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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!

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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
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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.

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Thank you very much~ –  athena Sep 25 '10 at 3:04
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