# Does every ideal class contains a prime ideal that splits?

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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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
@MattE: I added an answer to this question, and I'm pinging you in this comment because the question is so old. Hopefully I didn't do anything totally stupid. – Barry Smith Aug 8 '15 at 11:57

The answer to both A and B is no. Take the non-Galois field $L = \mathbb{Q} (\sqrt[4]{-5})$. Let $N = L(i)$, the Galois closure of $L$. Computing with PARI/GP tells me $L$ has cyclic class group of order $4$ and $N$ has class group of order $2$. Thus, the map $N \colon \mathrm{Cl}_N \rightarrow \mathrm{Cl}_L$ induced by the relative norm of ideals is not surjective.

Assume, to reach a contradiction, that one of the classes of order $4$ in $\mathrm{Cl}_L$ contains a prime $\mathfrak{p}$ lying over a prime $p \in \mathbb{Z}$ which is completely split in $L$. It is well-known that this happens if and only if $p$ is completely split in $N$. Then $\mathfrak{p}$ is the norm of an ideal $\mathfrak{P}$ in $N$. But then the class of $\mathfrak{p}$ is the norm of the class of $\mathfrak{P}$, contradicting the fact that the norm map on classes is not surjective.

A is true if we make the additional assumption that the normal closure of $L$ is linearly disjoint over $L$ from the Hilbert class field of $L$.

To see this, let $N$ be the normal closure of $L$ over $\mathbb{Q}$, let $H_N$ be the Hilbert class field of $N$, and let $H_L$ be the Hilbert class field of $L$. We must show that for each $\sigma \in G(H_L/L)$, there exists some prime ideal of $L$, completely split over $\mathbb{Q}$ and whose Frobenius for $H_L/L$ is $\sigma$.

We need some facts:

1. $H_N$ is Galois over $\mathbb{Q}$
2. $N H_L \subset H_N$

For 1, consider an embedding $\tau \colon H_N \hookrightarrow \overline{\mathbb{Q}}$. We have $\tau (N) = N$, so $\tau (H_N)$ is an abelian unramified extension of $N$, hence is contained in $H_N$ It follows that $\tau(H_N) = H_N$ and 1 follows.

For 2, consider a prime ideal $\tilde{\mathfrak{P}}$ in $N H_L$, which is Galois over $L$. Because $H_L/L$ is unramified, the inertia group for $\tilde{\mathfrak{P}}$ over $L$ is contained in $G(N H_L /H_L)$. But the definition of inertia groups also shows that the inertia group for $\tilde{\mathfrak{P}}$ over $N$ is contained in that for $\tilde{\mathfrak{P}}$ over $L$. Thus, the inertia group for $\tilde{\mathfrak{P}}$ over $N$ consists only of automorphisms that fix $N$ and $H_L$, so it is trivial and $N H_L/N$ is unramified. Since $G(N H_L/N)$ injects into $G(H_L/L)$ by restriction, $N H_L/N$ is abelian and 2 follows.

Now let us return to considering our $\sigma$ in $G(H_L/L)$. By 1 and the assumption that $N$ and $H_L$ are linearly disjoint over $L$, we can lift $\sigma$ to an automorphism $\tilde{\sigma}$ in $G(H_N/N)$. By Chebotarev, we can find some prime ideal $\mathfrak{P}$ in $H_N$, unramified over $\mathbb{Q}$ and whose Frobenius over $\mathbb{Q}$ is $\tilde{\sigma}$. The decomposition group of $\mathfrak{P}$ over $\mathbb{Q}$ is thus contained in $G(H_L/L)$, so the prime $p = \mathfrak{P} \cap \mathbb{Z}$ is completely split up to $L$.

Let $\mathfrak{p} = \mathfrak{P} \cap N$. By the formalism of the Artin symbol, the Frobenius for $\mathfrak{P}$ over $N$ is $\tilde{\sigma}$ and its restriction to $H_L$, namely $\sigma$ is itself the image under the Artin map of the ideal $N_{N/L} \mathfrak{p}$. But again, because $p$ is completely split up to $L$, it is also completely split up to $N$ and so $N_{N/L} \mathfrak{p}$ is a prime ideal of $L$ with degree $1$ and Frobenius equal to $\sigma$ as desired.

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