I am interested in proving and understanding the following statement:
If $K \ne \mathbb{Q}$, then the set of positive integers that are norms of ideals in $\mathcal{O}_K$ have density zero in $\mathbb{Z}$. That is, $$ \lim_{X \to \infty} \frac{\#\{n \le X : n = N(I), \, \, I \, \, \text{is an ideal of}\, \, \mathcal{O}_K\}}{X} = 0. $$
I am unsure if this is true because for any ideal $I$, we can factor it uniquely as the product of prime ideals $I = \mathfrak{p}_1 \cdots \mathfrak{p}_l$. Then taking norms we have $N(I) = N(\mathfrak{p}_1) \cdot \cdots \cdot N(\mathfrak{p}_k)$, where each term on the right is a power of a rational prime. If this is the case, then how can we not choose the ''right" prime ideals to produce an ideal $I$ with norm any integer we want? Is it because the prime ideals in $\mathcal{O}_k$ lie over only certain rational primes?
- Is it true that there are infinitely many prime ideals in $\mathcal{O}_K$ for $K \ne \mathbb{Q}$?
- Is it true that the prime ideals in $\mathbb{O}_K$ lie over finitely many rational primes? Given a prime $p \in \mathbb{Z}$ can we always find a prime ideal $\mathfrak{p} $ lying over $p$?
- How are the prime ideals distributed with respect to the rational primes they lie over? I.e. do prime ideals tends to cluster around certain primes as we increase the degree of the field extension?
I am self learning a lot of this material so any references, or clear expositions would be greatly appreciated. I realize that some of my questions may not have simple answers, so if that is the case then a heuristic/intuitive explanation will suffice.