TITLE: How to determine the prime ideals of a intermediate field above $p$?
QUESTION [5 upvotes]: For $\mathbb{Q}\subset \mathbb{Q}( \sqrt{5}, \sqrt{13})\subset \mathbb{Q}(\xi_{65})$. 
I am trying to the primes ideals in the ring of integers of $\mathbb{Q}( \sqrt{5}, \sqrt{13})$ above $2,3$. 
Since the ring of integers of $\mathbb{Q}( \sqrt{5}, \sqrt{13})$ is not quadratic, we can not apply the Dedekind-Kummer lemma. I also tried to do the factorization of $(2)$ in the ring of integers of $\mathbb{Q}(\xi_{65})$ first and try to push down to the ring of integers of $\mathbb{Q}( \sqrt{5}, \sqrt{13})$. But this way seems to be not doable. 
I can't figure out any methods to do that, any ideas please?

REPLY [4 votes]: You can apply Dedekind-Kummer twice, you just need to choose appropriate primitive elements.
Let $\alpha=\frac{1+\sqrt{5}}{2}$, $\beta=\frac{1+\sqrt{13}}{2}$, $F=\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{5})$ and $K=F(\beta)=\mathbb{Q}(\sqrt{5},\sqrt{13})$. We can verify that $\{1,\alpha \}$ and $\{1,\alpha,\beta, \alpha \beta\} $ are  integral basis of $\mathcal{O}_F$ and $\mathcal{O}_K$, in particular  $\mathcal{O}_F=\mathbb{Z}[\alpha]$ and $\mathcal{O}_K=\mathcal{O}_F[\beta]$.
Now for a extention of number fields $L \subset M=L(\gamma)$ where $\gamma \in \mathcal{O}_L$ Dedekind-Kummer applies to any prime $\mathfrak{p}$ in $\mathcal{O}_L$   as long as $[\mathcal{O}_{M}:\mathcal{O}_{L}[\gamma]] \notin \mathfrak{p}$. 
So if we apply it first to $L=\mathbb{Q}$,  $\gamma=\alpha$ and $\mathfrak{p}=2 \, \mathbb{Z}$ (which we can because $\mathcal{O}_F=\mathbb{Z}[\alpha]$ )  since the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is equal to $x^2-x-1$ 
 (irreducible mod 2) we have that $2 \,\mathcal{O}_F $ is prime.
If we now apply it to  $L=F$,  $\gamma=\beta$ and $\mathfrak{p}=2 \, \mathcal{O}_F$ (where again we use $\mathcal{O}_K=\mathcal{O}_F[\beta]$) , the minimal polynomial of $\beta$ over $F$ is $x^2-x-3$ which factor as $(x-\alpha)(x-\alpha')$ mod $\mathfrak{p}$  where $\alpha'=\frac{1-\sqrt{5}}{2}$. Thus $2 \, \mathcal{O}_K =\mathfrak{p} \, \mathcal{O}_K $ factors as $(2,\beta-\alpha)\cdot (2,\beta-\alpha')$ in $\mathcal{O}_K$.
In a similar way we  see that 3 is inert in $F$ and factor as  $(3,\beta)\cdot (3,\beta-1)$ in $\mathcal{O}_K$.