[PPT] - An Introduction to Integer Valued Polynomials Marie-Andre B.Langlois PowerPoint Presentation

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An Introduction to Integer Valued Polynomials

Marie-Andrée B.Langlois

Dalhousie University

February 2018

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What are Integer Valued Polynomials ?

For this talk Definition For any subset S of Z the ring of integer valued polynomials on S is defined to be Int(S) = {f (x) ∈ Q[x] | f (S) ⊆ Z}.

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What are Integer Valued Polynomials ?

For this talk Definition For any subset S of Z the ring of integer valued polynomials on S is defined to be Int(S) = {f (x) ∈ Q[x] | f (S) ⊆ Z}. The formal definition is for a domain D and field of fractions K. Definition For any subset S of D the ring of integer valued polynomials on S is defined to be Int(S, D) = {f (x) ∈ K[x] | f (S) ⊆ D}.

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What are Integer Valued Polynomials ?

Lets start with Int(Z) and find some examples :

25x5 − 13x3 + 7x − 23

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What are Integer Valued Polynomials ?

Lets start with Int(Z) and find some examples :

25x5 − 13x3 + 7x − 23
is a boring example, we want non-integer coefficients !

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What are Integer Valued Polynomials ?

Lets start with Int(Z) and find some examples :

25x5 − 13x3 + 7x − 23
is a boring example, we want non-integer coefficients !
Degree 1, can we do better than x ?

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What are Integer Valued Polynomials ?

Lets start with Int(Z) and find some examples :

25x5 − 13x3 + 7x − 23
is a boring example, we want non-integer coefficients !
Degree 1, can we do better than x ?
Degree 2,

x(x−1) 2

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What are Integer Valued Polynomials ?

Lets start with Int(Z) and find some examples :

25x5 − 13x3 + 7x − 23
is a boring example, we want non-integer coefficients !
Degree 1, can we do better than x ?
Degree 2,

x(x−1) 2

Degree 3,

x(x−1)(x−2) 2·3

Marie B.Langlois Dalhousie University Integer Valued Polynomials 3/ 37

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What are Integer Valued Polynomials ?

In general, for degree n : x n

= x(x − 1) · · · (x − n + 1)

n! Theorem A polynomial is integer valued on Z if and only if it can be written as a Z-linear combination of the polynomials x n

= x(x − 1) · · · (x − k + 1)

n! , for n = 0, 1, 2, . . ..

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Today’s Plan

We will go over the following :

Bases and IVPs on subsets of the integers.
p-orderings, p-sequences and invariants of Int(S).
Multivariable and homogeneous case.

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Int(S) is a Ring

Most of the axioms follow from Q[x] being a ring.
Int(S) is also closed under addition and multiplication.
Int(S) is a Z-module.

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Int(S) is a Module

Definition An R-module M, over the ring R consist of an abelian group (M, +) and an operation R × M → M (scalar multiplication). For all r, s ∈ R, x, y ∈ M we have

1 r(x + y) = rx + ry. 2 (r + s)x = rx + sx. 3 (rs)(x) = r(sx). 4 1Rx = x. Marie B.Langlois Dalhousie University Integer Valued Polynomials 7/ 37

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Int(S) is a Z-module

In this case R = Z, we want for all m, n ∈ Z and f (x), g(x) ∈ Int(S) :

1 m(f (x) + g(x)) = m · f (x) + m · g(x). 2 (m + n)f (x) = m · f (x) + n · f (x). 3 (mn)f (x) = m(n · f (x)). 4 1 · f (x) = f (x).

Multiply f (x) an IVP by an integer n will preserve its integer valued property.

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Bases

Definition A basis B of the R-module B is said to be a regular basis if it is formed by one and only one polynomial of each degree.

A regular basis for Int(Z) is {1}∪

x

n

= x(x−1)···(x−n+1)

n!

n≥1.

