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Mordell’s Theorem (I/II)
Justin Scarfy
The University of British Columbia
January 31, 2012
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 1 / 29
Introduction
Background
In the past few lectures we learned the definition of elliptic curves, absorbed theNagell-Lutz Theorem, were being introduced to the L-functions of elliptic curves,and got spoiled by a few mouth-watering conjectures.
Lecture PlanToday we shall start discussing a theorem that L. Mordell proved in 1922:
The group of rational points on a non-singular cubic elliptic curve is finitelygenerated.
The main source I follow is that of Silverman-Tate, in which they employ atechnique called heights, and cheerfully partition the π so I can serve them to youby slices.
DisclaimerI only have time to serve you a finite portion of the π today.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 2 / 29
Introduction
Background
In the past few lectures we learned the definition of elliptic curves, absorbed theNagell-Lutz Theorem, were being introduced to the L-functions of elliptic curves,and got spoiled by a few mouth-watering conjectures.
Lecture PlanToday we shall start discussing a theorem that L. Mordell proved in 1922:
The group of rational points on a non-singular cubic elliptic curve is finitelygenerated.
The main source I follow is that of Silverman-Tate, in which they employ atechnique called heights, and cheerfully partition the π so I can serve them to youby slices.
DisclaimerI only have time to serve you a finite portion of the π today.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 2 / 29
Introduction
Background
In the past few lectures we learned the definition of elliptic curves, absorbed theNagell-Lutz Theorem, were being introduced to the L-functions of elliptic curves,and got spoiled by a few mouth-watering conjectures.
Lecture PlanToday we shall start discussing a theorem that L. Mordell proved in 1922:
The group of rational points on a non-singular cubic elliptic curve is finitelygenerated.
The main source I follow is that of Silverman-Tate, in which they employ atechnique called heights, and cheerfully partition the π so I can serve them to youby slices.
DisclaimerI only have time to serve you a finite portion of the π today.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 2 / 29
Introduction
Background
In the past few lectures we learned the definition of elliptic curves, absorbed theNagell-Lutz Theorem, were being introduced to the L-functions of elliptic curves,and got spoiled by a few mouth-watering conjectures.
Lecture PlanToday we shall start discussing a theorem that L. Mordell proved in 1922:
The group of rational points on a non-singular cubic elliptic curve is finitelygenerated.
The main source I follow is that of Silverman-Tate, in which they employ atechnique called heights, and cheerfully partition the π so I can serve them to youby slices.
DisclaimerI only have time to serve you a finite portion of the π today.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 2 / 29
Review
Definition (Non-Singular Cubic Elliptic Curve)
Recall the precise definition of a non-singular cubic elliptic curve E is an equationof the form
y2 = f(x) = x3 + ax2 + bx+ c (1)
with non-zero discrimant:
∆ := −16(4a3 + 27b2) 6= 0
Nagell-Lutz Theorem
Let E be a non-singular cubic elliptic curve with the form y2 = x3 + ax+ b,a, b ∈ ZIf P ∈ E(Q) is a torsion point of order m ≥ 2, then,
1) x(P ), y(P ) ∈ Z2) Either 2P = 0 or y2|∆ := 4a2 + 27b2
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 3 / 29
Review
Definition (Non-Singular Cubic Elliptic Curve)
Recall the precise definition of a non-singular cubic elliptic curve E is an equationof the form
y2 = f(x) = x3 + ax2 + bx+ c (1)
with non-zero discrimant:
∆ := −16(4a3 + 27b2) 6= 0
Nagell-Lutz Theorem
Let E be a non-singular cubic elliptic curve with the form y2 = x3 + ax+ b,a, b ∈ ZIf P ∈ E(Q) is a torsion point of order m ≥ 2, then,
1) x(P ), y(P ) ∈ Z2) Either 2P = 0 or y2|∆ := 4a2 + 27b2
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 3 / 29
Heights
Definition (Heights)
For all x ∈ Q with x = mn where (m,n) = 1, we define H : Q→ Z+ by
H(x) := max{|m|, |n|}
The Finiteness Property of the Height
The set of all rational numbers whose height is less than a fixed number is finite.
Heights of Points on Elliptic Curves
For E : y2 = f(x) = x3 + ax2 + bx+ c with a, b, c ∈ Z and rational pointP = (x, y) on E,
H(P ) := H(x)
further we define “small h” height: h : E → R≥0 by:
h(P ) = logH(P )
and finally define the height of point O at infinity to be: H(O) = 1 or h(O) = 0
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 4 / 29
Heights
Definition (Heights)
For all x ∈ Q with x = mn where (m,n) = 1, we define H : Q→ Z+ by
H(x) := max{|m|, |n|}
The Finiteness Property of the Height
The set of all rational numbers whose height is less than a fixed number is finite.
Heights of Points on Elliptic Curves
For E : y2 = f(x) = x3 + ax2 + bx+ c with a, b, c ∈ Z and rational pointP = (x, y) on E,
H(P ) := H(x)
further we define “small h” height: h : E → R≥0 by:
h(P ) = logH(P )
and finally define the height of point O at infinity to be: H(O) = 1 or h(O) = 0
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 4 / 29
Heights
Definition (Heights)
For all x ∈ Q with x = mn where (m,n) = 1, we define H : Q→ Z+ by
H(x) := max{|m|, |n|}
The Finiteness Property of the Height
The set of all rational numbers whose height is less than a fixed number is finite.
Heights of Points on Elliptic Curves
For E : y2 = f(x) = x3 + ax2 + bx+ c with a, b, c ∈ Z and rational pointP = (x, y) on E,
H(P ) := H(x)
further we define “small h” height: h : E → R≥0 by:
h(P ) = logH(P )
and finally define the height of point O at infinity to be: H(O) = 1 or h(O) = 0
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 4 / 29
Partitioning the π
Theorem (Modrell 1922)
The group of rational points E(Q) is finitely generated.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4Not ready yet– will be ready and served next week!
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 5 / 29
Partitioning the π
Theorem (Modrell 1922)
The group of rational points E(Q) is finitely generated.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4Not ready yet– will be ready and served next week!
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 5 / 29
Partitioning the π
Theorem (Modrell 1922)
The group of rational points E(Q) is finitely generated.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4Not ready yet– will be ready and served next week!
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 5 / 29
Partitioning the π
Theorem (Modrell 1922)
The group of rational points E(Q) is finitely generated.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4Not ready yet– will be ready and served next week!
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 5 / 29
Partitioning the π
Theorem (Modrell 1922)
The group of rational points E(Q) is finitely generated.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4Not ready yet– will be ready and served next week!
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 5 / 29
Descent Theorem (1/6)
Descent TheoremLet Γ be an abelian group, and suppose that there is a function h : Γ→ R≥0 withthe following three properties:
a) For every M ∈ R, the set {P ∈ Γ : h(P ) ≤M} is finite.b) For every P0 ∈ Γ, there is a constant κ0 with h(P + P0) ≤ 2h(P ) + κ0.c) There is a constant κ so that h(2P ) ≥ 4h(P )− κ for all P ∈ Γ.
Oh, and further suppose that:d) The subgroup 2Γ has finite index in Γ
Then, Γ is finitely generated
Proof of the Descent Theorem (1/6)
First we take a representative for each coset of 2Γ in Γ, assume there are n ofthem, and Let Q1, Q2, . . . , Qn be representative of the cosets (i.e. For any P ∈ Γ,there exists an index i1, depending on P , such that P −Qi1 ∈ 2Γ).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 6 / 29
Descent Theorem (1/6)
Descent TheoremLet Γ be an abelian group, and suppose that there is a function h : Γ→ R≥0 withthe following three properties:
a) For every M ∈ R, the set {P ∈ Γ : h(P ) ≤M} is finite.b) For every P0 ∈ Γ, there is a constant κ0 with h(P + P0) ≤ 2h(P ) + κ0.c) There is a constant κ so that h(2P ) ≥ 4h(P )− κ for all P ∈ Γ.
Oh, and further suppose that:d) The subgroup 2Γ has finite index in Γ
Then, Γ is finitely generated
Proof of the Descent Theorem (1/6)
First we take a representative for each coset of 2Γ in Γ, assume there are n ofthem, and Let Q1, Q2, . . . , Qn be representative of the cosets (i.e. For any P ∈ Γ,there exists an index i1, depending on P , such that P −Qi1 ∈ 2Γ).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 6 / 29
Descent Theorem (1/6)
Descent TheoremLet Γ be an abelian group, and suppose that there is a function h : Γ→ R≥0 withthe following three properties:
a) For every M ∈ R, the set {P ∈ Γ : h(P ) ≤M} is finite.b) For every P0 ∈ Γ, there is a constant κ0 with h(P + P0) ≤ 2h(P ) + κ0.c) There is a constant κ so that h(2P ) ≥ 4h(P )− κ for all P ∈ Γ.
