The Polynomial method and cardinality of sum-sets in Zp

The Polynomial method and cardinality of
sum-sets in Z p
Dusan Milijancevic
Mathematical High School, Belgrade
dmilijancevic@gmail.com
Mihajlo Cekic
Mathematical High School, Belgrade
mcekic@sezampro.yu
December 6, 2009
Abstract
In this paper we prove two well known theorems Cauchy-Davenport
inequality and Heilbrunn-Erdos theorem and their extension. The proof
is based on a recently developed polynomial method, which in a relatively
simple way proves many facts that are considered to be very difficult to
solve.
Keywords: Caucy-Davenport, Erdos-Heilbrunn, polynomial, coefficient
1 Introduction
The first appearance of the Additive number theory is in the works of Cauchy
and Warring who sowed a couple of fundamental results in this area. Their work
is continued by Erdos, and later on by Alon, Ruzsa and others. Even today this
area contains many open problems that are partially related to the cardinality
of a sum of specific sets in abelian groups. We will focus on the group of
residues of a prime module equipped with the standard operation of addition +.
Let us define the sum of two sets in Z p as a set of all possible Sums of their
elements.
Definition 1 Denote by + the operation on subsets of Z p , where p is a prime
number, that is is defined as A + B = {a + b|a ∈ A, b ∈ B} for all subsets A and
B of Z p .
1

In a similar vein we define other operations on subsets. We will use only the
difference of sets whose definition is bellow.
Definition 2 Denote by − the operation on subsets of Z p , where p is a prime
number, that is defined as A − B = {a − b|a ∈ A, b ∈ B}, for all subsets A and
B of Z p .
The theorem below concerns with the simple estimates of sum of cardinalities
of two sets:
Theorem 1 If A and B are finite non empty subsets of Z p the following inequalities
hold:
a) |A + B| ≥ max{|A|, |B|}
b) |A + B| ≤ |A||B|.
Proof: a) Translate a set A (add one element on all its elements) so that it
contains 0 and keep its notation. Cardinalities of sets A, B and A + B are same
as before translation. Similarly we can translate B so that it contains 0. As
0 ∈ A, B both A and B are subsets of A + B, with |A|, |B| ≤ |A + B|, that is
max{|A|, |B|} ≤ |A + B|.
b) From the definition of A + B number |A + B| can not exceed |A||B|.
Many results and conjectures in this area are due to Erdos. All of them
were regarded as combinatorial until the new polynomial was created which
gave us a new point of view on these problems. In the following section we are
going the sow how this method applies to this kind of problem and a natural
generalization that follows from it.
2 Cauchy-Davenport inequality
Cauchy-Davenport inequality is probably the first result in this vein. We will
show two proofs of it. The first one is due to Terence Tao (reference [1]) that
uses methods of group theory while the other is quite elementary and easily
generalized (it appeared in [2] due to Noga Alon, Melvyn B. Nathanson and
Imre Ruzsa). Using the idea in the second proof we will derive two other results
in later sections.
Theorem 2 (Cauchy-Davenport inequality) If A and B are two non empty subsets
of Z p for some prime p then the inequality |A + B| ≥ min|A| + |B| − 1, p
is true.
2

