# p-torsion in the Picard group of a regular projective curve

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.

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 $$\mathrm{Pic}^0(C)=\mathrm{Pic}^0(C)$$. 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 $$K=\mathbb{F}_3(\!(t)\!)$$.
• 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? Jan 14, 2021 at 21:00
• 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. Jan 14, 2021 at 21:23