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The Polynomial Method Lecture 27

By Siegel’s lemma on integer solutions of linear integer equations (in...

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Diploma in civil engineering

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Winnacunnet High School - Hampton

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Polynomials that vanish to high order at a rational point Suppose that P ∈ Z[x 1 , x 2 ] has the special form

P (x 1 , x 2 ) = P 1 (x 1 )x 2 + P 0 (x 1 ). Suppose that r ∈ Q 2. If P vanishes to high order at a complicated point r, how big do the coefficients of P have to be? More precisely, we suppose that ∂ 1 j P (r) = 0 for 0 ≤ j ≤ l − 1. Last time we gave two examples. The polynomial q 2 x 2 − p 2 which has size ‖r 2 ‖, and the polynomial (q 1 x 1 − p 1 )l, which has size ‖r 1 ‖l. By parameter counting it is possible to do somewhat better.

Proposition 1. For any r ∈ Q 2 , and any l ≥ 0 , there is a polynomial P ∈ Z[x 1 , x 2 ] with the form P (x 1 , x 2 ) = P 1 (x 1 )x 2 + P 0 (x 1 ) obeying the following conditions.

∂ 1 j P (r) = 0 for j = 0, ..., l − 1.
|P | ≤ C(")l‖r 1 ‖

l +! 2 , for any " > 0.

The degree of P is! "− 1 l + log‖r 1 ‖ ‖r 2 ‖.

Proof. We will find our solution

(

by counting pa

)

rameters. We will choose a degree D, and let P 0 , P 1 be polynomials of degree ≤ D. The coefficients of P 0 and P 1 are ≥ 2 D integer variables at our disposal. We wish to satisfy the l equations

∂ 1 j P (r) = 0, j = 0, ..., l − 1. (1) After a minor rewriting, each of these equations is a linear equation in the coef- ficients of P with integer coefficients. If we write P 1 (x 1 ) = i

∑

∑ i bix 1 and P 0 (x 1 ) = i aix i, then

0 = qD

i i 1 q

j 2 (1/j!)∂ 1 P (r) = q 2 (

∑

( )

pi 1 − jqD 1 −i+j) + (

∑

( )

pi 1 − jq 1 D −i+jp 2 ). i j i j The size of the coefficients in the equations is ≤ 2 D‖r 1 ‖D‖r 2 ‖. By Siegel’s lemma on integer solutions of linear integer equations (in the last lecture), we find a non-zero integer solution of these equations with

l |P | ≤

[ l D 3 D · 2 D‖r 1 ‖D ‖r 2 ‖

] 2 D−l ≤ Cl‖r 1 ‖l 2 D −l ‖r 2 ‖ 2 D −l . 1

We choose D = 1000"− 1 l +1000"− 1 log‖r 1 ‖ ‖r 2 ‖. With this value of D, 2 DD− l≤ "/10, l and so the exponent of ‖r 1 ‖ is almost l/2. Also, the term ‖r 2 ‖ 2 D−l ≤ ‖r 1 ‖ !/ 10. "

Combining our parameter counting with the elementary example q 2 x 2 − p 2 , we can find P vanishing to order l at r with |P | on the order of min(‖r 1 ‖l/ 2 , ‖r 2 ‖). The following result shows that these examples are quite sharp. I believe it is a special case of a lemma of Schneider.

Proposition 1. (Schneider) If P (x 1 , x 2 ) = P 1 (x 1 )x 2 + P 0 (x 1 ) ∈ Z[x 1 , x 2 ], and r ∈ Q 2 , and ∂ 1 j P (r) = 0 for j = 0, ..., l − 1 , and if l ≥ 2 , then

|P | ≥ min((2DegP ) − 1 ‖r l 1 1 ‖

− 2 , ‖r 2 ‖).

Remark. We need to assume that l ≥ 2 to get any estimate. If we have vanishing only to order 1, then we could have P (x 1 , x 2 ) = 2x 1 − x 2 , which vanishes at (r 1 , 2 r 1 ) for any rational number r 1. As soon as l ≥ 2, the size of |P | constrains the complexity of r. It can still happen that one component of r is very complicated, but they can’t both be very complicated.

