# Tag Info

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Any curve of genus greater than two, whose Jacobian $J$ is simple, will do. If it were a divisor on an abelian surface $S$, then there would be a surjection $J\to S$ with positive dimensional kernel, contradicting the simplicity of $J$. Most curves of genus larger than two have this property; a randomly chosen example is $y^3 = x^4 - x$.

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An obvious class of counterexamples are uniruled varieties. In fact, abelian varieties contain no rational curves. More generally, and for the same reason, if $X$ is any algebraic variety that contain a (possibly singular) rational curve, then $X$ is not a subvariety of an abelian variety, in particular it is not a divisor there.

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Here's another answer using the Albanese that's of a slightly different flavor. Let $X$ be $n$-dimensional and suppose that $h^0(X,\Omega^1_X)<n$. Then any map $X\rightarrow A$ where $A$ is an abelian variety factors through the Albanese, which is of dimension less than $n$, so $X$ can't be a divisor on any abelian variety. So as an example you could take ...

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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, ...

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I just want to point out that "adjunction+translation" tells us quite a bit: Let $A$ be an abelian variety, say of dimension $n>1$ and let $D \subset A$ be a (let's say smooth) divisor. Since $\omega_A = \mathcal{O}_A$, the adjunction formula $$\omega_D = \omega_A(D)|_D = \mathcal{O}(D)|_D,$$ the normal bundle of $D$. By differentiating the ...

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Certainly it does not suffice to take the schematic closure if $S$ is nonreduced. For instance, let $S$ be $\text{Spec}\ k[x,y]/\langle x^2, xy \rangle$. Let $A$ be $E \times_{\text{Spec} k} S$, where $E$ is an elliptic curve over $k$ with specified zero point $z$. Let $p \in S$ be the closed point with maximal ideal $\langle x,y\rangle$. Let $U$ be the ...

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