y divides discriminant implies y^2 divides it

Multi tool use
$begingroup$
The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve
$$y^2 = x^3 + ax^2 + bx + c$$
with the cubic polynomial discriminant
$$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$
Later when explaining a particular case they state:
"... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."
I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?
elementary-number-theory cubic-equations
$endgroup$
add a comment |
$begingroup$
The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve
$$y^2 = x^3 + ax^2 + bx + c$$
with the cubic polynomial discriminant
$$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$
Later when explaining a particular case they state:
"... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."
I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?
elementary-number-theory cubic-equations
$endgroup$
add a comment |
$begingroup$
The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve
$$y^2 = x^3 + ax^2 + bx + c$$
with the cubic polynomial discriminant
$$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$
Later when explaining a particular case they state:
"... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."
I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?
elementary-number-theory cubic-equations
$endgroup$
The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve
$$y^2 = x^3 + ax^2 + bx + c$$
with the cubic polynomial discriminant
$$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$
Later when explaining a particular case they state:
"... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."
I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?
elementary-number-theory cubic-equations
elementary-number-theory cubic-equations
asked Jan 10 at 17:02
PugNosePugNose
112
112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
$$
D = Res_x(f(x), f'(x))
$$
and there exists some integer polynomials $g(x), h(x)$ such that
$$
D = g(x) f(x) + h(x)f'(x)
$$
Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.
Now the part you are missing is you can also write $D$ as
$$
D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
$$
Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.
$endgroup$
add a comment |
$begingroup$
In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
$$
It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
$$
begin{aligned}
kappa &= 2y +a_1 +a_3 ,\
kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
&qquadtext{ and one defines the following division polynomials:}\
psi_0 &= 0 ,\
psi_1 &= 1 ,\
psi_2 &= kappa ,\
psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
&qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
&qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
&qquadtext{ such that one has in case of $mPne0$ the relation}\
x(nP)
&=
frac{phi_m(P)}{psi_m(P)^2} ,\
Delta
&=
{color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
&qquadtext{ where}\[2mm]
sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
end{aligned}
$$
See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book
Ian Connel, Elliptic curve Handbook
(this is around page 150).
Proof of the implicit claim in the OP:
If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.
More general:
If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.
The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.
$square$
Usually it is good in such cases to have an explicit example, here we have one using sagemath:
sage: import random
sage: alfa = random.choice([10^4..10^5])
sage: alfa
40882
sage: c = alfa^2-alfa
sage: b = alfa*c
sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
sage: E
Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
sage: EE
Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
sage: EE.discriminant().factor()
2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
sage: EE.torsion_points()
[(0 : 1 : 0),
(8378881486178014515 : -7379204292472752 : 1),
(8378881486178014515 : 7379204292472752 : 1),
(8378881546344708027 : -301661871656786182296 : 1),
(8378881546344708027 : 301661871656786182296 : 1),
(8381341220942172099 : -12332842306323413283199584 : 1),
(8381341220942172099 : 12332842306323413283199584 : 1)]
sage: P = _[-1]
sage: P
(8381341220942172099 : 12332842306323413283199584 : 1)
sage: P.xy()
(8381341220942172099, 12332842306323413283199584)
sage: y = P.xy()[1]
sage: y
12332842306323413283199584
sage: ((2*y)^2).factor()
2^12 * 3^10 * 13627^4 * 20441^6
sage: ((2*y)^2).divides( EE.discriminant() )
True
In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
$$
D = Res_x(f(x), f'(x))
$$
and there exists some integer polynomials $g(x), h(x)$ such that
$$
D = g(x) f(x) + h(x)f'(x)
$$
Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.
Now the part you are missing is you can also write $D$ as
$$
D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
$$
Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.
$endgroup$
add a comment |
$begingroup$
Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
$$
D = Res_x(f(x), f'(x))
$$
and there exists some integer polynomials $g(x), h(x)$ such that
$$
D = g(x) f(x) + h(x)f'(x)
$$
Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.
Now the part you are missing is you can also write $D$ as
$$
D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
$$
Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.
$endgroup$
add a comment |
$begingroup$
Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
$$
D = Res_x(f(x), f'(x))
$$
and there exists some integer polynomials $g(x), h(x)$ such that
$$
D = g(x) f(x) + h(x)f'(x)
$$
Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.
