I'm trying to understand why the following claim is true (this is example $6.5.1$ from Hartshorne's Algebraic Geometry):
If $Y\subset \mathbb{P}^2$ is an irreducible curve of degree $d$, then $\text{Cl}(\mathbb{P}^2\setminus Y)\simeq \mathbb{Z}/d\mathbb{Z}$.
The author simply argues that this is a direct application of proposition $6.4$ (which basically says $\text{Cl}(\mathbb{P}^2)=\mathbb{Z}$) and proposition $6.5$, which says there is an exact sequence:
$$\mathbb{Z}\to \underbrace{\text{Cl}(\mathbb{P}^2)}_{\mathbb{Z}}\to\text{Cl}(\mathbb{P}^2\setminus Y)\to 0$$
where the first map is $1\mapsto [Y]$ and the second is $\sum_i n_i[Z_i]\mapsto \sum_i n_i[U\cap Z_i]$ with $U:=\mathbb{P}^2\setminus Y$.
Since the map $\mathbb{Z}\to\text{Cl}(\mathbb{P}^2\setminus Y)$ is surjective, we have that $\text{Cl}(\mathbb{P}^2\setminus Y)$ is cyclic. But how do we make $d$ show up?
Best Answer
The class group of $\mathbb P^2$ is generated by the class of a line, call it $L$. In this setting, $[Y] = dL$, so the (in fact injective) map on the left includes $\mathbb Z$ into $\operatorname{Cl}(\mathbb P^2)$ as $d\mathbb Z$.