Does this degree 12 genus 1 curve have only one point over infinitely many finite fields?How do the number of plane curves over a finite field of a fixed genus increase with the degree?Does a curve over a number field have a finite etale cover of given degreeFor how many primes does an elliptic curve over a totally imaginary field have supersingular reduction?Argument for unboundedness of integral points of elliptic curves over number fieldsIs it possible on an elliptic curve both $x,y$ to be arbitrary large powers infinitely often?Hyperelliptic curve of genus 2 over RWeierstrass form of genus one $y^10 z^30 - 8000 y^4 z^20 + 12800000 z^20 + 1600 y^2 z^10 - 64=0$What is the exact statement about uniform boundedness of rational points on curves of genus greater than one? Singular points can be unboundedConjecture: The number of points modulo $p$ of certain elliptic curve is $p$ or $p+2$ for $p$ of form $p=27a^2+27a+7$Faltings theorem and number of singularities
Does this degree 12 genus 1 curve have only one point over infinitely many finite fields?
How do the number of plane curves over a finite field of a fixed genus increase with the degree?Does a curve over a number field have a finite etale cover of given degreeFor how many primes does an elliptic curve over a totally imaginary field have supersingular reduction?Argument for unboundedness of integral points of elliptic curves over number fieldsIs it possible on an elliptic curve both $x,y$ to be arbitrary large powers infinitely often?Hyperelliptic curve of genus 2 over RWeierstrass form of genus one $y^10 z^30 - 8000 y^4 z^20 + 12800000 z^20 + 1600 y^2 z^10 - 64=0$What is the exact statement about uniform boundedness of rational points on curves of genus greater than one? Singular points can be unboundedConjecture: The number of points modulo $p$ of certain elliptic curve is $p$ or $p+2$ for $p$ of form $p=27a^2+27a+7$Faltings theorem and number of singularities
$begingroup$
Let $F(x,y,z)$ be the degree 12 homogeneous polynomial:
$$x^12 - x^9 y^3 + x^6 y^6 - x^3 y^9 + y^12 - 4 x^9 z^3 + 3 x^6 y^3 z^3 - 2 x^3 y^6 z^3 + y^9 z^3 + 6 x^6 z^6 - 3 x^3 y^3 z^6 + y^6 z^6 - 4 x^3 z^9 + y^3 z^9 + z^12$$
Over the rationals it is irreducible and $F=0$ is genus 1 curve.
Numerical evidence in Sagemath and Magma suggests that for infinitely
many primes $p$, the curve $F=0$ is irreducible over $mathbbF_p$ and $F=0$
has only one point over $mathbbF_p$, the singular point $(1 : 0 : 1)$.
Q1 Is this true?
Set $p=50033$. Then we have only one point over the finite field and the curve is irreducible of genus 1.
This appears to violate the bound on number of rational points over
finite fields given in the paper "The number of points on an algebraic curve over a finite field", J.W.P. Hirschfeld, G. Korchmáros and F. Torres ,p. 23.
Q2 What hypothesis am I missing for this violation?
Sagemath code:
def tesgfppoints2():
L1=5*10^4
L2=2*L1
for p in primes(L1,L2):
K.<x,y,z>=GF(p)[]
F=x^12 - x^9*y^3 + x^6*y^6 - x^3*y^9 + y^12 - 4*x^9*z^3 + 3*x^6*y^3*z^3 - 2*x^3*y^6*z^3 + y^9*z^3 + 6*x^6*z^6 - 3*x^3*y^3*z^6 + y^6*z^6 - 4*x^3*z^9 + y^3*z^9 + z^12
C=Curve(F)
ire=C.is_irreducible()
if not ire: continue
rp=len(C.rational_points())
print 'p=',p,';rp=',rp,'ir=',ire,'g=',C.genus()
ag.algebraic-geometry elliptic-curves finite-fields
edited May 27 at 10:14
Glorfindel
1,51641222
asked May 26 at 11:55
jorojoro
14.5k84794
$endgroup$
add a comment |
$begingroup$
Let $F(x,y,z)$ be the degree 12 homogeneous polynomial:
$$x^12 - x^9 y^3 + x^6 y^6 - x^3 y^9 + y^12 - 4 x^9 z^3 + 3 x^6 y^3 z^3 - 2 x^3 y^6 z^3 + y^9 z^3 + 6 x^6 z^6 - 3 x^3 y^3 z^6 + y^6 z^6 - 4 x^3 z^9 + y^3 z^9 + z^12$$
Over the rationals it is irreducible and $F=0$ is genus 1 curve.
Numerical evidence in Sagemath and Magma suggests that for infinitely
many primes $p$, the curve $F=0$ is irreducible over $mathbbF_p$ and $F=0$
has only one point over $mathbbF_p$, the singular point $(1 : 0 : 1)$.
