Is there an elliptic curve mod n with exactly one point?
up vote
3
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I have tried many elliptic curves $y^2 = x^3 + ax +b$ with no success. I know that for prime modules there exists a minimum number of points the elliptic curve has to have, and I couldn't satisfy this for the smallest primes. So I decided to try luck with modules with few quadratic residues such as 8. But again, no luck.
elliptic-curves
asked Nov 16 at 18:04
SlowerPhoton
408111
|
show 2 more comments
up vote
3
down vote
favorite
I have tried many elliptic curves $y^2 = x^3 + ax +b$ with no success. I know that for prime modules there exists a minimum number of points the elliptic curve has to have, and I couldn't satisfy this for the smallest primes. So I decided to try luck with modules with few quadratic residues such as 8. But again, no luck.
elliptic-curves
asked Nov 16 at 18:04
SlowerPhoton
408111
1
Do you know Hensel's lemma?
– Servaes
Nov 16 at 18:10
4
There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$.
– Lord Shark the Unknown
Nov 16 at 18:14
If $p | n, p^2 nmid n$ then let $e_p equiv 1 bmod p, e_p equiv 0 bmod frac{n}{p}$, the map $(x,y) mapsto (e_p x,e_p y)$ is an embedding $E/(mathbb{Z}/pmathbb{Z}) to E/(mathbb{Z}/nmathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius.
– reuns
Nov 16 at 19:30
@LordSharktheUnknown Why not? Is there a way to write it down then?
– SlowerPhoton
Nov 16 at 20:31
@Servaes No, reading it now.
– SlowerPhoton
Nov 16 at 20:31
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have tried many elliptic curves $y^2 = x^3 + ax +b$ with no success. I know that for prime modules there exists a minimum number of points the elliptic curve has to have, and I couldn't satisfy this for the smallest primes. So I decided to try luck with modules with few quadratic residues such as 8. But again, no luck.
elliptic-curves
asked Nov 16 at 18:04
SlowerPhoton
408111
I have tried many elliptic curves $y^2 = x^3 + ax +b$ with no success. I know that for prime modules there exists a minimum number of points the elliptic curve has to have, and I couldn't satisfy this for the smallest primes. So I decided to try luck with modules with few quadratic residues such as 8. But again, no luck.
elliptic-curves
elliptic-curves
asked Nov 16 at 18:04
SlowerPhoton
408111
asked Nov 16 at 18:04
SlowerPhoton
408111
asked Nov 16 at 18:04
SlowerPhoton
408111
asked Nov 16 at 18:04
SlowerPhoton
408111
asked Nov 16 at 18:04
SlowerPhoton
408111
408111
1
Do you know Hensel's lemma?
– Servaes
Nov 16 at 18:10
4
There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$.
– Lord Shark the Unknown
Nov 16 at 18:14
If $p | n, p^2 nmid n$ then let $e_p equiv 1 bmod p, e_p equiv 0 bmod frac{n}{p}$, the map $(x,y) mapsto (e_p x,e_p y)$ is an embedding $E/(mathbb{Z}/pmathbb{Z}) to E/(mathbb{Z}/nmathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius.
– reuns
Nov 16 at 19:30
@LordSharktheUnknown Why not? Is there a way to write it down then?
– SlowerPhoton
Nov 16 at 20:31
@Servaes No, reading it now.
– SlowerPhoton
Nov 16 at 20:31
|
show 2 more comments
1
Do you know Hensel's lemma?
– Servaes
Nov 16 at 18:10
4
There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$.
– Lord Shark the Unknown
Nov 16 at 18:14
If $p | n, p^2 nmid n$ then let $e_p equiv 1 bmod p, e_p equiv 0 bmod frac{n}{p}$, the map $(x,y) mapsto (e_p x,e_p y)$ is an embedding $E/(mathbb{Z}/pmathbb{Z}) to E/(mathbb{Z}/nmathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius.
– reuns
Nov 16 at 19:30
@LordSharktheUnknown Why not? Is there a way to write it down then?
– SlowerPhoton
Nov 16 at 20:31
@Servaes No, reading it now.
– SlowerPhoton
Nov 16 at 20:31
1
1
Do you know Hensel's lemma?
– Servaes
Nov 16 at 18:10
Do you know Hensel's lemma?
– Servaes
Nov 16 at 18:10
4
4
There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$.
– Lord Shark the Unknown
Nov 16 at 18:14
There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$.
– Lord Shark the Unknown
Nov 16 at 18:14
If $p | n, p^2 nmid n$ then let $e_p equiv 1 bmod p, e_p equiv 0 bmod frac{n}{p}$, the map $(x,y) mapsto (e_p x,e_p y)$ is an embedding $E/(mathbb{Z}/pmathbb{Z}) to E/(mathbb{Z}/nmathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius.
– reuns
Nov 16 at 19:30
If $p | n, p^2 nmid n$ then let $e_p equiv 1 bmod p, e_p equiv 0 bmod frac{n}{p}$, the map $(x,y) mapsto (e_p x,e_p y)$ is an embedding $E/(mathbb{Z}/pmathbb{Z}) to E/(mathbb{Z}/nmathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius.
– reuns
Nov 16 at 19:30
@LordSharktheUnknown Why not? Is there a way to write it down then?
– SlowerPhoton
Nov 16 at 20:31
@LordSharktheUnknown Why not? Is there a way to write it down then?
