number of prime ideals in algebraic integers Given an algebraic number field $K/\mathbb Q$, let $R = \mathbb Z_K$ be the ring the algebraic integers of that fields. Is it possible to say how many prime ideals there are in $R$? I suspect we always have infinitely many, as this is certainly the case for $K=\mathbb Q$ (because then $R = \mathbb Z$). 
I was already able to show that an Ideal $I\subseteq \mathbb R$ ($I\neq0$) is prime if and only if it is maximal. One idea would have been showing that you can generate infinitely many fields via $R/I$ but I was never able to show what $R/I$ would be isomorphic to, as I do not really know what the prime ideals $I$ look like. 
So have no idea how to go from there, and I'd be happy for every hint!
 A: There are infinitely many primes in $\mathbb{Z}_K$.  You can prove it by showing that for every prime $p$ of $\mathbb{Z}$, there exists a prime ideal $P$ of $\mathbb{Z}_K$ such that $P \cap \mathbb{Z} = p\mathbb{Z}$.  This is called the lying over theorem.  There are infinitely many prime ideals of $\mathbb{Z}$, hence infinitely many prime ideals of $\mathbb{Z}_K$.
A: In the specific case of rings of integers, you can mimic Euclid's proof over $\mathbb Q$. The key point is that every ideal $\mathfrak a \subset \mathcal O_K$ factors (uniquely) as a product of prime ideals.
Suppose that we have a finite set of non-zero prime ideals $\mathfrak p_1,\ldots,\mathfrak p_n\subset\mathcal O_K$. Then $\mathfrak p_1\cdots\mathfrak p_n\cap\mathbb Z$ is a non-trivial ideal of $\mathbb Z$, so is equal to $(n)$ for some $n\in\mathbb Z\setminus\{0,1\}$. In particular, $n\in \mathfrak p_i$ for all $i$
Now consider the ideal $(n+1)\mathcal O_K$. If $\mathfrak p_i\mid (n+1)\mathcal O_K$ for any $i$, then $n, n+1\in \mathfrak p_i$ and hence $1\in \mathfrak p_i$, a contradiction.
In particular, $(n+1)\mathcal O_K$ is divisible by a prime ideal not in $\{\mathfrak p_1,\ldots,\mathfrak p_n\}$ and the result follows.
For a different proof mimicking Euler's proof for $\mathbb Q$, see this question.
