathbb{Q}$ and $F_3$ is a cubic extension of $\mathbb{Q}$. Let $\{ 1, \omega \}$ be an integral basis for the ring of integers of $F_2$ and let $\{ 1, \alpha , \beta \}$ be an integral basis for the ring of integers of $F_3$. Under which conditions is $\{ 1, \alpha , \beta , \omega , \alpha \omega , \beta \omega \}$ an integral basis for the ring of integers of $E$?

Answer 1

This might not be the answer you are looking for, but I guess it will give you some insight.

First of all we must have that $[E:\mathbb{Q}] = 6$, as otherwise we can't get a basis of 6 elements. This means that $E=F_2F_3$. As Gal metioned in the comments if the two fields have coprime discriminants then we can conclude that $\mathcal O_{E} = \mathcal O_{F_2} \mathcal O_{F_3}$. However he missed another condition (which is satisfied in your case) and that is that we must have: $[EF:\mathbb{Q}]=[E:\mathbb{Q}][F:\mathbb{Q}]$. Note that having two disjoint field isn't enough to conclude this. For example take $E=\mathbb{Q}(\sqrt[6]{2})$ and $F=\mathbb{Q}(\zeta_8\sqrt[4]{2})$, which are extensions of degree $6$ and $4$ over $\mathbb{Q}$, but $EF = \mathbb{Q}(\sqrt[6]{2}+\zeta_8\sqrt[4]{2})$, an extension of degree $12$ over $\mathbb{Q}$. This happens because the fields aren't Galois extension. Nevertheless we don't have to worry about that in you case, as the extension degrees are coprime.

You can find the abovementioned result in Marcus' Number Theory book on page 33, listed as Theorem 12. It says that if $d=\gcd(d_E,d_F)$ and $[EF:\mathbb{Q}]=[E:\mathbb{Q}][F:\mathbb{Q}]$, then $\mathcal O_{EF} \subseteq \frac 1d \mathcal O_{E} \mathcal O_{F}$. By the way I would recommend reading Marcus' book, if you haven't done it yet.

Also I would recommend taking a look at Exercise 2.42, which considers the case of biquadratic extensions, i.e $\mathbb{Q}[\sqrt{m},\sqrt{n}]$. as you can see even in the most basic example it's hard to find the integral basis and it gets even harder for $\mathbb{Q}[\alpha,\sqrt{m}]$, where $\alpha$ is an element of degree $3$, as in you example. Moreover $\alpha$ doesn't have to be an element of the form $\sqrt[3]{n}$, which makes thing even harder. Indeed $\alpha = \zeta_7 + \zeta_7^{-1}$ isn't of that form, for instance.

To summirize, I doubt there is a general formula for determining the integral basis for the composition of the two fields. However the two discriminant being coprime is sufficient condition.