Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be an abelian variety over an algebraically closed field $K$, and $B$ an abelian subvariety. We know (for instance, from Milne course on AV) that there is an abelian "cofactor" $B'\subset A$ such that the addition $B\times B'\to A$ is an isogeny, which is the same as saying that $B\cap B'$ is a finite algebraic group. Nothing (to my knowledge) is said about the unicity of $B'$, although the proof of this result constructs such a $B'$ through a polarization of $A$ (so it may be that there aren't too many such $B'$'s).

My question is : is there such a $B'$ which satisfies $B\cap B'$ has cardinality prime to a given prime integer $p$ (in the exact problem I have, this $p$ is the characteristic of $K$) ?

If it is not the case, is there any way to control this finite subgroup ?

share|improve this question
2  
Take $B$ and $B'$ to be non-CM non-isogenous elliptic curves, pick a common $K$-subgroup $G$, and define $A = (B \times B')/G$. The only elliptic curves that are abelian subvarieties of $A$ are the evident copies of $B$ and $B'$, so there's nothing you can do to "fix" the problem in such cases. One way to avoid such a situation is to assume that $A$ admits a polarization of degree prime to $p$, but that's a rather severe assumption. –  grp Sep 16 '12 at 20:47

1 Answer 1

up vote 4 down vote accepted

Here is a (slightly more detailed) variant of what was pointed out by grp.

Let $E$ and $F$ be non-isogenous elliptic curves over $K$. Let $n$ be a positive integer. (If $p=char(K)>0$ and $p$ divides $n$ we assume additionally that both $E$ and $F$ are ordinary elliptic curves.) Then there are order $n$ cyclic subgroups $C_n \subset E(K)$ and $D_n \subset F(K)$. Fix a group isomorphism $\phi: C_n \cong D_n$. Let $$\Gamma(\phi)=[\{(x,\phi x) \mid x \in C_n \}] \subset C_n \times D_n \subset E(K) \times F(K)$$ be the graph of $\phi$; it is an order $n$ cyclic subgroup of $(E\times F)(K)$. Let us consider the quotient $A:= (E\times F)/\Gamma(\phi)$ and denote by $\pi: E\times F \to A$ the corresponding degree $n$ isogeny of abelian surfaces. Clearly, the restrictions of $\pi$ to $E \times \{e_F\}$ and $\{e_E\}\times F$ give us isomorphisms of elliptic curves $$E=E \times \{e_F\} \cong \pi(E \times \{e_F\})=: E^{\prime}\subset A,$$ $$F=\{e_E\}\times F \cong \pi(\{e_E\}\times F)=: F^{\prime} \subset A.$$

(Here $e_E$ (resp. $e_F$) is the zero of group law on $E$ (resp. on $F$).) It is also clear that the intersection of $E^{\prime}$ and $F^{\prime}$ (in $A$) is a cyclic order $n$ subgroup that is the image under $\pi$ of $$[\{(x,0) \mid x \in C_n \}] \subset C_n \times D_n \subset E(K) \times F(K).$$ Now let $Z$ be a 1-dim'l abelian subvariety of $A$ and let $Y$ be the identity component of its preimage $\pi^{-1}(Z)$ in $E\times F$. Clearly, $Y$ is a 1-dim'l abelian subvariety of $E\times F$ and $\pi(Y)=Z$. It is also clear that (at least) one of projection maps $$Y \to E, \ Y \to F$$ is non-constant. If $Y \to E$ is non-constant then it is an isogeny of elliptic curves. Since $E$ and $F$ are non-isogenous, $Y$ is non-isogenous to $F$ and therefore $Y \to F$ is the constant map to $e_F$. It follows that $Y=E\times \{e_F\}$ and therefore $Z=\pi(Y)=E^{\prime}\subset A$. The same arguments prove that if $Y \to F$ is non-constant then $Z=F^{\prime} \subset A$.

share|improve this answer
    
Thank you very much, this is very clear and helpful. –  Cyrille Corpet Sep 17 '12 at 9:18
    
You are welcome. –  Yuri Zarhin Sep 17 '12 at 19:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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