Now the part you are missing is you can also write $D$ as
$$
D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
$$
Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.
$endgroup$
Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
$$
D = Res_x(f(x), f'(x))
$$
and there exists some integer polynomials $g(x), h(x)$ such that
$$
D = g(x) f(x) + h(x)f'(x)
$$
Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.
Now the part you are missing is you can also write $D$ as
$$
D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
$$
Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.
answered Jan 10 at 19:17
Yong Hao NgYong Hao Ng
3,4941220
3,4941220
add a comment |
add a comment |
$begingroup$
In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
$$
It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
$$
begin{aligned}
kappa &= 2y +a_1 +a_3 ,\
kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
&qquadtext{ and one defines the following division polynomials:}\
psi_0 &= 0 ,\
psi_1 &= 1 ,\
psi_2 &= kappa ,\
psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
&qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
&qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
&qquadtext{ such that one has in case of $mPne0$ the relation}\
x(nP)
&=
frac{phi_m(P)}{psi_m(P)^2} ,\
Delta
&=
{color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
&qquadtext{ where}\[2mm]
sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
end{aligned}
$$
See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book
Ian Connel, Elliptic curve Handbook
(this is around page 150).
Proof of the implicit claim in the OP:
If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.
More general:
If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.
The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.
$square$
Usually it is good in such cases to have an explicit example, here we have one using sagemath:
sage: import random
sage: alfa = random.choice([10^4..10^5])
sage: alfa
40882
sage: c = alfa^2-alfa
sage: b = alfa*c
sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
sage: E
Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
sage: EE
Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
sage: EE.discriminant().factor()
2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
sage: EE.torsion_points()
[(0 : 1 : 0),
(8378881486178014515 : -7379204292472752 : 1),
(8378881486178014515 : 7379204292472752 : 1),
(8378881546344708027 : -301661871656786182296 : 1),
(8378881546344708027 : 301661871656786182296 : 1),
(8381341220942172099 : -12332842306323413283199584 : 1),
(8381341220942172099 : 12332842306323413283199584 : 1)]
sage: P = _[-1]
sage: P
(8381341220942172099 : 12332842306323413283199584 : 1)
sage: P.xy()
(8381341220942172099, 12332842306323413283199584)
sage: y = P.xy()[1]
sage: y
12332842306323413283199584
sage: ((2*y)^2).factor()
2^12 * 3^10 * 13627^4 * 20441^6
sage: ((2*y)^2).divides( EE.discriminant() )
True
In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.
$endgroup$
add a comment |
$begingroup$
In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
$$
It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
$$
begin{aligned}
kappa &= 2y +a_1 +a_3 ,\
kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
&qquadtext{ and one defines the following division polynomials:}\
psi_0 &= 0 ,\
psi_1 &= 1 ,\
psi_2 &= kappa ,\
psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
&qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
&qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
&qquadtext{ such that one has in case of $mPne0$ the relation}\
x(nP)
&=
frac{phi_m(P)}{psi_m(P)^2} ,\
Delta
&=
{color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
&qquadtext{ where}\[2mm]
sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
end{aligned}
$$
See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book
Ian Connel, Elliptic curve Handbook
(this is around page 150).
Proof of the implicit claim in the OP:
If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.
More general:
If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.
The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.
$square$
Usually it is good in such cases to have an explicit example, here we have one using sagemath:
sage: import random
sage: alfa = random.choice([10^4..10^5])
sage: alfa
40882
sage: c = alfa^2-alfa
sage: b = alfa*c
sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
sage: E
Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
sage: EE
Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
sage: EE.discriminant().factor()
2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
sage: EE.torsion_points()
[(0 : 1 : 0),
(8378881486178014515 : -7379204292472752 : 1),
(8378881486178014515 : 7379204292472752 : 1),
(8378881546344708027 : -301661871656786182296 : 1),
(8378881546344708027 : 301661871656786182296 : 1),
(8381341220942172099 : -12332842306323413283199584 : 1),
(8381341220942172099 : 12332842306323413283199584 : 1)]
sage: P = _[-1]
sage: P
(8381341220942172099 : 12332842306323413283199584 : 1)
sage: P.xy()
(8381341220942172099, 12332842306323413283199584)
sage: y = P.xy()[1]
sage: y
12332842306323413283199584
sage: ((2*y)^2).factor()
2^12 * 3^10 * 13627^4 * 20441^6
sage: ((2*y)^2).divides( EE.discriminant() )
True
In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.
