It's known that the class number of $\mathbb{Q}(\sqrt{p})$ is $1$ for all primes $p<229$.

Question: What would it be like for conceptual explanations of $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$ (equivalently, $\mathbb{Q}(\sqrt{p})$ having odd conductor)?

To clarify the question:

  • A conceptual explanation should treat the first few primes simutaneously, instead of a case-by-case analysis where a case is a single prime. (A case-by-case analysis with a finite number of cases that a priori covers the whole range of primes is allowed, e.g. the cases being p=1, 5 or 9 mod 12.)

  • For "the first few primes of form $4k+1$", I mean such continuous primes up to a bound, e.g. $5,13,17$ but not $5,17,29$. The argument should be able to cover primes in such a way.

  • To avoid trivialities, the conceptual explanations should cover at least $5, 13, 17$ and $29$.

An example of conceptual explanation would be like:

  • By Example 2.9 of Masley's paper Class numbers of real cyclic number fields with small conductor, the class number of such fields are odd.

  • The Minkowski bound gives $h(\mathbb{Q}(\sqrt{p}))<3$ for $p<36$. Thus we have established $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$: $5,13,17$ and $29$.

  • This explanation also works for cyclic cubic fields of conductor $7$ and $13$.

Bonus for explanations that are not specialized on real quadratic fields, e.g. the explanation presented above.

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    $\begingroup$ Tiny note that there are very few primes of the form p=9 mod 12. :) $\endgroup$ May 26, 2020 at 20:36

2 Answers 2


We give a uniform approach to $p \leq 61$ by applying analytic discriminant bounds to the Hilbert class field. To be sure this is not entirely "conceptual", but then some computation is needed even to deal with $p < 36$ using Minkowski.

If $p = 4k+1$ is prime then $K = {\bf Q}(\sqrt{p})$ has odd class number $h$, so either $h=1$ or $h \geq 3$. If $h \geq 3$ then the Hilbert class field $H_K$ is a totally real field of degree $2h \geq 6$ and discriminant $p^h$. We can now apply to $H_K$ the known lower bounds on the discriminants of totally number fields.

The Odlyzko bounds (see Table 4 on page 134 of his 1990 paper give a lower bound of $7.941$ on the root-discriminant $|{\rm disc}(F)|^{1/n}$ for any totally real field $F$ of degree $n \geq 6$. Hence $p > 7.941^2 > 63$. So we have accounted for all $p \leq 61$.

If the zeta function $\zeta_F$ satisfies the Riemann hypothesis, the lower bound improves to $8.143$. Unfortunately this does not account for any more primes because $8.143^2 = 66.3+$ and $65$ is not prime. Since there exists a totally real sextic field of discriminant $300125 = 5^3 7^4$ (see the LMFDB entry), such bounds can never get us past $300125^{1/3} = 66.95+$, so the primes $p \in [73, 197]$ must be dealt with in some other way.


Some complementary heuristics, too long for a comment.

Again start from the fact that the class number $h$ of $K = {\bf Q}(\sqrt{p})$ is odd if $p$ is a prime of the form $4k+1$. This time we compare with Dirichlet's class number formula, which here gives $$ L(1,\chi_p) = \frac{2\log \epsilon}{\sqrt p} h, $$ where the character $\chi_p$ is the Legendre symbol $\chi_p(n) = (n/p)$, and $\epsilon$ is the fundamental unit of $K$.

We expect that $L(1,\chi_p) \approx 1$, so large $h$ go with small $\epsilon$. A unit $\epsilon > 1$ in a real quadratic field of discriminant $D$ must be at least as large as $\frac12(m + \sqrt{m^2 \pm 4})$ for some odd integer $m$, with $D = m^2 \pm 4$. If $D$ is prime then we must use the plus sign (unless $m=3$, but then the fundamental unit is $(1+\sqrt5)/2$). $\epsilon > \sqrt{p} - O(1/\sqrt{p})$ and So, $2\log \epsilon / \sqrt{p} > \log p - O(1/p)$. Setting $L(1,\chi_p) \approx 1$ and $h=3$ in the class number formula gives $\sqrt{p} \approx 3 \log p$; the solution $p \approx 289$ is of about the right size for the minimal example of $h>1$.

We're actually closer here than we deserved to be: the solution of $L(1,\chi_p) \sqrt{p} = 3 \log p$ is quite sensitive to the size of $L(1,\chi_p)$, and $L(1,\chi_{229}^{\phantom.}) = 1.075+$ is unusually close to $1$; for example, $L(1,\chi_p) > 2$ for $p = 193, 241, 313, 337$, while $L(1,\chi_p) < 0.4$ for $p = 173, 293, 677, 773$. Most of the early examples of $h > 1$ have small $\epsilon$, either with $p=m^2+4$ as above or the next-smallest possibility, $\epsilon = m + \sqrt{p}$ with $p=m^2+1$. Indeed this LMFDB list of fields ${\bf Q}(\sqrt p)$ with $p<2000$ and $h>1$ begins with $$ 229 = 15^2 + 4,\ 257 = 16^2 + 1,\ 401 = 20^2 + 1,\ 577 = 24^2 + 1, $$ $733 = 27^2 + 4$, and then two exceptions $p=761$ and $p=1009$ and nine further $p$ of which all but $1429, 1489, 1901$ are not of the form $m^2+4$ or $m^2+1$. Moreover $229$ is the only second-smallest prime of the form $p = m^2 + 4$ that satisfies our analytic bound $p > 63$ --- and the smallest is $p = 173$, which was our example of an unusually small $L(1,\chi_p)$. Likewise the next two examples were $293 = 17^2 + 4$ and $677 = 26^2 + 1$, which are conjectured to be the largest discriminants $p = m^2+4$ and $p = m^2+1$ for which ${\bf Q}(\sqrt p)$ has class number $1$.

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    $\begingroup$ Noam: In 2 articles published in Acta Arithmetica in 2003, A. Biro claims to have established the truth of the conjectures (of Yokoi and Chowla) mentioned in your final sentence, using extensive computer calculation. Is his claim generally accepted? $\endgroup$ May 27, 2020 at 13:32
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    $\begingroup$ Thank you: I was not aware of this pair of theorems (and didn't know or remember the attribution of the conjectures to Yokoi and Chowla). I'll fix this in the next edit. $\endgroup$ May 27, 2020 at 16:32

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