# Rational points on towers of curves

Let $\ldots \to X_n \to X_{n-1} \to \ldots \to X_0$ be etale maps between smooth projective curves of genera $g(X_n)>1$, all defined over a fixed number field $K$. By Faltings' Theorem, we know that the sets $X_n(K)$ are finite.

Is the set $\bigcup_{n=0}^\infty X_n(K)$ always finite?

• Do you want the inverse limit of the sets of rational points to be infinite? That is stronger than what you wrote. – Jason Starr Dec 27 '15 at 20:21

No. Start with a tower of Galois covers for which $X_0(K) \ne \emptyset$. Then, inductively, replace each $X_n$ by a twist such that the point from the previous layer lifts to a rational point on the twist. In this way, all $X_n(K)$ are non-empty and their union is infinite.

• Perhaps you can achieve the same thing by considering the sequence $f_i:Y_{i+1}\to Y_i$ where each $Y_i$ is a fixed Abelian variety $(A,0)$ and each $f_i$ is the multiplication by $2$ morphism. Now let $X_0$ be a smooth, projective curve in $A$ containing $0$ and that is a complete intersection curve. Then the inverse image $X_i$ of $X_0$ in each $Y_i$ is a smooth, projective curve that is a complete intersection of ample divisors, thus it is geometrically connected. Finally, each $X_i$ contains the origin of $A$. – Jason Starr Dec 27 '15 at 20:20

Take a genus $>1$ curve $X_0$ with a rational point, equipped with a finite map to an elliptic curve $X_0\rightarrow E$, which has a point of order $n$. Let $X_i := X_{i-1}\times_{E} E$ be the base change by the multiplication by $n$ map $[n] : E\rightarrow E$. Since $[n]$ is etale (pick $n$ coprime to the characteristic of $k$), each $X_i$ is a smooth projective curve of genus $>1$, and by the universal property of fiber products, $|X_i(K)| = |E[n](K)|\cdot|X_{i-1}(K)|$.

EDIT: As Jason Starr mentioned, I should probably also arrange that each $X_i$ is connected (even though the OP didn't explicitly specify this!). This will be true if $X_0$ is a $G$-Galois branched cover of $E$ where $G$ is a finite perfect group.

Specifically, to see why this suffices, let $G$ be a finite 2-generated perfect group, ie a finite group generated by 2 elements such that $G = [G,G]$ (for example, any finite nonabelian simple group). Let $E^\circ$ denote a punctured elliptic curve (puncturing at the identity), then its etale fundamental group is isomorphic to $\widehat{F_2}$, the profinite free group on 2 generators. Let $p : \widehat{F_2}\rightarrow G$ be a surjection, which corresponds to a $G$-galois cover over $E^\circ$, which upon normalizing one obtains a ramified $G$-galois cover over $E$ which is a connected smooth projective curve of genus $>1$ (genus $>1$ because the map is ramified, ramified because $G$ is nonabelian). Take this curve to be $X_0$. I claim that the sequence $X_i := X_{i-1}\times_E E$, where the base change is via the map $$[n^i] : E\rightarrow E$$ is a sequence which gives a counterexample to your questions. By the above discussion each $X_i\rightarrow X_{i-1}$ is a finite etale map of smooth projective curves of genus $>1$. It remains to show that each $X_i$ is connected. This is true because $G$ is generated by commutators (ie, is perfect).

In more detail, the map $[n^i] : E\rightarrow E$ restricted to the preimage of $E^\circ$ gives a finite etale cover over $E^\circ$, corresponding to a subgroup $P_i\le\widehat{F_2}$ of index $n^{2i}$. Specifically $$P_i = \text{Ker}\left(\widehat{F_2}\rightarrow(\mathbb{Z}/n^i\mathbb{Z})^2\right)$$ and is by Galois theory the fundamental group of $[n^i]^{-1}(E^\circ)$. Hence, $P_i$ contains the commutator subgroup $[\widehat{F_2},\widehat{F_2}]\le\widehat{F_2} = \pi_1(E^\circ)$. Then, by Galois theory, the restriction of the pulled back $G$-Galois cover $X_i\rightarrow E$ to $E^\circ$ corresponds to the composition $$P_i\hookrightarrow\widehat{F_2}\stackrel{p}{\rightarrow} G$$ which is surjective since $G$ is perfect, and $P_i$ contains $[\widehat{F_2},\widehat{F_2}]$. By the Galois correspondence, this implies that $X_i$ is connected.

• You need to arrange that each $X_i$ is connected. For instance, if $X_0\to E$ factors through the multiplication by $n$ morphism $E\to E$, then $X_i$ factors through $E \times_E E$, which has the diagonal as one connected component. – Jason Starr Dec 27 '15 at 20:24
• @JasonStarr Thanks for the good point. I've fixed it now! – Will Chen Dec 27 '15 at 21:15
• Thanks! I don't understand why $|X_i(K)| = n\cdot |X_{i-1}(K)|$. For example, if $X_0 = E$ (forgetting $g(X_n)>1$ for now) is an elliptic curve with $E(K)$ finite, we get $X_i = E$ and so $|E(K)| = n\cdot |E(K)|$, contradiction. – Piotr Achinger Dec 27 '15 at 22:53
• (Though, of course, your construction shows that $X_i(K)$ is nonempty.) – Piotr Achinger Dec 27 '15 at 22:57
• @PiotrAchinger I made a mistake (fixed now). I should have assumed that $E$ has a point of order $n$, in which case $|X_i(K)| = |E[n](K)|\cdot|X_{i-1}(K)|$. This follows from the universal property of fiber products. A $K$-point $P_i\in X_i(K)$ is precisely the data of a $K$-point $P_{i-1}\in X_{i-1}(K)$, and a point $Q\in E(K)$ such that $[n]Q$ is the image of $P_{i-1}$. For every point $P_{i-1}\in X_{i-1}(K)$, there are precisely $|E[n](K)|$ such points $Q$ , corresponding to $|E[n](K)|$ such points of $X_i(K)$. – Will Chen Dec 28 '15 at 5:18