Relation between Picard group of affine curve and the projective closure

Question

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.

Answer 1

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.