Oh, and further suppose that:d) The subgroup 2Γ has finite index in Γ
Then, Γ is finitely generated
Proof of the Descent Theorem (1/6)
First we take a representative for each coset of 2Γ in Γ, assume there are n ofthem, and Let Q1, Q2, . . . , Qn be representative of the cosets (i.e. For any P ∈ Γ,there exists an index i1, depending on P , such that P −Qi1 ∈ 2Γ).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 6 / 29
Descent Theorem (1/6)
Descent TheoremLet Γ be an abelian group, and suppose that there is a function h : Γ→ R≥0 withthe following three properties:
a) For every M ∈ R, the set {P ∈ Γ : h(P ) ≤M} is finite.b) For every P0 ∈ Γ, there is a constant κ0 with h(P + P0) ≤ 2h(P ) + κ0.c) There is a constant κ so that h(2P ) ≥ 4h(P )− κ for all P ∈ Γ.
Oh, and further suppose that:d) The subgroup 2Γ has finite index in Γ
Then, Γ is finitely generated
Proof of the Descent Theorem (1/6)
First we take a representative for each coset of 2Γ in Γ, assume there are n ofthem, and Let Q1, Q2, . . . , Qn be representative of the cosets (i.e. For any P ∈ Γ,there exists an index i1, depending on P , such that P −Qi1 ∈ 2Γ).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 6 / 29
Descent Theorem (2/6)
Proof of the Descent Theorem (2/6)
After all, P has to be in one of the cosets (i.e. We can write P −Qi1 = 2P1 forsome P1 ∈ Γ).
Feeding this into the finite state automata yields:
P1 −Qi2 =2P2
P2 −Qi3 =2P3
...
Pm−1 −Qim =2Pm
where Qi1 , Qi2 , . . . , Qim are chosen from the coset representation Q1, Q2, . . . , Qn
and P1, P2, . . . , Pm are elements of Γ.
Moral: Pi is more-or-less equal to 2Pi+1, the height of Pi+1 is more-or-less onefourth the height of Pi. So the sequence of points P, P1, P2, . . . should havedecreasing height, and from property a), the set will be finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 7 / 29
Descent Theorem (2/6)
Proof of the Descent Theorem (2/6)
After all, P has to be in one of the cosets (i.e. We can write P −Qi1 = 2P1 forsome P1 ∈ Γ).
Feeding this into the finite state automata yields:
P1 −Qi2 =2P2
P2 −Qi3 =2P3
...
Pm−1 −Qim =2Pm
where Qi1 , Qi2 , . . . , Qim are chosen from the coset representation Q1, Q2, . . . , Qn
and P1, P2, . . . , Pm are elements of Γ.
Moral: Pi is more-or-less equal to 2Pi+1, the height of Pi+1 is more-or-less onefourth the height of Pi. So the sequence of points P, P1, P2, . . . should havedecreasing height, and from property a), the set will be finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 7 / 29
Descent Theorem (2/6)
Proof of the Descent Theorem (2/6)
After all, P has to be in one of the cosets (i.e. We can write P −Qi1 = 2P1 forsome P1 ∈ Γ).
Feeding this into the finite state automata yields:
P1 −Qi2 =2P2
P2 −Qi3 =2P3
...
Pm−1 −Qim =2Pm
where Qi1 , Qi2 , . . . , Qim are chosen from the coset representation Q1, Q2, . . . , Qn
and P1, P2, . . . , Pm are elements of Γ.
Moral: Pi is more-or-less equal to 2Pi+1, the height of Pi+1 is more-or-less onefourth the height of Pi. So the sequence of points P, P1, P2, . . . should havedecreasing height, and from property a), the set will be finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 7 / 29
Descent Theorem (3/6)
Proof of the Descent Theorem (3/6)
From the first equation we see P = Qi + 2P1, and substituting the secondequation P1 = Qi2 + 4P2 into the first gives P = Qi1 + 2Qi2 + 4P2
Continuing in this fashion, we obtain
P = Qi1 + 2Qi2 + 4Qi3 + . . .+ 2m−1Qim + 2mPm
In particular, this says that P is in the subgroup of Γ generated by the Qi’s andPm.
Next we show that by choosing m large enough, we can force Pm to have heightless than a certain fixed bound. Then the finite set of points with height less thanthat bound, together with the Q′is, will generate Γ.
Taking one of the P ′js in the sequence of points P, P1, P2, . . . and examine therelation between the height of Pj−1 and that of Pj . We want to show that theheight of Pj is considerably smaller.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 8 / 29
Descent Theorem (3/6)
Proof of the Descent Theorem (3/6)
From the first equation we see P = Qi + 2P1, and substituting the secondequation P1 = Qi2 + 4P2 into the first gives P = Qi1 + 2Qi2 + 4P2
Continuing in this fashion, we obtain
P = Qi1 + 2Qi2 + 4Qi3 + . . .+ 2m−1Qim + 2mPm
In particular, this says that P is in the subgroup of Γ generated by the Qi’s andPm.
Next we show that by choosing m large enough, we can force Pm to have heightless than a certain fixed bound. Then the finite set of points with height less thanthat bound, together with the Q′is, will generate Γ.
Taking one of the P ′js in the sequence of points P, P1, P2, . . . and examine therelation between the height of Pj−1 and that of Pj . We want to show that theheight of Pj is considerably smaller.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 8 / 29
Descent Theorem (3/6)
Proof of the Descent Theorem (3/6)
From the first equation we see P = Qi + 2P1, and substituting the secondequation P1 = Qi2 + 4P2 into the first gives P = Qi1 + 2Qi2 + 4P2
Continuing in this fashion, we obtain
P = Qi1 + 2Qi2 + 4Qi3 + . . .+ 2m−1Qim + 2mPm
In particular, this says that P is in the subgroup of Γ generated by the Qi’s andPm.
Next we show that by choosing m large enough, we can force Pm to have heightless than a certain fixed bound. Then the finite set of points with height less thanthat bound, together with the Q′is, will generate Γ.
Taking one of the P ′js in the sequence of points P, P1, P2, . . . and examine therelation between the height of Pj−1 and that of Pj . We want to show that theheight of Pj is considerably smaller.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 8 / 29
Descent Theorem (3/6)
Proof of the Descent Theorem (3/6)
From the first equation we see P = Qi + 2P1, and substituting the secondequation P1 = Qi2 + 4P2 into the first gives P = Qi1 + 2Qi2 + 4P2
Continuing in this fashion, we obtain
P = Qi1 + 2Qi2 + 4Qi3 + . . .+ 2m−1Qim + 2mPm
In particular, this says that P is in the subgroup of Γ generated by the Qi’s andPm.
Next we show that by choosing m large enough, we can force Pm to have heightless than a certain fixed bound. Then the finite set of points with height less thanthat bound, together with the Q′is, will generate Γ.
Taking one of the P ′js in the sequence of points P, P1, P2, . . . and examine therelation between the height of Pj−1 and that of Pj . We want to show that theheight of Pj is considerably smaller.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 8 / 29
Descent Theorem (4/6)
Proof of the Descent Theorem (4/6)
If we apply b) with −Qi in place of P0, we obtain a constant κi such thath(P −Qi) ≤ 2h(P ) + κi for all P ∈ Γ.
Now we do this for each Qi, 1 ≤ i ≤ n.Let κ′ be the largest of the κi’s, thenh(P −Qi) ≤ 2h(P ) +κ′ for all P ∈ Γ and all 1 ≤ i ≤ n, this can be done as thereare only Q′is, and is one place where property (d) that 2Γ has finite index in Γ.
Let κ be the constant from (c), then we deduce:
4h(Pj) ≤ h(2Pj) + κ = h(Pj−1 −Qij ) + κ ≤ 2h(Pj−1) + κ′ + κ
which can be rewritten as:
h(Pj) ≤1
2h(Pj−1) +
κ′ + κ
4=
3
4h(Pj−1)− 1
4
(h(Pj−1)− (κ′ + κ)
)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 9 / 29
Descent Theorem (4/6)
Proof of the Descent Theorem (4/6)
If we apply b) with −Qi in place of P0, we obtain a constant κi such thath(P −Qi) ≤ 2h(P ) + κi for all P ∈ Γ.
