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?



If a genus 2 curve has no $k$-rational points, can it have a $k$-rational divisor class of degree 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 $2x^3+3x-2 = y = 0$.


