# rank of Abelian schemes under ample hypersurface section

Let $k$ be an algebraic closure of a finite field, $\ell \neq \mathrm{Char}(k)$ be prime, $S/k$ a smooth projective geometrically connected surface and $C/k$ a smooth ample connected hypersurface section of $S$. Let $\mathscr{A}/S$ be an Abelian scheme.

Are there good cases when the injection $\mathscr{A}(S)/\ell^n \hookrightarrow \mathscr{A}(C) /\ell^n$ is surjective?

If $\mathrm{rk} \mathscr{A}(C) = 0$, this holds true since the $\ell^n$-torsion subgroups of $\mathscr{A}(S)$ and $\mathscr{A}(C)$ are (always) isomorphic. $\mathrm{rk} \mathscr{A}(C) = 0$ is e.g. the case if $C \cong \mathbf{P}^1_k$ and $\mathscr{A}/C$ is constant.

[Edit: Note that if $X$ is like $S$, but of arbitrary dimension, and $Y$ a smooth ample hypersurface section of dimension $\geq 2$, $\mathrm{rk} \mathscr{A}(X) = \mathrm{rk} \mathscr{A}(Y)$.]

I recommend that you look at my joint paper with Tom Graber.

MR3114946 Pending
Graber, Tom(1-CAIT); Starr, Jason Michael(1-SUNYS)
Restriction of sections for families of abelian varieties. (English summary) A celebration of algebraic geometry, 311–327,
Clay Math. Proc., 18, Amer. Math. Soc., Providence, RI, 2013.
14K12 (14C05)

In particular, surjectivity frequently fails. For instance, starting with $X$ the projective linear system of plane cubics containing one specified point, and denoting by $\mathcal{A}$ the universal plane cubic over $X$ (with its specified point as the "origin"), let $S$ be a general linear $2$-plane inside $X$, and let $C$ be a general line inside $S$. Then $C$ gives a pencil of plane cubics that has 9 base points (one of which is our specified point). These base points give lots of rational points in $\mathcal{A}(C)$, and these need not lift (of course your base field is "small", so this might take some work to completely justify).

What Tom and I do show is that, if you use "line pairs" rather than "lines", and if you allow $C$ to be a "sufficiently general" line pair, then the restriction map is a bijection.

• Are there problems as $k$ is not uncountable? – TKe Jul 7 '14 at 13:19
• @TimoKeller For the counterexample, I think there are no problems with $k$ being countable. However, you are quite correct that, in the last paragraph, "sufficiently general" almost certainly does require that $k$ is uncountable. – Jason Starr Jul 7 '14 at 17:52
• Do we need it for Lemma 5.1, 5.2 and 5.3; and Theorem 1.2(ii)? – TKe Jul 7 '14 at 18:18
• @TimoKeller: Yes, I think you need uncountability for all of the arguments in that paper (unfortunately). – Jason Starr Jul 9 '14 at 20:14