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$\underline{Pic}^0(X)$ denotes Picard scheme of $X$.

Let $X$ be a nonsingular projective surface over an algebraically closed field $k$ and let $D$ be an effective divisor on $X$.

Suppose there is a surjective homomorphism between tangent spaces to Picard schemes at the origin $H^1(\mathcal{O}_X)\to H^1(\mathcal{O}_D)$. Then

(a) $\underline{Pic}^0(X)\to \underline{Pic}^0(D)$ is surjective.

(b) if (a) holds and $\underline{Pic}^0(X)$ is an Abelian variety, then $\underline{Pic}^0(D)$ is also an Abelian variety.

I can't understand them well. Can anyone explain?

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I have serious doubts about (a), though I don't have explicit counter-examples. Suppose $\underline{Pic}^0(X)=\mu _p$, the group of $p$-th roots of unity ($p=\mathrm{char}(k)$), $\underline{Pic}^0(D)=\mathbb{G}_m$, and the restriction map $r$ is the inclusion: then $T_0(r)$ is the identity, but of course $r$ is not surjective. Could you say where you found this statement?

On the other hand if (a) holds and $\underline{Pic}^0(X)$ is smooth, then $\underline{Pic}^0(D)$ is smooth (use EGA IV, thm. 17.11.1) and proper, hence is an abelian variety.

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    $\begingroup$ I don't have Cossec-Dolgachev with me, but I bet you can get this behavior for $X$ an Enriques surface in characteristic $2$ with Picard group $\alpha_2$ and $D$ a cuspidal rational fiber of a quasielliptic fibration on $X$. $\endgroup$ – Jason Starr Mar 26 '14 at 8:53
  • $\begingroup$ Just a follow-up to my previous comment. For a supersingular Enrique surface $X$, there is always a morphism $\pi:X\to \mathbb{P}^1$ whose geometric generic fiber is either a (smooth) elliptic curve or else a cuspidal, integral curve of geometric genus $0$ and arithmetic genus $1$. In both cases, both $\pi_*\mathcal{O}_X$ and $R^1\pi_*\mathcal{O}_X$ equal $\mathcal{O}_{\mathbb{P}^1}$. Thus, for $D$ any but finitely many of the fibers of $\pi$, you get the behavior abx mentions (with $\alpha_2$ rather than $\mu_2$). $\endgroup$ – Jason Starr Mar 27 '14 at 9:16
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The point of this answer is to point out that (a) is true when $k$ is of characteristic zero, and that the issues being discussed by abx and Jason Starr are all about characteristic $p$. (Of course, abx and Jason realize this, but I'm not sure the original poster does.) In the analytic category, we have the short exact sequence $$0 \to \underline{\mathbb{Z}} \overset{2 \pi i}{\longrightarrow} \mathcal{O} \overset{\exp}{\longrightarrow} \mathcal{O}^{\ast} \to 0$$ where $\underline{\mathbb{Z}}$ is the sheaf of locally constant integer valued functions. So we have $$H^1(X_{an}, \underline{\mathbb{Z}}) \to H^1(X_{an}, \mathcal{O}) \to H^1(X_{an}, \mathcal{O}^{\ast}) \to H^2(X_{an}, \mathbb{Z}).$$ The kernel of the last map is $\mathrm{Pic}^0(X_{an})$ (whether or not this is obvious depends on how you define $\mathrm{Pic}^0$) so $H^1(X_{an}, \mathcal{O}) \to Pic^0(X_{an})$ is surjective. We then have a commutative diagram:

$$\begin{matrix} H^1(X_{an}, \mathcal{O}) & \longrightarrow & H^1(D_{an}, \mathcal{O}) \\ \downarrow && \downarrow \\ \mathrm{Pic}^0(X_{an}) & \longrightarrow & \mathrm{Pic}^0(D_{an}) \\ \end{matrix}$$

We have just shown that the right arrow is surjective so, if the top arrow is surjective, then the composite is surjective, implying that the bottom arrow is surjective.

When we work algebraically, the vertical maps can't be defined, but the objects and the horizontal maps makes sense, and GAGA tells us that they are the same. So the conclusion that "map on $H^1(\mathcal{O})$ surjective implies map on $\mathrm{Pic}^0$ surjective" is still correctly algebraically over $\mathbb{C}$ and, by the standard nonsense, also over any algebraically closed field of characteristic zero.

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    $\begingroup$ You don't need GAGA or Hodge theory for this, cf. "Lectures on Curves on an Algebraic Surface" by Mumford. $\endgroup$ – Jason Starr Mar 26 '14 at 19:24

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