Are there counterexamples of isogeny elliptic curves with non-isomorphic integral Tate modules?
$begingroup$
Let $K$ be a field and $G_K$ be its absolute Galois group. Let $E_1,E_2$ be two elliptic curves over $K$. Assume that there exists an isogeny $f:E_1rightarrow E_2$. Let $p$ be a prime number. Then $f$ induces an isomorphism of rational Tate module $V_p(E_1)cong V_p(E_2)$ as representations of $G_K$: let $f^{vee}:E_2rightarrow E_1$ be the dual isogeny, then $fcirc f^{vee}:E_1rightarrow E_1$ is $[mathrm{deg}(f)]$ and on the Tate module $T_p(E_1)$ $fcirc f^{vee}$ is just multiplicated by $mathrm{deg}(f)$. So after tensoring $mathbb{Q}_p$, $f$ induces an isomorphism and the isomorphism is $G_K$-equivariant.
My question is whether we can always get an isomorphism of integral Tate module $T_p(E_1)cong T_p(E_2)$ as $G_K$-modules over $mathbb{Z}_p$? Notice that the isomorphism doesn't need to be induced by $f$.
There is a possible way to find a counterexample. Tate's isogeny theorem tells that the isogeny classes of elliptic curves over a finite field $mathbb{F}_q$ is determined by the rational representations. If we could construct elliptic curves over a finite field for any integral representations, then we just need to find two non-isomorphic integral models for a rational representations then we get a counterexample for the original question.
If the counterexample exists for general fields, can it be true for some special fields?
elliptic-curves arithmetic-geometry galois-representations
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
$endgroup$
|
show 1 more comment
$begingroup$
Let $K$ be a field and $G_K$ be its absolute Galois group. Let $E_1,E_2$ be two elliptic curves over $K$. Assume that there exists an isogeny $f:E_1rightarrow E_2$. Let $p$ be a prime number. Then $f$ induces an isomorphism of rational Tate module $V_p(E_1)cong V_p(E_2)$ as representations of $G_K$: let $f^{vee}:E_2rightarrow E_1$ be the dual isogeny, then $fcirc f^{vee}:E_1rightarrow E_1$ is $[mathrm{deg}(f)]$ and on the Tate module $T_p(E_1)$ $fcirc f^{vee}$ is just multiplicated by $mathrm{deg}(f)$. So after tensoring $mathbb{Q}_p$, $f$ induces an isomorphism and the isomorphism is $G_K$-equivariant.
My question is whether we can always get an isomorphism of integral Tate module $T_p(E_1)cong T_p(E_2)$ as $G_K$-modules over $mathbb{Z}_p$? Notice that the isomorphism doesn't need to be induced by $f$.
There is a possible way to find a counterexample. Tate's isogeny theorem tells that the isogeny classes of elliptic curves over a finite field $mathbb{F}_q$ is determined by the rational representations. If we could construct elliptic curves over a finite field for any integral representations, then we just need to find two non-isomorphic integral models for a rational representations then we get a counterexample for the original question.
If the counterexample exists for general fields, can it be true for some special fields?
elliptic-curves arithmetic-geometry galois-representations
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
$endgroup$
$begingroup$
You mean $T_p(E_i) ={ Q in E_i, exists n, p^n Q = O} $ which is isomorphic to a subgroup of $mathbb{Q}_p^2 / mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) otimes_{mathbb{Z}_{p} } mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) cong deg(f) T_p(E_i) $. Did you try with $K $ a finite field ?
$endgroup$
– reuns
Dec 16 '18 at 20:32
$begingroup$
$T_p(E_i)=lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)otimes_{mathbb{Z}_p}mathbb{Q}_p$. The isomorphism of underlying $mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny.
$endgroup$
– wuzx
Dec 16 '18 at 20:34
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Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/ker(f)$ and $E_2$ are isomorphic as groups, $mathbb{Z}_p$ and Galois modules. Then what ?
$endgroup$
– reuns
Dec 16 '18 at 20:48
3
$begingroup$
The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question.
$endgroup$
– Brandon Carter
Dec 21 '18 at 3:36
$begingroup$
@BrandonCarter Thank you! It is very explicit!