Now we do this for each Qi, 1 ≤ i ≤ n.Let κ′ be the largest of the κi’s, thenh(P −Qi) ≤ 2h(P ) +κ′ for all P ∈ Γ and all 1 ≤ i ≤ n, this can be done as thereare only Q′is, and is one place where property (d) that 2Γ has finite index in Γ.
Let κ be the constant from (c), then we deduce:
4h(Pj) ≤ h(2Pj) + κ = h(Pj−1 −Qij ) + κ ≤ 2h(Pj−1) + κ′ + κ
which can be rewritten as:
h(Pj) ≤1
2h(Pj−1) +
κ′ + κ
4=
3
4h(Pj−1)− 1
4
(h(Pj−1)− (κ′ + κ)
)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 9 / 29
Descent Theorem (4/6)
Proof of the Descent Theorem (4/6)
If we apply b) with −Qi in place of P0, we obtain a constant κi such thath(P −Qi) ≤ 2h(P ) + κi for all P ∈ Γ.
Now we do this for each Qi, 1 ≤ i ≤ n.Let κ′ be the largest of the κi’s, thenh(P −Qi) ≤ 2h(P ) +κ′ for all P ∈ Γ and all 1 ≤ i ≤ n, this can be done as thereare only Q′is, and is one place where property (d) that 2Γ has finite index in Γ.
Let κ be the constant from (c), then we deduce:
4h(Pj) ≤ h(2Pj) + κ = h(Pj−1 −Qij ) + κ ≤ 2h(Pj−1) + κ′ + κ
which can be rewritten as:
h(Pj) ≤1
2h(Pj−1) +
κ′ + κ
4=
3
4h(Pj−1)− 1
4
(h(Pj−1)− (κ′ + κ)
)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 9 / 29
Descent Theorem (5/6)
Proof of the Descent Theorem (5/6)
From the previous equation we see that if h(Pj−1) ≥ κ′ + κ, then
h(Pj) ≤3
4h(Pj−1)
So in the sequence of points P, P1, P2, P3, . . . , as long as the points Pj satisfiesthe condition h(Pj) ≥ κ′ + κ, then the next point in the sequence has muchsmaller height, namely h(Pj+1) ≤ 3
4h(Pj).
N.B. If you start with a number and keep multiplying it by 3/4, it approacheszero. So eventually we will find an index m such that h(Pm) ≤ κ′ + κ
We have now shown that every element P ∈ Γ can be written in the form:
P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR
for certain integers a1, . . . , an and some point R ∈ Γ satisfying the inequalityh(R) ≤ κ′ + κ
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 10 / 29
Descent Theorem (5/6)
Proof of the Descent Theorem (5/6)
From the previous equation we see that if h(Pj−1) ≥ κ′ + κ, then
h(Pj) ≤3
4h(Pj−1)
So in the sequence of points P, P1, P2, P3, . . . , as long as the points Pj satisfiesthe condition h(Pj) ≥ κ′ + κ, then the next point in the sequence has muchsmaller height, namely h(Pj+1) ≤ 3
4h(Pj).
N.B. If you start with a number and keep multiplying it by 3/4, it approacheszero. So eventually we will find an index m such that h(Pm) ≤ κ′ + κ
We have now shown that every element P ∈ Γ can be written in the form:
P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR
for certain integers a1, . . . , an and some point R ∈ Γ satisfying the inequalityh(R) ≤ κ′ + κ
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 10 / 29
Descent Theorem (5/6)
Proof of the Descent Theorem (5/6)
From the previous equation we see that if h(Pj−1) ≥ κ′ + κ, then
h(Pj) ≤3
4h(Pj−1)
So in the sequence of points P, P1, P2, P3, . . . , as long as the points Pj satisfiesthe condition h(Pj) ≥ κ′ + κ, then the next point in the sequence has muchsmaller height, namely h(Pj+1) ≤ 3
4h(Pj).
N.B. If you start with a number and keep multiplying it by 3/4, it approacheszero. So eventually we will find an index m such that h(Pm) ≤ κ′ + κ
We have now shown that every element P ∈ Γ can be written in the form:
P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR
for certain integers a1, . . . , an and some point R ∈ Γ satisfying the inequalityh(R) ≤ κ′ + κ
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 10 / 29
Descent Theorem (5/6)
Proof of the Descent Theorem (5/6)
From the previous equation we see that if h(Pj−1) ≥ κ′ + κ, then
h(Pj) ≤3
4h(Pj−1)
So in the sequence of points P, P1, P2, P3, . . . , as long as the points Pj satisfiesthe condition h(Pj) ≥ κ′ + κ, then the next point in the sequence has muchsmaller height, namely h(Pj+1) ≤ 3
4h(Pj).
N.B. If you start with a number and keep multiplying it by 3/4, it approacheszero. So eventually we will find an index m such that h(Pm) ≤ κ′ + κ
We have now shown that every element P ∈ Γ can be written in the form:
P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR
for certain integers a1, . . . , an and some point R ∈ Γ satisfying the inequalityh(R) ≤ κ′ + κ
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 10 / 29
Descent Theorem (6/6)
Proof of the Descent Theorem (6/6)
Hence the set
{Q1, Q2, . . . , Qn} ∪ {R ∈ Γ : h(R) ≤ κ′ + κ}
generates Γ.
From a) and d), this set is finite, which completes the proof that Γ is finitelygenerated.
RemarkThis Theorem is called a Descent Theorem as the proof imitates the style ofFermat’s method of infinite descent: One starts with an arbitrary point P ∈ E(Q),and by manipulation descents to a smaller point [in terms of height, of course].
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 11 / 29
Descent Theorem (6/6)
Proof of the Descent Theorem (6/6)
Hence the set
{Q1, Q2, . . . , Qn} ∪ {R ∈ Γ : h(R) ≤ κ′ + κ}
generates Γ.
From a) and d), this set is finite, which completes the proof that Γ is finitelygenerated.
RemarkThis Theorem is called a Descent Theorem as the proof imitates the style ofFermat’s method of infinite descent: One starts with an arbitrary point P ∈ E(Q),and by manipulation descents to a smaller point [in terms of height, of course].
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 11 / 29
Descent Theorem (6/6)
Proof of the Descent Theorem (6/6)
Hence the set
{Q1, Q2, . . . , Qn} ∪ {R ∈ Γ : h(R) ≤ κ′ + κ}
generates Γ.
From a) and d), this set is finite, which completes the proof that Γ is finitelygenerated.
RemarkThis Theorem is called a Descent Theorem as the proof imitates the style ofFermat’s method of infinite descent: One starts with an arbitrary point P ∈ E(Q),and by manipulation descents to a smaller point [in terms of height, of course].
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 11 / 29
The Height of P + P0 (1/7)
First we recall that if P = (x, y) is a rational point on E, then x and y have theform
x =m
e2and y =
n
e3
for integers m,n, e with e > 0 and gcd(m, e) = gcd(n, e) = 1.This can be shown by mimicking the steps in the proof of Nagell-Lutz Theorem.
Now suppose we write
x =m
Mand y =
n
N
in lowest terms with M > 0 and N > 0.
Substituting into the equation (1) of E yields:
n2
N2=m3
M3+ a
m2
M2+ b
m
M+ c
Clearing the denominators gives:
M3n2 = N2m3 + aN2Mm2 + bN2M2m+ cN2M3 (2)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 12 / 29
The Height of P + P0 (1/7)
First we recall that if P = (x, y) is a rational point on E, then x and y have theform
x =m
e2and y =
n
e3
for integers m,n, e with e > 0 and gcd(m, e) = gcd(n, e) = 1.This can be shown by mimicking the steps in the proof of Nagell-Lutz Theorem.
Now suppose we write
x =m
Mand y =
n
N
in lowest terms with M > 0 and N > 0.
Substituting into the equation (1) of E yields:
n2
N2=m3
M3+ a
m2
M2+ b
m
M+ c
Clearing the denominators gives:
M3n2 = N2m3 + aN2Mm2 + bN2M2m+ cN2M3 (2)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 12 / 29
The Height of P + P0 (1/7)
First we recall that if P = (x, y) is a rational point on E, then x and y have theform
x =m
e2and y =
n
e3
for integers m,n, e with e > 0 and gcd(m, e) = gcd(n, e) = 1.This can be shown by mimicking the steps in the proof of Nagell-Lutz Theorem.
Now suppose we write
x =m
Mand y =
n
N
in lowest terms with M > 0 and N > 0.
