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.
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.
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.