Is every k-isogeny of abelian varieties given by polynomials over k?
$begingroup$
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.
abelian-varieties
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
$endgroup$
add a comment |
$begingroup$
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.
abelian-varieties
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
$endgroup$
add a comment |
$begingroup$
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.
abelian-varieties
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
$endgroup$
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
abelian-varieties
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
asked Dec 7 '18 at 16:59
A. GMA. GM
1189
1189
add a comment |
add a comment |
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