Teorema de Mordell - Beamer

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.

Introduction

Background

Introduction

Background

Introduction

Background

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

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

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

Heights

H(x) := max{|m|, |n|}

H(P ) := H(x)

h(P ) = logH(P )

Heights

H(x) := max{|m|, |n|}

H(P ) := H(x)

h(P ) = logH(P )

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!

Partitioning the π

Slice 1

Slice 3

Partitioning the π

Slice 1

Slice 3

Partitioning the π

Slice 1

Slice 3

Partitioning the π

Slice 1

Slice 3

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).

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.

P1 −Qi2 =2P2

P2 −Qi3 =2P3

Pm−1 −Qim =2Pm

P1 −Qi2 =2P2

P2 −Qi3 =2P3

Pm−1 −Qim =2Pm

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.

P = Qi1 + 2Qi2 + 4Qi3 + . . .+ 2m−1Qim + 2mPm

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

4h(Pj−1)− 1

(h(Pj−1)− (κ′ + κ)

h(Pj) ≤1

2h(Pj−1) +

κ′ + κ

4h(Pj−1)− 1

(h(Pj−1)− (κ′ + κ)

h(Pj) ≤1

2h(Pj−1) +

κ′ + κ

4h(Pj−1)− 1

(h(Pj−1)− (κ′ + κ)

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

h(Pj) ≤3

4h(Pj−1)

4h(Pj).

P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR

h(Pj) ≤3

4h(Pj−1)

4h(Pj).

P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR

h(Pj) ≤3

4h(Pj−1)

4h(Pj).

P = a1Q1 + a2Q2 + · · ·+ anQn + 2mR

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].

Hence the set

{Q1, Q2, . . . , Qn} ∪ {R ∈ Γ : h(R) ≤ κ′ + κ}

generates Γ.

Hence the set

{Q1, Q2, . . . , Qn} ∪ {R ∈ Γ : h(R) ≤ κ′ + κ}

generates Γ.

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

e2and y =

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

Mand y =

in lowest terms with M > 0 and N > 0.

Substituting into the equation (1) of E yields:

Clearing the denominators gives:

M3n2 = N2m3 + aN2Mm2 + bN2M2m+ cN2M3 (2)

e2and y =

Mand y =

e2and y =

Mand y =

e2and y =

Mand y =

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= M, and e3 =

N2= N.

Therefore x =m

e2and y =

e3have the desired form.

Therefore, M3 = N2.

e2 =N2

M2= M, and e3 =

N2= N.

Therefore x =m

e2and y =

Therefore, M3 = N2.

e2 =N2

M2= M, and e3 =

N2= N.

Therefore x =m

e2and y =

Therefore, M3 = N2.

e2 =N2

M2= M, and e3 =

N2= N.

Therefore x =m

e2and y =

Therefore, M3 = N2.

e2 =N2

M2= M, and e3 =

N2= N.

Therefore x =m

e2and y =

Therefore, M3 = N2.

e2 =N2

M2= M, and e3 =

N2= N.

Therefore x =m

e2and y =

Therefore, M3 = N2.

e2 =N2

M2= M, and e3 =

N2= N.

Therefore x =m

e2and y =

If the point P is given in lowest terms as P =(me2,n

), 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.

|n| ≤ KH(P )3/2

n2 = m3 + ae2m2 + be4m+ ce6

|n| ≤ KH(P )3/2

n2 = m3 + ae2m2 + be4m+ ce6

1 + |a|+ |b|+ |c|

Now we are ready to have Slice 2.

|n| ≤ KH(P )3/2

n2 = m3 + ae2m2 + be4m+ ce6

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)

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.

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).

ξ + x+ x0 = λ2 − a, λ =y − y0x− x0

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

ξ + x+ x0 = λ2 − a, λ =y − y0x− x0

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

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.

Em2 + Fme2 +Ge4

e ≤ H(P )1/2, n ≤ KH(P )3/2, m ≤ H(P )

where K depends on only a, b, c.

Em2 + Fme2 +Ge4

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

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).

It follows

h(P + P0) ≤ 2h(P ) + κ0

It follows

h(P + P0) ≤ 2h(P ) + κ0

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)

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)

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:

ξ =(f ′(x))2 − (8x+ 4a)f(x)

4f(x)=

x4 + · · ·4x3 + · · ·

h(ξ) ≥ 4h(x)− κ

ξ =(f ′(x))2 − (8x+ 4a)f(x)

4f(x)=

x4 + · · ·4x3 + · · ·

h(ξ) ≥ 4h(x)− κ

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

gcd(ndφ

), ndψ

))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,

)− κ1 ≤ h

(φ(m/n)

ψ(m/n)

)≤ dh

)+ κ2

Serving Slice 3i (1/8)

a) First we observe that since φ and ψ have degree at most d, both ndφ(

)are integers, so their g.c.d. makes sense.

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

gcd(ndφ

), ndψ

))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,

)− κ1 ≤ h

(φ(m/n)

ψ(m/n)

)≤ dh

)+ κ2

a) First we observe that since φ and ψ have degree at most d, both ndφ(

)are integers, so their g.c.d. makes sense.

