Mixing
Commutativity of $End(E)$ for $E$ elliptic curve
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If I consider an elliptic curve $E$ over a field of characteristic zero, is $End(E)$ always a commutative ring? If not, can someone give a counterexample?
algebraic-number-theory elliptic-curves
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
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add a comment |
$begingroup$
If I consider an elliptic curve $E$ over a field of characteristic zero, is $End(E)$ always a commutative ring? If not, can someone give a counterexample?
algebraic-number-theory elliptic-curves
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
$endgroup$
1
$begingroup$
To answer the original question, $mathrm{End}(E)$ is always commutative over a field of characteristic zero.
$endgroup$
– Brandon Carter
Feb 2 at 7:53
add a comment |
$begingroup$
If I consider an elliptic curve $E$ over a field of characteristic zero, is $End(E)$ always a commutative ring? If not, can someone give a counterexample?
algebraic-number-theory elliptic-curves
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
$endgroup$
If I consider an elliptic curve $E$ over a field of characteristic zero, is $End(E)$ always a commutative ring? If not, can someone give a counterexample?
algebraic-number-theory elliptic-curves
algebraic-number-theory elliptic-curves
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
asked Feb 1 at 15:35
Lei FeimaLei Feima
867
867
1
$begingroup$
To answer the original question, $mathrm{End}(E)$ is always commutative over a field of characteristic zero.
$endgroup$
– Brandon Carter
Feb 2 at 7:53
add a comment |
1
$begingroup$
To answer the original question, $mathrm{End}(E)$ is always commutative over a field of characteristic zero.
$endgroup$
– Brandon Carter
Feb 2 at 7:53
1
1
$begingroup$
To answer the original question, $mathrm{End}(E)$ is always commutative over a field of characteristic zero.
$endgroup$
– Brandon Carter
Feb 2 at 7:53
$begingroup$
To answer the original question, $mathrm{End}(E)$ is always commutative over a field of characteristic zero.
$endgroup$
– Brandon Carter
Feb 2 at 7:53
add a comment |
2 Answers
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It is classically known that the endomorphism ring of an elliptic curve $E/K$ is either $mathbf Z$, an order in an imaginary quadratic field, or an order in a quaternion algebra; see e.g. Silverman, " The Arithmetic of Elliptic Curves", III, 9. A complete description of $End(E)$ can be found in Deuring, "Abh. Math. Sem. Hamburg, 14, 1941. If char$(K)=0$, one says that $E$ has complex multiplication if $End(E) neq mathbf Z$; note that in characteristic $0$, then $End(E)otimes mathbf Q$ cannot be a quaternion algebra (the proof is analytic). In characteristic $pneq 0$, there is a surprising connection between $End(E)$ and the subgroup $E[p]$ of $p$-torsion (over $bar K$): if $K$ is perfect, $End(E)$ is an order in a quaternion algebra iff $E[p]=0$; $E$ is then called supersingular. If $K$ is a finite field, then $End(E) neq mathbf Z$, and there exist $E/bar K$ with non commutative $End(E)$; more precisely, for $p$ odd, there are simple criteria which allow to determine the supersingular $E$ up to isomorphism.
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
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$begingroup$
Let $E$ be a supersingular elliptic curve (over a field of positive characteristic). Then its endomorphism algebra is the (unique) quaternion algebra over $Bbb Q$ ramified at $p$ and at $infty$. So the endomorphism algebra need not be commutative in general.
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
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2 Answers
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2 Answers
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$begingroup$
It is classically known that the endomorphism ring of an elliptic curve $E/K$ is either $mathbf Z$, an order in an imaginary quadratic field, or an order in a quaternion algebra; see e.g. Silverman, " The Arithmetic of Elliptic Curves", III, 9. A complete description of $End(E)$ can be found in Deuring, "Abh. Math. Sem. Hamburg, 14, 1941. If char$(K)=0$, one says that $E$ has complex multiplication if $End(E) neq mathbf Z$; note that in characteristic $0$, then $End(E)otimes mathbf Q$ cannot be a quaternion algebra (the proof is analytic). In characteristic $pneq 0$, there is a surprising connection between $End(E)$ and the subgroup $E[p]$ of $p$-torsion (over $bar K$): if $K$ is perfect, $End(E)$ is an order in a quaternion algebra iff $E[p]=0$; $E$ is then called supersingular. If $K$ is a finite field, then $End(E) neq mathbf Z$, and there exist $E/bar K$ with non commutative $End(E)$; more precisely, for $p$ odd, there are simple criteria which allow to determine the supersingular $E$ up to isomorphism.
