Let $K$ be a real quadratic number field and $\mathcal O_K$ its ring of integers. Is it known whether for each element in the class group we have a representative $\mathfrak p \subset \mathcal O_K$ which is prime?

My conjecture: It is true.

In case it's true: How to prove it?

In case it's wrong: Which is the least discriminant $D$ that serves as counter-example?

Since each ideal can be written as product of prime ideals it's clear that the prime ideals at least generate the class group. Further it's clear that the neutral element of the class group is always represented by a prime ideal since there are always inert primes (namely for $\chi_D(p)=-1$). So this implies that my conjecture is at least true for those $K$ with class number $h_K \le 3$.

  • 2
    $\begingroup$ This is the analog of Dirichlet's Theorem on primes in arithmetic progressions; the proof is analytic. $\endgroup$
    – user23365
    Apr 30, 2021 at 12:46
  • $\begingroup$ @franzlemmermeyer Can you give a reference? $\endgroup$ Apr 30, 2021 at 12:48

1 Answer 1


Yes, it’s true.

Let $H/K$ be the Hilbert class field. There is a bijection $Gal(H/K) \rightarrow Cl(O_K)$ that maps the Frobenius above a prime $\mathfrak{p}$ of $K$ ($H/K$ is abelian so that element is well-defined) to the class of the prime $\mathfrak{p}$.

By Cebotarev, every element of $Gal(H/K)$ is the Frobenius of some prime (and some more but never mind), and thus every class in $Cl(O_K)$ comes from some prime.

  • $\begingroup$ Is is true as well that given two primes $p,q$ in $\mathbb Z$ with $p \equiv q \pmod D$ that for the case that they are split (which is equivalent to $\chi_D(p)=1$), so $(p)=PP'$ and $q=QQ'$, we have $[P]=[Q]$ or $[P]=[Q']$ in $\operatorname{Cl}(\mathcal O_K)$? So in particular in order to get all different types of prime ideals you have only to check the splitting of primes $1<p\le D$. $\endgroup$ May 1, 2021 at 5:27
  • $\begingroup$ I’m pretty sure that it’s not the case in general, because we’re talking about Frobenius repartition in a “higher degree extension” setting. But I don’t know any particular counterexample – maybe you can find some with software, I don’t think you’d need to check it very far. $\endgroup$
    – Aphelli
    May 1, 2021 at 8:07
  • $\begingroup$ I'll try that. When I checked $\mathbb Q(\sqrt{10})$ which has class number $2$ I was surprised that that seems to be the case. There are eight values mod $D=40$ where we have $\chi_D(x)=1$. For exactely four of them your prime splits always into two principal ideals, for the other four your prime spilts into two non-principal ideals. Since the class number is $2$, that's all you need for my question. $\endgroup$ May 1, 2021 at 8:15
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    $\begingroup$ Yes! Class field theory yields an isomorphism $Gal(H/K) \rightarrow Cl(K)$ mapping the Frobenius at a prime $q$ of $K$ (well-defined as $Gal(H/K)$ is abelian) to the class of $q$ in the class group. By Cebotarev, every element in $Gal(H/K)$ is the Frobenius at infinitely many primes – QED. $\endgroup$
    – Aphelli
    Nov 10, 2022 at 18:29
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    $\begingroup$ @principal-ideal-domain by Chebotarev's density theorem for the abelian extension $H/K$, the density of prime ideals in $K$ having a particular ideal class in the class group of $K$ is $1/h(K)$. $\endgroup$
    – KCd
    Nov 10, 2022 at 22:36

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