Mixing
How to prove that $text{End}_{mathbb{F}_p}(E)$ is commutative for a given elliptic curve E?
$begingroup$
Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:
$mathbb{Z}$
An order in an imaginary quadratic field
An order in a quaternion algebra
The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.
Any help on how to approach this will be appreciated, thanks.
number-theory analytic-number-theory elliptic-curves
asked Jan 26 at 16:18
AiYagamiAiYagami
635
$endgroup$
add a comment |
$begingroup$
Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:
$mathbb{Z}$
An order in an imaginary quadratic field
An order in a quaternion algebra
The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.
Any help on how to approach this will be appreciated, thanks.
number-theory analytic-number-theory elliptic-curves
asked Jan 26 at 16:18
AiYagamiAiYagami
635
$endgroup$
$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05
add a comment |
$begingroup$
Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:
$mathbb{Z}$
An order in an imaginary quadratic field
An order in a quaternion algebra
The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.
Any help on how to approach this will be appreciated, thanks.
number-theory analytic-number-theory elliptic-curves
asked Jan 26 at 16:18
AiYagamiAiYagami
635
$endgroup$
Given a prime $p$ and considering the finite field $mathbb{F}_p$, I need to see that $text{End}_{mathbb{F}_p}$(E) is commutative using orders. It is known that $text{End}_{mathbb{F}_p} subseteq text{End}(E)$, and I have seen how $text{End}(E)$ is one of the following:
$mathbb{Z}$
An order in an imaginary quadratic field
An order in a quaternion algebra
The key is, if I show that $text{End}_{mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $text{End}_{mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $text{End}_{mathbb{F}_p}subseteq text{End}(E)$ is strict or not.
Any help on how to approach this will be appreciated, thanks.
number-theory analytic-number-theory elliptic-curves
number-theory analytic-number-theory elliptic-curves
asked Jan 26 at 16:18
AiYagamiAiYagami
635
asked Jan 26 at 16:18
AiYagamiAiYagami
635
edited Jan 26 at 16:24
AiYagami
asked Jan 26 at 16:18
AiYagamiAiYagami
635
asked Jan 26 at 16:18
AiYagamiAiYagami
635
asked Jan 26 at 16:18
AiYagamiAiYagami
635
635
$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05
add a comment |
$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05
$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05
$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05
add a comment |
1 Answer
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$begingroup$
$text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $Bbb Z[F]$ is a quadratic imaginary order.
If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.
If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
$Bbb Q(F)$.
In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
$Bbb Z [F]$.
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
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add a comment |
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1 Answer
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$begingroup$
$text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $Bbb Z[F]$ is a quadratic imaginary order.
If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.
If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
$Bbb Q(F)$.
In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
$Bbb Z [F]$.
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
$text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $Bbb Z[F]$ is a quadratic imaginary order.
If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.
If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
$Bbb Q(F)$.
In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
$Bbb Z [F]$.
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
$text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $Bbb Z[F]$ is a quadratic imaginary order.
If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.
If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
$Bbb Q(F)$.
In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
$Bbb Z [F]$.
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$text{End}_{Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $Bbb Z[F]$ is a quadratic imaginary order.
If $E$ is ordinary, $text{End}(E)$ is a quadratic imaginary order
and so $text{End}_{Bbb F_p}(E)=text{End}(E)$.
If $E$ is supersingular, then $text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$Bbb Q(F)captext{End}(E)$, which form an order in the quadratic field
$Bbb Q(F)$.
In both cases, $text{End}_{Bbb F_p}(E)$ may be an order strictly containing
$Bbb Z [F]$.
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 26 at 16:35
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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$begingroup$
In AEC $deg_i(phi)=p$ and $p=phi^*phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^infty]$ is infinite procyclic $cong mathbb{Z}[p^{-1}]/mathbb{Z}$ and $End(E)$ injects in $End(E[p^infty])$.The harder case $E[p^infty]=O$ is obtained from considering the dual isogeny map $fto f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $phi^{2n}-t_nphi^n +p^n=0$ lets us conclude about the subring commuting with $phi^n$.
$endgroup$
– reuns
Jan 26 at 18:05