Elliptic curve with only one point
$begingroup$
Is there an elliptic curve $E$ over an infinite field $K$ such that $E(K)={infty}$?
My original task was to find an elliptic curve over some field $K$ with only one point, which I did for $K=mathbb{F}_2.$ Now, I'm curious about the case of infinite cardinality, which I am not able to handle.
field-theory elliptic-curves
asked Nov 10 '18 at 13:59
byk7byk7
326110
$endgroup$
add a comment |
$begingroup$
Is there an elliptic curve $E$ over an infinite field $K$ such that $E(K)={infty}$?
My original task was to find an elliptic curve over some field $K$ with only one point, which I did for $K=mathbb{F}_2.$ Now, I'm curious about the case of infinite cardinality, which I am not able to handle.
field-theory elliptic-curves
asked Nov 10 '18 at 13:59
byk7byk7
326110
$endgroup$
add a comment |
$begingroup$
Is there an elliptic curve $E$ over an infinite field $K$ such that $E(K)={infty}$?
My original task was to find an elliptic curve over some field $K$ with only one point, which I did for $K=mathbb{F}_2.$ Now, I'm curious about the case of infinite cardinality, which I am not able to handle.
field-theory elliptic-curves
asked Nov 10 '18 at 13:59
byk7byk7
326110
$endgroup$
Is there an elliptic curve $E$ over an infinite field $K$ such that $E(K)={infty}$?
My original task was to find an elliptic curve over some field $K$ with only one point, which I did for $K=mathbb{F}_2.$ Now, I'm curious about the case of infinite cardinality, which I am not able to handle.
field-theory elliptic-curves
field-theory elliptic-curves
asked Nov 10 '18 at 13:59
byk7byk7
326110
asked Nov 10 '18 at 13:59
byk7byk7
326110
asked Nov 10 '18 at 13:59
byk7byk7
326110
asked Nov 10 '18 at 13:59
byk7byk7
326110
asked Nov 10 '18 at 13:59
byk7byk7
326110
326110
add a comment |
add a comment |
1 Answer
1
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$begingroup$
According to this database,
the elliptic curve
$$E : y^2z=x^3−108z^3$$
has only $[0:1:0]$ as rational point, i.e. $E(Bbb Q)$ is the trivial group.
Further examples are given here.
(Notice that if $K$ is an algebraically closed field (hence infinite), then $E(K)$ is always infinite, since it contains $(Bbb Z/n Bbb Z)^2$ for every $n geq 1$, coprime to $mathrm{char}(K)$ if $K$ has positive characteristic.)
I may add the following related and interesting result, by Mazur and Rubin (theorem 1.1 here):
if $K$ is a number field, then there is an elliptic curve $E$ over $K$ with $E(K) = {0}$.
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
$endgroup$
1
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
3
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
According to this database,
the elliptic curve
$$E : y^2z=x^3−108z^3$$
has only $[0:1:0]$ as rational point, i.e. $E(Bbb Q)$ is the trivial group.
Further examples are given here.
(Notice that if $K$ is an algebraically closed field (hence infinite), then $E(K)$ is always infinite, since it contains $(Bbb Z/n Bbb Z)^2$ for every $n geq 1$, coprime to $mathrm{char}(K)$ if $K$ has positive characteristic.)
I may add the following related and interesting result, by Mazur and Rubin (theorem 1.1 here):
if $K$ is a number field, then there is an elliptic curve $E$ over $K$ with $E(K) = {0}$.
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
$endgroup$
1
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
3
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
add a comment |
$begingroup$
According to this database,
the elliptic curve
$$E : y^2z=x^3−108z^3$$
has only $[0:1:0]$ as rational point, i.e. $E(Bbb Q)$ is the trivial group.
Further examples are given here.
(Notice that if $K$ is an algebraically closed field (hence infinite), then $E(K)$ is always infinite, since it contains $(Bbb Z/n Bbb Z)^2$ for every $n geq 1$, coprime to $mathrm{char}(K)$ if $K$ has positive characteristic.)
I may add the following related and interesting result, by Mazur and Rubin (theorem 1.1 here):
if $K$ is a number field, then there is an elliptic curve $E$ over $K$ with $E(K) = {0}$.
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
$endgroup$
1
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
3
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
add a comment |
$begingroup$
According to this database,
the elliptic curve
$$E : y^2z=x^3−108z^3$$
has only $[0:1:0]$ as rational point, i.e. $E(Bbb Q)$ is the trivial group.
Further examples are given here.
(Notice that if $K$ is an algebraically closed field (hence infinite), then $E(K)$ is always infinite, since it contains $(Bbb Z/n Bbb Z)^2$ for every $n geq 1$, coprime to $mathrm{char}(K)$ if $K$ has positive characteristic.)
I may add the following related and interesting result, by Mazur and Rubin (theorem 1.1 here):
if $K$ is a number field, then there is an elliptic curve $E$ over $K$ with $E(K) = {0}$.
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
$endgroup$
According to this database,
the elliptic curve
$$E : y^2z=x^3−108z^3$$
has only $[0:1:0]$ as rational point, i.e. $E(Bbb Q)$ is the trivial group.
Further examples are given here.
(Notice that if $K$ is an algebraically closed field (hence infinite), then $E(K)$ is always infinite, since it contains $(Bbb Z/n Bbb Z)^2$ for every $n geq 1$, coprime to $mathrm{char}(K)$ if $K$ has positive characteristic.)
I may add the following related and interesting result, by Mazur and Rubin (theorem 1.1 here):
if $K$ is a number field, then there is an elliptic curve $E$ over $K$ with $E(K) = {0}$.
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
edited Nov 30 '18 at 10:28
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
answered Nov 10 '18 at 14:22
WatsonWatson
15.9k92970
15.9k92970
1
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
3
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
add a comment |
1
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
3
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
1
1
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
$begingroup$
$y^2=x^3-5$ seems to work as well
$endgroup$
– byk7
Nov 10 '18 at 15:08
3
3
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
$begingroup$
If $K$ has characteristic $p$, the $p$-torsion of $E(K)$ is never $(mathbf Z/pmathbf Z)^2$ even when $K$ is algebraically closed.
$endgroup$
– KCd
Nov 10 '18 at 16:09
add a comment |
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