question about Picard scheme

Question

$\underline{Pic}^0(X)$ 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?

Answer 1

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.

Answer 2

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.