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Suppose that $X\subset \Bbb P^n$ is a nondegenerate smooth projective curve of degree $n$. Let $H$ be a hyperplane in $\Bbb P^n$, $D=\text{div}(H)$, and $Q\subset |D|$ be the linear subsystem of hyperplane divisors. Then we have the equalities $$ n=\dim (Q)\leq \dim |D|=\dim L(D)-1\leq n$$ Therefore $Q=|D|$ and $\dim L(D)=1+\deg (D)$. The latter equality implies that $X$ has genus zero, and the first equality implies that any divisor of degree $n$ is a hyperplane divisor.

This is a paragraph in p.217 of Miranda's book Algebraic Curves and Riemann Surfaces, and I can't understand the last sentence.

  1. How do we know that the genus of $X$ is zero? I can only see that by Riemann-Roch $g=\dim L(K-D)$.

  2. How do we know that any divisor of degree $n$ is a hyperplane divisor? Is any positive divisor of degree $n$ contained in $|D|$?

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1 Answer 1

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I’m not sure at all because my knowledge of curves and divisors is a little bit rusty. But here’s what I came up with.

For 1, if $g>0$, then $\ell(K-D) >0$ so that $D$ is special and thus by Clifford’s theorem $\ell(D) \leq 1+\frac{\deg(D)}{2}$. It follows that $\deg(D) \leq 0$, a contradiction.

For 2, if $R$ is a positive divisor of degree $n$ on $X$ with multiplicities $1$ everywhere (let’s say the field is algebraically closed to keep everything simple), its image in $\mathbb{P}^n$ is a formal positive linear combination of $n$ distinct points, so there is one hyperplane $H$ going through all of them. Then $L(R) \subset |D|$ (at least I think so) and thus $L(R)=|D|$.

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  • $\begingroup$ Thanks for 1! Also in 2, I can see that if $R$ is a positive divisor of degree $n$ of the form $p_1+\cdots+p_n$ with distinct $p_i$'s, then $R$ is a hyperplane divisor. But how do we know that $R$ is a hyperplane divisor if there is a point $p$ with $R(p)>1$? $\endgroup$
    – blancket
    Jul 16, 2021 at 9:42
  • $\begingroup$ Not entirely sure, but maybe consider a hyperplane with enough intersection multiplicity with the curve at each $p$? $\endgroup$
    – Aphelli
    Jul 16, 2021 at 10:36

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