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What if S ⊂ Z

We will look at Int(S) for S ⊂ Z, to motivate why we need better tools to find bases. Here are examples of sets we can consider

Even/Odd integers
Prime numbers
Fibonacci Numbers
Sum of ℓ d-th powers

x = xd

1 + xd 2 + · · · + xd ℓ

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Even/Odd Integers

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Even/Odd Integers

For S = 2Z, a basis for Int(S) is made of the polynomials

x/2

n

1, x

2, x(x − 2) 8 , x(x − 2)(x − 4) 48 , . . .

Marie B.Langlois

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Even/Odd Integers

For S = 2Z, a basis for Int(S) is made of the polynomials

x/2

n

1, x

2, x(x − 2) 8 , x(x − 2)(x − 4) 48 , . . .

For S = 1 + 2Z, a basis for Int(S) is made of the polynomials

(x−1)/2

n

1, (x − 1)

2 , (x − 1)(x − 3) 8 , (x − 1)(x − 3)(x − 5) 48 , . . .

Marie B.Langlois

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Prime Numbers

The beginning of a basis for Int(P) : f0 = 1, f1 = (x − 1), f2 = (x − 1)(x − 2) 2 , f3 = (x − 1)(x − 2)(x − 3) 24 , f4 = (x − 1)2(x − 2)(x − 3) 48 , f5 = (x − 1)(x − 2)(x − 3)(x − 5)(x − 79) 5760 , . . . and f3(4) = 1

4, f4(4) = 3 4 and f5(4) = 5 64.

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Manjul Bhargava

We will go over some work that Bhargava did during his undergraduate degree and define

p-orderings
p-sequences.

For the next part

f

the presentation we will work locally at a prime p.

Source

f

Image : https://opc.mfo.de/ detail?photo_id=7108 Marie B.Langlois Dalhousie University Integer Valued Polynomials 13/ 37

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A Game Called p-orderings

Fix a prime p. Definition (Bhargava) A p-ordering of S a subset of Z is a sequence (an)n≥0, such that a0 is arbitrarily chosen and for each n > 0, an ∈ S is chosen to minimize νp((a0 − an) · · · (an−1 − an)). where νp(n) = max{ν ∈ N: pν|n} n ≥ 0 ∞ n = 0

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Lets play !

Let p = 2 and S = {0, 1, 2, 3, 4}

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Lets play !

Let p = 2 and S = {0, 1, 2, 3, 4}

a0 = 0

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Lets play !

Let p = 2 and S = {0, 1, 2, 3, 4}

a0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1

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Lets play !

Let p = 2 and S = {0, 1, 2, 3, 4}

a0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1

a2

(0 − a2)(1 − a2) pick 2 or 3, a2 = 2

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Lets play !

Let p = 2 and S = {0, 1, 2, 3, 4}

a0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1

a2

(0 − a2)(1 − a2) pick 2 or 3, a2 = 2

a3

(0− a3)(1− a3)(2− a3), need to pick an odd value a3 = 3

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Lets play !

Let p = 2 and S = {0, 1, 2, 3, 4}

a0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1

a2

(0 − a2)(1 − a2) pick 2 or 3, a2 = 2

a3

(0− a3)(1− a3)(2− a3), need to pick an odd value a3 = 3

a4 = 4.

Our p-ordering of S is {0, 1, 2, 3, 4}.

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In General

Proposition (Bhargava) The natural ordering of Z≥0 with ai = i is a p-ordering of Z for all primes p. Are p-orderings unique ?

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In General

Proposition (Bhargava) The natural ordering of Z≥0 with ai = i is a p-ordering of Z for all primes p. Are p-orderings unique ? No, we made some choices in the previous example.

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p-sequences

Definition (Bhargava) Given (an)n≥0 a p-ordering of S a subset of Z with α0 = 0 and αn(S, p) = νp((a0 − an) · · · (an−1 − an)), {αn(S, p)} is the associated p-sequence of S. A nice property for a set S is that the p-sequence is independent of the choice of p-ordering. Example : {αn(Z, p)} = {νp(n!)}.

