What is the line bundle on $\mathbb{P}^1_\mathbb{C}$ whose transition function is $e^z$ Let $U_0,U_\infty$ be the two affine patches of $\mathbb{P}^1_\mathbb{C}$, neighborhoods of "0" and "$\infty$" respectively. Let $L$ be the line bundle on $\mathbb{P}^1_\mathbb{C}$ constructed by gluing the trivial bundles over $U_0$ and $U_\infty$ via the function $e^z$, which is a nowhere vanishing holomorphic function on $U_0\cap U_\infty$.
I've never taken complex geometry (my background is in algebraic geometry, and $e^z$ isn't an algebraic function), so my question is - does $L$ exist in the algebraic world? Which one is it? (for which $n\in\mathbb{Z}$ is $L\cong\mathcal{O}(n)$?). If it isn't any of them, what is the "right statement" to say that it doesn't exist? For example, is $e^z$ somehow not in $\mathcal{O}_{\mathbb{P}^1}(U_0\cap U_\infty)$?
 A: Topologically there are $\Bbb Z$-many line bundles, determined by the first Chern class, and one can calculate this by calculating what element of $\pi_1 \Bbb C^*$ the transition function represents. So it suffices to see how $e^z$ winds around $0$ as $z$ winds once counterclockwise around zero. But the very fact that $e^z$ extends to all of $\Bbb C$ means that this winding number is zero. So you've (topologically!) given the trivial line bundle. If I remember my Riemann surfaces, for $\Bbb P^1$, each complex line bundle has one and only one holomorphic structure up to isomorphism (because $H^i(\Bbb P^n,\mathcal O)=0$ for $i>0$, yes?) - so you've just written down the trivial line bundle. 
A: Since $e^z$ extends as a non-zero function over $U_0$, you can change basis on $U_0$ so that your gluing function is now just $1$.  This makes it clear that you have the trivial bundle.
(Incidentally, by GAGA, any holomorphic line bundle on $\mathbb P^1$ has to be algebraic, so what you wrote down had to be $\mathscr O(n)$ for some $n$.)
