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Timeline for $p$-divisibility of Picard groups

Current License: CC BY-SA 4.0

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Aug 22, 2023 at 8:36 vote accept Boaz Moerman
Aug 19, 2023 at 4:20 answer added naf timeline score: 3
Aug 18, 2023 at 5:23 comment added naf If $\widetilde{C}$ is smooth, it suffices to know that $\widetilde{C} \setminus C$ supports a divisor of degree prime to $p$, which always holds by your assumption on $k$. This implies that $\mathrm{Jac}(\widetilde{C})$ is $p$-divisible (see the answer by R. van Dobben de Bruyn below). It follows that the Picard group is also $p$-divisible, since if $L$ is a line bundle whose class in the Jacobian is divisible by $p$, then the obstruction to finding a $p$-th root in the Picard group is an element of $\mathrm{Br}(k)[p]$, but $\mathrm{Br}(k)[p]=0$ (by the same cohomological argument).
Aug 18, 2023 at 0:14 answer added R. van Dobben de Bruyn timeline score: 1
Aug 17, 2023 at 19:55 history edited Boaz Moerman CC BY-SA 4.0
Added brackets in exponent for clarity (otherwise l^p might look like p if zoomed out)
Aug 17, 2023 at 17:39 comment added Piotr Achinger I thought you meant $k$ is algebraically closed in the last sentence, sorry. In fact I misunderstood your question and thought $k$ has characteristic $p$, and anyway your assumption on $k$ disallows the removal of a point of degree $p$ from $\tilde C$.
Aug 17, 2023 at 15:33 history edited Boaz Moerman CC BY-SA 4.0
Added examples of such fields $k$.
Aug 17, 2023 at 15:04 comment added Boaz Moerman Could you explain why $A(k)$ is $p$-divisible? I know that it is true for separably closed fields, but it is not clear to me why it should hold for the fields I described.
Aug 17, 2023 at 14:18 history edited Boaz Moerman CC BY-SA 4.0
added 22 characters in body
Aug 17, 2023 at 13:34 comment added Piotr Achinger Yes, it’s true that $A(k)$ is always divisible. What worries me is the situation when you’re removing a point of degree $p$ from the compact curve.
Aug 17, 2023 at 10:40 history asked Boaz Moerman CC BY-SA 4.0