Curves with many torsion points

(Jit here.)

I am going to write about some recent ideas in diophantine geometry. Consider a curve $C$ inside $\mathbb{C}^2$ given by some polynomial $P(x,y) = 0$. Can there be infinitely many points $(x,y)$ lying on our curve $C$ such that both $x$ and $y$ are roots of unity? If we just want one of the coordinates to be a root of unity, it is clear that there are infinitely such points, although they should be "sparse". The idea is then that if we impose this condition on both coordinates, there should somehow only be finitely many such points, unless $C$ is a special curve. For example, if $C$ was of the form $\{\omega \} \times \mathbb{C}$ where $\omega$ is a root of unity, then clearly it has infinitely many such points on it. In general, if our curve $C$ were given by $x^i y^j = \omega$, where $\omega$ is a root of unity, then $y$ will be a root of unity if $x$ is and so we get infinitely many points with both coordinates roots of unity.

We can understand curves of these form more conceptually. We call points with both coordinates being roots of unity "torsion" points. If our curve is given by $x^i y^j = 1$, then its points actually form a group under multiplication. Indeed clearly if $x^i y^j = 1$ and $(x')^i (y')^j = 1$ then so do we have $(xx')^i (yy')^j = 1$. Let's call this curve $G$ since it forms a group. Then if we consider $\omega \cdot G$, the curve formed by multiplying all points of $G$ by $\omega$, then certainly we still have infinitely many torsion points on it. We call $\omega \cdot G$ a "torsion coset". It turns out that any curve with infinitely many torsion points must be a union of torsion cosets.

Let's give an argument as to why is this the case. We will assume that $P(x,y)$ is given by $\mathbb{Q}$ coefficients and let's just assume it is irreducible for simplicity. Then if a torsion point $(\omega_1,\omega_2)$ satisfies $P(\omega_1,\omega_2) = 0$, then so do any conjugate $(\omega'_1,\omega'_2)$ of $(\omega_1,\omega_2)$. We can check that if $m$ is the order of $(\omega_1,\omega_2)$, i.e. $m$ is the smallest positive integer such that $\omega_1^m = \omega_2^m = 1$, then we have at least $\varphi(m)$ many conjugates. Now let $p$ be a prime number coprime to $m$. Then $(\omega_1^p, \omega_2^p)$ is a conjugate of $(\omega_1,\omega_2)$ and we may also choose $p$ to be $O(\log m)$ in size.

Now consider the curve $C^{(p)}$ which is given by the image of $C$ under the map $(x,y) \mapsto (x^p,y^p)$. Assume that $C^{(p)} \not = C$. Since $C$ is irreducible, so is $C^{(p)}$ and so it is given by some irreducible polynomial $Q(x,y) = 0$ with $\mathbb{Q}$-coefficients. Now we know that $(\omega_1^p, \omega_2^p)$ lie on $C^{(p)}$, but it also lies on the original curve $C$ because it is a conjugate of $(\omega_1,\omega_2)$. This gives us $\varphi(m)$ many torsion points in the intersection of $C$ and $C^{(p)}$. If $C$ has degree $d$, then the degree of $C^{(p)}$ is at most $p^2 d$, which is $d O(\log m)^2$. Hence there should only be at most $d^2 O(\log m)^2$ common intersection points between $C$ and $C^{(p)}$, but we showed that there is at least $\varphi(m)$ many of them, which is a contradiction for $m$ sufficiently large. Hence there can only be finitely many torsion points on $C$ unless $C = C^{(p)}$. It is then not so hard to show that if this is the case, then $C$ must be a union of torsion cosets.

The theorem I just proved is known as Ihara-Serre-Tate and was a question of Serge Lang that was proven by Ihara and Serre-Tate independently in the 1960s. In general, you can replace $C$ with a subvariety $V$ and the multiplicative group $\mathbb{G}_m^2$ with an abelian variety $A$. Then we can ask when is it possible that the torsion points of $A$ lying on $V$ is Zariski dense in $V$, and again it turns out this is only possible if $V$ is an union of torsion cosets of subgroups of $A$. This is known as the Manin-Mumford conjecture, which was proven by Raynaud.

Recently, there have been alot of interest in uniformity results. For example in our original setting of a curve $C$ inside the multiplicative group $\mathbb{G}_m^2$, could we obtain an uniform bound on the number of torsion points in $C$ if we fix the degree $d$? Of course we will have many curves of degree $d$ with infinitely many torsion points, but if we exclude those, then the question makes sense. In the argument given, we can see that our bound only depends on the degree $d$, but with an important assumption that our curve $C$ was defined over the rationals $\mathbb{Q}$. If it was defined over a number field $K$ of higher degree, then the number of conjugates will certainly decrease (we may still lower bound it by $[K:\mathbb{Q}]^{-1} \varphi(m)$) and so our bound gets worse as the degree $[K:\mathbb{Q}]$ increases.

It turns out that we can actually prove an uniform bound over all curves $C$ with $\mathbb{C}$ coefficients of degree $d$. The idea is to somehow use the higher dimensional version, that a subvariety $V$ of $\mathbb{G}_m^n$ has a dense set of torsion points if and only if it is a torsion coset. Consider the Chow variety $X_d$ that parametrizes all curve $C$ of $\mathbb{G}_m$ of degree $d$. It is itself a variety of dimension $\frac{d(d+3)}{2}$. There is then an universal family of curves $C_d \to X_d$, where the fiber above each point $x \in X_d$ is the curve $C_x$ that the point $x$ represents.

We now consider the the fiber powers $C_d^m$, where again it is a family over $X_d$ and each fiber is given by $C_x^m$. We then consider the map $C_d^m \to (\mathbb{C}^2)^m$, where we do the natural map $C_x^m \to (\mathbb{C}^2)^m$ by mapping each copy of the curve $C$ into $\mathbb{C}^2$. Now when $m = 1$, the universal family of curves $C_d$ is of very high dimension and the image $\mathbb{C}^2$ is just dimension $1$. But for each $m$ we increase by, the domain $C_d^m$ only increases in dimension by one, but the image $(\mathbb{C}^2)^m$ increases by $2$. Hence if $m$ is large enough, our map $C_d^m \to (\mathbb{C}^2)^m$ will necessarily be not surjective. Hence if we look at the image of all special points, its Zariski closure will be some special subvariety of $(\mathbb{C}^2)^m$.

Now we can use the fact that it has to be a torsion coset to argue that this Zariski closure will not be the same as the image of $C_d^m$ and so must be one dimension less as the image is necessarily irreducible. We then look at its preimage and this will be a variety of codimension one inside $C_d^m$. Hence on some Zariski open $U$ of the base, its fiber must given by a hypersurface $H_x \subseteq C_x^m$ with a well-defined degree by projection onto each of its coordinates. In particular, there is some $d$ such that if we fix $x_2,\ldots,x_m \in H_x$, then there are at most $d$ points of the form $(x,x_2,\ldots,x_m)$ in $H$. But if we had a generic sequence of curves $\{C_n\}$ with $C_n$ having at least $n$ torsion points, then by the symmetry of $C^m$ this gives a contradiction if $n > d$. Hence for some Zariski open we do have an uniform bound $d$ and we then proceed by repeating the argument on the complement of this Zariski open. We only go down finitely many times which gives us our uniform bound as desred, but an ineffective one.