Let $C \subset \mathbb{P}^2$ be a smooth projective plane curve of degree 2 defined over $\mathbb{Q}$ with no rational points. Also the genus of $C$ is $0$.
I want to show that every rational divisor D can only have even degree.
Here a rational divisor is a divisor on C that is stable under the action of the absolute galois group $G_\mathbb{Q}$ of $\mathbb{Q}$.
I already managed to show that there are divisors of even degree now the last thing I want to show is that any rational divisor of degree one cannot exist:
Let $Q \in C$ be a point such that $\sigma(Q)=Q$ for all $\sigma \in G_\mathbb{Q}$, by riemann-roch I obtain that $\mathcal{L}(Q)$ has dimension $2$. Hence there exists a nonconstant f such that div$(f)+Q$ has degree one. Now I want to say that there exists a rational point P such that div$(f)+Q=P$, so we obtain a contradiction. How do I manage to do that? and where do I use the assumption of $Q$ being stable under galois action?
Best Answer
$Q$ is invariant under the Galois action is irrelevant. Since $C$ is a quadric, a line intersects it in a degree 2 divisor, so it has plenty of degree 2 divisors. If there is a divisor of odd degree, then by adding or subtracting degree two divisors, you may assume that it has a divisor of degree one. By Riemann-Roch (as you said), this divisor is linearly equivalent to an effective divisor. But an effective divisor of degree one is just a rational point.