Very ample divisors and the Riemann-Roch theorem

Question

What is the easiest way to prove that a divisor $D$ is very ample if and only if $l(D - P - Q) = l(D) - 2$ for all points $P, Q \in C$. It seems like it might be a consequence of the Riemann-Roch theorem, but I am not sure how to deduce this from the said theorem.

Answer 1

Hint: Show first that it's globally generated. This should be simple since

$$\ell(D-P)\leqslant \ell(D-P-Q)+1=\ell(D)-1$$

so $\ell(D)-\ell(D-P)\ne 0$. Show that this implies that $\mathcal{L}$ is globally generated (Hint: think about what this means in terms of 'having a section which doesn't vanish there').

Then, you want to show that the map $\varphi_D:C\to \mathbb{P}^m$ is a closed embedding. You want to show that it separates tangent vectors and points. Think, onc again, about what this means (similar to the parenthetical hint of the last paragraph) and show that your condition implies this.

Then use the fact in, say Hartshorne, that separating tangent vectors and points is enough to be a closed embedding (morally this is because your morphism is then a proper monomorphism, which is a closed embedding).

Remark: The last step of the above (if you look at Hartshorne's proof) uses the fact that $k=\bar{k}$. This isn't necessary since you can just tensor up to $\bar{k}$, prove it's a closed embedding (since separating tangent vectors and points isn't effected by field extension) and then conclude your original morphism was a closed embedding.