I want to prove that a specific object is a line bundle.
Consider a normal variety $X$ and let $E$ be a line bundle on $X$. Denote by $s:X\to E$ the zero section, and consider
$$F=(E\setminus s(X))\times(\mathbb{P}^1\setminus \{0\}))/\mathbb{C}^*,$$
where $\mathbb{C}^*$ act as $t\cdot (x,y)=(tx,t^{-1}y)$, for $x\in E\setminus s(X)$, $y\in \mathbb{P}^1\setminus \{0\}$.
So in my head $F$ is obtained by replacing $0_p\in E_p$, for any $p\in X$, with another point (since $\mathbb{P}^1\setminus\{0\}\simeq \mathbb{C}$, and then I quotient by $\mathbb{C}^*$-action)… I suspect that on the left side, picking an open subset $U$ of $X$, I have $\pi^{-1}(U)\simeq U\times \mathbb{C}$ and I'm considering the action only on $\mathbb{C}$, becuase in principle $X$ may not admit the structure of $\mathbb{C}^*$-variety.
Claim: $F$ is a line bundle on $X$.
Proof: I don' know how to proceed: I should first define $\phi$, the canonical projection $F\to X$, but even here I'm quite stuck. Moreover I quite don't see how the quotient by the $\mathbb{C}^*$-action is used in order to make $F$ a line bundle.
Edit, second doubt: Later on in the article I linked in the comments it pick the "infinity section divisor" on $s_{\infty}\in\mathcal{O}_X(F)$: now, I know in a line bundle we can consider for istance the zero section $s:X\rightarrow E$, sending $p\mapsto s(p)=0_p\in E_p$, but since we work in a line bundle (i.e., something locally ismorphic to $U\times \mathbb{C}$, we don't have a notion of $\infty$, thus I was wondering what is that iinfinity section divisor they're talking about. I thought we justify it by considering $\mathbb{P}^1\setminus \{0\}$, but after the quotient by $\mathbb{C}^*$ is just $\mathbb{C}$, thus I'm a bit confused.
I know I should post some attempts, but I really don't know where to put my hands. I'd highly appreciate some help.
Best Answer
Let $U\subset X$ be an open set on which $E$ is trivial. Then $E|_U\cong U\times \Bbb C$, so $(E\setminus s(X))|_U\cong U\times \Bbb C^*$, and so $F|_U\cong (U\times \Bbb C^*\times \Bbb C)/\Bbb C^*$, where the action of $\Bbb C^*$ is $t\cdot(u,x,y)=(u,tx,t^{-1}y)$. This quotient is isomorphic to $U\times \Bbb C$: we can represent any point $(u,x,y)$ as $(u,1,\frac{y}{x})$ up to our action of $\Bbb C^*$. This is compatible with transition functions on $E$, so our collection of isomorphisms $F|_U\cong U\times \Bbb C$ patch together in to an honest line bundle.
In response to edit: My guess here is that the infinity section divisor is the map which sends $u$ to $(u,1,\infty)$. Note that the value $1$ is arbitrary: any value in $\Bbb C^*$ will do because the action of $\Bbb C^*$ fixes $\infty$ but can move that second value around to anything you want. Under the isomorphism detailed above ($\Bbb P^1\setminus\{0\} \cong \Bbb C$ by sending the coordinate $z\mapsto \frac1z$), this turns out to be the zero section divisor of $F$.