Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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
1  
If you make this an answer, I'll accept it. –  Justin Campbell Jan 1 '12 at 1:23
add comment

1 Answer

up vote 3 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 $Q\sim P_0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.