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I'm having troubles solving exercise K on page 167 of the book "Algebraic curves and Riemann surfaces" of Miranda.

The question is the following one : Let Q be a base-point-free linear system, let p_1, ...,p_m be some points on the Riemann Surface X. Show that there is a divisor D in Q without any p_i in its support.

I tried the following :

Since Q is a base-point-free linear system, it corresponds to some linear system |phi| associated to some holomorphic map phi:X \rightarrow P^n: x \mapsto [f_0(x): \dots : f_n(x)]. Here the f_i's denote some meromorphic functions. We have a very clear discription of such a system, namely any divisor in it can be written as

div(g)+D,

where D is defined as -min_i{div(f_i)} and g is a linear combination of the f_i's.

Since we're dealing with a base-point-free system, for any i, there is a divisor E_i in Q such that E_i(p_i)=0.

But now I'm stuck, I tried some combinations of the E_i's to get the desired divisor, but I don't find it.

Remarks : I don't know much about sheaves and schemes, so I can't use it. And I'm very sorry for the notation above, I'm new to this site and I haven't figured out yet how to get nice notations.

Thank you in advance.

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This may be cheating but here goes, since $D$ is base point free $\dim L(D) < \dim L(D+p)$ for all $p$. You want to show that $L(D) \neq \cup_{i} L(D+p_i)$ but this should be obvious since a vector space cannot be the union of a finite number of spaces of smaller dimension.

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  • $\begingroup$ I'm not really convinced that I want to show that, how do you get the union in there? $\endgroup$ – Mathematician 42 Jun 8 '14 at 13:46
  • $\begingroup$ Wnat I mean to say, I am not sure I am using the right notation, but if you consider all divisors which contain $p$ (that is you add $p$ as a base point) then you get a system of lower dimension. The question is for a divisor which does not lie in a finite number of such subsystems. Whats the notation for adding $p$ as base point ? $D+p$ ? $\endgroup$ – Rene Schipperus Jun 8 '14 at 13:58
  • $\begingroup$ I think I get what you mean, it seems like a useful idea. $\endgroup$ – Mathematician 42 Jun 8 '14 at 14:08
  • $\begingroup$ I think I found a proof, thanks, I'm going to check it again tomorrow, I feel my head spinning today, I've been stuck on this for far too long. $\endgroup$ – Mathematician 42 Jun 8 '14 at 14:27
  • $\begingroup$ If you have a different proof, please post it. $\endgroup$ – Rene Schipperus Jun 8 '14 at 14:29

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