2
$\begingroup$

Let $A$ be an Abelian variety. Let $L$ and $C$ be an effective divisor and a curve on $A$, respectively. If $C$ is not including in $L$, can we conclude that $L+C=A$? Thanks.

$\endgroup$

2 Answers 2

9
$\begingroup$

Let us try to come up with a criterion for $L+C\ne A$. Let us assume both $L$ and $C$ are irreducible. (Otherwise consider separate irreducible components.) Let us also shift both $L$ and $C$ so that they pass through the origin. (I guess here I am implicitly assuming the field is algebraically closed, so $L$ and $C$ have points.)

Note that $L+C$ is an irreducible closed variety containing $L$. If $L+C\ne A$, then $L+C=L$. This implies $C+C+C+\dots+C\subset L$ for any number of summands. Clearly, the chain $C\subset C+C\subset\dots$ must stabilize; denote the limit by $B$. It is a semigroup: $B+B=B$. Therefore, it is an abelian variety. It has the following properties: $C\subset B$ and $B+L\subset L$. We thus arrive at the following criterion:

$L+C\ne A$ if and only if there is an abelian subvariety $B$ such that $C$ lies in a translate of $B$ and $L$ is the preimage of a divisor under $A\to A/B$.

P.S. Note that for such $L$ and $C$, we clearly have $L.C=0$, as suggested by Jack Huizenga in his answer.

$\endgroup$
7
$\begingroup$

Not necessarily. For instance, if $L\subset A$ is a subgroup and $C$ is contained in a translate $L + x$ of $L$, then $L + C$ would be contained in $L+x$.

Additional hypotheses should make this work out, though. For instance, it should be true if $C$ intersects $L$ transversely at some point $x$. For then the differential of the map $L\times C \to A$ at the point $(x,x)$ will be surjective, and so the map is dominant, hence surjective in light of projectivity. I doubt the intersection actually has to be transverse (in particular, I believe $L.C >0$ should be enough), but somebody else can fill in the details.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.