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.



p-torsion in the Picard group of a regular projective curve
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 $K=\mathbb{F}\_3(\!(t)\!)$.