The First Proof: If |A| + |B| − 1 ≥ p sets A and {x} − B have a non empty
intersection for every x ∈ Z p with A + B = Z p . Therefore |A + B| = p and our
inequality is true (it reduces to equality). Hence we may take that |A|+|B|1 < p.
Suppose that the inequality is not true and from all of its counterexamples
(A, B) select the one for which |A| is the least. Let it be (A, B). Now we can
assume that |A| ≥ 2 as in the case of |A| = 1 the inequality is obviously valid.
Note that by replacing the set B with the set B + {x} for arbitrary x ∈ Z p
(translation) both the left and right sides of inequality have the same value as
before transformation. So translate B for an element such that |A ∩ B| ≥ 1.
Let A ′ = A ∩ B and B ′ = A cupB. Therefore |A| + |B| = |A ′ | + |B ′ |, and so
is A ′ + B ′ ⊂ A + B. It implies |A ′ | + |B ′ | ≤ |A| + |B|. Hence (A ′ , B ′ ) is an additional
counterexample to the given inequality. However |A| is minimal with this
property so we must have |A ′ | ≥ |A|, and consequently A ′ ⊂ A, that is A ′ = A.
It follows that for every minimal counterexample {A, B} such that |A ∩ B| ≥ 1
it holds A ⊂ B. However, if (A, B) is a counterexample then (A + {x}, B) is
counterexample too for each x ∈ Z p so x ∈ B −A. Therefore for each x ∈ B −A
is A + {x} ⊂ B, i.e. A + (B − A) ⊂ B and B + (A − A) ⊂ B, because of
associative and commutativity of addition. As set A − A contains 0 it must be
B subsetB + AA, and B = B + (A − A), ie. |B + C| = |B|, for C = A − A. We
will use the following lemma that is essential for the characterization of B and
C :
Lemma 1 Let A and B be two nonempty subsets of G, where G = (G, +) is a
finite abelian group. The equality |A + B| = |a| holds if and only if A is union
of a finite number of translates of G ′ and B is a subset of one of translates of
G ′ where G ′ = (G, +) is a subgroup of the group G.
Proof: Let us prove the first direction of lemma. Assume that A = ⊎ k i=1 G′ + {a i }
and B ⊂ G ′ +b for some a i 1 ≤ i ≤ k and b from G. We want to prove |A+B| =
|A|. Note that the intersection of two translates of subgroup G ′ is the empty set
or those the two translates coincide. Hence A = {a 1 , a 2 , ..., a k } + G ′ = A ′ + G,
and
A + B ⊂ A ′ + G ′ + b + G ′ = A ′ + b + G ′ = {a 1 + b, a 2 + b + b, ..., a k } + G ′ .
If some of two translates of form {a i +b}+G ′ and {a j }+{b}+G ′ have some
common element for i ≠ j, then two translates of the form {a i } + G and a j + G
will have a common element too, which is impossible for i ≠ j. Therefore any
two of the translates are disjoint and |A + B| ≤ k|G ′ |. However as |A| = k|G ′ |,
and |A + B| ≤ |a| it is |A + B| = |A| what we wanted to prove.
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Evidence of a different direction of lemma. Let A, B ⊂ G and let |A + B| =
|A|. Consider the set G ′ = {g ∈ G|{±g} + A = A}. Obviously, 0 ∈ G ′ . For G ′
to be a group it is enough to prove that for all g 1 , g 2 ∈ G ′ is g 1 − g 2 ∈ G ′ . As
{g 1 − g 2 } + G ′ = {g 1 } + {−g 2 } + G ′ = {g 1 } + G ′ = G ′ so it is g 1 − g 2 ∈ G ′ .
It implies that G ′ is a group with respect to operation +. Hence A is a union
of translates of G ′ as A = G ′ + A. Consider an element b from B. Then
|A + b| = |A|, and so it is A + b ⊂ A + B. Therefore it is A + B = A + b. As
b is an arbitrary element of B the equality A + b = A + b ′ is true for any two
elements b and b ′ from B, i.e. b ′ − b + A = A, which means that b ′ − b ∈ G ′ , i.e.
B is a subset of the translates of B. Thus the proof of lemma is completed.
Using this lemma on the equality |B +C| = |B|, we derive that B is an union
of translates of a subgroup of Z p and C is a subset of the translate of the same
subgroup. As the only subgroups of Z p are the whole group and {0}, it is either
B = Z p or |C| = 1 which means that |A| = 1. However, this is a contradiction
because in both cases we have a valid inequality. Hence the proof of the whole
theorem is completed.
The second proof: As in the first proof we can assume that |A| + |B| −
1 < p. Let |A| = k, |B| = l and |A + B| = n. Suppose the opposite, i.e.
n < minp, k + l − 1 and n < p and n < k + l− 1. Consider the polynomial
fx, y =
∏
∑
(x + y − c) =
f i,j x i y j ,
c∈A+B
i+j≤n,i,j∈N
where f i,j are some integers. This polynomial can be written in the following
format:
fx, y =
∑
f i,j x i y j +
∑
f i,j x i y j +
∑
f i,j x i y j +
∑
f i,j x i y j ,
i

∏
The determinant of this system is the Vandermonde i.e. it is equal to
a i − a j . Hence it is different from 0, since a i ≠ a j for i ≠ j. Therefore
1≤i