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

a0 = 0 and α0 = 0

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

a0 = 0 and α0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1 and α1 = 0

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

a0 = 0 and α0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1 and α1 = 0

a2

(0 − a2)(1 − a2), a2 = 27 is the best choice and α2 = 1.

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

a0 = 0 and α0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1 and α1 = 0

a2

(0 − a2)(1 − a2), a2 = 27 is the best choice and α2 = 1.

a3

(0 − a3)(1 − a3)(27 − a3), we can choose between 8 and 125, a3 = 8 and α3 = 3.

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

a0 = 0 and α0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1 and α1 = 0

a2

(0 − a2)(1 − a2), a2 = 27 is the best choice and α2 = 1.

a3

(0 − a3)(1 − a3)(27 − a3), we can choose between 8 and 125, a3 = 8 and α3 = 3.

a4

(0 − a4)(1 − a4)(27 − a4)(8 − a4) start checking with 125, since we hope for α < 6, we obtain a4 = 125 and α4 = 3

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A More Interesting Version of the Game

Let p = 2 and S = {0, 1, 8, 27, 64, 125}

a0 = 0 and α0 = 0
for a1 we want to minimize the power of 2 dividing (0 − a1)

take any odd number, a1 = 1 and α1 = 0

a2

(0 − a2)(1 − a2), a2 = 27 is the best choice and α2 = 1.

a3

(0 − a3)(1 − a3)(27 − a3), we can choose between 8 and 125, a3 = 8 and α3 = 3.

a4

(0 − a4)(1 − a4)(27 − a4)(8 − a4) start checking with 125, since we hope for α < 6, we obtain a4 = 125 and α4 = 3

a5 = 64 and (0 − 64)(1 − 64)(27 − 64)(8 − 64)(125 − 64) =

26(−63)(−37)(−56)(61) and α5 = 9. Our p-ordering of S is {0, 1, 27, 8, 125, 64} and the p-sequence is {0, 0, 1, 3, 3, 9}.

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Factorial Function

The factorial function is very important when working with IVPs. It was in the denominators of the basis elements of Int(Z) and in the p-sequence of Z. What is the factorial function when working with S not Z ? Proposition

For any non-negative k and ℓ, (k +ℓ)! is still a multiple of k!ℓ!.
Let f be a primitive polynomial with integer coefficients of

degree k, and let d(Z, f ) = gcd{f (a) | a ∈ Z}. Then d(Z, f ) divides k!. This is called the fixed divisor of f .

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Generalized Factorial Function

Definition (Bhargava) Let S be any subset of Z. the the factorial function on S denoted k!S is defined by k!S =

p pαk(S,p).

This definition preserves many properties of the factorial function

n Z :
For any non-negative k and ℓ, (k + ℓ)!S is still a multiple of

k!Sℓ!S.

Let f be a primitive polynomial of degree k, and let

d(S, f ) = gcd{f (a) | a ∈ S}. Then d(S, f ) divides k!S.

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Generalized Factorial is the Same on Z

We know that 6! = 1 · 2 · 3 · 4 · 5 · 6, we use the generalized factorial instead : αn(Z, 2) = {0, 0, 1, 1, 3, 3, 4, 4, 7, 8, . . .} αn(Z, 3) = {0, 0, 0, 1, 1, 1, 2, 2, 2, 4, . . .} αn(Z, 5) = {0, 0, 0, 0, 0, 1, 1, 1, 1, 1, . . .} 6! = 24 · 32 · 5

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Generalized Factorial is the Same on Z

We know that 6! = 1 · 2 · 3 · 4 · 5 · 6, we use the generalized factorial instead : αn(Z, 2) = {0, 0, 1, 1, 3, 3, 4, 4, 7, 8, . . .} αn(Z, 3) = {0, 0, 0, 1, 1, 1, 2, 2, 2, 4, . . .} αn(Z, 5) = {0, 0, 0, 0, 0, 1, 1, 1, 1, 1, . . .} 6! = 24 · 32 · 5

