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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?

near
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  • Have you worked it out for $K=\Bbb Q(i)$ ? – Lubin Apr 08 '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 – near Apr 08 '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 08 '18 at 20:31
  • Right, I definitely agree! But I still don't see clearly how to move forward... – near Apr 08 '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 08 '18 at 20:54
  • You're right, that may be the point, I'll try to see if I figure this out via the Čebotarev Density Theorem. Anyway, thank you very much for your help! – near Apr 08 '18 at 21:06
  • First you should try to use the Cebotarev Density Theorem to reduce the problem to the following purely theoretic exercise: Suppose that a finite group G acts transitively on a finite set X (say the roots of an irreducible polynomial). Then the average number of fixed points in X for elements g in G is 1. You can prove this by showing that the probability any chosen element $x \in X$ is fixed is $1/|H|$, where $H$ is the stabilizer of $x$, and then use the Orbit-Stabilizer Theorem. – Infinity Apr 10 '18 at 13:11
  • @Infinity thanks for the hint! However I still think I'm missing something (maybe trivial)...I agree that you can reduce the problem to that of showing the average number of fixed points in X for elements g in G is 1, but I don't really understand how to prove this. How do I show that the probability is $\frac{1}{|H|}$? – near Apr 10 '18 at 15:13
  • Sorry, I meant the probability that $x \in X$ is fixed is $|H|/|G|$, where $H = H_x$ is the stabilizer of $x$, simply because the elements that fix $x$ are precisely those in $H$. Then the the average number of fixed points is just the sum $\sum_{x \in X} |H_x|/|G|$. But if $G$ is transitive, then all the $H_x$ are conjugate and are of the same order, so this sum is $|H||X|/|G| = 1$ by the Orbit-Stabilizer theorem. – Infinity Apr 11 '18 at 01:17
  • Ok, I think I have it clearer now. Just a question, is this (I mean this proof that the average is 1) related in some way to this approach to the problem? https://math.stackexchange.com/questions/349118/probability-permutations By the way, thank you very much for your help! – near Apr 11 '18 at 13:58

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