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

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 $E$?

• There is a theorem which says that if you have two number fields $E,F$ whose intersection is trivial their discriminants are relatively prime then the rings of integers satisfy $\mathcal{O}_{EF}=\mathcal{O}_E\mathcal{O}_F$, although I can't seem to have a reference for this fact. This theorem is often used when one wants to compute the ring of integers of the cyclotomic extension $\mathbb{Q}(\zeta_m)/\mathbb{Q}$. Commented Aug 8, 2018 at 11:44
• Thanks Gal, that's somewhere to start. Commented Aug 8, 2018 at 11:47
• I'm not sure if this was clear from what I wrote above, but note that this implies that if the discrminants of $F_2$ and $F_3$ are relatively prime (and also, I presume that you mean to have $E=F_1F_2$?), then the set you wrote above will be an integral basis. Thus the discrminants being relatively prime is a sufficient condition. Commented Aug 8, 2018 at 12:00

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.
• Thanks very much Stefan4024 ! I was thinking about the field $E = \mathbb{Q}(\zeta , \sqrt{D})$, where $\zeta$ is the root of a cubic polynomial, and the discriminant of the cubic field $F_3$ is $D f^2$, where $D$ is the fundamental discriminant of the quadratic field $F_2$. In this case by your and Gal's answer it would suggest that $\{ 1, \alpha , \beta , \omega , \alpha \omega , \beta \omega \}$ is likely not an integral basis of $\mathcal{O}_{E}$ since the field discs are not coprime. Commented Aug 8, 2018 at 22:45