# A question about a nondegenerate smooth curve in $\Bbb P^n$ of degree $n$

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 $$|D|$$?

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 $$L(R)=|D|$$.
• 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 $R(p)>1$? Jul 16, 2021 at 9:42
• Not entirely sure, but maybe consider a hyperplane with enough intersection multiplicity with the curve at each $p$? Jul 16, 2021 at 10:36