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Dec 31, 2022 at 12:01 comment added Tom Goodwillie @abx Yes, I did not write what I meant. I've changed it to $P(L\oplus 1)$ now.
Dec 31, 2022 at 12:00 history edited Tom Goodwillie CC BY-SA 4.0
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Dec 31, 2022 at 9:58 comment added user283487 Thank you for the counterexample.
Dec 31, 2022 at 9:57 vote accept user283487
Dec 31, 2022 at 7:46 comment added abx A notation remark: for a standard algebraic geometer, what you call $P(L)$ is $\mathbb{P}(\mathscr{O}_E\oplus L)$ (or $\mathbb{P}(\mathscr{O}_E\oplus L^{-1})$, it doesn't matter).
Dec 31, 2022 at 3:13 history answered Tom Goodwillie CC BY-SA 4.0