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The Polynomial Method Lecture 28

By using parameter counting, we constructed polynomials P with integer...
Subject

Diploma in civil engineering

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Academic year: 2023/2024
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  1. Outline of the proof of Thue’s theorem

Theorem 1. (Thue) If β is an irrational algebraic number, and γ > deg(β 2 )+2 , then there are only finitely many integer solutions to the inequality

p |β − | ≤ |q|−γ . q By using parameter counting, we constructed polynomials P with integer coeffi- cients that vanish to high order at (β, β). The degree of P and the size of P are controlled. If r 1 , r 2 are rational numbers with large height, then we proved that P cannot vanish to such a high order at r = (r 1 , r 2 ). For some j of controlled size, we have ∂j 1 P (r) = 0. Since P has integer coefficients, and r is rational, |∂ 1 j P (r)| is bounded below. Since P vanishes to high order at (β, β), we can use Taylor’s theorem to bound |∂ 1 j P (r)| from above in terms of |β − r 1 | and |β − r 2 |. So we see that |β − r 1 | or |β − r 2 | needs to be large. Here is the framework of the proof. We suppose that there are infinitely many rational solutions to the inequality |β − r| ≤ ‖r‖−γ . Let $ > 0 be a small parameter we will play with. We let r 1 be a solution with very large height, and we let r 2 be a solution with much larger height. Using these, we will prove that γ ≤ deg(β 2 )+2 +C(β)$.

  1. The polynomials For each integer m ≥ 1, we proved that there exists a polynomial P = Pm ∈ Z[x 1 , x 2 ] with the following properties:

(1) We have ∂ 1 j P (β, β) = 0 for j = 0, ..., m − 1. (2) We have Deg 2 P ≤ 1 and Deg 1 P ≤ (1 + $) deg 2 (β) m. (3) We have |P | ≤ C(β, $)m.

  1. The rational point Suppose that r 1 , r 2 are good rational approximations to β in the sense that

‖β − ri‖ ≤ ‖r 1 ‖−γ . 1

&

Also, we will suppose that ‖r 1 ‖ is sufficiently large in terms of β, $, and that ‖r 2 ‖ is sufficiently large in terms of β, $, and ‖r 1 ‖. If l ≥ 2 and ∂j 1 P (r) = 0 for j = 0, ..., l − 1, then we proved the following estimate:

l 1 |P | ≥ min((2degP )− 1 ‖r 1 ‖

− 2 , ‖r 2 ‖). Given our bound for |P |, we get

l 1 C(β, $)m ≥ min(‖r 1 ‖

− 2 , ‖r 2 ‖). From now on, we only work with m small enough so that

C(β, $)m < ‖r 2 ‖. Assumption l 1 Therefore, ‖r ‖

− 1 2 ≤ C(β, $)m. We assume that ‖r 1 ‖ is large enough so that ‖r 1 ‖# > C(β, $), and this implies that l ≤ $m. Therefore, there exists some j ≤ $m so that ∂ 1 j P (r) = 0. Let P ̃ = (1/j !)∂ 1 j P. The polynomial P ̃ has integer coefficients, and |P ̃ | ≤ 2 degP |P |. Therefore, P ̃ obeys essentially all the good properties of P above:

(1) We have ∂ 1 j P ̃ (β, β) = 0 for j = 0, ..., (1 − $)m − 1. (2) We have Deg 2 P ̃ ≤ 1 and Deg 1 P ̃ ≤ (1 + $) deg(β) m. 2 (3) We have |P ̃| ≤ C(β, $)m. (4) We also have P ̃ (r) = 0. Since P ̃ has integer coefficients, we can write P ̃ (r) as a fraction with a known

denominator: q Deg P 1 ̃ q Deg P 2 ̃ 1 2. Therefore, ̃ ̃ deg(β) |P ̃(r)| ≥ ‖r 1 ‖−Deg 1 P ‖r 2 ‖−Deg 2 P ≥ ‖r 1 ‖−(1+#) 2 m‖r 2 ‖− 1. We make some notation to help us focus on what’s important. In our problem, terms like ‖r 1 ‖m or ‖r 2 ‖ are substantial, but terms like ‖r 1 ‖#m or ‖r 1 ‖ are minor in comparison. Therefore, we write A! B to mean A ≤ ‖r 1 ‖a#m‖r 1 ‖b , for some constants a, b depending only on β. Recall that ‖r 1 ‖# is bigger than C(β, $), so C(β, $)m! 1. Our main inequality for this section is

deg(β) |P ̃(r)| " ‖r 1 ‖ − 2 m‖r 2 ‖ − 1. (1)

  1. Taylor’s theorem estimates We recall Taylor’s theorem.

&

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We see that ‖r 2 ‖ ≥ ‖r 1 ‖m > C(β, $)m, so the assumption about r 2 and m above is satisfied. The inequality becomes

deg(β) ‖r 1 ‖− 2 m−m! ‖r 1 ‖−γm. Multiplying through to make everything positive, we get deg(β)+ ‖r 1 ‖γm ! ‖r 1 ‖ 2 m. Unwinding the !, this actually means deg(β)+ ‖r 1 ‖γm ≤ ‖r 1 ‖b+a#m+ 2 m. (If we had been more explicit, we could have gotten specific values for a, b, but it doesn’t matter much.) Taking the logarithm to base ‖r 1 ‖ and dividing by m, we get

deg(β) + 2 γ ≤ (b/m) + a$ +. 2 If ‖r 2 ‖ is large enough compared to ‖r 1 ‖, then (1/m) ≤ $, and we have γ ≤ (a + b)$ + deg(β 2 )+2 . Taking $ → 0 finishes the proof.

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The Polynomial Method Lecture 28

Subject: Diploma in civil engineering

32 Documents
Students shared 32 documents in this course

School: Winnacunnet High School - Hampton

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PROOF OF THUE’S THEOREM – PART III
1. Outline of the proof of Thue’s theorem
Theorem 1.1. (Thue) If βis an irrational algebraic number, and γ>deg(β)+2 ,then
2
there are only finitely many integer solutions to the inequality
p
|β−|≤|q|−γ.
q
By using parameter counting, we constructed polynomials Pwith integer coeffi-
cients that vanish to high order at (β,β). The degree of Pand the size of Pare
controlled.
If r1,r2are rational numbers with large height, then we proved that Pcannot
vanish to such a high order at r=(r1,r
2). For some jof controlled size, we have
∂j
1P(r)=0. SincePhas integer coefficients, and ris rational, |∂j
1P(r)|is bounded
below.
Since Pvanishes to high order at (β,β), we can use Taylor’s theorem to bound
|∂j
1P(r)|from above in terms of |β−r1|and |β−r2|.Soweseethat|β−r1|or
|β−r2|needs to be large.
Here is the framework of the proof. We suppose that there are infinitely many
rational solutions to the inequality |β−r|≤#r#−γ. Let $>0beasmallparameter
we will play with. We let r1be a solution with very large height, and we let r2be a
solution with much larger height. Using these, we will prove that γ≤deg(β)+2 +C(β)$.
2
2. The polynomials
For each integer m≥1, we proved that there exists a polynomial P=Pm∈
Z[x1,x
2]withthefollowingproperties:
(1) We have ∂j
1P(β,β)=0forj=0,...,m−1.
(2) We have Deg2P≤1andDeg1P≤(1 + $)deg(β)m.
2
(3) We have |P|≤C(β,$)m.
3. The rational point
Suppose that r1,r
2are good rational approximations to βin the sense that
#β−ri#≤#r1#−γ.
1
&

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