How to determine the prime ideals of a intermediate field above $p$? 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?
 A: 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 $\mathcal{O}_K$.
A: In biquadratic extensions, one may completely determine the splitting of a prime by examining its splitting in the quadratic subfields. Here, the quadratic subfields are $ \mathbf Q(\sqrt{5}) $, $ \mathbf Q(\sqrt{13}) $, $ \mathbf Q(\sqrt{65}) $ respectively. $ 2 $ remains inert in the first two and splits in the third, $ 3 $ remains inert in the first and the third, and splits in the second. Compositums of unramified extensions are unramified, therefore $ 2, 3 $ are both unramified in $ K = \mathbf Q(\sqrt{5}, \sqrt{13}) $. We thus have enough information to conclude that both $ 2 $ and $ 3 $ split as the product of two primes in $ K $.
To find these primes, we first find them downstairs, in the quadratic extensions where splitting occurs. We find the factorizations
$$ (2) = \left( 2, \frac{1 + \sqrt{65}}{2} \right) \left(2, \frac{-1 + \sqrt{65}}{2} \right) $$
$$ (3) = \left( 3, \frac{1 + \sqrt{13}}{2} \right) \left( 3, \frac{-1 + \sqrt{13}}{2} \right) $$
using Dedekind's criterion in $ \mathbf Q(\sqrt{65}) $ and $ \mathbf Q(\sqrt{13}) $ respectively. By what we know above; these primes lying in the quadratic fields must remain inert when going upstairs to $ K $, therefore, these factorizations are also prime ideal factorizations of $ 2 $ and $ 3 $, valid in $ K $.
A: If it helps you, that's what Sagemaths says :
 Q.<x> = QQ[]; K.<a,b> = NumberField([x^2-5,x^2-13])
 OK = K.ring_of_integers(); OK.basis()
 >> [3/2*a - b + 1/2, 3*a - 2*b, (-1/4*b + 11/4)*a - 7/4*b + 9/4, 11/2*a - 7/2*b]
 I = OK.ideal(2); I.factor()
 >> (1/2*a + 1/2*b + 1) * ((-1/4*b + 3/4)*a - 1/4*b + 7/4)
 J = OK.ideal(3); J.factor()
 >> (1/2*b - 1/2) * (1/2*b + 1/2)

