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Jul 18, 2020 at 18:53 history edited Georges Elencwajg CC BY-SA 4.0
added 74 characters in body
Jul 18, 2020 at 18:34 history edited Georges Elencwajg CC BY-SA 4.0
Distinguished the algebraic from the holomorphic case, which I had mixed up in the previous version.
Jul 18, 2020 at 18:28 comment added Georges Elencwajg Dear @abx, you are completely right, of course: thank you for your attention. I had shamefully mixed up the holomorphic and algebraic case, which resulted in the absurd answer I posted. I have corrected this mess, and I hope that everything is OK now.
Jul 18, 2020 at 17:14 comment added abx All these open subsets $U$ have $H^2(U,\mathbb{Z})=0$ and $H^1(U,\mathscr{O}_U)=0$ because they are Stein as you observe, therefore $\operatorname{Pic}(U)=0 $.
Jul 18, 2020 at 15:24 history answered Georges Elencwajg CC BY-SA 4.0