Substituting into the equation (1) of E yields:
n2
N2=m3
M3+ a
m2
M2+ b
m
M+ c
Clearing the denominators gives:
M3n2 = N2m3 + aN2Mm2 + bN2M2m+ cN2M3 (2)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 12 / 29
The Height of P + P0 (1/7)
First we recall that if P = (x, y) is a rational point on E, then x and y have theform
x =m
e2and y =
n
e3
for integers m,n, e with e > 0 and gcd(m, e) = gcd(n, e) = 1.This can be shown by mimicking the steps in the proof of Nagell-Lutz Theorem.
Now suppose we write
x =m
Mand y =
n
N
in lowest terms with M > 0 and N > 0.
Substituting into the equation (1) of E yields:
n2
N2=m3
M3+ a
m2
M2+ b
m
M+ c
Clearing the denominators gives:
M3n2 = N2m3 + aN2Mm2 + bN2M2m+ cN2M3 (2)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 12 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (2/7)
Since N2 is a factor of all terms on the R.H.S. of (2), it follows that N2|M3n2.But gcd(n,N) = 1, so N2|M3.
Conversely, we show M3|N2, in three steps:
From (2) we immediately see that M |N2m3, and since gcd(m,M) = 1, wefind M |N2.
Using the above back to (2), we find M2|N2m3, so M |N .
Again by (2), we see that this implies M3|N2m3, so M3|N2.
Therefore, M3 = N2.
Further, during the proof we showed that M |N .
So if we let e = NM ,we find that
e2 =N2
M2=M3
M2= M, and e3 =
N2
M3=N3
N2= N.
Therefore x =m
e2and y =
n
e3have the desired form.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 13 / 29
The Height of P + P0 (3/7)
If the point P is given in lowest terms as P =(me2,n
e3
), then |e2| ≤ H(P ) and
|m| ≤ H(P ), and we claim there is a constant K > 0, depending on a, b, c suchthat
|n| ≤ KH(P )3/2
To prove this we just use the fact that the point satisfies the equation:substituting into equation (1) and multiplying by e6 to clear denominator:
n2 = m3 + ae2m2 + be4m+ ce6
Taking the absolute value and using the triangle inequality yields:
|n2| ≤ |m3|+ |ae2m2|+ |be4m|+ |ce6|≤ H(P )3 + |a|H(P )3 + |b|H(P )3 + |c|H(P )3
So we can take K =√
1 + |a|+ |b|+ |c|Now we are ready to have Slice 2.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 14 / 29
The Height of P + P0 (3/7)
If the point P is given in lowest terms as P =(me2,n
e3
), then |e2| ≤ H(P ) and
|m| ≤ H(P ), and we claim there is a constant K > 0, depending on a, b, c suchthat
|n| ≤ KH(P )3/2
To prove this we just use the fact that the point satisfies the equation:substituting into equation (1) and multiplying by e6 to clear denominator:
n2 = m3 + ae2m2 + be4m+ ce6
Taking the absolute value and using the triangle inequality yields:
|n2| ≤ |m3|+ |ae2m2|+ |be4m|+ |ce6|≤ H(P )3 + |a|H(P )3 + |b|H(P )3 + |c|H(P )3
So we can take K =√
1 + |a|+ |b|+ |c|Now we are ready to have Slice 2.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 14 / 29
The Height of P + P0 (3/7)
If the point P is given in lowest terms as P =(me2,n
e3
), then |e2| ≤ H(P ) and
|m| ≤ H(P ), and we claim there is a constant K > 0, depending on a, b, c suchthat
|n| ≤ KH(P )3/2
To prove this we just use the fact that the point satisfies the equation:substituting into equation (1) and multiplying by e6 to clear denominator:
n2 = m3 + ae2m2 + be4m+ ce6
Taking the absolute value and using the triangle inequality yields:
|n2| ≤ |m3|+ |ae2m2|+ |be4m|+ |ce6|≤ H(P )3 + |a|H(P )3 + |b|H(P )3 + |c|H(P )3
So we can take K =√
1 + |a|+ |b|+ |c|
Now we are ready to have Slice 2.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 14 / 29
The Height of P + P0 (3/7)
If the point P is given in lowest terms as P =(me2,n
e3
), then |e2| ≤ H(P ) and
|m| ≤ H(P ), and we claim there is a constant K > 0, depending on a, b, c suchthat
|n| ≤ KH(P )3/2
To prove this we just use the fact that the point satisfies the equation:substituting into equation (1) and multiplying by e6 to clear denominator:
n2 = m3 + ae2m2 + be4m+ ce6
Taking the absolute value and using the triangle inequality yields:
|n2| ≤ |m3|+ |ae2m2|+ |be4m|+ |ce6|≤ H(P )3 + |a|H(P )3 + |b|H(P )3 + |c|H(P )3
So we can take K =√
1 + |a|+ |b|+ |c|Now we are ready to have Slice 2.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 14 / 29
The Height of P + P0 (4/7)
Slice 2For P0 a fixed rational point on E, there is a constant κ0, depending on P0 andon a, b, c such that h(P + P0) ≤ 2h(P ) + κ0 for all P ∈ E(Q)
Serving Slice 2 (1/4)
First we remark that if P0 = O, the slice is trivial.So we take P0 6= O, say P0 = (x0, y0).
To prove the existence of κ0, it suffices to prove that the inequality holds for all Pexcept those in a some fixed set; this holds because, for any finite number of P ,we just looking at the difference h(P + P0)− 2h(P ) and take κ0 larger than thefinite number of values that occur. Hence, it suffices to prove for all points Pwith P /∈ {P0,−P0,O}.We write P = (x, y), the reason for avoiding both P0 and −P0 is to have x 6= x0.
We also denoteP + P0 = (ξ, η)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 15 / 29
The Height of P + P0 (4/7)
Slice 2For P0 a fixed rational point on E, there is a constant κ0, depending on P0 andon a, b, c such that h(P + P0) ≤ 2h(P ) + κ0 for all P ∈ E(Q)
Serving Slice 2 (1/4)
First we remark that if P0 = O, the slice is trivial.So we take P0 6= O, say P0 = (x0, y0).
To prove the existence of κ0, it suffices to prove that the inequality holds for all Pexcept those in a some fixed set; this holds because, for any finite number of P ,we just looking at the difference h(P + P0)− 2h(P ) and take κ0 larger than thefinite number of values that occur. Hence, it suffices to prove for all points Pwith P /∈ {P0,−P0,O}.We write P = (x, y), the reason for avoiding both P0 and −P0 is to have x 6= x0.
We also denoteP + P0 = (ξ, η)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 15 / 29
The Height of P + P0 (4/7)
Slice 2For P0 a fixed rational point on E, there is a constant κ0, depending on P0 andon a, b, c such that h(P + P0) ≤ 2h(P ) + κ0 for all P ∈ E(Q)
Serving Slice 2 (1/4)
First we remark that if P0 = O, the slice is trivial.So we take P0 6= O, say P0 = (x0, y0).
To prove the existence of κ0, it suffices to prove that the inequality holds for all Pexcept those in a some fixed set; this holds because, for any finite number of P ,we just looking at the difference h(P + P0)− 2h(P ) and take κ0 larger than thefinite number of values that occur. Hence, it suffices to prove for all points Pwith P /∈ {P0,−P0,O}.
We write P = (x, y), the reason for avoiding both P0 and −P0 is to have x 6= x0.
We also denoteP + P0 = (ξ, η)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 15 / 29
The Height of P + P0 (4/7)
Slice 2For P0 a fixed rational point on E, there is a constant κ0, depending on P0 andon a, b, c such that h(P + P0) ≤ 2h(P ) + κ0 for all P ∈ E(Q)
Serving Slice 2 (1/4)
First we remark that if P0 = O, the slice is trivial.So we take P0 6= O, say P0 = (x0, y0).
To prove the existence of κ0, it suffices to prove that the inequality holds for all Pexcept those in a some fixed set; this holds because, for any finite number of P ,we just looking at the difference h(P + P0)− 2h(P ) and take κ0 larger than thefinite number of values that occur. Hence, it suffices to prove for all points Pwith P /∈ {P0,−P0,O}.We write P = (x, y), the reason for avoiding both P0 and −P0 is to have x 6= x0.
We also denoteP + P0 = (ξ, η)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 15 / 29
The Height of P + P0 (5/7)
Serving Slice 2 (2/4)
Now H(P + P0) = ξ, so we need a formula for ξ in terms of (x, y) and (x0, y0):
ξ + x+ x0 = λ2 − a, λ =y − y0x− x0
.
