Conceptual explanations of the class numbers for the first few $\mathbb{Q}(\sqrt{p})$ with odd conductor

Question

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.

Answer 1

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.

Answer 2

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 $1$.