$endgroup$
– wuzx
Dec 22 '18 at 19:51
|
show 1 more comment
$begingroup$
Let $K$ be a field and $G_K$ be its absolute Galois group. Let $E_1,E_2$ be two elliptic curves over $K$. Assume that there exists an isogeny $f:E_1rightarrow E_2$. Let $p$ be a prime number. Then $f$ induces an isomorphism of rational Tate module $V_p(E_1)cong V_p(E_2)$ as representations of $G_K$: let $f^{vee}:E_2rightarrow E_1$ be the dual isogeny, then $fcirc f^{vee}:E_1rightarrow E_1$ is $[mathrm{deg}(f)]$ and on the Tate module $T_p(E_1)$ $fcirc f^{vee}$ is just multiplicated by $mathrm{deg}(f)$. So after tensoring $mathbb{Q}_p$, $f$ induces an isomorphism and the isomorphism is $G_K$-equivariant.
My question is whether we can always get an isomorphism of integral Tate module $T_p(E_1)cong T_p(E_2)$ as $G_K$-modules over $mathbb{Z}_p$? Notice that the isomorphism doesn't need to be induced by $f$.
There is a possible way to find a counterexample. Tate's isogeny theorem tells that the isogeny classes of elliptic curves over a finite field $mathbb{F}_q$ is determined by the rational representations. If we could construct elliptic curves over a finite field for any integral representations, then we just need to find two non-isomorphic integral models for a rational representations then we get a counterexample for the original question.
If the counterexample exists for general fields, can it be true for some special fields?
elliptic-curves arithmetic-geometry galois-representations
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
$endgroup$
Let $K$ be a field and $G_K$ be its absolute Galois group. Let $E_1,E_2$ be two elliptic curves over $K$. Assume that there exists an isogeny $f:E_1rightarrow E_2$. Let $p$ be a prime number. Then $f$ induces an isomorphism of rational Tate module $V_p(E_1)cong V_p(E_2)$ as representations of $G_K$: let $f^{vee}:E_2rightarrow E_1$ be the dual isogeny, then $fcirc f^{vee}:E_1rightarrow E_1$ is $[mathrm{deg}(f)]$ and on the Tate module $T_p(E_1)$ $fcirc f^{vee}$ is just multiplicated by $mathrm{deg}(f)$. So after tensoring $mathbb{Q}_p$, $f$ induces an isomorphism and the isomorphism is $G_K$-equivariant.
My question is whether we can always get an isomorphism of integral Tate module $T_p(E_1)cong T_p(E_2)$ as $G_K$-modules over $mathbb{Z}_p$? Notice that the isomorphism doesn't need to be induced by $f$.
There is a possible way to find a counterexample. Tate's isogeny theorem tells that the isogeny classes of elliptic curves over a finite field $mathbb{F}_q$ is determined by the rational representations. If we could construct elliptic curves over a finite field for any integral representations, then we just need to find two non-isomorphic integral models for a rational representations then we get a counterexample for the original question.
If the counterexample exists for general fields, can it be true for some special fields?
elliptic-curves arithmetic-geometry galois-representations
elliptic-curves arithmetic-geometry galois-representations
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
edited Dec 16 '18 at 20:57
wuzx
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
asked Dec 16 '18 at 20:12
wuzxwuzx
1163
1163
$begingroup$
You mean $T_p(E_i) ={ Q in E_i, exists n, p^n Q = O} $ which is isomorphic to a subgroup of $mathbb{Q}_p^2 / mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) otimes_{mathbb{Z}_{p} } mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) cong deg(f) T_p(E_i) $. Did you try with $K $ a finite field ?
$endgroup$
– reuns
Dec 16 '18 at 20:32
$begingroup$
$T_p(E_i)=lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)otimes_{mathbb{Z}_p}mathbb{Q}_p$. The isomorphism of underlying $mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny.
$endgroup$
– wuzx
Dec 16 '18 at 20:34
$begingroup$
Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/ker(f)$ and $E_2$ are isomorphic as groups, $mathbb{Z}_p$ and Galois modules. Then what ?
$endgroup$
– reuns
Dec 16 '18 at 20:48
3
$begingroup$
The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question.
$endgroup$
– Brandon Carter
Dec 21 '18 at 3:36
$begingroup$
@BrandonCarter Thank you! It is very explicit!
$endgroup$
– wuzx
Dec 22 '18 at 19:51
|
show 1 more comment
$begingroup$
You mean $T_p(E_i) ={ Q in E_i, exists n, p^n Q = O} $ which is isomorphic to a subgroup of $mathbb{Q}_p^2 / mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) otimes_{mathbb{Z}_{p} } mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) cong deg(f) T_p(E_i) $. Did you try with $K $ a finite field ?
$endgroup$
– reuns
Dec 16 '18 at 20:32
$begingroup$
$T_p(E_i)=lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)otimes_{mathbb{Z}_p}mathbb{Q}_p$. The isomorphism of underlying $mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny.