After writing out this a little bit:
ξ =(y − y0)2
(x− x0)2− a− x− x0
=(y − y0)2 − (x− x0)2(x+ x0 + a)
(x− x0)2
=Ay +Bx2 + Cx+D
Ex2 + Fx+G
where A,B,C,D,E, F,G are certain rational numbers which can be expressed interms of a, b, c and (x0, y0).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 16 / 29
The Height of P + P0 (5/7)
Serving Slice 2 (2/4)
Now H(P + P0) = ξ, so we need a formula for ξ in terms of (x, y) and (x0, y0):
ξ + x+ x0 = λ2 − a, λ =y − y0x− x0
.
After writing out this a little bit:
ξ =(y − y0)2
(x− x0)2− a− x− x0
=(y − y0)2 − (x− x0)2(x+ x0 + a)
(x− x0)2
=Ay +Bx2 + Cx+D
Ex2 + Fx+G
where A,B,C,D,E, F,G are certain rational numbers which can be expressed interms of a, b, c and (x0, y0).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 16 / 29
The Height of P + P0 (5/7)
Serving Slice 2 (2/4)
Now H(P + P0) = ξ, so we need a formula for ξ in terms of (x, y) and (x0, y0):
ξ + x+ x0 = λ2 − a, λ =y − y0x− x0
.
After writing out this a little bit:
ξ =(y − y0)2
(x− x0)2− a− x− x0
=(y − y0)2 − (x− x0)2(x+ x0 + a)
(x− x0)2
=Ay +Bx2 + Cx+D
Ex2 + Fx+G
where A,B,C,D,E, F,G are certain rational numbers which can be expressed interms of a, b, c and (x0, y0).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 16 / 29
The Height of P + P0 (6/7)
Serving Slice 2 (3/4)
Further, we are able to multiply the numerator and denominator by the l.c.d. ofA, . . . , G, and hence we may assume that A, . . . , G ∈ Z, which depend only ona, b, c and (x0, y0), After substituting x = m
e2 and y = ne3 and clearing out the
denominators by multiplying the numerator and denominator by e4, we find:
ξ =Ane+Bm2 + Cme2 +De4
Em2 + Fme2 +Ge4
Notice we have an expression for ξ as an integer divided by an integer: althoughwe are uncertain that it is in the lowest terms, but cancellation will only make theheight smaller:
H(ξ) ≤ max{|Ane+Bm2 + Cme2 +De4|, |Em2 + Fme2 +Ge4|}
Further, from above we have the estimates
e ≤ H(P )1/2, n ≤ KH(P )3/2, m ≤ H(P )
where K depends on only a, b, c.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 17 / 29
The Height of P + P0 (6/7)
Serving Slice 2 (3/4)
Further, we are able to multiply the numerator and denominator by the l.c.d. ofA, . . . , G, and hence we may assume that A, . . . , G ∈ Z, which depend only ona, b, c and (x0, y0), After substituting x = m
e2 and y = ne3 and clearing out the
denominators by multiplying the numerator and denominator by e4, we find:
ξ =Ane+Bm2 + Cme2 +De4
Em2 + Fme2 +Ge4
Notice we have an expression for ξ as an integer divided by an integer: althoughwe are uncertain that it is in the lowest terms, but cancellation will only make theheight smaller:
H(ξ) ≤ max{|Ane+Bm2 + Cme2 +De4|, |Em2 + Fme2 +Ge4|}
Further, from above we have the estimates
e ≤ H(P )1/2, n ≤ KH(P )3/2, m ≤ H(P )
where K depends on only a, b, c.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 17 / 29
The Height of P + P0 (6/7)
Serving Slice 2 (3/4)
Further, we are able to multiply the numerator and denominator by the l.c.d. ofA, . . . , G, and hence we may assume that A, . . . , G ∈ Z, which depend only ona, b, c and (x0, y0), After substituting x = m
e2 and y = ne3 and clearing out the
denominators by multiplying the numerator and denominator by e4, we find:
ξ =Ane+Bm2 + Cme2 +De4
Em2 + Fme2 +Ge4
Notice we have an expression for ξ as an integer divided by an integer: althoughwe are uncertain that it is in the lowest terms, but cancellation will only make theheight smaller:
H(ξ) ≤ max{|Ane+Bm2 + Cme2 +De4|, |Em2 + Fme2 +Ge4|}
Further, from above we have the estimates
e ≤ H(P )1/2, n ≤ KH(P )3/2, m ≤ H(P )
where K depends on only a, b, c.Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 17 / 29
The Height of P + P0 (7/7)
Serving Slice 2 (4/4)
Using the above and triangle inequality gives
|Ane+Bm2 + Cme2 +De4| ≤ |Ane|+ |Bm2|+ |Cme2|+ |De4|≤ (|AK|+ |B|+ |C|+ |D|)H(P )2
|Em2 + Fme2 +Ge4| ≤ |Em2|+ |Fme2|+ |Ge4|≤ (|E|+ |F |+ |G|)H(P )2
It follows
H(P + P0) = H(ξ) ≤ max{|AK|+ |B|+ |C|+ |D|, |E|+ |F |+ |G|}H(P )2
Taking the logarithm of both sides yields
h(P + P0) ≤ 2h(P ) + κ0
where the constant κ0 = log max{|AK|+ |B|+ |C|+ |D|, |E|+ |F |+ |G|}depends only on a, b, c and (x0, y0) and does NOT depend on P = (x, y).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 18 / 29
The Height of P + P0 (7/7)
Serving Slice 2 (4/4)
Using the above and triangle inequality gives
|Ane+Bm2 + Cme2 +De4| ≤ |Ane|+ |Bm2|+ |Cme2|+ |De4|≤ (|AK|+ |B|+ |C|+ |D|)H(P )2
|Em2 + Fme2 +Ge4| ≤ |Em2|+ |Fme2|+ |Ge4|≤ (|E|+ |F |+ |G|)H(P )2
It follows
H(P + P0) = H(ξ) ≤ max{|AK|+ |B|+ |C|+ |D|, |E|+ |F |+ |G|}H(P )2
Taking the logarithm of both sides yields
h(P + P0) ≤ 2h(P ) + κ0
where the constant κ0 = log max{|AK|+ |B|+ |C|+ |D|, |E|+ |F |+ |G|}depends only on a, b, c and (x0, y0) and does NOT depend on P = (x, y).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 18 / 29
The Height of P + P0 (7/7)
Serving Slice 2 (4/4)
Using the above and triangle inequality gives
|Ane+Bm2 + Cme2 +De4| ≤ |Ane|+ |Bm2|+ |Cme2|+ |De4|≤ (|AK|+ |B|+ |C|+ |D|)H(P )2
|Em2 + Fme2 +Ge4| ≤ |Em2|+ |Fme2|+ |Ge4|≤ (|E|+ |F |+ |G|)H(P )2
It follows
H(P + P0) = H(ξ) ≤ max{|AK|+ |B|+ |C|+ |D|, |E|+ |F |+ |G|}H(P )2
Taking the logarithm of both sides yields
h(P + P0) ≤ 2h(P ) + κ0
where the constant κ0 = log max{|AK|+ |B|+ |C|+ |D|, |E|+ |F |+ |G|}depends only on a, b, c and (x0, y0) and does NOT depend on P = (x, y).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 18 / 29
The Height of 2P (1/10)
Slice 3There is a constant κ, depending on a, b, c so that
h(2P ) ≥ 4h(P )− κ for all P ∈ E(Q)
Serving Slice 3 (1/2)
Mimicking the Serving of Slice 2, we ignore the finite set of points satisfying2P = O since we can always take κ larger than 4h(P ) for all points in that finiteset.
Let P = (x, y) and write 2P = (ξ, η).The duplication formula we derived earlier states that
ξ + 2x = λ2 − a, where λ =f ′(x)
2y
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 19 / 29
The Height of 2P (1/10)
Slice 3There is a constant κ, depending on a, b, c so that
h(2P ) ≥ 4h(P )− κ for all P ∈ E(Q)
Serving Slice 3 (1/2)
Mimicking the Serving of Slice 2, we ignore the finite set of points satisfying2P = O since we can always take κ larger than 4h(P ) for all points in that finiteset.
