Suppose $C$ is a curve of genus $g$. Let $K$ be a canonical divisor on $C$, let $n\geq 3$ be an integer, and let $C\to\mathbb{P}^N$ be the map determined by the linear system $|nK|$, where $N$ can be determined by a Riemann Roch calculation. In this question, the selected answer proves why the degree of the image of $C$ is $n\deg K$. I have a couple of questions about the answer.
- By definition, the degree of $C$ is the intersection number of $C$ with a hyperplane $H$. The author writes that this is equivalent to the degree (as a divisor on $C$) of the restriction of the hyperplane to $C$. Why is this so?
- In terms of invertible sheaves, a hyperplane $H$ corresponds to $\mathcal{O}(1)$, and "it's restriction to $C$ is by construction $nK$". Why is the part in quotations here true?
EDIT: For 2, if we call our map $f$, is this just saying that we have $\mathcal{O}_C(nK)=f^*\mathcal{O}_{\mathbb{P}^N}(1)$ and calling the pullback a "restriction" even if $f$ isn't a literal inclusion?
Let $C$ be a smooth curve over an algebraically closd field, and suppose we have some closed immersion $C\hookrightarrow \Bbb P^n$. If $D$ is a hypersurface in $\Bbb P^n$, then the intersection multiplicity of $D$ and $C$ at some point $c\in C$ is equal to the valuation of a local equation for $D$ in the local ring $\mathcal{O}_{C,c}$, because $\mathcal{O}_{C,c}$ is a DVR. On the other hand, this quantity is exactly the multiplicity of $D\cap C$ as a divisor on $C$ at the point $c$. Summing over all closed points, we have the result.
Yes, you're exactly correct. If we have a map $f$ from $X$ to $\Bbb P^n$ given by some sections of some base-point free line bundle $\mathcal{L}$, then $f^*\mathcal{O}(1)\cong\mathcal{L}$ by construction.