Let $C$ be a smooth projective geometrically connected curve of genus 2 defined over a number field $k$. Here are some definitions:

The index $I$ of a curve $C$ is the greatest common divisor of all effective divisors $D \in \mathrm{Div}(C)$. Equivalently, it is the greatest common divisor of the degrees $[L:k]$, where $[L:k]$ ranges over algebraic extensions such that $C(L) \neq \emptyset$. The period $P$ is the smallest positive degree of rational divisor classes, i.e., those given by divisors linearly equivalent to their Galois conjugates.

Let us list 2 lemmas:

For a hyperelliptic curve (which includes all genus 2 curves), we have $I=P$ (see Lemma 5 of Brauer groups of local elliptic and hyperelliptic curves and central division algebras over their function fields by Yanchevski and Margolin).

Let $C$ be a smooth projective curve of genus $g \geq 2$. Then $C$ has a closed point of degree at most $2g-2$.

So here is the issue. If $C(k)= \emptyset$, then all closed points are of degree 2. In other words, for any closed point $P$, we can find a degree 2 extension $L/k$ such that $P \in C(L)$. By the definition of the index of a curve, we must have $I = 2$ since for any finite extension $K/k$ such that $C(K) \neq \emptyset$, $K$ must contain a degree 2 extension and thus $[K:k]$ must be even $\implies I = 2$. Then $P = I = 2$ and so the smallest positive degree of $k$-rational divisor classes is 2.

However, in the paper The Hasse Principle and the Brauer-Manin obstruction for curves by Flynn, Corollary 4 states that for $k = \mathbb{Q}$, there are genus 2 curves with no rational points but has a degree 1 rational divisor class. What is going on?

  • $\begingroup$ Careful: a genus $2$ curve need not be hyperelliptic over the ground field, just like a genus $1$ curve need not be elliptic or a genus $0$ curve need not be rational. $\endgroup$ Feb 7, 2023 at 1:47
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    $\begingroup$ It's not true that every closed point $p$ of $C$ has residue field $k(p)$ having degree 2 over $k$. $\endgroup$
    – Eoin
    Feb 7, 2023 at 2:01
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    $\begingroup$ @Eoin Right I read the second lemma wrongly. It simply says that if $C(k) = \emptyset$, then there exists a closed point of degree 2. But that does not imply the nonexistence of closed points of odd degree, and hence the index (gcd) can still be 1. Thanks for pointing that out. $\endgroup$
    – oleout
    Feb 7, 2023 at 2:06
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    $\begingroup$ @R.vanDobbendeBruyn Doesn't the canonical linear series show it's hyperelliptic? $\endgroup$ Feb 7, 2023 at 3:53
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    $\begingroup$ Likewise for hyperelliptic curves of every even genus: the canonical map goes 2:1 to a rational normal curve of odd degree, so we have an odd-degree divisors which together with the canonical divisor (degree -2) gets us to degree 1. (I think I learned this from Bjorn Poonen.) $\endgroup$ Feb 7, 2023 at 16:21

1 Answer 1


Rummaging a bit through the LMFDB turns up the curve https://www.lmfdb.org/Genus2Curve/Q/129600/b/129600/1 with equation $y^2 = -(2x^3+3x-2)(2x^3+4x^2+x-2)$ with no rational points (indeed trivial Mordell-Weil group) but a degree-3 divisor $2x^3+3x-2 = y = 0$.


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