Mixing
Torsion of elliptic curves and abelian extensions
1
$begingroup$
Let $L/K$ be an abelian $p$-extension of number fields and $E$ be an elliptic curve over $Bbb Q$.
If $E[p](K)=0$, does it follow that $E[p](L)=0$ ? The converse is obviously true, but I don't have any reference for my problem.
galois-theory elliptic-curves arithmetic-geometry
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
$endgroup$
add a comment |
1
$begingroup$
Let $L/K$ be an abelian $p$-extension of number fields and $E$ be an elliptic curve over $Bbb Q$.
If $E[p](K)=0$, does it follow that $E[p](L)=0$ ? The converse is obviously true, but I don't have any reference for my problem.
galois-theory elliptic-curves arithmetic-geometry
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
$endgroup$
1
$begingroup$
Not sure but you may try : replace $L$ by $ K(E[p]) cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(mathbb{F}_p)$. Does $g in G$ of order $p$ mean $langle g rangle = u T u^{-1}$ where $T = {pmatrix{1 & b \ 0 & 1 }}$ ? Then only $DT = {pmatrix{ a & b \ 0 & a }}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element.
$endgroup$
– reuns
Jan 21 at 15:03
$begingroup$
The case of $char K=p$ is easier since $rank E[p](bar{K})$ is $0$ or $1$. The group $Aut(mathbb F_p)$ has no order $p$ elements.
$endgroup$
– eduard
Jan 21 at 15:47
add a comment |
1
1
1
$begingroup$
Let $L/K$ be an abelian $p$-extension of number fields and $E$ be an elliptic curve over $Bbb Q$.
If $E[p](K)=0$, does it follow that $E[p](L)=0$ ? The converse is obviously true, but I don't have any reference for my problem.
galois-theory elliptic-curves arithmetic-geometry
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
$endgroup$
Let $L/K$ be an abelian $p$-extension of number fields and $E$ be an elliptic curve over $Bbb Q$.
If $E[p](K)=0$, does it follow that $E[p](L)=0$ ? The converse is obviously true, but I don't have any reference for my problem.
galois-theory elliptic-curves arithmetic-geometry
galois-theory elliptic-curves arithmetic-geometry
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
asked Jan 21 at 12:58
AlphonseAlphonse
2,208724
2,208724
1
$begingroup$
Not sure but you may try : replace $L$ by $ K(E[p]) cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(mathbb{F}_p)$. Does $g in G$ of order $p$ mean $langle g rangle = u T u^{-1}$ where $T = {pmatrix{1 & b \ 0 & 1 }}$ ? Then only $DT = {pmatrix{ a & b \ 0 & a }}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element.
$endgroup$
– reuns
Jan 21 at 15:03
$begingroup$
The case of $char K=p$ is easier since $rank E[p](bar{K})$ is $0$ or $1$. The group $Aut(mathbb F_p)$ has no order $p$ elements.
$endgroup$
– eduard
Jan 21 at 15:47
add a comment |
1
$begingroup$
Not sure but you may try : replace $L$ by $ K(E[p]) cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(mathbb{F}_p)$. Does $g in G$ of order $p$ mean $langle g rangle = u T u^{-1}$ where $T = {pmatrix{1 & b \ 0 & 1 }}$ ? Then only $DT = {pmatrix{ a & b \ 0 & a }}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element.
$endgroup$
– reuns
Jan 21 at 15:03
$begingroup$
The case of $char K=p$ is easier since $rank E[p](bar{K})$ is $0$ or $1$. The group $Aut(mathbb F_p)$ has no order $p$ elements.
$endgroup$
– eduard
Jan 21 at 15:47
1
1
$begingroup$
Not sure but you may try : replace $L$ by $ K(E[p]) cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(mathbb{F}_p)$. Does $g in G$ of order $p$ mean $langle g rangle = u T u^{-1}$ where $T = {pmatrix{1 & b \ 0 & 1 }}$ ? Then only $DT = {pmatrix{ a & b \ 0 & a }}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element.
$endgroup$
– reuns
Jan 21 at 15:03
$begingroup$
Not sure but you may try : replace $L$ by $ K(E[p]) cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(mathbb{F}_p)$. Does $g in G$ of order $p$ mean $langle g rangle = u T u^{-1}$ where $T = {pmatrix{1 & b \ 0 & 1 }}$ ? Then only $DT = {pmatrix{ a & b \ 0 & a }}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element.
$endgroup$
– reuns
Jan 21 at 15:03
$begingroup$
The case of $char K=p$ is easier since $rank E[p](bar{K})$ is $0$ or $1$. The group $Aut(mathbb F_p)$ has no order $p$ elements.
$endgroup$
– eduard
Jan 21 at 15:47
$begingroup$
The case of $char K=p$ is easier since $rank E[p](bar{K})$ is $0$ or $1$. The group $Aut(mathbb F_p)$ has no order $p$ elements.
$endgroup$
– eduard
Jan 21 at 15:47
add a comment |
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1
$begingroup$
Not sure but you may try : replace $L$ by $ K(E[p]) cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(mathbb{F}_p)$. Does $g in G$ of order $p$ mean $langle g rangle = u T u^{-1}$ where $T = {pmatrix{1 & b \ 0 & 1 }}$ ? Then only $DT = {pmatrix{ a & b \ 0 & a }}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element.
$endgroup$
– reuns
Jan 21 at 15:03
$begingroup$
The case of $char K=p$ is easier since $rank E[p](bar{K})$ is $0$ or $1$. The group $Aut(mathbb F_p)$ has no order $p$ elements.
$endgroup$
– eduard
Jan 21 at 15:47