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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?

  1. Is it true that there are infinitely many prime ideals in $\mathcal{O}_K$ for $K \ne \mathbb{Q}$?
  2. 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 $p$?
  3. 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.

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    $\begingroup$ I always find it helpful to consider the simplest example, $K = \mathbb Q(i)$. In this case, there are three types of primes in $\mathcal O_K = \mathbb Z[i]$. First, if $p \in \mathbb Z$ is a prime congruent to 3 mod 4, then $p$ remains prime in $\mathcal O_K$, and has norm $p^2$. Second, if $p \equiv 1 \pmod 4$, then $p$ splits as a product of two complex primes $\pi \overline \pi$, each of norm $p$. Finally, if $p = 2$, then $p$ factors as $(1+i)^2$ up to units, where $1+i$ has norm 2. These cases are called "inert", "split", and "ramified" respectively. $\endgroup$ – Ravi Fernando Oct 15 '15 at 3:55
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    $\begingroup$ There certainly are always primes $\mathfrak p$ lying over any given $p$. But in the example of $\mathbb Q(i)$, the norm of every ideal must be divisible by each prime $p \equiv 3 \pmod 4$ with even multiplicity, because the primes $\mathfrak p$ over such $p$ have norm $p^2$. So what you'd want to prove in this case is that almost all positive integers are divisible by some $p \equiv 3 \pmod 4$ with odd multiplicity. $\endgroup$ – Ravi Fernando Oct 15 '15 at 3:58
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    $\begingroup$ In answer to your numbered questions: 2 and hence 1 are both yes. For 3, there are always primes over each $p$: namely, think of $(p)$ as an ideal in $\mathcal O_K$, and look at its prime factors. The questions are how many prime factors there are, what their norms are, and whether they are repeated. These three questions correspond exactly to the three cases we saw for $K = \mathbb Q(i)$: splitting, inertia, and ramification. For more complicated $K$, these can occur in combination with each other, and there is lots of beautiful theory about them. $\endgroup$ – Ravi Fernando Oct 15 '15 at 4:04
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    $\begingroup$ Possibly relevant: en.wikipedia.org/wiki/… $\endgroup$ – Ravi Fernando Oct 15 '15 at 4:04
  • $\begingroup$ Thank you for the detailed answer and reference. $\endgroup$ – Kevin Sheng Oct 15 '15 at 18:31

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