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 fieldsby 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?

evengenus: 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$2more comments