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Unicity of subgroup-scheme of a supersingular elliptic curve of a given rank
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Let $E/k$ be a supersingular elliptic curve over an algebraically closed field $k$ of characteristic $p>0$. Then for any positive integer $r$, there is a unique closed subgroup-scheme of $E$ of rank $p^r$ (which can be realized as the kernel of the $r$-fold Froebenius morphism $F^r:Erightarrow E^(r)$).
This statement is made in some point in Katz and Mazur's book. The unicity is important as it allows us to deduce that $operatornameKer(F^2r)=operatornameKer(p^r)$. However, I totally fail to see why it is true. Actually, I fail to relate this to the hypothesis that $E$ is supersingular, as the only definition of this that is given in the book is the fact that the formal completion $widehatE$ of $E$ along its zero section has height $2$.
Could someone please provide a reference or an explanation for this result?
I thank you very much for your help.
EDIT:
I found a proof of this statement in the book "Fermat's Last Theorem: the Proof" by Takeshi Saito. Here, a supersingular elliptic curve is defined as an elliptic curve $E/S$ where $S$ is a scheme over $mathbbF_p$ such that $E[p]=operatornameKer(F^2)$, where $F^2:Erightarrow E^(p^2)$ is the Froebenius morphism (composed twice).
Given a supersingular elliptic curve $E$ over an algebraically closed field $k$ of characteristic $p$, we have that $E[p](k)=left(operatornamekerF^2right)(k)=0$ (as it can be seen, for instance, by considering a description of $E$ by a Weierstraß equation). Hence, the abelian group $ E[p](k)$ is of order $1$.
Then, he writes down the following:
Let $G$ be a closed subgroup scheme of $E$ of order $p^e$. It is then a closed subgroup scheme of $E[p^e]$ of degree $p^e$, thus it is connected. Therefore if $mathfrakm_0$ is the maximal ideam of the local ring $mathcalO_E,0$, we must have $mathbfG=operatornameSpec(mathcalO_E,0/mathfrakm_0^p^e)=operatornameKer(F^e)$.
I am having a bad time understand pretty much every step in this last proof, however... More precisely, I do not understand the parts written in bold letters. Could someone please explain these parts to me?
algebraic-geometry elliptic-curves
asked Aug 6 at 7:28
Suzet
2,216527
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up vote
2
down vote
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Let $E/k$ be a supersingular elliptic curve over an algebraically closed field $k$ of characteristic $p>0$. Then for any positive integer $r$, there is a unique closed subgroup-scheme of $E$ of rank $p^r$ (which can be realized as the kernel of the $r$-fold Froebenius morphism $F^r:Erightarrow E^(r)$).
This statement is made in some point in Katz and Mazur's book. The unicity is important as it allows us to deduce that $operatornameKer(F^2r)=operatornameKer(p^r)$. However, I totally fail to see why it is true. Actually, I fail to relate this to the hypothesis that $E$ is supersingular, as the only definition of this that is given in the book is the fact that the formal completion $widehatE$ of $E$ along its zero section has height $2$.
Could someone please provide a reference or an explanation for this result?
I thank you very much for your help.
EDIT:
I found a proof of this statement in the book "Fermat's Last Theorem: the Proof" by Takeshi Saito. Here, a supersingular elliptic curve is defined as an elliptic curve $E/S$ where $S$ is a scheme over $mathbbF_p$ such that $E[p]=operatornameKer(F^2)$, where $F^2:Erightarrow E^(p^2)$ is the Froebenius morphism (composed twice).
Given a supersingular elliptic curve $E$ over an algebraically closed field $k$ of characteristic $p$, we have that $E[p](k)=left(operatornamekerF^2right)(k)=0$ (as it can be seen, for instance, by considering a description of $E$ by a Weierstraß equation). Hence, the abelian group $ E[p](k)$ is of order $1$.
Then, he writes down the following:
Let $G$ be a closed subgroup scheme of $E$ of order $p^e$. It is then a closed subgroup scheme of $E[p^e]$ of degree $p^e$, thus it is connected. Therefore if $mathfrakm_0$ is the maximal ideam of the local ring $mathcalO_E,0$, we must have $mathbfG=operatornameSpec(mathcalO_E,0/mathfrakm_0^p^e)=operatornameKer(F^e)$.
I am having a bad time understand pretty much every step in this last proof, however... More precisely, I do not understand the parts written in bold letters. Could someone please explain these parts to me?
algebraic-geometry elliptic-curves
asked Aug 6 at 7:28
Suzet
2,216527
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $E/k$ be a supersingular elliptic curve over an algebraically closed field $k$ of characteristic $p>0$. Then for any positive integer $r$, there is a unique closed subgroup-scheme of $E$ of rank $p^r$ (which can be realized as the kernel of the $r$-fold Froebenius morphism $F^r:Erightarrow E^(r)$).