Proof. Our assumption is that

∂j P 1 (r 1 )r 2 + ∂j P 0 (r 1 ) = 0, 0 ≤ j ≤ l − 1. Let V (x) be the vector (P 1 (x), P 0 (x)). Our assumption is that for 0 ≤ j ≤ l − 1, the derivatives ∂j V (r 1 ) all lie on the line V · (r 2 , 1) = 0. In particular, any two of these derivatives are linearly dependent. This tells us that many determinants vanish. If V and W are two vectors in R 2 , we write [V, W ] for the 2 × 2 matrix with first column V and second column W. Therefore,

det[∂j 1 V, ∂j 2 V ](r 1 ) = 0, for any 0 ≤ j 1 , j 2 ≤ l − 1. Now it follows by the Liebniz rule that

∂j det[V, ∂V ](r 1 ) = 0, for any 0 ≤ j ≤ l − 2. Remark: Because the determinant is multilinear, we have the Leibniz rule ∂det[V, W ] = det[∂V, W ]+ det[V, ∂W ], which holds for any vector-valued functions V, W : R → R 2. Now det[V, ∂V ] is a polynomial in one variable with integer coefficients. If this polynomial is non-zero, then by Gauss’s lemma (see last lecture) we conclude that

number, we will check that 1, β, ..., βdeg(β)− 1 form a basis for the vector space Q[β] over the field Q. In particular, any power βi can be expanded as a rational combination of 1, β, ..., βdeg(β)− 1. Substituting in, we can rewrite equation (1) in the form:

deg(β)− 1 0 = βk b A k=

[

iBik + ai ik 0 i i

]

∑ ∑ ∑

= 0,

where Aik and Bik are rational numbers. Since 1, β, ..., βdeg(β)− 1 are linearly indepen- dent over Q, this list of equations is equivalent to the deg(β) equations

∑ biBik +

∑

aiAik = 0, for all 0 ≤ k ≤ deg(β) − 1. (2) i i After multiplying by a large constant to clear the denominators, we get deg(β) equations with integer coefficients. In total, our original l equations ∂ 1 j P (r) = 0 for j = 0, ..., l − 1 are equivalent to deg(β)l integer linear equations in the coefficients of P. Since we have > 2 D coefficients, we can find a non-trivial integer solution as long as D ≥ (1/2)deg(β)l. Our next task is to estimate the size of the solution. To do this, we need to estimate the heights of the coefficients Aik, Bik. Also we get a much better estimate by taking D slightly larger than (1/2)deg(β)l, and for this reason we choose D to be the least integer ≥ (1 + ")(1/2)deg(β)l. To estimate the heights of Aik, Bik, we consider more carefully how to expand βd in terms of 1, β, ..., βd− 1.

Lemma 2. Suppose Q(β) = 0, where Q ∈ Z[x] with degree deg(Q) = deg(β) and leading coefficient qdeg(beta). Then for any d ≥ 0 , we can write

deg(β)− 1 qdegd (β)βd = Akdβ k, k= where A kd ∈ Z and |Akd| ≤ [2|Q|]d.

∑

Proof. We have 0 = Q(β) = ∑ deg(β) k=0 qkβ

k. We do the proof by induction on d,

starting with d = deg(β). For d = deg(β), the equation Q(β) = 0 directly gives

deg(β)− 1 deg(β) qdeg(be a t )βdeg(β)=

∑

(−qk)βk. (∗) k= If we multiply both sides by q eg(β)− 1 deg(β) , we get a good expansion for the case d = deg(β)− 1 deg(β). Now we proceed by induction. Suppose that qd gde ( β )βd =

∑ k k=0 Akdβ . Multiplying by qdeg(β)β, we get

deg(β)− 1 deg(β)− 1 deg(β)− 1

q deg (β)+1 deg(β)+ deg(β) β =

∑

A kdqdeg(β)βk+ 1 =

∑

A k− 1 ,dq deg(β)β k+ k

∑

Adeg(β)− 1 ,d(−qk)β. k=0 k=1 k= " Plugging in this lemma, we see that qdegD (β)Aik, qdegD(β)Bik are integers of size ≤

