# A conjecture of Lang

Of course, all errors (mathematical and literary) are my fault. Please email me with your suggestions and corrections.

Instead of writing the same thing but worse, I will steal the introduction to the paper *Torsion points on modular curves and Galois theory*, by Minhyong Kim and Ken Ribet.

"In elementary terms, the arithmetic theory of a curve X is concerned with solutions to a geometrically irreducible polynomial equation in two variables:

\begin{equation} \label{eq:f} f(x, y) = 0 \end{equation}In contrast to the geometric theory, where the different kinds of number pairs \((x, y)\) that can occur as solutions are viewed as homogeneous, the arithmetic classifies more carefully the structure of solutions of specific type. That is, one tries to understand the solutions to the Equation (\ref{eq:f}) where \((x, y)\) are constrained to lie in some arithmetically defined set. One common case is that of rational solutions or, more generally, the case of solutions where \(x, y\) are constrained to lie inside a fixed number field \(F\). For example, when \(f(x, y)\) has genus 1 (that is, the smooth points of the complex solution set form a genus one Riemann surface with punctures), the Mordell-Weil theorem says the solution set, in conjunction with a few additional points, acquires the natural structure of a finitely generated abelian group. For another example, when the genus is greater than one, Faltings proved that the solution set is finite. In both cases, one derives finite-type structures for the solution set from finiteness contraints on the ‘type’ of the solution. A theorem of Ihara-Serre-Tate gives an example of finiteness theorems deriving from a different kind of arithmetic constraint. Namely, one considers solutions that are roots of unity of arbitrary order. Then as soon as the genus is at least one, there are again only finitely many solutions to (\ref{eq:f})."

The theorem of Ihara, Serre, and Tate mentioned here is the solution to the following conjecture proposed by Lang in the sixties.

Let \(f(x,y)\) be an irreducible polynomial with complex coefficients. Suppose that there exist infinitely many pairs of roots of unity \((\theta, \eta)\) such that \(f(\theta, \eta) = 0\). Then, \(f(x,y)\) is of the form:

\begin{equation*} ax^n + by^m,\quad cx^ny^m + d. \end{equation*}The proofs were found independently and communicated privately to Lang. Ihara proved a special case, and Serre and Tate gave very similar proofs. For the original source, check out Lang's paper.

## Split tori

Before discussing the proof of Serre and Tate, lets recall some generalities about the multiplicative group \(\mathbf{G}_m\) and its powers.

Let \(k\) be some field. The *multiplicative group* \(\mathbf{G}_m\) is the \(k\)-variety obtained by removing the origin from the affine line. It is an algebraic group, with functor of points given by \(\mathbf{G}_m(R) = R^\times\) for any \(k\)-algebra \(R\). Indeed, any morphism of \(k\)-algebras \(k[t, t^{-1}] \to R\) is determined by the image of the variable \(t\), which has to be a unit in \(R\).

A *split torus* is simply a power of the multiplicative group \(T = \mathbf{G}_m^n\). It's functor of points is given by \(T(R) = (R^\times)^n\) for any \(k\)-algebra \(R\), with coordinate-wise multiplication.

For our applications, we restrict to the case where \(k\) has characteristic zero, and we fix \(\overline{k}\) an algebraic closure. Given an algebraic group \(G\) over \(k\), denote by \(\overline{G} = G\times_k \overline{k}\) the base-change of \(G\) to the algebraic closure. A *geometric character* is a morphism \(\chi\colon \overline{G} \to \mathbf{G}_{m, \overline{k}}\) of algebraic groups over \(\overline{k}\). It is customary (and sometimes confusing) to denote by \(X^*(G)\) the abstract group of geometric characters of \(G\). This group comes with a continuous action of the *absolute Galois group* \(\mathrm{Gal}_k\). Moreover, \(X^*\) is a contravariant equivalence from the category of algebraic groups of multiplicative type over \(k\), to the category of finitely generated abelian groups equipped with a continuous action of \(\mathrm{Gal}_k\). I learned this from Milne's book on algebraic groups (Theorem 12.23 in page 240). He also has a preliminary version of the book available for free here.

The reason I mentioned this is that in the case of maximal tori \(T = \mathbf{G}_m^n\), the functor \(X^*\) tells us everything we need to know about \(T\) and its subgroups:

- Algebraic subgroups \(G \subset T\) correspond to lattices \(X^*(G) := \Lambda_G \subset \mathbf{Z}^n\). In particular, \(G\) is defined by equations \(\mathbf{t^a} -1\), where \(\mathbf{t^a} := t_{1}^{a_1}\cdots t_n^{a_n}\) and \(\mathbf{a} = (a_1, \dots, a_n)\) runs through \(\Lambda_G\).
- It is enough to have \(\mathbf{a}\) run through a basis of \(\Lambda_G\) to get all the equations that cut out \(G\).
- The correspondence \(G \leftrightarrow \Lambda_G\) is one-to-one.
- If \(G\) has dimension \(n-r\), then \(\Lambda_G\) has rank \(r\).
- \(G \subset T\) is irreducible if and only if \(\Lambda_G\) is a factor of \(\mathbf{Z}^n\). Irreducible algebraic subgroups of \(T\) are called
*subtori*.

Given an integer \(N\), the *multiplication-by-N* map \([N]\colon T \to T\) is given in coordinates by \(\mathbf{x} \mapsto \mathbf{x}^N\). Therefore, the torsion points of \(T\) are those whose coordinates are roots of unity!

As you may have guessed, we may rephrase Lang's conjecture in terms of curves in \(\mathbf{G}_m^2\) containing infinitely many torsion points. But before we do that, we will state (and prove in dimension \(1\)) an important lemma which will be the final nail in the coffin in the proof of Serre and Tate.

