Which schemes are divisors of an abelian variety?

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?

• The moduli space of smooth genus $g$ curves which are divisors on an abelian surface is an open subset of a $\mathbb P^{g-2}$-bundle on the moduli space of abelian surfaces with a polarization of degree $g-1$, hence has dimension $\leq 3 + (g-2) = g+1$, while the moduli space of all curves has dimension $3g-3$. Aug 19, 2020 at 18:45
• Any rational variety, e.g. $\Bbb{P}^n$ for $n≥1$...
– abx
Aug 19, 2020 at 18:46
• @WillSawin Wow, thats a cool way of thinking about it! Aug 19, 2020 at 18:46

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

• Thanks. Apologies for the dumb question. Aug 19, 2020 at 18:46

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.

Here's another answer using the Albanese that's of a slightly different flavor. Let $$X$$ be $$n$$-dimensional and suppose that $$h^0(X,\Omega^1_X). Then any map $$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.

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.