the simplest non-trivial line bundle over Riemann sphere We define Riemann sphere as $S=\mathbb{C}^2-\{0\}/\sim$. Given a point $p$ over $S$, I have seen somewhere there exists a line bundle $L_p$ associated to $p$, and $L_p$ has a non-zero holomorphic section with only one zero at $p$.
I think this construction is well-known to most people, I want to know how to construct the line bundle associated to $p$, by definition of the line bundle, how to define the space $E$ and the map $E\rightarrow S$ such that each fiber is a complex vector space of dimension $1$?
Thanks!

For simplicity let's take $P = [0: 1]$.
Cover $S$ with the two opens $U_0 = S \setminus [1: 0]$ and $U_\infty = S \setminus [0:1]$.
Now take the disjoint union of two copies of the trivial bundle:
$M = (U_0 \times \mathbb{C}) \sqcup(U_\infty \times \mathbb{C})$
We impose an equivalence relation on $M$ by the rule $([a:b], z)_0 \sim ([a:b], bz/a)_\infty$ when neither $a$ nor $b$ are zero, and set $L = M/\sim$.
Since $\sim$ respects the projection to $S$, there is a natural map $L \to S$. The fiber over a point is $\mathbb{C}$, and local triviality is clear (every point is in $U_0$ or $U_\infty$).
To give a section of $L$ is to give a map $f_0: U_0 \to \mathbb{C}$ and a map $f_\infty: U_\infty \to \mathbb{C}$ such that
$$
bf_0([a: b]) = af_\infty([a:b])
$$ whenever neither $a$ nor $b$ are zero.
There is such a section: $f_0([a: b]) = a/b$ and $f_\infty([a: b]) = 1$. This is locally holomorphic, so it's holomorphic, and it has a unique zero of order one at $P$ as desired.
