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?



Which schemes are divisors of an abelian variety?
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 $y^3 = x^4 - x$.


