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.
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.
By Hartshorne, Chapter 2 (same below) Proposition 6.15, $\mathcal O(1)$ can be expressed as a subsheaf of $\mathcal K$ by natural maps $\mathcal O(1) \to \mathcal O(1) \otimes \mathcal K \cong \mathcal K$. On opens $D_+(x_i)$, $\Gamma(D_+(x_i), \mathcal O(1)) = k[\frac{x_0}{x_i}, \cdots, \hat{\frac{x_i}{x_i}}, \cdots \frac{x_n}{x_i}] \cdot x_i$ and $\Gamma(D_+(x_i), \mathcal K) = k[\frac{x_j}{x_k}]_{j\neq k}$. Then $\Gamma(D_+(x_i), \mathcal O(1) \otimes \mathcal K) = k[\frac{x_j}{x_k}]_{j\neq k} \cdot x_i$.
Fix $l$, one can get the isomorphism $O(1) \otimes \mathcal K \cong \mathcal K$ by multiplying $\frac{1}{x_l}$. Then as a subsheaf of $\mathcal K$, $\Gamma(D_+(x_i), \mathcal O(1)) = k[\frac{x_0}{x_i}, \cdots, \hat{\frac{x_i}{x_i}}, \cdots \frac{x_n}{x_i}] \cdot \frac{x_i}{x_l}$, corresponding to Cartier divisor $D = \left\{ \left(D_+(x_i), \frac{x_l}{x_i} \right) \right\}$. By the proof of Proposition 6.11, this corresponds to a Weil divisor, which is exactly prime divisor $H=\{x_l = 0\}$ for the reasons below:
Since $D_+(x_l) \cap H = \emptyset$, we discuss only $j\neq l$ here. $H$ has generic point $\eta=(x_l)$ and $\mathcal O_{\eta} = k[x_0, \cdots , x_n]_{(\eta)}$. As $x_l$ is the generator of $\eta$ and $x_j\notin \eta$, $v_H(\frac{x_l}{x_i}) = 1$ on each $D_+(x_i)$.
Moreover, for any other prime divisor $Y$, since $k[x_0,\cdots, x_n]$ is a UFD, the generic point of $Y$ is of the form $\eta^\prime = (f)$ and $f\neq x_l$. Hence for each $x_i \neq f$(i.e. $D_+(x_i) \cap Y \neq \emptyset$), $x_l, x_i \notin \eta^\prime$, i.e. $v_Y(\frac{x_l}{x_i}) = 0$.
As Fredrik Meyer said, all hyperplanes are equal in $\mathrm{Cl} X$, choices of the isomorphism $O(1) \otimes \mathcal K \cong \mathcal K$ do only affect the hyperplane $H$ it corresponds to.
