Prime decomposition in $\mathbb{Q}(\sqrt2 + \sqrt3)$ I am having trouble finding detailed examples of prime factorization in number fields. Here is the question I was working:
Find the prime ideal decomposition of $p\mathcal{O}_L$ for each prime $p \in \{2,3,5\}$ where $L=\mathbb{Q}(\sqrt2, \sqrt3)$. Determine the inertial degree and ramication index for each case.
I find that $L/\mathbb{Q}$ is normal and $Gal(L/ \mathbb{Q})$ is $\mathbb{Z}/\mathbb{2Z} \times \mathbb{Z}/\mathbb{2Z}$. Also since $d_L=2^8.3^2$, primes 2 and 3 ramifies in $L$, but 5 does not.
I couldn't go any forward proberly. Any help would be much appreciated. Or any other concrete examples would be nice.
 A: Since you already know that $\mathbb{Q}(\sqrt{2} + \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, it's obvious that $2$ and $3$ ramify. The only real difficulty here is $5$.
But like hunter said, we can avail ourselves to the "intermediate" rings. We see that $5$ is prime in $\mathbb{Z}[\sqrt{2}]$ and also prime in $\mathbb{Z}[\sqrt{3}]$. Okay, so no help there.
Did I mention I had a raw vegan meal for lunch today? I'm having trouble deciding which photo of it to post to Instagram. It's not that I'm vegetarian, far from it.
Okay, I hope I lost a few people with the preceding paragraph. Anyway, $(1 - \sqrt{6})(1 + \sqrt{6}) = -5$, which means that $5$ can't be inert in $\mathbb{Q}(\sqrt{2} + \sqrt{3})$.
Now notice that $$\frac{5}{\sqrt{2} + \sqrt{3}} = -5 \sqrt{2} + 5 \sqrt{3}.$$
You know what, I can't concentrate with this hunger. I gotta go get myself something more substantial to eat. Bye!
A: It shouldn't be too hard to prove that the ring of integers is $\mathbb{Z}[\alpha]$, where $\alpha = \frac{\sqrt{2}+\sqrt{6}}{2}$ and $\min(\alpha, \mathbb{Q}) = x^4 - 4x^2 + 1$. Now you can reduce it modulo $2,3,5$ to get:
$$x^4 - 4x^2 + 1 = (x+1)^4 \text{ in } \mathbb{F}_2[x] \implies 2O_k = (2,\alpha+1)^4$$
$$x^4 - 4x^2 + 1 = (x^2+1)^2 \text{ in } \mathbb{F}_3[x] \implies 3O_k = (3,\alpha^2+1)^2$$
$$x^4 - 4x^2 + 1 = (x^2+x+1)(x^2+4x+1) \text{ in } \mathbb{F}_5[x] \implies 5O_k = (5,\alpha^2 + \alpha+1)(5,\alpha^2 + 4\alpha+1)$$
