Dimension of linear system of divisor of two points on curve of genus greater than 2

Question

This should not be hard, but I am stuck on it nonetheless, so I would much appreciate a solution.

Suppose $C$ is a projective non-singular curve of genus $g\geq 2$ and $P,Q$ are distinct points on $C$.

The notes of my algebraic geometry course now claim: "since $g\geq 2$, the dimension of the linear system $|P+Q|$ must be equal to 0 or 1."

Why is this?

Answer 1

Assertion:
If $D$ is an effective divisor of degree $d\geq 1$ on a smooth projective curve $C$ of genus $g\geq 1$, then $l(D)\leq d$.

The proof is by induction on $d$:

Initial step d=1
If $d=1$, then $D=1. P$ for some $P\in C$ and $L(1.P)=k$ (= base field) so that the assertion is true. [Here we have used that $g\geq 1$, i.e. that $C$ is not isomorphic to $\mathbb P^1$]

Inductive step $d-1\to d$
Choose some $P\in \operatorname {supp} D$ and consider the exact sequence of $k$-vector spaces $$0\to L(D-P)\to L(D)\to k$$ It immediately implies $l(D)\leq l(D-P)+1\leq (d-1)+1=d$

The assertion, which is much stronger than what you ask about, is thus proved.
Note carefully that this proof is from first principles and does definitely not involve grandiose concepts like Riemann-Roch, sheaf cohomology or Serre duality...

Answer 2

I think that you can reason like this: you know in general that (in Hartshorne notation) $\ell(P+Q)\leq \deg(P+Q)+1 = 3$ so that you just have to rule out the case $\ell(P+Q)=3$. Suppose that this is the case, then $\ell(P)\geq \ell(P+Q)-1 \geq 2$, so that by the previous inequality, $\ell(P)=2$. But since this divisor has degree $1$, the induced map to $\mathbb{P}^1$ is an isomorphism.