Let $d \geq 2$ be a positive integer, and let $K_d$ denote the compositum of all fields of degree $d$ over $\mathbb{Q}$.

Let $Y$ be an algebraic curve defined over the rationals and has genus $g \geq 2$. It is a well-known theorem of Faltings that asserts that for any number field $K$, the set of $K$-points of $Y$ is finite. However, if $Y$ is given by a plane model of the form $F(x,y,z) = 0$ for some homogeneous polynomial $F$ with rational coefficients and degree $d$, then certainly $Y$ has infinitely many points over $K_d$. Indeed, one can choose $x,y$ to be any two rational numbers, and the corresponding equation in $z$ will surely have a solution over $K_d$.

Sometimes $Y$ has infinitely many points over $K_d$ for a much smaller $d$. For instance if $Y$ is hyperelliptic, then $Y$ always has infinitely many points over $K_2$.

The gonality of $Y$, given as a plane curve, is the smallest positive integer $k$ for which $Y$ admits a degree $k$, non-constant map to the projective line $\mathbb{P}^1$. Hyperelliptic curves have gonality equal to two, which explains the behaviour above. In general, if $m$ is the gonality of $Y$, then $Y(K_m)$ will be infinite.

In general, if $Y$ is given as a plane curve by a homogeneous polynomial $F(x,y,z) = 0$ of gonality $m$, do we expect $Y(K_s)$ to be finite or infinite for $s < m$?

Edit: thanks to Jason Starr for pointing out the notion of gonality.

  • $\begingroup$ Are you asking about the gonality of a plane curve? $\endgroup$ – Jason Starr Jan 3 '19 at 19:04
  • $\begingroup$ I was not aware of the notion of gonality before; I will edit the question to reflect this new notion. $\endgroup$ – Stanley Yao Xiao Jan 3 '19 at 19:38
  • $\begingroup$ It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0. $\endgroup$ – Xarles Jan 3 '19 at 20:46
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    $\begingroup$ This paper of Smith and Vogt might be of interest to you. math.harvard.edu/~gsmith/lowdeg_v7.pdf $\endgroup$ – Jackson Morrow Jan 4 '19 at 17:31
  • $\begingroup$ @JacksonMorrow thank you very much for the reference! It is indeed very helpful to me. $\endgroup$ – Stanley Yao Xiao Jan 4 '19 at 22:01

In the first place, there's also the possibility that there is a map of degree $m$ from $Y$ to an elliptic curve $E$ such that $E(\mathbb Q)$ is infinite, in which case $Y(K_m)$ will clearly be infinite. There's actually a bunch of literature on this subject, including for example:

  • MR1055774 Harris, Joe; Silverman, Joe; Bielliptic curves and symmetric products. Proc. Amer. Math. Soc. 112 (1991), no. 2, 347–356. (This deals with quadratic points, so in your notation, with $K_2$. The answer is that in order for $Y(K_2)$ to be infinite, the curve $Y$ must be hyperelliptic or admit a degree 2 map to an elliptic curve.)

  • MR11047 Abramovich, Dan; Harris, Joe; Abelian varieties and curves in $W_d(C)$. Compositio Math. 78 (1991), no. 2, 227–238. (This deals with the general case. The answer is a bit more complicated than just looking at gonality.)

In all cases, the results rely on Faltings' theorem describing subvarieties of $A(K)$ having having a Zariski dense set of $K$-rational points.

  • $\begingroup$ I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=\mathbb{Q}$. $\endgroup$ – Xarles Jan 3 '19 at 21:09
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    $\begingroup$ @Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields. $\endgroup$ – Joe Silverman Jan 4 '19 at 0:15

I will consider only the case $d=2$ (and for simplicity $K=\mathbb{Q}$ if you wish).

As Silverman says in his answer and his comment, there are two distinct but related problems: one is to consider the points of the curve $Y$ (that I assume smooth and projective of genus $\ge 2$) defined on some quadratic extension of $K$, and the other the points defined over $K_2$, the compositum of all quadratic extensions of $K$. The first case is already answered in the paper by Harris and Silverman cited above (the answer being the hyperelliptic curves and the bielliptic curves with an elliptic quotient with infinitely many points over $K$, so their gonality is bounded by $4$), so I will say same words on the second case.

First, the field $K_2$ can be written as the union of all fields $L/K$ which are abelian extensions with Galois group $(\mathbb{Z}/2\mathbb{Z})^r$ for some $r\ge 1$. So you are considering now which curves have infinitely many points defined over this fields. Since there are much more fields you expect to have other examples than the ones above.

For example, all curves that have a map $\pi$ to $\mathbb{P}^1_K$ which is Galois with Galois group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^r$ for some $r\ge 1$ (with this I mean that the corresponding extension between the function fields is Galois, etc, but not that the map is unramified). This curves can have gonality as big as you wish, so you get much more examples.

To be more explicit, you can consider $f_1(x),\dots, f_r(x)\in K[x]$ pairwise coprime polynomials of degree $\ge 3$, and the curve $Y$ to be the projectivization of the affine curve in the affine space $\mathbb{A}_K^{r+1}$ given by the equations $y_i^2=f_i(x)$ for $i=1,\dots, r$. It is clear that this curve has infinitely many points in $K_2$ just taking the points with coordinate $x\in K$. On the other hand this curve has (at least generically, but I think always under the given conditions) gonality $2^r$, with gonal map given by sending a point $(x,y_1,\dots,y_r)$ to $x$. More general cases with not necessarily pairwise coprime polynomials, or with degrees $\le 2$, or with terms of degree $1$ in the $y_i$ can be taken, but the gonality is not so easy to control.

Of course you can consider also the curves $Y$ with a map $\pi:Y\to E$ to an elliptic curve $E$ defined over $K$, with $\sharp E(K)=\infty$ and which is Galois with Galois group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^r$ for some $r\ge 1$. I don't know if there other other examples of curves that these two types.

Finally, you talk in the message of plane curves, but I assume you are considering plane singular models of the curves. If you really only want to consider curves with non-singular plane projective models, then one should study if these type of curves but with high degree can have a map $\pi$ as described above.


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