Let $K$ be a number field. For each ideal $I$ of the ring of integers $\mathcal{O}\_K$ let $N\_K(I)$ denote the norm of $I$. For a prime $\mathfrak{p}\subset \mathcal{O}\_K$ above the rational prime $p\in \mathbb{Z}$, set $\deg(\mathfrak p):=[\mathcal{O}\_K/\mathfrak{p}:\mathbb{Z}/p]$. Consider the following limit:
$$
\delta\_n:=\lim\_{B\to\infty} \frac{\#\{\text{primes } \mathfrak{p}\subset \mathcal{O}\_K \text{ with } \deg(\mathfrak{p})=n \text{ and } N\_K(\mathfrak{p})\leq B \}}{\#\{\text{primes } \mathfrak{p}\subset \mathcal{O}\_K \text{ with } N\_K(\mathfrak{p})\leq B\}}.
$$


**Question:** Does this limit exist? If so, is there a nice formula for it?


**Remark:** This is sort of like a 'backwards' Chebotarev density theorem. Instead of considering a density of primes in the base of the extension we are considering a density of primes in the extension itself.



Density of prime ideals of a given degree
This is fairly classical, and the answer is zero for any $n>1$, and $1$ for $n=1$. The reason basically comes down to the fact that asymptotic-wise, almost all prime powers are prime.


Here is the proper proof. Firstly, by Chebotarev, a positive density of primes in $\Bbb Z$ split completely in $O\_K$. This alone implies that your denominator is $\gg \frac{B}{\log B}$. On the other hand, the norm a prime of degree $n$ must be of the form $p^n$ for some prime $p$, and for each $p$ you have at most $[K:\Bbb Q]$ primes dividing $p$. Therefore your numerator is $\leq[K:\Bbb Q]B^{1/n}$, which immediately gives $\delta\_n=0$ for $n>1$.


It also follows quite easily that $\delta\_1=1$ - we just have to bound the sum of the relevant terms over all possible $n>1$. But since $p^n\leq B$ for a prime $p$ implies $n\leq\log\_2 B$, we get $\ll B^{1/2}\log B$ primes of degree more than $1$ in total.