D[2|Q|]D. The integer matrix that we are solving has coefficients of size ≤ D[2|Q|]D. It is a matrix with dimensions (2D + 2) × deg(β)l, and so it has operator norm ≤ (2D + 2)D[2|Q|]D ≤ C(β)D. Now applying Siegel’s lemma, we see that we can find an integer solution P with |P | bounded by

deg(β)l C(β)D 2 D−deg(β)l ≤ C(β)D/!. Since D ≤ C(β)l, we can redefine C(β) so that |P | ≤ C(β)l/!. "

Summary Suppose that β is an algebraic number, and that r 1 , r 2 are two very good rational approximations of β. We may suppose that ‖r 1 ‖ is very large and ‖r 2 ‖ is (much) larger. Say ‖r 2 ‖ ∼ ‖r 1 ‖m. We consider polynomials P ∈ Z[x 1 , x 2 ] of the simple form P (x 1 , x 2 ) = P 1 (x 1 )x 2 + P 0 (x 1 ). We can arrange that ∂ 1 j P (β, β) = 0 for 0 ≤ j ≤ m − 1 with |P | ≤ C(β)m. On the other hand, if ∂ 1 j P (r) = 0 for 0 ≤ j ≤ l − 1, then we must have |P | # ‖r 1 ‖l/ 2. Since ‖r 1 ‖ is much larger than C(β), it follows that l must be much smaller than m. This creates a certain tension. As we will see, if r was too close to (β, β), than P would have to vanish too much at r, giving a contradiction.

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The Polynomial Method Lecture 27

Subject: Diploma in civil engineering

32 Documents

Students shared 32 documents in this course

School: Winnacunnet High School - Hampton

Was this document helpful?

PROOF OF THUE’S THEOREM – PART II

1. Polynomials that vanish to high order at a rational point

Suppose that P∈Z[x1,x

2]hasthespecialform

P(x1,x

2)=P1(x1)x2+P0(x1).

Suppose that r∈Q2.IfPvanishes to high order at a complicated point r,how

big do the coefficients of Phave to be? More precisely, we suppose that ∂j

1P(r)=0

for 0 ≤j≤l−1. Last time we gave two examples. The polynomial q2x2−p2which

has size $r2$,andthepolynomial(q1x1−p1)l,whichhassize$r1$l.

By parameter counting it is possible to do somewhat better.

Proposition 1.1. For any r∈Q2,andanyl≥0,thereisapolynomialP∈Z[x1,x

with the form P(x1,x

2)=P1(x1)x2+P0(x1)obeying the following conditions.

•∂j

1P(r)=0for j=0,...,l−1.

•|P|≤C(")l$r1$l+!

2,forany">0.

•The degree of Pis !"−1l+log

"r1"$r2$.

Proof. We will find our solution !by counting pa

"rameters. We will choose a degree

D,andletP0,P

1be polynomials of degree ≤D.ThecoefficientsofP0and P1are

≥2Dinteger variables at our disposal. We wish to satisfy the lequations

∂j

1P(r)=0,j =0,...,l−1.(1)

After a minor rewriting, each of these equations is a linear equation in the coef-

ficients of Pwith integer coefficients. If we write P1(xi

1)=#ibix1and P0(x1)=

#iaixi,then

0=qDi i

1qj

2(1/j!)∂1P(r)=q2($bi%&

pi−j

1qD−i+j

1)+(

$ai%&

pi−j

1qD−i+j

1p2).

j j

i i

The size of the coefficients in the equations is ≤2D$r1$D$r2$.

By Siegel’s lemma on integer solutions of linear integer equations (in the last

lecture), we find a non-zero integer solution of these equations with

|P|≤'lD

3D·2D$rD

1$$r2$(2D−l≤Cl$rl2Dl 2Dl

1$−$r2$−.

The Polynomial Method Lecture 27

Diploma in civil engineering

Winnacunnet High School - Hampton

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)

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∑

( )

∑

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[

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∑ ∑ ∑

= 0,

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The Polynomial Method Lecture 27

Subject: Diploma in civil engineering

School: Winnacunnet High School - Hampton

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