Riffaut's conjecture: no singular modulus of degree at least 3 is a root of a rational trinomial (unconditional status)

Riffaut's conjecture asserts that no singular modulus — the $j$-invariant of an elliptic curve with complex multiplication — of degree at least 3 over $\mathbb{Q}$ can be a root of a trinomial with rational coefficients. It arose as the last unsettled case of Riffaut's classification of the equations $Ax^m+By^n=C$ in singular moduli and as an obstruction in effective approaches to the André–Oort conjecture. Bilu, Luca and Pizarro-Madariaga proved the conjecture under the Generalized Riemann Hypothesis (GRH) for Dirichlet $L$-functions, and showed unconditionally that the set of 'trinomial discriminants' — imaginary quadratic discriminants $\Delta$ with class number $h(\Delta)\ge 3$ whose singular moduli are roots of rational trinomials — is finite, consists of odd fundamental discriminants of the form $-p$ or $-pq$, satisfies $|\Delta|>10^{11}$ with at most one exception with $|\Delta|\ge 10^{160}$, and forces the exponent gap $m-n\le 2$ whenever $|\Delta|\ge 10^{40}$. The open problem is the unconditional status: prove without GRH that the set of trinomial discriminants is empty, or exhibit a counterexample.

Background

A singular modulus is the value of the modular $j$-function (the standard function on the complex upper half-plane parametrizing elliptic curves up to isomorphism) at an imaginary quadratic point; equivalently, it is the $j$-invariant of an elliptic curve with complex multiplication (CM), meaning that its endomorphism ring is larger than $\mathbb{Z}$. Singular moduli are algebraic integers. Two singular moduli are conjugate over $\mathbb{Q}$ exactly when they correspond to the same discriminant $\Delta<0$ (the discriminant of the underlying imaginary quadratic order), and there are exactly $h(\Delta)$ of them, where $h(\Delta)$ is the class number of that order (the number of equivalence classes of its invertible ideals); hence a singular modulus of discriminant $\Delta$ has degree $h(\Delta)$ over $\mathbb{Q}$. There are 13 rational singular moduli and 29 $\mathbb{Q}$-conjugate pairs of degree 2.

A trinomial, in the convention of the main source, is a monic three-term polynomial $t^m+At^n+B$ with rational coefficients, integers $m>n>0$ and $B\ne 0$. This normalization loses no generality: any trinomial with rational coefficients becomes monic after dividing by its leading coefficient, and if $B=0$ then a root $x$ satisfies $x^{m-n}\in\mathbb{Q}$, which for a singular modulus forces $x\in\mathbb{Q}$ (because $\mathbb{Q}(x^k)=\mathbb{Q}(x)$ for a singular modulus $x$ and $k\ge 1$), i.e. degree 1. An imaginary quadratic discriminant $\Delta$ with $h(\Delta)\ge 3$ is called a trinomial discriminant, of signature $(m,n)$, if the singular moduli of discriminant $\Delta$ are roots of such a trinomial. Riffaut's conjecture is exactly the assertion that trinomial discriminants do not exist.

The question comes from the study of Diophantine equations whose unknowns are singular moduli. Linear relations $Ax+By=C$ and multiplicative relations $xy\in\mathbb{Q}^{\times}$ between two singular moduli had been completely classified (Allombert–Bilu–Pizarro-Madariaga for the linear case, Bilu–Luca–Pizarro-Madariaga for products). Riffaut then conjectured a full classification of the equations $Ax^m+By^n=C$ with unknown positive exponents (his Conjecture 1.4) and proved it for distinct singular moduli, the two exceptional degree-3 pairs being eliminated subsequently by Luca–Riffaut. The one case his method could not treat is $x=y$: then $Ax^m+Bx^n=C$ says precisely that the singular modulus $x$ is a root of a trinomial, and he conjectured the negative answer. The problem also has an independent motivation in the effective André–Oort conjecture (the André–Oort conjecture, proved non-effectively, states that an algebraic curve containing a Zariski-dense set of special, i.e. CM, points is a special curve; effective versions seek computable height bounds for the remaining special points), where possible trinomial relations among powers of one singular modulus are an obstruction.

