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Let $K$ be a field of characteristic $p>0$ and $C$ a regular projective geometrically integral curve over $K$.

If $C$ is smooth, then the connected component ${\rm Pic}^0_C$ of the Picard scheme of $C$ is isomorphic to the Jacobian $J_C$, so in particular the $n$-torsion of the class group of $C$, ${\rm Cl}(C)[n]={\rm Pic}^0_C(K)[n]=J_C(K)[n]$, is finite for every $n$, including $n=p$.

Question: Is ${\rm Cl}(C)[p]={\rm Pic}^0_C(K)[p]$ finite also when $C$ is regular but not smooth?

Of course, $K$ is then necessarily imperfect. From what I understand (e.g. from Chapters 8 and 9 of the book by Bosch-Lütkebohmert-Raynaud), in this case ${\rm Pic}^0_C$ is still a group scheme, possibly with a unipotent part, but not containing a copy of $\mathbb{G}_a$ (in particular non-split). However, that in itself does not seem to be sufficient to conclude finiteness of the $p$-torsion.

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Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, putting $L:=K(t^{1/3})$, $C_L$ is isomorphic to the usual cuspidal cubic (explicitly, the equation becomes $y^2 z=(x - t^{1/3} z)^3$).

Thus, putting $J:=\mathrm{Pic}^0_{C/K}$, it follows that $J_L=\mathrm{Pic}^0_{C_L/L}$ is isomorphic to $\mathbb{G}_{a,L}$. In other words, $J$ is a form of $\mathbb{G}_{a,K}$, in particular smooth, one-dimensional and killed by $p$.

The natural map $\mathrm{Pic}^0(C)\to J(K)$ is injective, so $\mathrm{Pic}^0(C)=\mathrm{Pic}^0(C)[3]$. It is even bijective since $C$ has a rational point, namely $(0:1:0)$, so we get a counterexample if $J(K)$ is infinite, which is certainly the case if $K$ is large (aka fertile or ample). So, explicitly, we can take for instance $K=\mathbb{F}_3(\!(t)\!)$.

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  • $\begingroup$ Thanks a lot, Laurent. To help my ignorance, could you please explain how ${\rm Pic}^0(C_{K(t^{1/3})})$ and ${\rm Pic}^0(C)$ are related here? $\endgroup$
    – Arno Fehm
    Jan 14 at 21:00
  • $\begingroup$ Apologies, maybe it wasn't clear that I meant $p$-torsion in the Picard group of $C$, so really $p$-torsion in the $K$-rational points of the Picard scheme. I'll edit the question to try to make this clear. $\endgroup$
    – Arno Fehm
    Jan 14 at 21:23
  • $\begingroup$ Arno, you are right, I was not careful enough. I have edited. $\endgroup$ Jan 15 at 9:21
  • $\begingroup$ Great, thank you very much, Laurent. That indeed clarifies things for me. $\endgroup$
    – Arno Fehm
    Jan 15 at 16:01

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