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
$endgroup$
add a comment |
$begingroup$
It is classically known that the endomorphism ring of an elliptic curve $E/K$ is either $mathbf Z$, an order in an imaginary quadratic field, or an order in a quaternion algebra; see e.g. Silverman, " The Arithmetic of Elliptic Curves", III, 9. A complete description of $End(E)$ can be found in Deuring, "Abh. Math. Sem. Hamburg, 14, 1941. If char$(K)=0$, one says that $E$ has complex multiplication if $End(E) neq mathbf Z$; note that in characteristic $0$, then $End(E)otimes mathbf Q$ cannot be a quaternion algebra (the proof is analytic). In characteristic $pneq 0$, there is a surprising connection between $End(E)$ and the subgroup $E[p]$ of $p$-torsion (over $bar K$): if $K$ is perfect, $End(E)$ is an order in a quaternion algebra iff $E[p]=0$; $E$ is then called supersingular. If $K$ is a finite field, then $End(E) neq mathbf Z$, and there exist $E/bar K$ with non commutative $End(E)$; more precisely, for $p$ odd, there are simple criteria which allow to determine the supersingular $E$ up to isomorphism.
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
$endgroup$
add a comment |
$begingroup$
It is classically known that the endomorphism ring of an elliptic curve $E/K$ is either $mathbf Z$, an order in an imaginary quadratic field, or an order in a quaternion algebra; see e.g. Silverman, " The Arithmetic of Elliptic Curves", III, 9. A complete description of $End(E)$ can be found in Deuring, "Abh. Math. Sem. Hamburg, 14, 1941. If char$(K)=0$, one says that $E$ has complex multiplication if $End(E) neq mathbf Z$; note that in characteristic $0$, then $End(E)otimes mathbf Q$ cannot be a quaternion algebra (the proof is analytic). In characteristic $pneq 0$, there is a surprising connection between $End(E)$ and the subgroup $E[p]$ of $p$-torsion (over $bar K$): if $K$ is perfect, $End(E)$ is an order in a quaternion algebra iff $E[p]=0$; $E$ is then called supersingular. If $K$ is a finite field, then $End(E) neq mathbf Z$, and there exist $E/bar K$ with non commutative $End(E)$; more precisely, for $p$ odd, there are simple criteria which allow to determine the supersingular $E$ up to isomorphism.
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
$endgroup$
It is classically known that the endomorphism ring of an elliptic curve $E/K$ is either $mathbf Z$, an order in an imaginary quadratic field, or an order in a quaternion algebra; see e.g. Silverman, " The Arithmetic of Elliptic Curves", III, 9. A complete description of $End(E)$ can be found in Deuring, "Abh. Math. Sem. Hamburg, 14, 1941. If char$(K)=0$, one says that $E$ has complex multiplication if $End(E) neq mathbf Z$; note that in characteristic $0$, then $End(E)otimes mathbf Q$ cannot be a quaternion algebra (the proof is analytic). In characteristic $pneq 0$, there is a surprising connection between $End(E)$ and the subgroup $E[p]$ of $p$-torsion (over $bar K$): if $K$ is perfect, $End(E)$ is an order in a quaternion algebra iff $E[p]=0$; $E$ is then called supersingular. If $K$ is a finite field, then $End(E) neq mathbf Z$, and there exist $E/bar K$ with non commutative $End(E)$; more precisely, for $p$ odd, there are simple criteria which allow to determine the supersingular $E$ up to isomorphism.
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
answered Feb 2 at 19:06
nguyen quang donguyen quang do
9,0691724
9,0691724
add a comment |
add a comment |
$begingroup$
Let $E$ be a supersingular elliptic curve (over a field of positive characteristic). Then its endomorphism algebra is the (unique) quaternion algebra over $Bbb Q$ ramified at $p$ and at $infty$. So the endomorphism algebra need not be commutative in general.
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
add a comment |
$begingroup$
Let $E$ be a supersingular elliptic curve (over a field of positive characteristic). Then its endomorphism algebra is the (unique) quaternion algebra over $Bbb Q$ ramified at $p$ and at $infty$. So the endomorphism algebra need not be commutative in general.
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
add a comment |
$begingroup$
Let $E$ be a supersingular elliptic curve (over a field of positive characteristic). Then its endomorphism algebra is the (unique) quaternion algebra over $Bbb Q$ ramified at $p$ and at $infty$. So the endomorphism algebra need not be commutative in general.
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
Let $E$ be a supersingular elliptic curve (over a field of positive characteristic). Then its endomorphism algebra is the (unique) quaternion algebra over $Bbb Q$ ramified at $p$ and at $infty$. So the endomorphism algebra need not be commutative in general.
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
answered Feb 1 at 15:55
Dietrich BurdeDietrich Burde
82k649107
82k649107
add a comment |
add a comment |
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To answer the original question, $mathrm{End}(E)$ is always commutative over a field of characteristic zero.
$endgroup$
– Brandon Carter
Feb 2 at 7:53