Q1 Is this true?
Set $p=50033$. Then we have only one point over the finite field and the curve is irreducible of genus 1.
This appears to violate the bound on number of rational points over
finite fields given in the paper "The number of points on an algebraic curve over a finite field", J.W.P. Hirschfeld, G. Korchmáros and F. Torres ,p. 23.
Q2 What hypothesis am I missing for this violation?
Sagemath code:
def tesgfppoints2():
L1=5*10^4
L2=2*L1
for p in primes(L1,L2):
K.<x,y,z>=GF(p)[]
F=x^12 - x^9*y^3 + x^6*y^6 - x^3*y^9 + y^12 - 4*x^9*z^3 + 3*x^6*y^3*z^3 - 2*x^3*y^6*z^3 + y^9*z^3 + 6*x^6*z^6 - 3*x^3*y^3*z^6 + y^6*z^6 - 4*x^3*z^9 + y^3*z^9 + z^12
C=Curve(F)
ire=C.is_irreducible()
if not ire: continue
rp=len(C.rational_points())
print 'p=',p,';rp=',rp,'ir=',ire,'g=',C.genus()
ag.algebraic-geometry elliptic-curves finite-fields
edited May 27 at 10:14
Glorfindel
1,51641222
asked May 26 at 11:55
jorojoro
14.5k84794
$endgroup$
1
$begingroup$
You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find.
$endgroup$
– Gerry Myerson
May 26 at 12:36
$begingroup$
@GerryMyerson Thanks. I edited with link and authors.
$endgroup$
– joro
May 26 at 13:07
6
$begingroup$
Perhaps the curve is not geometrically irreducible over the rationals.
$endgroup$
– ulrich
May 26 at 13:30
$begingroup$
@ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:30
add a comment |
$begingroup$
Let $F(x,y,z)$ be the degree 12 homogeneous polynomial:
$$x^12 - x^9 y^3 + x^6 y^6 - x^3 y^9 + y^12 - 4 x^9 z^3 + 3 x^6 y^3 z^3 - 2 x^3 y^6 z^3 + y^9 z^3 + 6 x^6 z^6 - 3 x^3 y^3 z^6 + y^6 z^6 - 4 x^3 z^9 + y^3 z^9 + z^12$$
Over the rationals it is irreducible and $F=0$ is genus 1 curve.
Numerical evidence in Sagemath and Magma suggests that for infinitely
many primes $p$, the curve $F=0$ is irreducible over $mathbbF_p$ and $F=0$
has only one point over $mathbbF_p$, the singular point $(1 : 0 : 1)$.
Q1 Is this true?
Set $p=50033$. Then we have only one point over the finite field and the curve is irreducible of genus 1.
This appears to violate the bound on number of rational points over
finite fields given in the paper "The number of points on an algebraic curve over a finite field", J.W.P. Hirschfeld, G. Korchmáros and F. Torres ,p. 23.
Q2 What hypothesis am I missing for this violation?
Sagemath code:
def tesgfppoints2():
L1=5*10^4
L2=2*L1
for p in primes(L1,L2):
K.<x,y,z>=GF(p)[]
F=x^12 - x^9*y^3 + x^6*y^6 - x^3*y^9 + y^12 - 4*x^9*z^3 + 3*x^6*y^3*z^3 - 2*x^3*y^6*z^3 + y^9*z^3 + 6*x^6*z^6 - 3*x^3*y^3*z^6 + y^6*z^6 - 4*x^3*z^9 + y^3*z^9 + z^12
C=Curve(F)
ire=C.is_irreducible()
if not ire: continue
rp=len(C.rational_points())
print 'p=',p,';rp=',rp,'ir=',ire,'g=',C.genus()
ag.algebraic-geometry elliptic-curves finite-fields
edited May 27 at 10:14
Glorfindel
1,51641222
asked May 26 at 11:55
jorojoro
14.5k84794
$endgroup$
Let $F(x,y,z)$ be the degree 12 homogeneous polynomial:
$$x^12 - x^9 y^3 + x^6 y^6 - x^3 y^9 + y^12 - 4 x^9 z^3 + 3 x^6 y^3 z^3 - 2 x^3 y^6 z^3 + y^9 z^3 + 6 x^6 z^6 - 3 x^3 y^3 z^6 + y^6 z^6 - 4 x^3 z^9 + y^3 z^9 + z^12$$
Over the rationals it is irreducible and $F=0$ is genus 1 curve.
Numerical evidence in Sagemath and Magma suggests that for infinitely
many primes $p$, the curve $F=0$ is irreducible over $mathbbF_p$ and $F=0$
has only one point over $mathbbF_p$, the singular point $(1 : 0 : 1)$.
Q1 Is this true?
Set $p=50033$. Then we have only one point over the finite field and the curve is irreducible of genus 1.