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

Ψ(m,n) := ndψ(mn

)= b0m

end−e + b1me−1nd−n+1 + · · ·+ ben

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)

Then we can write:

Φ(m,n) := ndφ(mn

)= a0m

d + a1md−1 + · · ·+ adn

Ψ(m,n) := ndψ(mn

)= b0m

end−e + b1me−1nd−n+1 + · · ·+ ben

F (X)φ(X) +G(X)ψ(X) = 1 (3)

Then we can write:

Φ(m,n) := ndφ(mn

)= a0m

d + a1md−1 + · · ·+ adn

Ψ(m,n) := ndψ(mn

)= b0m

end−e + b1me−1nd−n+1 + · · ·+ ben

F (X)φ(X) +G(X)ψ(X) = 1 (3)

Then we can write:

Φ(m,n) := ndφ(mn

)= a0m

d + a1md−1 + · · ·+ adn

Ψ(m,n) := ndψ(mn

)= b0m

end−e + b1me−1nd−n+1 + · · ·+ ben

F (X)φ(X) +G(X)ψ(X) = 1 (3)

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

)+ nDAG

)· ndψ

)= AnD+d

Let γ = γ(m,n) := gcd(

Φ(m,n),Ψ(m,n))

We have: {nDAF

)}Φ(m,n) +

)}Ψ(m,n) = AnD+d

Since the quantities in braces are integers, we see that γ|AnD+d

nDAF(mn

)· ndφ

)+ nDAG

)· ndψ

)= AnD+d

Φ(m,n),Ψ(m,n))

We have: {nDAF

)}Φ(m,n) +

)}Ψ(m,n) = AnD+d

nDAF(mn

)· ndφ

)+ nDAG

)· ndψ

)= AnD+d

Φ(m,n),Ψ(m,n))

We have: {nDAF

)}Φ(m,n) +

)}Ψ(m,n) = AnD+d

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).

d−1nD+d + · · ·+AadnD+2d−1.

divides Aa0nD+d−1

d−1nD+d + · · ·+AadnD+2d−1.

divides Aa0nD+d−1

d−1nD+d + · · ·+AadnD+2d−1.

divides Aa0nD+d−1

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

). 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(

) =ndφ

) =Φ(m,n)

Ψ(m,n)

except when |Φ(m,n)| and |Ψ(m,n)| have common factors.

). So reverting the

ξ =φ(

) =ndφ

) =Φ(m,n)

Ψ(m,n)

). So reverting the

ξ =φ(

) =ndφ

) =Φ(m,n)

Ψ(m,n)

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)|}

{∣∣∣ndφ(mn

)∣∣∣, ∣∣∣ndψ(mn

)∣∣∣}≥ 1

(∣∣∣ndφ(mn

)∣∣∣+∣∣∣ndψ(m

)∣∣∣)

In terms of multiplicative notation, we want to compare H(ξ) to

)d= max{|m|d, |n|d}, so we consider the quotient:

H(ξ) ≥ 1

{∣∣∣ndφ(mn

)∣∣∣, ∣∣∣ndψ(mn

)∣∣∣}≥ 1

(∣∣∣ndφ(mn

)∣∣∣+∣∣∣ndψ(m

)∣∣∣)

H(ξ) ≥ 1

{∣∣∣ndφ(mn

)∣∣∣, ∣∣∣ndψ(mn

)∣∣∣}≥ 1

(∣∣∣ndφ(mn

)∣∣∣+∣∣∣ndψ(m

)∣∣∣)

H(m/n)d≥ 1

(∣∣∣ndφ(mn

)∣∣∣+∣∣∣ndψ(m

)∣∣∣)max{|m|d, |n|d}

(∣∣∣φ(mn

)∣∣∣+∣∣∣ψ(m

)∣∣∣)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

H(m/n)d≥ 1

(∣∣∣ndφ(mn

)∣∣∣+∣∣∣ndψ(m

)∣∣∣)max{|m|d, |n|d}

(∣∣∣φ(mn

)∣∣∣+∣∣∣ψ(m

)∣∣∣)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|+ |b0| if deg(φ) = d

H(m/n)d≥ 1

(∣∣∣ndφ(mn

)∣∣∣+∣∣∣ndψ(m

)∣∣∣)max{|m|d, |n|d}

(∣∣∣φ(mn

)∣∣∣+∣∣∣ψ(m

)∣∣∣)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|+ |b0| if deg(φ) = d

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

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.

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

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

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

Slice 3

Slice 4 (SPOILER)

The index (E(Q) : 2E(Q) is finite.

Summary

Slice 1

Slice 3

Slice 4 (SPOILER)

Summary

Slice 1

Slice 3

Slice 4 (SPOILER)

Summary

Slice 1

Slice 3

Slice 4 (SPOILER)

Summary

Slice 1

Slice 3

Slice 4 (SPOILER)

Teorema de Mordell - Beamer

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