# Existence question on rational points on a curve

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.

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It may very well be that C contains no further point other than O. Thus the answer to the question could be "no" simply because C contains no more points. Are you assuming that there are $n>g$ points on $C$, or that you can enlarge the ground field? If you only know that there are $n>g$ points, then I suspect that what you want will be false in general; if you are allowed to increase the size of the field, then, for any $n>g$, you can certainly find $n$ points on C with the required property. – damiano Sep 21 '10 at 11:27
Firstly, I'd like to say that there are $n>g$ points, If the algebraic curve is chosen properly. For example, consider the Hermitian curves over $GF(q^2)$, the total number of GF(q)-rational points is $q^3+1$, while the genus of such curves is equal to $\frac{1}{2}(q^2-q)$. Secondly, I want to ask if it is allowed to increase the size of the field, then, how to fix the $n(>g)$ points on C with the required property. Thank you~ – athena Sep 21 '10 at 12:35
I do not know what Hermitian curves are. If the points are general, then the condition that you want is satisfied (see Angelo's answer to mathoverflow.net/questions/21816). Thus, if you are allowed to increase the size of the field, you will find points with the required property. If you are stuck with a bunch of points that you cannot choose, then these points may fail to be general and thus the condition that you want might fail. As a general expectation, the more points you have to choose from, the likelier it is that you will be able to extract a subset of general points from them. – damiano Sep 21 '10 at 13:11

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.