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Let $C$ be a curve over $\mathbb{Q}_p$ and let $\mathcal{C}$ be a regular model of $C$ over $\mathbb{Z}_p$, with $\mathcal{J}$ the Neron model of the Jacobian of $C$. Raynaud's theorem asserts that $\mathcal{J}^0(\mathbb{F}_p) \cong \mathrm{Pic}^0(\mathcal{C}_{\mathbb{F}_p})$, that is, the $\mathbb{F}_p$ points of the connected component is the Picard group of the special fibre of $\mathcal{C}$.

How would one actually compute with this group? For example, given a reasonably nice Weil divisor (e.g. supported on reduced components, away from crossing points) on the special fibre, how can one test if it is the identity? When the special fibre is reduced, but not necessarily irreducible, this can be done by considering the rational functions on the components which agree on crossing points. When it is no longer reduced, it is less clear what can be expected. My intuition from cuspidal cubics is that crossing a non-reduced component should impose an order of vanishing condition on the function at the crossing point. This seems to hold in examples when the divisor is supported on a single component, but not when moving points between components.

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    $\begingroup$ Let $I$ be any ideal sheaf on $\mathcal{C}_{\mathbb{F}_p}$ such that $I^2$ is the zero ideal sheaf. Denote by $\mathcal{C}'_{\mathbb{F}_p}$ the closed subscheme of $\mathcal{C}_{\mathbb{F}_p}$ with ideal sheaf $I$. There is a long exact sequence: $H^1(\mathcal{C}_{\mathbb{F}_p},I)\to \text{Pic}(\mathcal{C}_{\mathbb{F}_p})\to \text{Pic}(\mathcal{C}'_{\mathbb{F}_p})\to 0.$ $\endgroup$ Commented Dec 16, 2024 at 15:32
  • $\begingroup$ Thanks for this suggestion! I had been trying to compute this directly via linear equivalences etc, but the cohomological viewpoint is much simpler to use in this case. $\endgroup$ Commented Dec 17, 2024 at 11:22
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    $\begingroup$ Depending on what "compute" means, this thesis might help in the non-reduced case: oops.uni-oldenburg.de/5467/1/junasy22.pdf It also contains a lot of theoretical results that could be useful. $\endgroup$ Commented Jan 3 at 14:28

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