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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 ?

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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
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1 Answer

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 $Z=F^{\prime} \subset A$.

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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
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