# Relation between Picard group of affine curve and the projective closure

Suppose that we have an smooth affine curve $$C$$ over an algebraically closed field of finite characteristic $$p$$, defined by $$f(x,y) = 0$$, and let $$D$$ be the smooth projective curve defined by the homogenisation of $$f$$; $$F(X,Y,Z) = 0$$.

I know that there is an exact sequence $$\mathbb{Z}^{|S|} \rightarrow \renewcommand{\Pic}{\operatorname{Pic}} \Pic(D) \rightarrow \Pic(C) \rightarrow 0,$$ where $$S$$ is the finite set of points $$D \setminus C$$.

I am wondering if I know that the $$p$$-torsion of $$\Pic(D)$$ vanishes, can I conclude immediately that $$\Pic(C)[p] = 0$$ too?

If not, are there standard techniques for establishing this?

Thanks.

I want to talk about two special case, and prove that you can't necessarily say $$\renewcommand{\Pic}{\operatorname{Pic}} \Pic C$$ does not have any $$p$$-torison but I think the question about what you can say about $$p$$-torison is interesting.
First if $$S$$ consists of a single point $$P$$ then $$\Pic C$$ doesn't have any $$p$$-torison: if we have $$p\bar x=0$$ in $$\Pic C$$,then we should have $$px=nP$$ in $$\Pic D$$ for some $$n$$, by looking at the degree you get that $$p|n$$ and because you don't have $$p$$-torison we have $$x=n'P$$ hence $$\bar x=0$$.
Second if $$S$$ consists of points $$P,Q$$ such that $$P-Q$$ is not torison, then take a $$p^{th}$$ root of $$P-Q$$, which exists because $$p$$ is an isogeny, and call it $$T$$. this point is clearly a $$p$$ torsion in $$C$$, but if $$T=aP+bQ$$ in $$\Pic D$$ you get a relation $$mP=nQ$$ and by looking at the degree you get that $$m(P-Q)=0$$ a contradiction. hence $$T$$ is a nontrivial $$p$$-torison in this case.