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Generalized Factorial on S

Let S = {0, 1, 8, 27, 64, 125}, we will calculate 3!S, the factorial of an element with index 3 in our p-ordering, in out case this was 8 : αn(Z, 2) = {0, 0, 1, 3, 3, 9} αn(Z, 3) = {0, 0, 0, 2, 2, 3} αn(Z, 5) = {0, 0, 0, 0, 0, 3} αn(Z, 7) = {0, 0, 0, 1, 2, 2} 3!S = 23 · 32 · 50 · 71

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Generalized Factorial on S

Let S = {0, 1, 8, 27, 64, 125}, we will calculate 3!S, the factorial of an element with index 3 in our p-ordering, in out case this was 8 : αn(Z, 2) = {0, 0, 1, 3, 3, 9} αn(Z, 3) = {0, 0, 0, 2, 2, 3} αn(Z, 5) = {0, 0, 0, 0, 0, 3} αn(Z, 7) = {0, 0, 0, 1, 2, 2} 3!S = 23 · 32 · 50 · 71

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Basis for IVPs on a subset of Z

Theorem (Bhargava) A polynomial is integer valued on a subset S of Z if and only if it can be written as a Z-linear combination of the polynomials Bk,S k!S = (x − a0,k)(x − a1,k) · · · (x − ak−1,k) k!S , for k = 0, 1, 2, . . . where the Bk,S(x) are the polynomials defined by (x − a0,k)(x − a1,k) · · · (x − ak−1,k), where {ai,k}∞

i=0 is a sequence in Z that, for each prime p dividing

k!S, is term-wise congruent modulo αk(S, p) to some p-ordering of S.

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Characteristic Ideal

Definition (Chabert) The characteristic ideal of index n of Z is the set Jn(Z) formed by 0 and the leading coefficients of the polynomials in Int(S, Z) of degree n. For example, when S = Z, Jn(Z) = 1

n!Z.

Definition (Chabert) The characteristic sequence of S with respect to a fixed prime p is the sequence of negatives of the p-adic valuations of these ideals, denoted by αn(S, p).

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The Valuative Capacity

Definition (Chabert) For S a subset of Z and p a fixed prime, the valuative capacity of S with respect to the prime p is the following limit : LS,p = lim

n→∞

αn(S, p) n . The positive integers in increasing order are a p-ordering of Z and we have that αn(Z, p) = νp(n!). By Legendre’s formula νp(n!) = n− ni

p−1 , we can compute

LZ,p = lim

n→∞

αn(Z, p) n = 1 p − 1.

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Sums of Two and Three Squares

Let S be the set of perfect squares in Z and E = S + S F = S + S + S Theorem (Fares, Johnson) LE,p =     

1 p−1

if p ≡ 1 (mod 4) −1 +

1 +

2p (p−1)2

if p ≡ 3 (mod 4)

−1+ √ 13 2

if p = 2 LF,p =

1

p−1

if p > 2

−25+3 √ 705 52

if p = 2

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Sums of ℓ Elements to the Power of d

For D = {xd | x ∈ Z} and we let ℓD = D + · · · + D, for ℓ terms in the sum. Theorem (B.) Suppose p is a prime and d = pjd′ a positive integer not equal to 4, where p ∤ d′ and let e = 2j + 1. Then, LℓD,p is an algebraic number of degree at most 2. When 0 can be written non-trivially as a sum of ℓ elements to the power of d (mod pe), LℓD,p is a rational number. Corollary (B.) For a fixed ℓ, if d is odd and p is a prime, then LℓD,p ∈ Q.

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Timeline of IVPs

Pólya in 1915
Cahen and Chabert wrote a textbook on the subject, mid 90’s
Bhargava, p-sequences, late 90’s
Chabert, valuative capacity 2001
Chabert, survey article 2014

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Generalization

Recall that the official definition is Int(S, D).
IVPs over quaternions.

