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Let $X$ be a smooth projective curve over $k=\bar{k}$ and denote its Picard group by $\operatorname{Pic}(X)$, with the usual scheme structure coming from the representability of the relative Picard functor.

It's well known that $\operatorname{Pic}(X)$ is smooth of dimension $g$ everywhere in the case of characteristic $0$.

For positive characteristic, Igusa and Serre constructed examples of smooth surfaces presenting singular Picard groups.

What can be said about smooth curves in positive characteristic? Is the Picard group always smooth?

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  • $\begingroup$ Is this really standard terminology? The Picard group is a thing: the group of divisors modulo linear equivalence. It is thus a group. It doesn't make sense to call a group "smooth." I assume you mean Picard scheme? The definition of "the" Picard scheme is a bit more subtle and so one should be careful to define what is exactly meant here. $\endgroup$
    – Matt
    Feb 14, 2014 at 19:38
  • $\begingroup$ You are right on the fact that it's more subtle to define the Picard scheme (it can be done by showing that the relative Picard functor is representable, for instance). I will add this remark to the question $\endgroup$
    – Abramo
    Feb 15, 2014 at 9:30

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The Picard group splits as the product of $\mathbb{Z}$ and the Jacobian variety of $X$, and so each connected component of $\mbox{Pic}(X)$ is (non-canonically) isomorphic to the Jacobian of $X$ which is smooth.

Edit: To see that the Picard group splits, consider the exact sequence $$0\to\mbox{Pic}^0(X)\to\mbox{Pic}(X)\stackrel{\deg}{\to}\mathbb{Z}\to0$$ where $\mbox{Pic}^0(X)=\{\mathcal{O}_X(D):\deg(D)=0\}$. This sequence splits since if $p_0\in X$, we have a section $\mathbb{Z}\to\mbox{Pic}(X)$ where $m\mapsto\mathcal{O}_X(mp_0)$. It is well-known (and most of the time defined this way) that the Jacobian of $X$ is isomorphic to $\mbox{Pic}^0(X)$.

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  • $\begingroup$ To prove the Jacobian is smooth, one has to prove that $\mathrm{Pic}^0_{X/k}$ is smooth. And here we need the hypothesis that $X$ is a curve because then $H^2(X, O_X)=0$. $\endgroup$
    – Cantlog
    Feb 15, 2014 at 18:45
  • $\begingroup$ Dear Cantlog: Yes, one criterion for $\mbox{Pic}(X)$ to be smooth is when $H^2(X,\mathcal{O}_X)=0$. Now is smoothness of $\mbox{Pic}^0(X)$ used to prove the existence of the Jacobian variety (as an abelian variety)? $\endgroup$
    – rfauffar
    Feb 15, 2014 at 22:11
  • $\begingroup$ Dear Robert, the Jacobian is Pic$^0$ by definition. $\endgroup$
    – Cantlog
    Feb 17, 2014 at 15:20
  • $\begingroup$ Dear Cantlog: Yes, and that's why smoothness of $\mbox{Pic}(X)$ is easy, assuming that the Jacobian variety exists (that is, assuming that the functor $\mbox{Pic}^0$ is represented by an abelian variety). Now if you assume that the Jacobian variety exists, it is trivially smooth since it is a group variety. I was just wondering if maybe my answer uses circular logic in the sense that perhaps smoothness of $\mbox{Pic}^0$ was used to prove the existence of the Jacobian (as an abelian variety). $\endgroup$
    – rfauffar
    Feb 17, 2014 at 20:51
  • $\begingroup$ OK I see. Non, the smoothness of Pic$^0$ is not need to prove the existence (representability) of Pic. $\endgroup$
    – Cantlog
    Feb 17, 2014 at 21:08

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