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fixed problems with LaTeX ; added 2 characters in body

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edited Sep 17 at 0:15

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

(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 $${(x,0) $[{(x,0) \mid x \in C_n }] \subset C_n \times D_n \subset E(K) \times F(K).$$ 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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edited Sep 17 at 0:00

Yuri Zarhin
2,234●5●11

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

(Here $e_E$ (resp. $e_F$) is the zero of group law on $E$ (resp. on $F$.) 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 $\pi(Y)=Z$. Clearly, 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 $Z=F^{\prime} \subset A$.

added 2 characters in body

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edited Sep 16 at 23:54

Yuri Zarhin
2,234●5●11

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.$$ (Here $e_E$ (resp. $e_F$) is the zero of group law on $E$ (resp. on $F$.) It is also clear that the intersection of $E^{\prime}$ and $F^{\prime}$ (in $A$) is a cyclic order $n$ subgroup that is the image under $\pi$ of {(x,0) $${(x,0) \mid x \in C_n } \subset C_n \times D_n \subset E(K) \times F(K).$$ Now let $Z$ be a 1-dim'l abelian subvariety of $A$ and let $Y$ be the identity component of its preimage $\pi^{-1}(Z)$ in $E\times F$. Clearly, $Y$ is a 1-dim'l abelian subvariety of $E\times F$ and $\pi(Y)=Z$. Clearly, one of projection maps $$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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answered Sep 16 at 23:41

Yuri Zarhin
2,234●5●11

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 $\pi(Y)=Z$. Clearly, 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 $Z=F^{\prime} \subset A$.