holds in this case. We can take that |A| + |B| − 2 < p. Let |A| = k, |B| = l and
|A + B| = n. Assume the opposite i.e. that n < p and n < k + l − 2. Consider
the polynomial
f(x, y) = (x − y) ∏ c∈A+ 1B x + y − c = ∑ i+j≤n,i,j∈N 0
f i,j x i y j ,
for some integers f i,j . Hence f(x, y) = 0 for all (x, y) ∈ A × B. Applying
the lemma shown in the second proof of Cauchy-Davenport inequality we can
transform f(x, y) into polynomial p(x, y) whose degree in x is less than k and
degree in y is less than l, which has the same value as f(x, y) (of course in Z p ).
Thus p(x, y) ≡ 0 . Consider the coefficient of x u y v where u + v = deg(f) and
u < k, v < l. For these numbers is deg(f) < k + l. Therefore the coefficient of
x u y v in f(x, y) equals
( ) ( )
u + v − 1 u + v − 1
−
= u + v − 1! 1
u − 1 v − 1 u − 1!v − 1! v − 1 u − vu + v − 1!
=
u uvu − 1!v − 1! ,
which can not be divisible by p since u + v − 1 < p. This is a contradiction
because the coefficient of x u y v in the polynomial p equals 0.
Using this claim we are able to profe Erdos-Heilbrunn theorem.
Theorem 4 (Erdos-Heilbrunn theorem) Let A be a non empty subset of Z p . It
holds |A + 1 A| ≥ min{p, 2|A| − 3}.
Proof: Applying the previous theorem to the sets A and B where B = A {x}
and x is an arbitrary element of A we will derive this theorem. As A + 1 B =
A + 1 A because all numbers of a form a + x (a ≠ x and a ∈ A) belong to A + 1 B
for x ∈ A, a ∈ B and a ≠ x. Then |A+ 1 A| = |A+ 1 B| ≥ min{p, |A|+|B|−2} =
min{p, 2|A| − 3} what we wanted to prove.
4 Generalization Cauchy-Davenport inequality
and Erdos-Heilbrunn theorem
In this section we will show one way for generalization of the previous theorems
continuing with the use of the polynomial method and the lemma in the
second prove of the Cauchy-Davenport inequality. This result is due to Noga
Alon, Melvyn B. Nathanson and Imre Ruzsa and was published in [3]. We are
starting with a definition that will generalize sum of sets.
Definition 3 For a polynomial h = h(x 0 , x 1 , ..., x k ) on Z p where p is a prime
number and sets A 0 , A 1 , ..., A k define its addition by
⊕ h
i=0
k∑
A i = {a 0 + a 1 + ... + a k |a i ∈ A i , h(a 0 , a 1 , ..., a k ) ≠ 0}.
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As we will show later the former two theorems are special cassis of this
result. We will perform a little change in the proof concerning the selection of
the charachteristic polynomial in order to be able to choose specific coefficients.
Theorem 5 Let p be a prime number and h = h(x 0 , x 1 , ..., x k ) a polynomial
over Z p . Let A 0 , A 1 , .., A k be a non empty subsets of Z p where |A i | = c i + 1.
Moreover let m = ∑ k
i=0 (c i)−deg(h) and m ≥ 0. If the coefficient of x c 0
0 xc 1
1 ...xc k
k
in polynomial
is non zero then
(x 0 + x 1 + ... + x k ) m h(x 0 , x 1 , ..., x k )
k∑
| ⊕ h A i | ≥ m + 1.
i=0
Proof:
Suppose that the statement is false, and let E be a set of m elements of Z p
containing the set ⊕ h
∑ k
i=0 A i. Let Q = Q(x 0 , ..., x k ) be a polynomial defined
as follows:
Note that
Q(x 0 , ..., x k ) = h(x 0 , ..., x k ) · ∏
(x 0 + ... + x k − e).
e∈E
Q(x 0 , ..., x k ) = 0 for all (x 0 , ..., x k ) ∈ A 0 × A 1 × ... × A k .
∑ k
This is because for any (x 0 , ..., x k ) is h(x 0 , ..., x k ) = 0 or x 0 , ..., x k ∈ ⊕ h i=0 A i ⊂
E, and x 0 + x 1 + ... + x n = e for the corresponding e ∈ E. For the degree of
the polynomial Q we have deg(Q) = m + deg(h) = ∑ k
i=0 c i and therefore the
coefficient of monomial x c0 · x k ...x c k
k
in the polynomial Q is the same as in the
polynomial (x 0 + ... + x k ) m h(x 0 , ..., x k ), which is a nonzero, according to the
assumption.
By the lemma shown in the second proof of Cauchy-Davenport theorem we can
reduce the degree of every x i in Q to up to c i , so that it does not change its
value on A i . The transformed polynomial for has degree in x i less than c i and
is 0 on A 0 × ...A 1 × A k then it is a zero polynomial in Z p .
The coefficient of monomial ∏ k
i=0 xc i
i in the initial polynomial Q is equal to
the coefficients in the transformed polynomial and is equal to 0, because there
were no changes in this coefficient, because otherwise the degree of polynomial Q
would be larger than ∑ k
i=0 c i, which is impossible. Hence the theorem is proved.
Note that for m < 0 the same inequality is valid
k∑
| ⊕ h A i | ≥ 0 ≥ m + 1,
i=0
7

which means that a given inequality is valid for negative m.
Under the given conditions of the theorem it is also true that
so that m < p.
k∑
p ≥ | ⊕ h A i | ≥ m + 1,
i=0
Alternative proof of Cauchy-Davenport inequality can be made if the inequality
|A| + |B|1 < p is obtained and this theorem applied to the case k =
1 and h ≡ 1. Similarly, an alternative proof of Theorem 3. can be made if
after obtaining the inequality |A| + |B| − 2 < p and eliminating the trivial case
|A| = |B| = 1 we apply this theorem to the case k = 1 and hx 0 , x 1 ≡ x 0 − x 1 .
By choosing different polynomials for h we can make additional assertions. The
problem with many variables occurs when the coefficient that should be zero
needs to calculated, that in some cases may represent some of the open combinatorial
problems. One of the interesting results in this context was proved by
Dias de Silva and Hamidoune:
Theorem 6 Let p be a prime number , A a subset of Z p and k ∈ N. If the
polynomial h is defined with h =
∏ (x i − x j ) i, j ∈ N 0 then
0≤i

[5] Noga Alon, Combinatorial Nullstellensatz, Combinatorics, Probability and
Computing 8 1-2: 7-29
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