This statement is made in some point in Katz and Mazur's book. The unicity is important as it allows us to deduce that $operatornameKer(F^2r)=operatornameKer(p^r)$. However, I totally fail to see why it is true. Actually, I fail to relate this to the hypothesis that $E$ is supersingular, as the only definition of this that is given in the book is the fact that the formal completion $widehatE$ of $E$ along its zero section has height $2$.
Could someone please provide a reference or an explanation for this result?
I thank you very much for your help.
EDIT:
I found a proof of this statement in the book "Fermat's Last Theorem: the Proof" by Takeshi Saito. Here, a supersingular elliptic curve is defined as an elliptic curve $E/S$ where $S$ is a scheme over $mathbbF_p$ such that $E[p]=operatornameKer(F^2)$, where $F^2:Erightarrow E^(p^2)$ is the Froebenius morphism (composed twice).
Given a supersingular elliptic curve $E$ over an algebraically closed field $k$ of characteristic $p$, we have that $E[p](k)=left(operatornamekerF^2right)(k)=0$ (as it can be seen, for instance, by considering a description of $E$ by a Weierstraß equation). Hence, the abelian group $ E[p](k)$ is of order $1$.
Then, he writes down the following:
Let $G$ be a closed subgroup scheme of $E$ of order $p^e$. It is then a closed subgroup scheme of $E[p^e]$ of degree $p^e$, thus it is connected. Therefore if $mathfrakm_0$ is the maximal ideam of the local ring $mathcalO_E,0$, we must have $mathbfG=operatornameSpec(mathcalO_E,0/mathfrakm_0^p^e)=operatornameKer(F^e)$.
I am having a bad time understand pretty much every step in this last proof, however... More precisely, I do not understand the parts written in bold letters. Could someone please explain these parts to me?
algebraic-geometry elliptic-curves
asked Aug 6 at 7:28
Suzet
2,216527
Let $E/k$ be a supersingular elliptic curve over an algebraically closed field $k$ of characteristic $p>0$. Then for any positive integer $r$, there is a unique closed subgroup-scheme of $E$ of rank $p^r$ (which can be realized as the kernel of the $r$-fold Froebenius morphism $F^r:Erightarrow E^(r)$).
This statement is made in some point in Katz and Mazur's book. The unicity is important as it allows us to deduce that $operatornameKer(F^2r)=operatornameKer(p^r)$. However, I totally fail to see why it is true. Actually, I fail to relate this to the hypothesis that $E$ is supersingular, as the only definition of this that is given in the book is the fact that the formal completion $widehatE$ of $E$ along its zero section has height $2$.
Could someone please provide a reference or an explanation for this result?
I thank you very much for your help.
EDIT:
I found a proof of this statement in the book "Fermat's Last Theorem: the Proof" by Takeshi Saito. Here, a supersingular elliptic curve is defined as an elliptic curve $E/S$ where $S$ is a scheme over $mathbbF_p$ such that $E[p]=operatornameKer(F^2)$, where $F^2:Erightarrow E^(p^2)$ is the Froebenius morphism (composed twice).
Given a supersingular elliptic curve $E$ over an algebraically closed field $k$ of characteristic $p$, we have that $E[p](k)=left(operatornamekerF^2right)(k)=0$ (as it can be seen, for instance, by considering a description of $E$ by a Weierstraß equation). Hence, the abelian group $ E[p](k)$ is of order $1$.
Then, he writes down the following:
Let $G$ be a closed subgroup scheme of $E$ of order $p^e$. It is then a closed subgroup scheme of $E[p^e]$ of degree $p^e$, thus it is connected. Therefore if $mathfrakm_0$ is the maximal ideam of the local ring $mathcalO_E,0$, we must have $mathbfG=operatornameSpec(mathcalO_E,0/mathfrakm_0^p^e)=operatornameKer(F^e)$.
I am having a bad time understand pretty much every step in this last proof, however... More precisely, I do not understand the parts written in bold letters. Could someone please explain these parts to me?
algebraic-geometry elliptic-curves
asked Aug 6 at 7:28
Suzet
2,216527
edited Aug 7 at 0:54
asked Aug 6 at 7:28
Suzet
2,216527
asked Aug 6 at 7:28
Suzet
2,216527
asked Aug 6 at 7:28
Suzet
2,216527
2,216527
add a comment |Â
add a comment |Â
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