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Suppose you have a number field $L$, and a non-zero ideal $I$ of the ring of integers $O$ of $L$.

Question part A: Is there prime ideal $\mathcal{P} \subseteq O$ in the ideal class of $I$ such that $p =\mathcal{P} \cap \mathbb{Z}$ splits completely in $O$?

If the extension $L/\mathbb{Q}$ is Galois I think one can show that the answer is yes. In fact I think it is possible to show that for every $I$ there are infinitely many such $\mathcal{P}$'s. The necessary tools for this are class field theory and Chebotarev's density theorem. On the other hand I'm not asking for infinitely many primes, only for one. So to be concrete:

Question part B: If the answer to A is yes, is it possible to give a proof that avoids showing that there are infinitely many $\mathcal{P}$'s?

Thank you.

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1 Answer 1

Yes, every ideal class contains a split prime. Here is one possible proof, which however doesn't satisfy the conditions of part B of your question: the analogue of Dirichlet's argument, using ideal class characters, shows that $\sum_{\mathfrak p \in [I]} N\mathfrak p^{-1}$ (the sum being taken over prime ideals in the class of $I$) diverges. On the other hand, the sum over non-split primes converges (because a non-split prime lying over $p$ has norm at least $p^2$, and so the sum over non-split primes is majorized by (some constant times) $\sum_p p^{-2}$, which converges). Thus there must be infinitely many split primes contributing to this sum.

Note: As the OP points out in a comment below, this argument applies only in the Galois case, and so doesn't actually address the question.

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Dear prof Emerton: Thank you for the nice answer. I'm unsure if you are only thinking in the Galois case. In the general case I don't see why "a non-split prime...has norm at least $p^2$" In fact I think that's false, e.g. $x^3-2=(x-3)((x-1)^2+3))$ on $F_{5}[x]$, hence over $\mathbb{Q}(\sqrt[3]{2})$ there is a prime lying over 5 with norm equal to 5. However 5 is not split. –  user8317 Mar 16 '11 at 3:54
@user8317: Dear user8317, You are right. I wasn't conscious of it, but indeed my argument applies as written only in the Galois case. Let me see if I an fix it ... . Best wishes, –  Matt E Mar 16 '11 at 4:03

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