$endgroup$
add a comment |
$begingroup$
In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
$$
It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
$$
begin{aligned}
kappa &= 2y +a_1 +a_3 ,\
kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
&qquadtext{ and one defines the following division polynomials:}\
psi_0 &= 0 ,\
psi_1 &= 1 ,\
psi_2 &= kappa ,\
psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
&qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
&qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
&qquadtext{ such that one has in case of $mPne0$ the relation}\
x(nP)
&=
frac{phi_m(P)}{psi_m(P)^2} ,\
Delta
&=
{color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
&qquadtext{ where}\[2mm]
sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
end{aligned}
$$
See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book
Ian Connel, Elliptic curve Handbook
(this is around page 150).
Proof of the implicit claim in the OP:
If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.
More general:
If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.
The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.
$square$
Usually it is good in such cases to have an explicit example, here we have one using sagemath:
sage: import random
sage: alfa = random.choice([10^4..10^5])
sage: alfa
40882
sage: c = alfa^2-alfa
sage: b = alfa*c
sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
sage: E
Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
sage: EE
Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
sage: EE.discriminant().factor()
2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
sage: EE.torsion_points()
[(0 : 1 : 0),
(8378881486178014515 : -7379204292472752 : 1),
(8378881486178014515 : 7379204292472752 : 1),
(8378881546344708027 : -301661871656786182296 : 1),
(8378881546344708027 : 301661871656786182296 : 1),
(8381341220942172099 : -12332842306323413283199584 : 1),
(8381341220942172099 : 12332842306323413283199584 : 1)]
sage: P = _[-1]
sage: P
(8381341220942172099 : 12332842306323413283199584 : 1)
sage: P.xy()
(8381341220942172099, 12332842306323413283199584)
sage: y = P.xy()[1]
sage: y
12332842306323413283199584
sage: ((2*y)^2).factor()
2^12 * 3^10 * 13627^4 * 20441^6
sage: ((2*y)^2).divides( EE.discriminant() )
True
In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.
$endgroup$
In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
$$
It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
$$
begin{aligned}
kappa &= 2y +a_1 +a_3 ,\
kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
&qquadtext{ and one defines the following division polynomials:}\
psi_0 &= 0 ,\
psi_1 &= 1 ,\
psi_2 &= kappa ,\
psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
&qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
&qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
&qquadtext{ such that one has in case of $mPne0$ the relation}\
x(nP)
&=
frac{phi_m(P)}{psi_m(P)^2} ,\
Delta
&=
{color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
&qquadtext{ where}\[2mm]
sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
end{aligned}
$$
See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book
Ian Connel, Elliptic curve Handbook
(this is around page 150).
Proof of the implicit claim in the OP:
If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.
More general:
If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.
The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.
$square$
Usually it is good in such cases to have an explicit example, here we have one using sagemath:
sage: import random
sage: alfa = random.choice([10^4..10^5])
sage: alfa
40882
sage: c = alfa^2-alfa
sage: b = alfa*c
sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
sage: E
Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
sage: EE
Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
sage: EE.discriminant().factor()
2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
sage: EE.torsion_points()
[(0 : 1 : 0),
(8378881486178014515 : -7379204292472752 : 1),
(8378881486178014515 : 7379204292472752 : 1),
(8378881546344708027 : -301661871656786182296 : 1),
(8378881546344708027 : 301661871656786182296 : 1),
(8381341220942172099 : -12332842306323413283199584 : 1),
(8381341220942172099 : 12332842306323413283199584 : 1)]
sage: P = _[-1]
sage: P
(8381341220942172099 : 12332842306323413283199584 : 1)
sage: P.xy()
(8381341220942172099, 12332842306323413283199584)
sage: y = P.xy()[1]
sage: y
12332842306323413283199584
sage: ((2*y)^2).factor()
2^12 * 3^10 * 13627^4 * 20441^6
sage: ((2*y)^2).divides( EE.discriminant() )
True
In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.
answered Jan 15 at 13:01
dan_fuleadan_fulea
6,4281312
6,4281312
add a comment |
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