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Let $K/\mathbb{Q}$ be a Galois Number field. Let $p$ be an unramified rational prime. In this extension, for any $P,Q | p\mathcal{O}_K$ then the relative degrees $f(P) := [\mathcal{O}_K/P : \mathbb{Z}/p\mathbb{Z}]$ satisfy $f(P) = f(Q)$. The Chebotarev density theorem gives statistical information on the size of sets of rational primes $p$ such that the primes lying above $p$ have a specific Frobenius conjugacy class. On the other hand, it does not provide information on how many rational primes in Galois extensions have a fixed $f(P)$ (aside from primes that split i.e. $f(P) = 1$, which will have trivial conjugacy class). Are there any general results about the statistics of primes with specific values $f(P)$? In the Abelian case? I would appreciate any references about this.

Dead-End
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    The Frobenius conjugacy class determines $f(P)$ -- it's just the order in the group of any element of the conjugacy class -- so you can read this off immediately from Chebotarev. – David Loeffler Mar 22 '14 at 18:49
  • @David Loeffler: This is certainly true. But how many conjugacy classes will occur with that value of $f(P)$? How large must they be? – Dead-End Mar 22 '14 at 19:17
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    That is simply a question about the number of elements of each order in the Galois group: the density of rational primes for which $f(P) = r$ for some (and hence all) $P \mid p$ is exactly $ {$# of elts of G of order r $}/$#G. – David Loeffler Mar 23 '14 at 09:10
  • Note $f(P)=f(Q)$ isn't guaranteed to be true unless $p$ is unramified. Indeed Chebotarev density can be leveraged to predict the density of any kind of splitting type of a prime, even in finite non-Galois extensions. I explain how that works in this answer. – anon Mar 24 '14 at 18:23
  • @sea turtles. Edited, thank you for catching that – Dead-End Mar 25 '14 at 03:22

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