Let P = (x, y) and write 2P = (ξ, η).The duplication formula we derived earlier states that
ξ + 2x = λ2 − a, where λ =f ′(x)
2y
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 19 / 29
The Height of 2P (1/10)
Slice 3There is a constant κ, depending on a, b, c so that
h(2P ) ≥ 4h(P )− κ for all P ∈ E(Q)
Serving Slice 3 (1/2)
Mimicking the Serving of Slice 2, we ignore the finite set of points satisfying2P = O since we can always take κ larger than 4h(P ) for all points in that finiteset.
Let P = (x, y) and write 2P = (ξ, η).The duplication formula we derived earlier states that
ξ + 2x = λ2 − a, where λ =f ′(x)
2y
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 19 / 29
The Height of 2P (2/10)
Serving Slice 3 (2/2)
Using y2 = f(x), we obtain an explicit formula for ξ in terms of x :
ξ =(f ′(x))2 − (8x+ 4a)f(x)
4f(x)=
x4 + · · ·4x3 + · · ·
Note that f(x) 6= 0 because 2P 6= O.
Thus, ξ is the quotient of two polynomials in x with integer coefficients. Since thecubic y2 = f(x) is non-singular by assumption, we know that f(x) and f ′(x) haveNO common (complex) roots, and thus the polynomials in the numerator and thedenominator of ξ also have NO common roots.
Since h(P ) = h(x) and h(2P ) = h(ξ), we are trying to prove that
h(ξ) ≥ 4h(x)− κ
Hence we reduced our task to proving the following slice about heights andquotients of polynomials:
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 20 / 29
The Height of 2P (2/10)
Serving Slice 3 (2/2)
Using y2 = f(x), we obtain an explicit formula for ξ in terms of x :
ξ =(f ′(x))2 − (8x+ 4a)f(x)
4f(x)=
x4 + · · ·4x3 + · · ·
Note that f(x) 6= 0 because 2P 6= O.
Thus, ξ is the quotient of two polynomials in x with integer coefficients. Since thecubic y2 = f(x) is non-singular by assumption, we know that f(x) and f ′(x) haveNO common (complex) roots, and thus the polynomials in the numerator and thedenominator of ξ also have NO common roots.
Since h(P ) = h(x) and h(2P ) = h(ξ), we are trying to prove that
h(ξ) ≥ 4h(x)− κ
Hence we reduced our task to proving the following slice about heights andquotients of polynomials:
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 20 / 29
The Height of 2P (2/10)
Serving Slice 3 (2/2)
Using y2 = f(x), we obtain an explicit formula for ξ in terms of x :
ξ =(f ′(x))2 − (8x+ 4a)f(x)
4f(x)=
x4 + · · ·4x3 + · · ·
Note that f(x) 6= 0 because 2P 6= O.
Thus, ξ is the quotient of two polynomials in x with integer coefficients. Since thecubic y2 = f(x) is non-singular by assumption, we know that f(x) and f ′(x) haveNO common (complex) roots, and thus the polynomials in the numerator and thedenominator of ξ also have NO common roots.
Since h(P ) = h(x) and h(2P ) = h(ξ), we are trying to prove that
h(ξ) ≥ 4h(x)− κ
Hence we reduced our task to proving the following slice about heights andquotients of polynomials:
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 20 / 29
The Height of 2P (3/10)
Slice 3i
Let φ(X) and ψ(X) be polynomials with integer coefficients and NO common(complex) roots. Let d = max{deg(φ),deg(ψ)}a) There is an integer R ≥ 1, depending on φ and ψ, so that for all rationalnumbers m
n ,
gcd(ndφ
(mn
), ndψ
(mn
))divides R
b) There are constants κ1 and κ2, depending on φ and ψ, so that for all rationalnumbers m
n which are NOT roots of ψ,
dh(mn
)− κ1 ≤ h
(φ(m/n)
ψ(m/n)
)≤ dh
(mn
)+ κ2
Serving Slice 3i (1/8)
a) First we observe that since φ and ψ have degree at most d, both ndφ(
mn
)and
ndψ(
mn
)are integers, so their g.c.d. makes sense.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 21 / 29
The Height of 2P (3/10)
Slice 3i
Let φ(X) and ψ(X) be polynomials with integer coefficients and NO common(complex) roots. Let d = max{deg(φ),deg(ψ)}a) There is an integer R ≥ 1, depending on φ and ψ, so that for all rationalnumbers m
n ,
gcd(ndφ
(mn
), ndψ
(mn
))divides R
b) There are constants κ1 and κ2, depending on φ and ψ, so that for all rationalnumbers m
n which are NOT roots of ψ,
dh(mn
)− κ1 ≤ h
(φ(m/n)
ψ(m/n)
)≤ dh
(mn
)+ κ2
Serving Slice 3i (1/8)
a) First we observe that since φ and ψ have degree at most d, both ndφ(
mn
)and
ndψ(
mn
)are integers, so their g.c.d. makes sense.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 21 / 29
The Height of 2P (4/10)
Serving Slice 3i (2/8)
Next we note that φ and ψ are interchangeable, so for correctness, we will takedeg(φ) := d and deg(ψ):=e ≤ d.
Then we can write:
Φ(m,n) := ndφ(mn
)= a0m
d + a1md−1 + · · ·+ adn
d,
Ψ(m,n) := ndψ(mn
)= b0m
end−e + b1me−1nd−n+1 + · · ·+ ben
d
So we need to find an estimate for gcd(Φ(m,n),Ψ(m,n)) which does NOTdepend on m OR n.
Since φ(X) and ψ(X) have NO common roots, they are relative prime in theEuclidean ring Q[X], so they generate the unit ideal:
So we can find polynomials F (X) and G(X) with rational coefficients satisfying
F (X)φ(X) +G(X)ψ(X) = 1 (3)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 22 / 29
The Height of 2P (4/10)
Serving Slice 3i (2/8)
Next we note that φ and ψ are interchangeable, so for correctness, we will takedeg(φ) := d and deg(ψ):=e ≤ d.
Then we can write:
Φ(m,n) := ndφ(mn
)= a0m
d + a1md−1 + · · ·+ adn
d,
Ψ(m,n) := ndψ(mn
)= b0m
end−e + b1me−1nd−n+1 + · · ·+ ben
d
So we need to find an estimate for gcd(Φ(m,n),Ψ(m,n)) which does NOTdepend on m OR n.
Since φ(X) and ψ(X) have NO common roots, they are relative prime in theEuclidean ring Q[X], so they generate the unit ideal:
So we can find polynomials F (X) and G(X) with rational coefficients satisfying
F (X)φ(X) +G(X)ψ(X) = 1 (3)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 22 / 29
The Height of 2P (4/10)
Serving Slice 3i (2/8)
Next we note that φ and ψ are interchangeable, so for correctness, we will takedeg(φ) := d and deg(ψ):=e ≤ d.
Then we can write:
Φ(m,n) := ndφ(mn
)= a0m
d + a1md−1 + · · ·+ adn
d,
Ψ(m,n) := ndψ(mn
)= b0m
end−e + b1me−1nd−n+1 + · · ·+ ben
d
So we need to find an estimate for gcd(Φ(m,n),Ψ(m,n)) which does NOTdepend on m OR n.
Since φ(X) and ψ(X) have NO common roots, they are relative prime in theEuclidean ring Q[X], so they generate the unit ideal:
So we can find polynomials F (X) and G(X) with rational coefficients satisfying
F (X)φ(X) +G(X)ψ(X) = 1 (3)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 22 / 29
The Height of 2P (4/10)
Serving Slice 3i (2/8)
Next we note that φ and ψ are interchangeable, so for correctness, we will takedeg(φ) := d and deg(ψ):=e ≤ d.
Then we can write:
Φ(m,n) := ndφ(mn
)= a0m
d + a1md−1 + · · ·+ adn
d,
Ψ(m,n) := ndψ(mn
)= b0m
end−e + b1me−1nd−n+1 + · · ·+ ben
d
So we need to find an estimate for gcd(Φ(m,n),Ψ(m,n)) which does NOTdepend on m OR n.
Since φ(X) and ψ(X) have NO common roots, they are relative prime in theEuclidean ring Q[X], so they generate the unit ideal:
So we can find polynomials F (X) and G(X) with rational coefficients satisfying
F (X)φ(X) +G(X)ψ(X) = 1 (3)
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 22 / 29
The Height of 2P (5/10)
Servicing Slice 3i (3/8)
Now let A be a large enough integer so that AF (X) and AG(X) have integercoefficients, and let D = max{deg(F ),deg(G)}.N.B. A and D do NOT depend on m or n.