$endgroup$
– wuzx
Dec 16 '18 at 20:34
$begingroup$
Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/ker(f)$ and $E_2$ are isomorphic as groups, $mathbb{Z}_p$ and Galois modules. Then what ?
$endgroup$
– reuns
Dec 16 '18 at 20:48
3
$begingroup$
The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question.
$endgroup$
– Brandon Carter
Dec 21 '18 at 3:36
$begingroup$
@BrandonCarter Thank you! It is very explicit!
$endgroup$
– wuzx
Dec 22 '18 at 19:51
$begingroup$
You mean $T_p(E_i) ={ Q in E_i, exists n, p^n Q = O} $ which is isomorphic to a subgroup of $mathbb{Q}_p^2 / mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) otimes_{mathbb{Z}_{p} } mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) cong deg(f) T_p(E_i) $. Did you try with $K $ a finite field ?
$endgroup$
– reuns
Dec 16 '18 at 20:32
$begingroup$
You mean $T_p(E_i) ={ Q in E_i, exists n, p^n Q = O} $ which is isomorphic to a subgroup of $mathbb{Q}_p^2 / mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) otimes_{mathbb{Z}_{p} } mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) cong deg(f) T_p(E_i) $. Did you try with $K $ a finite field ?
$endgroup$
– reuns
Dec 16 '18 at 20:32
$begingroup$
$T_p(E_i)=lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)otimes_{mathbb{Z}_p}mathbb{Q}_p$. The isomorphism of underlying $mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny.
$endgroup$
– wuzx
Dec 16 '18 at 20:34
$begingroup$
$T_p(E_i)=lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)otimes_{mathbb{Z}_p}mathbb{Q}_p$. The isomorphism of underlying $mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny.
$endgroup$
– wuzx
Dec 16 '18 at 20:34
$begingroup$
Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/ker(f)$ and $E_2$ are isomorphic as groups, $mathbb{Z}_p$ and Galois modules. Then what ?
$endgroup$
– reuns
Dec 16 '18 at 20:48
$begingroup$
Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/ker(f)$ and $E_2$ are isomorphic as groups, $mathbb{Z}_p$ and Galois modules. Then what ?
$endgroup$
– reuns
Dec 16 '18 at 20:48
3
3
$begingroup$
The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question.
$endgroup$
– Brandon Carter
Dec 21 '18 at 3:36
$begingroup$
The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question.
$endgroup$
– Brandon Carter
Dec 21 '18 at 3:36
$begingroup$
@BrandonCarter Thank you! It is very explicit!
$endgroup$
– wuzx
Dec 22 '18 at 19:51
$begingroup$
@BrandonCarter Thank you! It is very explicit!
$endgroup$
– wuzx
Dec 22 '18 at 19:51
|
show 1 more comment
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$begingroup$
You mean $T_p(E_i) ={ Q in E_i, exists n, p^n Q = O} $ which is isomorphic to a subgroup of $mathbb{Q}_p^2 / mathbb{Z}_p^2$ and $V_p(E_i) = T_p(E_i) otimes_{mathbb{Z}_{p} } mathbb{Q}_{p}$ ? Then it depends if $T_p(E_i) cong deg(f) T_p(E_i) $. Did you try with $K $ a finite field ?
$endgroup$
– reuns
Dec 16 '18 at 20:32
$begingroup$
$T_p(E_i)=lim_n E[p^{n}]$ and $V_p(E_i)=T_p(E_i)otimes_{mathbb{Z}_p}mathbb{Q}_p$. The isomorphism of underlying $mathbb{Z}_p$-modules is automatic. The problem is about Galois representations. We don't need to restrict to $f$ since if $E_1=E_2$ and $f=[p]$, then the isomorphic of Tate module as Galois representation is trivial and has no relationship to the isogeny.
$endgroup$
– wuzx
Dec 16 '18 at 20:34
$begingroup$
Can you put clearly what you think you know about $V_p(E_i)$. ? $E_1/ker(f)$ and $E_2$ are isomorphic as groups, $mathbb{Z}_p$ and Galois modules. Then what ?
$endgroup$
– reuns
Dec 16 '18 at 20:48
3
$begingroup$
The answer is no (i.e. there are counterexamples). Explicit examples can be found in this question.
$endgroup$
– Brandon Carter
Dec 21 '18 at 3:36
$begingroup$
@BrandonCarter Thank you! It is very explicit!
$endgroup$
– wuzx
Dec 22 '18 at 19:51