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I'm having trouble understanding a remark in Hartshorne: Let $X$ be the nonsingular projective cubic defined by $y^2z = x^3 - xz^2$ and put $P_0 = (0,1,0)$. The claim is that $\mathscr{L}(P_0)$ is not very ample because it is not generated by its global sections, and that this in turn is because $X$ is not rational, so $P_0$ cannot be equivalent to another point of $X$.

But why would $\mathscr{L}(P_0)$ being generated by its global sections imply $P_0$ is equivalent to another point? This is not clear to me. A less precise question: to what extent does this remark apply to $\mathscr{L}(P)$ for other points $P \in X$, or even other nonsingular projective curves of positive genus?

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Take a global section with a pole at $P_0$, say $f$. Since $X$ is nonsingular projective the divisor $(f)$ must have exactly one zero somewhere since the degree of $(f)$ is zero, say $Q$. By definition of linear equivalence $Q-P_0 = (f) \sim 0$, so $Q\sim P_0$. – Matt Dec 31 '11 at 22:55
Thanks Matt! It did not occur to me that there must be a global section with a pole at $P_0$. Even after you said this I had to think about it... – Justin Campbell Jan 1 '12 at 1:20
If you make this an answer, I'll accept it. – Justin Campbell Jan 1 '12 at 1:23
up vote 5 down vote accepted

Let $f$ be a global section of the sheaf with a pole at $P_0$. We now consider the divisor $(f)$. Since $X$ is a nonsingular projective curve the divisor has degree $0$ which means that there is exactly one zero somewhere else. Call this point $Q$. Now we see that $Q-P_0 = (f)\sim 0$, so $Q\sim P_0$.

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