Is every k-isogeny of abelian varieties given by polynomials over k?












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Given an abelian variety $A$ over the rational integers $mathbb{Q}$, for every finite group $Gsubset A(bar{mathbb{Q}})$ consider the field $mathbb{Q}(G)$ obtained by adjoining to $mathbb{Q}$ the coordinates of the points in $G$.



When $G=A[p]$, the $p$-torsion points of $A$, $mathbb{Q}(A[p])$ is the $p$-division field of $A$, it is a normal extension of $mathbb{Q}$.



Is this just because $A[p]$ is the kernel of the multiplication by $p$ and this is given by polynomials in $mathbb{Q}$?



If yes, that means that, for every finite group $G$, $mathbb{Q}(G)$ is normal?, since every finite group is the kernel of some separable isogeny.










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    0












    $begingroup$


    Given an abelian variety $A$ over the rational integers $mathbb{Q}$, for every finite group $Gsubset A(bar{mathbb{Q}})$ consider the field $mathbb{Q}(G)$ obtained by adjoining to $mathbb{Q}$ the coordinates of the points in $G$.



    When $G=A[p]$, the $p$-torsion points of $A$, $mathbb{Q}(A[p])$ is the $p$-division field of $A$, it is a normal extension of $mathbb{Q}$.



    Is this just because $A[p]$ is the kernel of the multiplication by $p$ and this is given by polynomials in $mathbb{Q}$?



    If yes, that means that, for every finite group $G$, $mathbb{Q}(G)$ is normal?, since every finite group is the kernel of some separable isogeny.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Given an abelian variety $A$ over the rational integers $mathbb{Q}$, for every finite group $Gsubset A(bar{mathbb{Q}})$ consider the field $mathbb{Q}(G)$ obtained by adjoining to $mathbb{Q}$ the coordinates of the points in $G$.



      When $G=A[p]$, the $p$-torsion points of $A$, $mathbb{Q}(A[p])$ is the $p$-division field of $A$, it is a normal extension of $mathbb{Q}$.



      Is this just because $A[p]$ is the kernel of the multiplication by $p$ and this is given by polynomials in $mathbb{Q}$?



      If yes, that means that, for every finite group $G$, $mathbb{Q}(G)$ is normal?, since every finite group is the kernel of some separable isogeny.










      share|cite|improve this question









      $endgroup$




      Given an abelian variety $A$ over the rational integers $mathbb{Q}$, for every finite group $Gsubset A(bar{mathbb{Q}})$ consider the field $mathbb{Q}(G)$ obtained by adjoining to $mathbb{Q}$ the coordinates of the points in $G$.



      When $G=A[p]$, the $p$-torsion points of $A$, $mathbb{Q}(A[p])$ is the $p$-division field of $A$, it is a normal extension of $mathbb{Q}$.



      Is this just because $A[p]$ is the kernel of the multiplication by $p$ and this is given by polynomials in $mathbb{Q}$?



      If yes, that means that, for every finite group $G$, $mathbb{Q}(G)$ is normal?, since every finite group is the kernel of some separable isogeny.







      abelian-varieties






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 16:59









      A. GMA. GM

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