[Werner] [Johnson and Pavlovski]

IVPs over matrix rings. [Evrard, Fares and Johnson] [Frisch]

[Werner]

Multivariable case. [Bhargava]

[Evrard]

Homogeneous 2-variable case. [Johnson and Patterson]

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Multivariable

Let n be a positive integer, let S be a subset of Zn and let m0, m1, . . . be an ordering of the monomials of Zn[X] , and consider the Z-algebra Int(S) = {f (X1, . . . , Xn) ∈ Q[X1, . . . , Xn] | f (S) ⊆ Z}. Using matrices, p-orderings and p-sequences can be generalized to the multivariable case.

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Multivariable - A Basis Int(Zn)

Proposition The polynomials x1

r1

x2

r2

· · ·

xn

rn

| r1, . . . , rn ∈ Z, r1, . . . , rn ≥ 0
form a basis of the Z-module of Int(Zn).

In the three variable case, the elements of degree 3 are x 3

,

y 3

,

z 3

,

x 2 y 1

,

x 2 z 1

,

x 1 y 2

,

y 2 z 1

x

1 z 2

,

y 1 z 2

,

x 1 y 1 z 1

Marie B.Langlois

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Homogeneous Polynomials

A homogeneous polynomial is one of the form f (x1, . . . , xn) =

i1+i2+···+in=m

c1xi1

1 xi2 2 · · · xin n .

For degree m they have the property that for a constant h : f (hx1, . . . , hxn) = hmf (x1, . . . , xn)

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Homogeneous as a Product of Linears

When written as a product of linear factors each term must

contain a variable, we can’t have (x − 2) we would need (x − 2y).

Start by focusing on how big of a power of p = 2 we can get.

How can we construct 2-variable IVPs ?

In degree 1 and 2 we need integer coefficients : x, y, xy.
Can we obtain a denominator in degree 3 ?

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Homogeneous as a Product of Linears

When written as a product of linear factors each term must

contain a variable, we can’t have (x − 2) we would need (x − 2y).

Start by focusing on how big of a power of p = 2 we can get.

How can we construct 2-variable IVPs ?

In degree 1 and 2 we need integer coefficients : x, y, xy.
Can we obtain a denominator in degree 3 ?
Yes ! xy(x−y)

2

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Basis for Degree 3, 3 Variables

In degree m we will have (m+1)(m+2)

2

basis elements.

xyz, xy(x − y)

2 , xz(x − z) 2 , yz(y − z) 2 , x2(x − y), x2(x − z), y2(y − x), y2(y − z), z2(z − x), z3 .

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Another Example

For example when m = 6, we have 7 elements with a 20, 14

with a 21, 4 with a 22 and 3 with a 23 in their denominators.

The polynomials with a 23 in their denominator are :

f = 1 4x5y + 7 8x4y2 + 1 8x2y4 + 3 4xy5 g = 3 4x5z + 1 8x4z2 + 7 8x2z4 + 1 4xz5 h = 1 4x5y + 1 2x5z + 3 8x4y2 + 3 4x4z2 + 5 8x2y4 + 1 4x2z4 + 3 4xy5 + 1 2xz5 + 3 4y5z + 1 8y4z2 + 7 8y2z4 + 1 4yz5

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Counting the Basis Elements

20 21 22 23 24 25 26 27 28 29                                               1 3 2 6 3 7 3 4 7 8 5 7 14 6 7 14 4 3 7 7 14 6 9 8 7 14 7 14 3 9 7 14 7 14 13 10 7 14 7 14 21 3 11 7 14 7 14 28 8 12 7 14 7 14 28 14 4 3 13 7 14 7 14 28 14 6 15 14 7 14 7 14 28 14 7 25 3 1

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SLIDE 59

Counting the Basis Elements

20 21 22 23 24 25 26 27 28 29                                               1 3 2 6 3 7 3 4 7 8 5 7 14 6 7 14 4 3 7 7 14 6 9 8 7 14 7 14 3 9 7 14 7 14 13 10 7 14 7 14 21 3 11 7 14 7 14 28 8 12 7 14 7 14 28 14 4 3 13 7 14 7 14 28 14 6 15 14 7 14 7 14 28 14 7 25 3 1

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SLIDE 60

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