Now we evaluate the identity (3) at X = mn and multiply both sides by AnD+d.
nDAF(mn
)· ndφ
(mn
)+ nDAG
(mn
)· ndψ
(mn
)= AnD+d
Let γ = γ(m,n) := gcd(
Φ(m,n),Ψ(m,n))
We have: {nDAF
(mn
)}Φ(m,n) +
{nDAG
(mn
)}Ψ(m,n) = AnD+d
Since the quantities in braces are integers, we see that γ|AnD+d
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 23 / 29
The Height of 2P (5/10)
Servicing Slice 3i (3/8)
Now let A be a large enough integer so that AF (X) and AG(X) have integercoefficients, and let D = max{deg(F ),deg(G)}.N.B. A and D do NOT depend on m or n.
Now we evaluate the identity (3) at X = mn and multiply both sides by AnD+d.
nDAF(mn
)· ndφ
(mn
)+ nDAG
(mn
)· ndψ
(mn
)= AnD+d
Let γ = γ(m,n) := gcd(
Φ(m,n),Ψ(m,n))
We have: {nDAF
(mn
)}Φ(m,n) +
{nDAG
(mn
)}Ψ(m,n) = AnD+d
Since the quantities in braces are integers, we see that γ|AnD+d
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 23 / 29
The Height of 2P (5/10)
Servicing Slice 3i (3/8)
Now let A be a large enough integer so that AF (X) and AG(X) have integercoefficients, and let D = max{deg(F ),deg(G)}.N.B. A and D do NOT depend on m or n.
Now we evaluate the identity (3) at X = mn and multiply both sides by AnD+d.
nDAF(mn
)· ndφ
(mn
)+ nDAG
(mn
)· ndψ
(mn
)= AnD+d
Let γ = γ(m,n) := gcd(
Φ(m,n),Ψ(m,n))
We have: {nDAF
(mn
)}Φ(m,n) +
{nDAG
(mn
)}Ψ(m,n) = AnD+d
Since the quantities in braces are integers, we see that γ|AnD+d
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 23 / 29
The Height of 2P (6/10)
Serving Slice 3i (4/8)
We also need to show γ|AaD+d0 , where a0 is the leading coefficient of φ(X). We
observe that since γ divides Φ(m,n), it certainly divides:
AnD+d−1Φ(m,n) = Aa0mdnD+d−1 +Aa1m
d−1nD+d + · · ·+AadnD+2d−1.
But in the sum, every term after the first one contains AnD+d as a factor; andwe just proved that γ divides AnD+d.
It follows that γ also divides the first term Aa0mdnD+d−1. Thus, γ divides
gcd(AnD+d, Aa0mdnD+d−1); and because (m,n) = 1, we conclude that γ
divides Aa0nD+d−1
Now using the fact that γ divides Aa0nD+d−2Φ(m,n) and repeating the above
argument shows that γ divides Aa20nD+d−2; eventually, we reach the conclusion
that γ divides AaD+d0 , finishing the serving of a).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 24 / 29
The Height of 2P (6/10)
Serving Slice 3i (4/8)
We also need to show γ|AaD+d0 , where a0 is the leading coefficient of φ(X). We
observe that since γ divides Φ(m,n), it certainly divides:
AnD+d−1Φ(m,n) = Aa0mdnD+d−1 +Aa1m
d−1nD+d + · · ·+AadnD+2d−1.
But in the sum, every term after the first one contains AnD+d as a factor; andwe just proved that γ divides AnD+d.
It follows that γ also divides the first term Aa0mdnD+d−1. Thus, γ divides
gcd(AnD+d, Aa0mdnD+d−1); and because (m,n) = 1, we conclude that γ
divides Aa0nD+d−1
Now using the fact that γ divides Aa0nD+d−2Φ(m,n) and repeating the above
argument shows that γ divides Aa20nD+d−2; eventually, we reach the conclusion
that γ divides AaD+d0 , finishing the serving of a).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 24 / 29
The Height of 2P (6/10)
Serving Slice 3i (4/8)
We also need to show γ|AaD+d0 , where a0 is the leading coefficient of φ(X). We
observe that since γ divides Φ(m,n), it certainly divides:
AnD+d−1Φ(m,n) = Aa0mdnD+d−1 +Aa1m
d−1nD+d + · · ·+AadnD+2d−1.
But in the sum, every term after the first one contains AnD+d as a factor; andwe just proved that γ divides AnD+d.
It follows that γ also divides the first term Aa0mdnD+d−1. Thus, γ divides
gcd(AnD+d, Aa0mdnD+d−1); and because (m,n) = 1, we conclude that γ
divides Aa0nD+d−1
Now using the fact that γ divides Aa0nD+d−2Φ(m,n) and repeating the above
argument shows that γ divides Aa20nD+d−2; eventually, we reach the conclusion
that γ divides AaD+d0 , finishing the serving of a).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 24 / 29
The Height of 2P (6/10)
Serving Slice 3i (4/8)
We also need to show γ|AaD+d0 , where a0 is the leading coefficient of φ(X). We
observe that since γ divides Φ(m,n), it certainly divides:
AnD+d−1Φ(m,n) = Aa0mdnD+d−1 +Aa1m
d−1nD+d + · · ·+AadnD+2d−1.
But in the sum, every term after the first one contains AnD+d as a factor; andwe just proved that γ divides AnD+d.
It follows that γ also divides the first term Aa0mdnD+d−1. Thus, γ divides
gcd(AnD+d, Aa0mdnD+d−1); and because (m,n) = 1, we conclude that γ
divides Aa0nD+d−1
Now using the fact that γ divides Aa0nD+d−2Φ(m,n) and repeating the above
argument shows that γ divides Aa20nD+d−2; eventually, we reach the conclusion
that γ divides AaD+d0 , finishing the serving of a).
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 24 / 29
The Height of 2P (7/10)
Serving Slice 3i (5/8)
b) Two inequalities to be proven:- The upper bound is similar to Slice 2 so it is left as an exercise.- For the lower bound: as usual, we are cool to exclude some finite set of rationalnumbers when we prove prove the inequality of this sort: so we assume that therational number m
n is NOT a root of φ.
For 0 6= r ∈ Q, it is clear from the definition that h(r) = h(1
r
). So reverting the
role of φ and ψ if necessary, we may assume that deg(φ) = d and deg(ψ) = e withe ≤ d.
Continuing from a), the rational number whose height we want to estimate is
ξ =φ(
mn
)ψ(
mn
) =ndφ
(mn
)ndψ
(mn
) =Φ(m,n)
Ψ(m,n)
except when |Φ(m,n)| and |Ψ(m,n)| have common factors.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 25 / 29
The Height of 2P (7/10)
Serving Slice 3i (5/8)
b) Two inequalities to be proven:- The upper bound is similar to Slice 2 so it is left as an exercise.- For the lower bound: as usual, we are cool to exclude some finite set of rationalnumbers when we prove prove the inequality of this sort: so we assume that therational number m
n is NOT a root of φ.
For 0 6= r ∈ Q, it is clear from the definition that h(r) = h(1
r
). So reverting the
role of φ and ψ if necessary, we may assume that deg(φ) = d and deg(ψ) = e withe ≤ d.
Continuing from a), the rational number whose height we want to estimate is
ξ =φ(
mn
)ψ(
mn
) =ndφ
(mn
)ndψ
(mn
) =Φ(m,n)
Ψ(m,n)
except when |Φ(m,n)| and |Ψ(m,n)| have common factors.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 25 / 29
The Height of 2P (7/10)
Serving Slice 3i (5/8)
b) Two inequalities to be proven:- The upper bound is similar to Slice 2 so it is left as an exercise.- For the lower bound: as usual, we are cool to exclude some finite set of rationalnumbers when we prove prove the inequality of this sort: so we assume that therational number m
n is NOT a root of φ.
For 0 6= r ∈ Q, it is clear from the definition that h(r) = h(1
r
). So reverting the
role of φ and ψ if necessary, we may assume that deg(φ) = d and deg(ψ) = e withe ≤ d.
Continuing from a), the rational number whose height we want to estimate is
ξ =φ(
mn
)ψ(
mn
) =ndφ
(mn
)ndψ
(mn
) =Φ(m,n)
Ψ(m,n)
except when |Φ(m,n)| and |Ψ(m,n)| have common factors.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 25 / 29
The Height of 2P (8/10)
Serving Slice 3i (6/8)
We showed in a) that there is some R ≥ 1, independent of m and n, so that theg.c.d. of Φ(m,n) and Ψ(m,n) divides R.
