# Smooth projective plane curve only has rational divisors of even degree

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?

$$Q$$ 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.