Suppose that \(X \subset T = \mathbf{G}_m^n\) is a non-empty irreducible \(\overline{\mathbf{Q}}\)-subvariety such that \([N]X \subseteq X\) for some positive integer \(N >1\). Then, \(X\) is the translate of a subvariety by a torsion point.

We consider the case of curves. \(X\) is given by an irreducible polynomial \(f(x,y)\), and since \([N]X \subset X\), we see that \(f(\theta x, \eta y)\) divides \(f(x^N, y^N)\) for all \(N\)-th roots of unity \(\theta, \eta\). Since \(\deg f(x^N, y^N) = (\deg f(x,y))N\) and \((\theta, \eta)\) varies in a set of more than \(N\) elements, we conclude that \(X\) is invariant by at least \(N\) multiplicative translations \((x,y) \mapsto (\theta x, \eta y)\), where \((\theta, \eta)\) has order dividing \(N\). Since \(m\)-fold composition of \([N]\) corresponds to \([N^m]\), we see that \([N^m]X \subset X\), and we can run the same argument for every \([N^m]\) where \(m\) is a positive integer. We deduce that \(X\) is translation invariant by an infinite set of torsion points \(\Sigma\), and furthermore, it will be translation invariant by the Zariski closure of \(\Sigma\). This closure must contain a curve \(Z\), and \(Z\cdot X \subset X\), so that \(X\) must be a translate of \(Z\). From this, we see that \(Z\cdot Z \subset \tau Z\) for some root of unity \(\tau\), and the conclusion follows.

## The proof of Serre and Tate

We can restate Lang's conjecture in geometric terms as follows.

Suppose that \(X \subset \mathbf{G}_m^2\) is an irreducible complex curve containing infinitely many torsion points. Then, \(X\) is the translation of a subgroup \(G \subset \mathbf{G}_m^2\) by a root of unity.

If \(G \subset \mathbf{G}_m^2\) is a one-dimensional subgroup, then it is given by a polynomial equation \(x^ny^m-1\), since it corresponds to a rank one sublattice of \(\mathbf{Z}^2\). Translating \(G\) by a root of unity we get a variety defined by an equation of the form \(x^ny^m - \rho\), for some root of unity \(\rho\). We refer to these translates of subgroups by roots of unity as *torsion cosets*.

### First observation

Even though our curve is apriori defined over the complex numbers, **the condition that it vanishes on infinitely many roots of unity actually forces the curve to have coefficients in some cyclotomic field!** The argument goes as follows:

We may assume, after dividing by a non-zero coefficient, that the polynomial \(f(x,y)\) defining out curve \(X\) has some coefficient equal to \(1\). Let \(\mathbf{Q}(\mu_\infty) \subset \mathbf{C}\) be the field extension of the rationals generated by all roots of unity. Let \(K \supset \mathbf{Q}(\mu_\infty)\) be the field extension obtained by adjoining the coefficients of the polynomial \(f(x,y) \in \mathbf{C}[x,y]\). Take some field automorphism \(\sigma \in \mathrm{Aut}(K/\mathbf{Q}(\mu_\infty))\). By hypothesis, both \(f\) and \(f^\sigma\) have infinitely many zeros in common; so they divide each other. Since both \(f\) and \(f^\sigma\) have a coefficient equal to one, we conclude that \(f = f^\sigma\). This shows that \(K = \mathbf{Q}(\mu_\infty)\), so that every coefficient of \(f\) is contained in \(\mathbf{Q}(\mu_\infty)\).

Let's call \(k = \mathbf{Q}(\zeta_m)\) the cyclotomic number field over which our curve is defined. This will be the base field of our varieties going forward.

### The final step

Let \(\boldsymbol{\zeta}=(\theta. \eta)\) be a torsion point on \(X\), say of order \(n\). Let \(N\) be a positive integer coprime to \(n\), and let \(\sigma \in \mathrm{Gal}(\mathbf{Q}(\boldsymbol{\zeta})/\mathbf{Q})\) be an automorphism with \(\sigma \boldsymbol{\zeta} = \boldsymbol{\zeta}^N\). Choosing \(N\) such that in addition \(N \equiv 1 \, (\mathrm{mod} \, m)\), \(\sigma\) can be extended to a field automorphism of \(k(\boldsymbol{\zeta})/k\). This implies that \(\boldsymbol{\zeta}^N \in X\), and \(X \cap [N]X\) contains every conjugate of \(\boldsymbol{\zeta}\). If \(X\) and \([N]X\) don't have a common component, Bezout's theorem tells us that the number of points in the intersection does not excede the product of the degrees of the curves. We then have the following inequalities:

\begin{equation} \label{ineq:1} \frac{\varphi(n)}{m} \leq [k(\boldsymbol{\zeta}):k] \leq \# (X\cap [N]X) \leq N(\deg f)^2. \end{equation}By Dirichlet's theorem of primes in arithmetic progressions, for \(n\) large enough, we can find a prime number \(N\) which is congruent to \(1\) modulo \(m\) for which Inequality (\ref{ineq:1}) does not hold. Thus, for this particular \(N\), we have that \([N]X \subset X\) and the Lemma implies that \(X\) is a torsion coset.

## References

I wrote this blog post in preparation for a talk at our reading seminar in preparation for the Arizona Winter School 2023.

- Lang, Serge. "
*Division points on curves*." Annali di Matematica Pura ed Applicata 70.1 (1965): 229-234. - Lang, Serge. "
*Fundamentals of Diophantine geometry*". Springer Science & Business Media, 2013. - Zannier, Umberto. "
*Some Problems of Unlikely Intersections in Arithmetic and Geometry*".(AM-181). Princeton University Press, 2012.