# average number of primes of degree one which lie above p

I'm reading P.J. Weinberger's article "Finding the Number of Factors of a Polynomial" (Journal of Algorithms, 5, 1984) and I can't fully understand an assertion on primes of (inertial) degree one in the ring of algebraic integers of a number field. What is stated is that, given a number field $K$ and a prime $p$, the average number of primes of degree one of $\mathcal{O}_{K}$ which lie above $p$ is one, due to prime number theorem for number fields. Why is this true?

• Have you worked it out for $K=\Bbb Q(i)$ ? – Lubin Apr 8 '18 at 20:14
• Unfortunately not yet...I have found some results in "A conversational introduction to algebraic number theory: arithmetic beyond $\mathbb{Z}$" by Paul Pollack, from which it seems that this follows in some way from class number formula, but it still remains quite unclear to me – thanks_emmy Apr 8 '18 at 20:25
• Well, you do know that half the primes of $\Bbb Z$ split, and half remain prime, in $\Bbb Z[i]$, right? And that the split ones contribute two primes $\mathfrak p$ of degree one, right? And that above a nonsplit prime of $\Bbb Z$, the prime is of degree two, right? – Lubin Apr 8 '18 at 20:31
• Right, I definitely agree! But I still don't see clearly how to move forward... – thanks_emmy Apr 8 '18 at 20:44
• This is not my meat at all, but it looks to me like the Čebotarev Density Theorem, which you can find a Wikipedia article for. I think that this is what takes the place of the Prime Number Theorem for number fields. – Lubin Apr 8 '18 at 20:54