This bounds the possible cancellation, and we find that
H(ξ) ≥ 1
Rmax{|Φ(m,n)|, |Ψ(m,n)|}
=1
Rmax
{∣∣∣ndφ(mn
)∣∣∣, ∣∣∣ndψ(mn
)∣∣∣}≥ 1
2R
(∣∣∣ndφ(mn
)∣∣∣+∣∣∣ndψ(m
n
)∣∣∣)
In terms of multiplicative notation, we want to compare H(ξ) to
H(mn
)d= max{|m|d, |n|d}, so we consider the quotient:
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 26 / 29
The Height of 2P (8/10)
Serving Slice 3i (6/8)
We showed in a) that there is some R ≥ 1, independent of m and n, so that theg.c.d. of Φ(m,n) and Ψ(m,n) divides R.
This bounds the possible cancellation, and we find that
H(ξ) ≥ 1
Rmax{|Φ(m,n)|, |Ψ(m,n)|}
=1
Rmax
{∣∣∣ndφ(mn
)∣∣∣, ∣∣∣ndψ(mn
)∣∣∣}≥ 1
2R
(∣∣∣ndφ(mn
)∣∣∣+∣∣∣ndψ(m
n
)∣∣∣)
In terms of multiplicative notation, we want to compare H(ξ) to
H(mn
)d= max{|m|d, |n|d}, so we consider the quotient:
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 26 / 29
The Height of 2P (8/10)
Serving Slice 3i (6/8)
We showed in a) that there is some R ≥ 1, independent of m and n, so that theg.c.d. of Φ(m,n) and Ψ(m,n) divides R.
This bounds the possible cancellation, and we find that
H(ξ) ≥ 1
Rmax{|Φ(m,n)|, |Ψ(m,n)|}
=1
Rmax
{∣∣∣ndφ(mn
)∣∣∣, ∣∣∣ndψ(mn
)∣∣∣}≥ 1
2R
(∣∣∣ndφ(mn
)∣∣∣+∣∣∣ndψ(m
n
)∣∣∣)
In terms of multiplicative notation, we want to compare H(ξ) to
H(mn
)d= max{|m|d, |n|d}, so we consider the quotient:
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 26 / 29
The Height of 2P (9/10)
Serving Slice 3i (7/8)
H(ξ)
H(m/n)d≥ 1
2R·
(∣∣∣ndφ(mn
)∣∣∣+∣∣∣ndψ(m
n
)∣∣∣)max{|m|d, |n|d}
=1
2R·
(∣∣∣φ(mn
)∣∣∣+∣∣∣ψ(m
n
)∣∣∣)max
{∣∣∣mn ∣∣∣d, 1}
This suggests that we look at the function p of a real variable t defined by
p(t) =
∣∣φ(t)∣∣+∣∣ψ(t)
∣∣max
{|t|d, 1
}Since max{deg(φ),deg(ψ)} ≤ d, we see that lim|t|→∞ p(t) 6= 0 and
lim|t|→∞
p(t) =
{|a0| if deg(φ) < d
|a0|+ |b0| if deg(φ) = d
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 27 / 29
The Height of 2P (9/10)
Serving Slice 3i (7/8)
H(ξ)
H(m/n)d≥ 1
2R·
(∣∣∣ndφ(mn
)∣∣∣+∣∣∣ndψ(m
n
)∣∣∣)max{|m|d, |n|d}
=1
2R·
(∣∣∣φ(mn
)∣∣∣+∣∣∣ψ(m
n
)∣∣∣)max
{∣∣∣mn ∣∣∣d, 1}This suggests that we look at the function p of a real variable t defined by
p(t) =
∣∣φ(t)∣∣+∣∣ψ(t)
∣∣max
{|t|d, 1
}
Since max{deg(φ),deg(ψ)} ≤ d, we see that lim|t|→∞ p(t) 6= 0 and
lim|t|→∞
p(t) =
{|a0| if deg(φ) < d
|a0|+ |b0| if deg(φ) = d
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 27 / 29
The Height of 2P (9/10)
Serving Slice 3i (7/8)
H(ξ)
H(m/n)d≥ 1
2R·
(∣∣∣ndφ(mn
)∣∣∣+∣∣∣ndψ(m
n
)∣∣∣)max{|m|d, |n|d}
=1
2R·
(∣∣∣φ(mn
)∣∣∣+∣∣∣ψ(m
n
)∣∣∣)max
{∣∣∣mn ∣∣∣d, 1}This suggests that we look at the function p of a real variable t defined by
p(t) =
∣∣φ(t)∣∣+∣∣ψ(t)
∣∣max
{|t|d, 1
}Since max{deg(φ),deg(ψ)} ≤ d, we see that lim|t|→∞ p(t) 6= 0 and
lim|t|→∞
p(t) =
{|a0| if deg(φ) < d
|a0|+ |b0| if deg(φ) = d
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 27 / 29
The Height of 2P (10/10)
Serving Slice 3i (8/8)
So there is a constant C > 0 so that p(t) > C for all t ∈ R
Use the fact in the inequality we derived above, we find that
H(ξ) ≥ C
2RH(mn
)d
Finally we see that the constants C and R do NOT depend on m and n, sotaking logarithms gives the desired inequality:
h(ξ) ≥ dh(mn
)− κ1 with κ1 = log(2R/C).
RemarksNotice that there are two ideas in the above proof:1) to bound the amount of cancellation;
2) to look atH(φ(x)/ψ(x))
H(x)das a function on something compact.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 28 / 29
The Height of 2P (10/10)
Serving Slice 3i (8/8)
So there is a constant C > 0 so that p(t) > C for all t ∈ R
Use the fact in the inequality we derived above, we find that
H(ξ) ≥ C
2RH(mn
)dFinally we see that the constants C and R do NOT depend on m and n, so
taking logarithms gives the desired inequality:
h(ξ) ≥ dh(mn
)− κ1 with κ1 = log(2R/C).
RemarksNotice that there are two ideas in the above proof:1) to bound the amount of cancellation;
2) to look atH(φ(x)/ψ(x))
H(x)das a function on something compact.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 28 / 29
The Height of 2P (10/10)
Serving Slice 3i (8/8)
So there is a constant C > 0 so that p(t) > C for all t ∈ R
Use the fact in the inequality we derived above, we find that
H(ξ) ≥ C
2RH(mn
)dFinally we see that the constants C and R do NOT depend on m and n, so
taking logarithms gives the desired inequality:
h(ξ) ≥ dh(mn
)− κ1 with κ1 = log(2R/C).
RemarksNotice that there are two ideas in the above proof:1) to bound the amount of cancellation;
2) to look atH(φ(x)/ψ(x))
H(x)das a function on something compact.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 28 / 29
Summary
Today we initiated the proof of Modrell’s Theorem by partitioning the π into fourslices, picked up the definition of heights on the road, and proved the DescendTheorem which connects the π to Modrell.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4 (SPOILER)
The index (E(Q) : 2E(Q) is finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 29 / 29
Summary
Today we initiated the proof of Modrell’s Theorem by partitioning the π into fourslices, picked up the definition of heights on the road, and proved the DescendTheorem which connects the π to Modrell.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4 (SPOILER)
The index (E(Q) : 2E(Q) is finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 29 / 29
Summary
Today we initiated the proof of Modrell’s Theorem by partitioning the π into fourslices, picked up the definition of heights on the road, and proved the DescendTheorem which connects the π to Modrell.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4 (SPOILER)
The index (E(Q) : 2E(Q) is finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 29 / 29
Summary
Today we initiated the proof of Modrell’s Theorem by partitioning the π into fourslices, picked up the definition of heights on the road, and proved the DescendTheorem which connects the π to Modrell.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4 (SPOILER)
The index (E(Q) : 2E(Q) is finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 29 / 29
Summary
Today we initiated the proof of Modrell’s Theorem by partitioning the π into fourslices, picked up the definition of heights on the road, and proved the DescendTheorem which connects the π to Modrell.
Slice 1
For every M ∈ R, the set {P ∈ E(Q) : h(P ) ≤M} is finite.
Slice 2Let P0 ∈ Q on E be fixed, then there exists a constant κ0 depending on P0 anda, b, c such that h(P + P0) < 2h(P ) + κ0 for all P ∈ E(Q).
Slice 3
There is a constant κ, depending on a, b, c so that h(2P ) ≥ 4h(P )− κ for allP ∈ E(Q).
Slice 4 (SPOILER)
The index (E(Q) : 2E(Q) is finite.
Justin Scarfy (UBC) Mordell’s Theorem (I/II) January 31, 2012 29 / 29