Bilu–Luca–Pizarro-Madariaga proved the conjecture under the Generalized Riemann Hypothesis for Dirichlet $L$-functions (GRH; Dirichlet characters are the completely multiplicative arithmetic functions modulo an integer, their $L$-functions generalize the Riemann zeta function, and GRH asserts that all non-trivial zeros of these $L$-functions lie on the line $\mathrm{Re}(s)=1/2$), and obtained far-reaching unconditional structure theorems. Their argument shows that a trinomial discriminant $\Delta$ cannot admit a small 'suitable' integer: a positive integer $a$ is called suitable for $\Delta$ if there exist integers $b,c$ such that the triple $(a,b,c)$ is a primitive reduced binary quadratic form of discriminant $\Delta$, meaning $\gcd(a,b,c)=1$, $b^2-4ac=\Delta$ and $|b|\le a\le c$ up to the standard boundary conventions (the singular modulus attached to this form is $j((b+\sqrt{\Delta})/2a)$). Suitable integers of a trinomial discriminant are forced to be large, of order $|\Delta|^{1/2}/\log|\Delta|$, and densely spaced; but a prime $p\le 4|\Delta|^{1/2}/\log|\Delta|$ that splits in $\mathbb{Q}(\sqrt{\Delta})$, i.e. with Kronecker symbol $(\Delta/p)=1$, is automatically suitable when $|\Delta|\ge 4p^2$, so such a prime would contradict the lower bound on suitable integers. Under GRH such a prime always exists, by an adaptation of conditional bounds for the least quadratic non-residue. Unconditionally, the classical bound $p\ll_\varepsilon m^{1/4+\varepsilon}$ for the least prime with $\chi(p)=1$ is ineffective because of a possible Siegel zero (a real zero of some $L(s,\chi)$ very close to $1$); replacing Siegel's ineffective lower bound for $L(1,\chi)$ by Tatuzawa's version, which permits at most one exceptional discriminant to escape, they obtain: every trinomial discriminant satisfies $|\Delta|>10^{11}$; at most one satisfies $|\Delta|\ge 10^{160}$, so the set of trinomial discriminants is finite; every trinomial discriminant is of the form $-p$ or $-pq$ with $p,q$ distinct odd primes, hence odd and fundamental (not a proper multiple of another discriminant); and every signature $(m,n)$ of a trinomial discriminant with $|\Delta|\ge 10^{40}$ satisfies $m-n\le 2$. The trinomial side is controlled by Bilu–Luca's theorem that, up to obvious exceptions, only boundedly many trinomials vanish at given algebraic numbers, with effectively bounded degrees and heights.

What remains open is the unconditional emptiness of the set of trinomial discriminants: the possible single exceptional discriminant with $|\Delta|\ge 10^{160}$ has not been excluded, and the finite window $10^{11}<|\Delta|<10^{160}$ (over discriminants $-p$ and $-pq$) is far beyond direct computation.

Problem Statement

Determine, without unproved hypotheses, whether a singular modulus of degree at least 3 over $\mathbb{Q}$ can be a root of a trinomial with rational coefficients (this is Problem 1.1 of Bilu–Luca–Pizarro-Madariaga; Riffaut's conjecture asserts that the answer is negative).

Specifically: prove, without assuming GRH or any other unproved hypothesis, that the set of trinomial discriminants is empty, where a trinomial discriminant is an imaginary quadratic discriminant $\Delta$ with class number $h(\Delta)\ge 3$ such that the singular moduli of discriminant $\Delta$ are roots of a monic trinomial $t^m+At^n+B$ with $A,B\in\mathbb{Q}$, $m>n>0$, $B\ne 0$ (a normalization that loses no generality, as explained above). Within the framework of the known unconditional results, this means closing the two remaining gaps: (i) ruling out the at most one possible exceptional trinomial discriminant with $|\Delta|\ge 10^{160}$ whose existence is permitted by the effective form of Siegel's lower bound for $L(1,\chi)$; and (ii) excluding trinomial discriminants in the remaining finite range $10^{11}<|\Delta|<10^{160}$, over discriminants of the form $-p$ and $-pq$. A solution need not follow this route: any unconditional proof that no singular modulus of degree at least 3 is a root of a rational trinomial, by whatever method, resolves the problem.

A complete resolution may alternatively take the form of an explicit counterexample: a specific fundamental imaginary quadratic discriminant $\Delta$ with $h(\Delta)\ge 3$ and a specific trinomial $t^m+At^n+B\in\mathbb{Q}[t]$, $m>n>0$, $B\ne 0$, having a singular modulus of discriminant $\Delta$ among its roots; such a witness would refute the conjecture.

References

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