This appears to violate the bound on number of rational points over
finite fields given in the paper "The number of points on an algebraic curve over a finite field", J.W.P. Hirschfeld, G. Korchmáros and F. Torres ,p. 23.
Q2 What hypothesis am I missing for this violation?
Sagemath code:
def tesgfppoints2():
L1=5*10^4
L2=2*L1
for p in primes(L1,L2):
K.<x,y,z>=GF(p)[]
F=x^12 - x^9*y^3 + x^6*y^6 - x^3*y^9 + y^12 - 4*x^9*z^3 + 3*x^6*y^3*z^3 - 2*x^3*y^6*z^3 + y^9*z^3 + 6*x^6*z^6 - 3*x^3*y^3*z^6 + y^6*z^6 - 4*x^3*z^9 + y^3*z^9 + z^12
C=Curve(F)
ire=C.is_irreducible()
if not ire: continue
rp=len(C.rational_points())
print 'p=',p,';rp=',rp,'ir=',ire,'g=',C.genus()
ag.algebraic-geometry elliptic-curves finite-fields
ag.algebraic-geometry elliptic-curves finite-fields
edited May 27 at 10:14
Glorfindel
1,51641222
asked May 26 at 11:55
jorojoro
14.5k84794
edited May 27 at 10:14
Glorfindel
1,51641222
asked May 26 at 11:55
jorojoro
14.5k84794
edited May 27 at 10:14
Glorfindel
1,51641222
edited May 27 at 10:14
Glorfindel
1,51641222
edited May 27 at 10:14
Glorfindel
1,51641222
1,51641222
asked May 26 at 11:55
jorojoro
14.5k84794
asked May 26 at 11:55
jorojoro
14.5k84794
asked May 26 at 11:55
jorojoro
14.5k84794
14.5k84794
1
$begingroup$
You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find.
$endgroup$
– Gerry Myerson
May 26 at 12:36
$begingroup$
@GerryMyerson Thanks. I edited with link and authors.
$endgroup$
– joro
May 26 at 13:07
6
$begingroup$
Perhaps the curve is not geometrically irreducible over the rationals.
$endgroup$
– ulrich
May 26 at 13:30
$begingroup$
@ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:30
add a comment |
1
$begingroup$
You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find.
$endgroup$
– Gerry Myerson
May 26 at 12:36
$begingroup$
@GerryMyerson Thanks. I edited with link and authors.
$endgroup$
– joro
May 26 at 13:07
6
$begingroup$
Perhaps the curve is not geometrically irreducible over the rationals.
$endgroup$
– ulrich
May 26 at 13:30
$begingroup$
@ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:30
1
1
$begingroup$
You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find.
$endgroup$
– Gerry Myerson
May 26 at 12:36
$begingroup$
You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find.
$endgroup$
– Gerry Myerson
May 26 at 12:36
$begingroup$
@GerryMyerson Thanks. I edited with link and authors.
$endgroup$
– joro
May 26 at 13:07
$begingroup$
@GerryMyerson Thanks. I edited with link and authors.
$endgroup$
– joro
May 26 at 13:07
6
6
$begingroup$
Perhaps the curve is not geometrically irreducible over the rationals.
$endgroup$
– ulrich
May 26 at 13:30
$begingroup$
Perhaps the curve is not geometrically irreducible over the rationals.
$endgroup$
– ulrich
May 26 at 13:30
$begingroup$
@ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:30
$begingroup$
@ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.
answered May 26 at 13:39
dinamodinamo
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
add a comment |
Your Answer
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$begingroup$
The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.
answered May 26 at 13:39
dinamodinamo
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
add a comment |
$begingroup$
The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.
answered May 26 at 13:39
dinamodinamo
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
add a comment |
$begingroup$
The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.
answered May 26 at 13:39
dinamodinamo
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.
answered May 26 at 13:39
dinamodinamo
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered May 26 at 13:39
dinamodinamo
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered May 26 at 13:39
dinamodinamo
15613
answered May 26 at 13:39
dinamodinamo
15613
15613
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dinamo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
add a comment |
5
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
5
5
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $mathbbF_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$.
$endgroup$
– dhy
May 26 at 14:18
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:29
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
$begingroup$
Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G equiv F pmodp$, so this is still not absolutely irreducible modulo $p$.
$endgroup$
– dinamo
May 27 at 19:17
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1
$begingroup$
You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find.
$endgroup$
– Gerry Myerson
May 26 at 12:36
$begingroup$
@GerryMyerson Thanks. I edited with link and authors.
$endgroup$
– joro
May 26 at 13:07
6
$begingroup$
Perhaps the curve is not geometrically irreducible over the rationals.
$endgroup$
– ulrich
May 26 at 13:30
$begingroup$
@ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right?
$endgroup$
– joro
May 27 at 12:30