I am puzzled about the following question:


Let C be a smooth, projective, absolutely irreducible curve defined over GF(q) and let g denote the genus of C. O is a rational point on C, and the divisor D is defined as D = (2g-1)O.


Question:
Whether there exist rational points $P\_1,P\_2,\cdots,P\_n~(n>g)$ on C such that for any g rational points $P\_{i\_1},P\_{i\_2},\cdots,P\_{i\_g}$ of them, $l(D-\sum^g\_{j=1}P\_{i\_j})=0$ (i.e. the dimension of $L(D-\sum^g\_{j=1}P\_{i\_j})$).


I know it is easy to find g such points. But the existence of further points really confused me.


Thank you for your help!



Existence question on rational points on a curve
Damiano's comments are spot-on and maybe you should clarify your hypotheses. Here is an inductive argument which shows that, if you have enough rational points on the curve, you can grow your set. 


First, by using the linear system $|(2g-1)O|$ you can embed your curve in $\mathbb{P}^{g-1}$ as a curve of degree $2g-1$ and you want $n$ points such that no $g$ of them are in a hyperplane. Suppose you have $n$ such points and that your curve has more than $n + g{n \choose g-1}$ rational points. I claim you can choose a further point $P\_{n+1}$ and have $n+1$ points, no $g$ in a hyperplane. Indeed, for each subset of $g-1$ points of your $n$ points, there are at most $g$ other points on the intersection of the curve and the hyperplane spanned by them. If $P\_{n+1}$ is in none of these hyperplanes, you are done. My hypothesis ensures that there is at least one such point. 


