Which schemes are divisors of an abelian variety?

Question

Let $X$ be a smooth, projective ireducible scheme over an algebraically closed field $k$. I'm trying to understand when there exists an abelian variety $A$ such that $X$ is isomorphic to a prime divisor on $A$.

There are some simple cases, of course. If $X$ is zero-dimensional, i.e. a point, then it is isomorphic to the identity of any elliptic curve $E$ over $k$, hence it is a divisor of $E$. If $X$ is of genus $1$, then if we choose a $k$-point, then $X$ is an elliptic curve. Then $X$ is isomorphic to the diagonal $\Delta\subset X\times X$, which is a divisor. Since $X$ is an elliptic curve, $X\times X$ is also an abelian variety. If $X$ is a curve of genus $2$, then the Jacobian of $X$ is 2-dimensional, and thus $X$ is of codimension one and thus the embedding $X\rightarrow \text{Jac}(X)$ lets us identify $X$ with a divisor of $\text{Jac}(X)$.

However, these simple cases do not give me an idea for the general case. The Jacobian only works for the genus $2$ case etc. The Albanse Variety also doesn't help, as the codimension might be to big. Are there any counter-examples of a smooth, projective ireducible scheme over an algebraically closed field which is not a divisor of an abelian variety?

Answer 1

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 $y^3 = x^4 - x$.

Answer 2

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.

Answer 3

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 any simply connected variety. Of course, $\mathbb{P}^1$ does the trick.

Answer 4

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 translation action of $A$, we can obtain non-0 global sections $0 \neq \sigma \in H^0(D,\omega_D)$, in which case the powers $\sigma^d$ show $H^0(D, \omega_D^d) \neq 0$ for all $d>0$. This shows that $D$ has non-negative Kodaira dimension: $\kappa(D) \geq 0$.

Remark: it's known that $D$ uniruled $\implies$ $H^0(D, \omega_D^d)=0$ for all $d > 0$ (and the converse is a conjecture), so the above is more-or-less an elaboration of Polizzi's observation that $D$ can't be uniruled.