– SlowerPhoton
Nov 16 at 20:31
@Servaes No, reading it now.
– SlowerPhoton
Nov 16 at 20:31
@Servaes No, reading it now.
– SlowerPhoton
Nov 16 at 20:31
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The answer is: $y^2 + y = x^3 + x +1 pmod{2}$. The only element here is the point in infinity: the neutral element.
answered Nov 18 at 19:32
SlowerPhoton
408111
add a comment |
up vote
2
down vote
You can also construct an answer mod 3 (which is in short weierstrass form) as follows:
For all $x in mathbf Z/3mathbf Z$ we know by Fermat's little theorem that $x^3 = x$, therefore the polynomial $x^3 - x + 2$ always takes the value $2$ on elements of $mathbf Z/3 mathbf Z$, as 2 is not a square in this ring the curve
$$y^2= x^3 - x + 2$$
has no non-infinite points.
answered Nov 22 at 14:08
Alex J Best
1,91711222
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The answer is: $y^2 + y = x^3 + x +1 pmod{2}$. The only element here is the point in infinity: the neutral element.
answered Nov 18 at 19:32
SlowerPhoton
408111
add a comment |
up vote
2
down vote
accepted
The answer is: $y^2 + y = x^3 + x +1 pmod{2}$. The only element here is the point in infinity: the neutral element.
answered Nov 18 at 19:32
SlowerPhoton
408111
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The answer is: $y^2 + y = x^3 + x +1 pmod{2}$. The only element here is the point in infinity: the neutral element.
answered Nov 18 at 19:32
SlowerPhoton
408111
The answer is: $y^2 + y = x^3 + x +1 pmod{2}$. The only element here is the point in infinity: the neutral element.
answered Nov 18 at 19:32
SlowerPhoton
408111
answered Nov 18 at 19:32
SlowerPhoton
408111
answered Nov 18 at 19:32
SlowerPhoton
408111
answered Nov 18 at 19:32
SlowerPhoton
408111
408111
add a comment |
add a comment |
up vote
2
down vote
You can also construct an answer mod 3 (which is in short weierstrass form) as follows:
For all $x in mathbf Z/3mathbf Z$ we know by Fermat's little theorem that $x^3 = x$, therefore the polynomial $x^3 - x + 2$ always takes the value $2$ on elements of $mathbf Z/3 mathbf Z$, as 2 is not a square in this ring the curve
$$y^2= x^3 - x + 2$$
has no non-infinite points.
answered Nov 22 at 14:08
Alex J Best
1,91711222
add a comment |
up vote
2
down vote
You can also construct an answer mod 3 (which is in short weierstrass form) as follows:
For all $x in mathbf Z/3mathbf Z$ we know by Fermat's little theorem that $x^3 = x$, therefore the polynomial $x^3 - x + 2$ always takes the value $2$ on elements of $mathbf Z/3 mathbf Z$, as 2 is not a square in this ring the curve
$$y^2= x^3 - x + 2$$
has no non-infinite points.
answered Nov 22 at 14:08
Alex J Best
1,91711222
add a comment |
up vote
2
down vote
up vote
2
down vote
You can also construct an answer mod 3 (which is in short weierstrass form) as follows:
For all $x in mathbf Z/3mathbf Z$ we know by Fermat's little theorem that $x^3 = x$, therefore the polynomial $x^3 - x + 2$ always takes the value $2$ on elements of $mathbf Z/3 mathbf Z$, as 2 is not a square in this ring the curve
$$y^2= x^3 - x + 2$$
has no non-infinite points.
answered Nov 22 at 14:08
Alex J Best
1,91711222
You can also construct an answer mod 3 (which is in short weierstrass form) as follows:
For all $x in mathbf Z/3mathbf Z$ we know by Fermat's little theorem that $x^3 = x$, therefore the polynomial $x^3 - x + 2$ always takes the value $2$ on elements of $mathbf Z/3 mathbf Z$, as 2 is not a square in this ring the curve
$$y^2= x^3 - x + 2$$
has no non-infinite points.
answered Nov 22 at 14:08
Alex J Best
1,91711222
answered Nov 22 at 14:08
Alex J Best
1,91711222
answered Nov 22 at 14:08
Alex J Best
1,91711222
answered Nov 22 at 14:08
Alex J Best
1,91711222
1,91711222
add a comment |
add a comment |
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1
Do you know Hensel's lemma?
– Servaes
Nov 16 at 18:10
4
There is an elliptic curve modulo $2$ with just one point defined over the integers modulo $2$. Being of characteristic two, one cannot write its equation as $y^2=x^3+ax+b$.
– Lord Shark the Unknown
Nov 16 at 18:14
If $p | n, p^2 nmid n$ then let $e_p equiv 1 bmod p, e_p equiv 0 bmod frac{n}{p}$, the map $(x,y) mapsto (e_p x,e_p y)$ is an embedding $E/(mathbb{Z}/pmathbb{Z}) to E/(mathbb{Z}/nmathbb{Z})$, the former being well-understood in term of the characteristic polynomial of the Frobenius.
– reuns
Nov 16 at 19:30
@LordSharktheUnknown Why not? Is there a way to write it down then?
– SlowerPhoton
Nov 16 at 20:31
@Servaes No, reading it now.
– SlowerPhoton
Nov 16 at 20:31