Reference: Hartshorne,Chapter 2, Proposition 6.17
$X= \mathbb P ^n _k$ for some field k. Then the generator of the $Cl
(X)$ (which is the group of weil divisors modulo principal divisor) is generated by a hyperplane which corresponds to the invertible sheaf $\mathcal O(1)$.
I don't understand how does a hyperplane corresponds to the invertible sheaf $\mathcal O(1)$
Can someone please help.
Best Answer
It is not true that a hyperplane corresponds to the invertible sheaf $\mathcal O(1)$. What is true, is that the class of a hyperplane corresponds to the invertible sheaf $\mathcal O(1)$.
Here's (a sketch of) the correspondence. The sheaf $\mathcal O(1)$ is generated by global sections, and the sections correspond to hyperplanes in $X$. Recall that the sheaf is generated by $x_0,\cdots,x_n$. Now any two hyperplanes represent the same non-trivial class in $\mathrm{Cl}(X)$, because if $H_1$ is the zero set of $a_0x_0+\cdots+a_nx_n$, and $H_2$ is the zero set of $b_0x_0+\cdots+b_nx_n$, then $H_2-H_1=((a_0x_0+\cdots+a_nx_n)/(b_0x_0+\cdots+b_nx_n))$, so they are equal in $\mathrm{Cl}(X)$.
So the class of a hyperplane corresponds to all global sections of $\mathcal O(1)$, and hence we can identify these two sets (for our purposes). And since $\mathcal O(1)$ is generated by global sections, they determine $\mathcal (X)$.
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Here's an example of what is ment by "generated by a hyperplane section". Let $D$ be the divisor defined by the ideal sheaf $(x^3) \subseteq \mathcal O_X$. This corresponds to the divisor $(x^3)=3(x)$, i.e. 3 times a hyperplane. Similarly, every divisor on $\mathbb P^n$ is determined by its degree.