An approach to the Gauß class number one problem for imaginary quadratic fields is to determine the integral points on the modular curve $Y_{nonsplit}(n)$ for a suitable $n$. Here follows a quick summary, but see Appendix A in Serre's book "Lectures on the Mordell-Weil Theorem" for more details:

Given an imaginary quadratic $K$ of class number 1, one considers an elliptic curve $E$ with CM by $O_K$, which is unique up to $\mathbb{C}$-isomorphism. Its $j$-invariant lies in $\mathbb{Z}$. Given any integer $n$, all of whose prime divisors are inert in $K$, our $E$ yields a unique integral point on $Y_{nonsplit}(n)$. That means, fixing an $n$ and letting $p$ be its largest prime divisor, any imaginary quadratic $K$ of class number 1 whose discriminant is larger than $4p$ in absolute value will furnish a unique integral point on $Y_{nonsplit}(n)$. Hence, determining the integral points on $Y_{nonsplit}(n)$ for one $n$ for which there are finitely many integral points solves the problem. Heegner used $n=24$, as did Stark.

My question is whether a similar approach has been, or can be, used in the real quadratic case;

the goal being to prove that there are infinitely many such fields of class number one.

For what it's worth, here's what would happen in my pipe dream:

  1. To a real quadratic $K$ of class number 1 one attaches uniquely an abelian surface (or some other kind of object) with real multiplication by $O_K$;

  2. One shows that this object gives rise to a unique rational point on some moduli space, and conversely, all such rational points would come from a $K$; ($Y_{nonsplit}$ plays this role in the imaginary case...)

  3. One shows that the moduli space has infinitely many rational points.

(Since this is a well-known open problem, I'll follow the advice of the FAQ and make it community-wiki. I am a little embarrassed asking such a speculative question, but I also feel that speculation drives a lot of mathematical research.)


1 Answer 1


I would say there are two issues: first of all, the good thing making the above machinery work is that the moduli problem you look at, represented by $Y_\text{nonsplit}(n)$, works for all quadratic fields at once. In other words, you are interested in looking at all possible curves with CM by some imaginary quadratic field and you end up looking at points always in the same moduli space. This is not the case for Hilbert-Blumenthal surfaces with real multiplication by a real quadratic $F$, who do possess a moduli space (stack, indeed) $\mathfrak{M}_F$ which heavily depends upon $F$.

But suppose this is only a technical issue, either by constructing some monstrous $\mathfrak{M}$ parametrizing abelian surfaces with al multiplication by some quadratic field not specified in the moduli problem; or by hoping, in your "pipe dream" that one proves that there are infinitely many $F$ with at least one element in $\mathfrak{M}_F(\mathbb{Q})$. Then a major (and, to my knowledge and understanding, not a merely technical one) problem is the following: the crucial step in the CM procedure, as you said, is the implication (now $K$ is imaginary quadratic) $$ h(K)=1\Leftrightarrow j(E_K)\in\mathbb{Q} $$ (it is indeed integral, but that won't matter, here) where $E_K$ is some/any elliptic curve with CM by $K$. This is false for real multiplication: to convince yourself of the failure of one implication, observe that there are elliptic curves with real multiplication by $\mathbb{Q}$ (what else?) which are not defined over the rationals although $h(\mathbb{Q})=1$. Likewise, the field of definition of the abelian surface tells you almost nohing on the arithmetic of the field of real multiplication, unlike the CM case, preventing you from reducing the class number problem to counting varieties (with bonus structure) defined over $\mathbb{Q}$ - i.e. rational points on some moduli space.


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