Adding Distinct Congruence Classes Modulo a Prime

TLDR: The Erdős-Heilbronn conjecture was recently proven by Dias da Silva and Hamidoune as mentioned in this paper using linear algebra and the representation theory of the symmetric group.

Abstract: The Cauchy-Davenport theorem states that if A and B are nonempty sets of congruence classes modulo a prime p, and if |A| = k and |B| = l, then the sumset A + B contains at least min(p, k + l − 1) congruence classes. It follows that the sumset 2A contains at least min(p, 2k − 1) congruence classes. Erdős and Heilbronn conjectured 30 years ago that there are at least min(p, 2k − 3) congruence classes that can be written as the sum of two distinct elements of A. Erdős has frequently mentioned this problem in his lectures and papers (for example, Erdős-Graham [4, p. 95]). The conjecture was recently proven by Dias da Silva and Hamidoune [3], using linear algebra and the representation theory of the symmetric group. The purpose of this paper is to give a simple proof of the Erdős-Heilbronn conjecture that uses only the most elementary properties of polynomials. The method, in fact, yields generalizations of both the Erdős-Heilbronn conjecture and the Cauchy-Davenport theorem.

Full text

Adding distinct congruence classes
modulo a prime
Noga Alon
∗
Melvyn B. Nathanson
†
Imre Ruzsa
‡
1 The Erd˝os-Heilbronn conjecture
The Cauchy-Davenport theorem states that if A and B are nonempty sets of
congruence classes modulo a prime p, and if |A| = k and |B| = l, then the
sumset A + B contains at least min(p, k + l − 1) congruence classes. It follows
that the sumset 2A contains at least min(p, 2k − 1) congruence classes. Erd˝os
and Heilbronn conjectured 30 years ago that there are at least min(p, 2k − 3)
congruence classes that can be written as the sum of two distinct elements of
A. Erd˝os has frequently mentioned this problem in his lectures and papers (for
example, Erd˝os-Graham [4, p. 95]). The conjecture was recently proven by
Dias da Silva and Hamidoune [3], using linear algebra and the representation
theory of the symmetric group. The purpose of this paper is to give a simple
proof of the Erd˝os-Heilbronn conjecture that uses only the most elementary
properties of polynomials. The method, in fact, yields generalizations of both
the Erd˝os-Heilbronn conjecture and the Cauchy-Davenport theorem.
2 The polynomial method
Lemma 1 (Alon-Tarsi [2]) Let A and B be nonempty subsets of a field F
with |A| = k and |B| = l. Let f(x, y) be a polynomial with coefficients in F and
∗
Institute for Advanced Study, Princeton, NJ 08540, and Department of Mathematics,
Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv, Israel.
E-mail: noga@math.tau.ac.il. Research supported in part by the Sloan Foundation, Grant No.
93-6-6. Alon also wishes to thank Doron Zeilberger for helpful discussions.
†
Department of Mathematics, Lehman College (CUNY), Bronx, New York 10468. E-mail:
nathansn@dimacs.rutgers.edu. Research supported in part by grants from the PSC-CUNY
Research Award Program
‡
Mathematical Institute of the Hungarian Academy of Sciences, Budapest, P.O.B. 127, H-
1364, Hungary. E-mail: h1140ruz@ella.hu. Research supported in part by DIMACS, Rutgers
University, and by the Hungarian National Foundation for Scientific Research, Grant No.
1901.
1

of degree at most k − 1 in x and l − 1 in y. If f (a, b) = 0 for all a ∈ A and
b ∈ B, then f(x, y) is identically zero.
Proof. This follows immediately from the fact that a nonzero polynomial
p(x) ∈ F [x] of degree at most k − 1 cannot have k distinct roots in F . We can
write
f(x, y) =
k−1
X
i=0
l−1
X
j=0
f
i,j
x
i
y
j
=
k−1
X
i=0
v
i
(y)x
i
,
where
v
i
(y) =
l−1
X
j=0
f
i,j
y
j
is a polynomial of degree at most l − 1 in y. Fix b ∈ B. Then
u(x) =
k−1
X
i=0
v
i
(b)x
i
is a polynomial of degree at most k−1 in x such that u(a) = 0 for all a ∈ A. Since
u(x) has at least k distinct roots, it follows that u(x) is the zero polynomial,
and so v
i
(b) = 0 for all b ∈ B. Since deg(v
i
(y)) ≤ l − 1 and |B| = l, it follows
that v
i
(y) is the zero polynomial, and so f
i,j
= 0 for all i and j. This completes
the proof. 2
Lemma 2 Let A be a finite subset of a field F , and let |A| = k. For every
m ≥ k there exists a polynomial g
m
(x) ∈ F [x] of degree at most k − 1 such that
g
m
(a) = a
m
for all a ∈ A.
Proof. Let A = {a
0
, a
1
, . . . , a
k−1
}. We must show that there exists a
polynomial u(x) = u
0
+ u
1
x + · · · + u
k−1
x
k−1
∈ F [x] such that
u(a
i
) = u
0
+ u
1
a
i
+ u
2
a
2
i
+ · · · + u
k−1
a
k−1
i
= a
m
i
for i = 0, 1, . . . , k − 1. This is a system of k linear equations in the k unknowns
u
0
, u
1
, . . . , u
k−1
, and it has a solution if the determinant of the coefficients of
the unknowns is nonzero. The Lemma follows immediately from the observation
that this determinant is the Vandermonde determinant









1 a
0
a
2
0
· · · a
k−1
0
1 a
1
a
2
1
· · · a
k−1
1
.
.
.
1 a
k−1
a
2
k−1
· · · a
k−1
k−1









=
Y
0≤i<j≤k−1
(a
j
− a
i
) 6= 0.
2
2

Theorem 1 Let p be a prime number, and let F = Z/pZ. Let A and B be
nonempty subsets of the field F , and let
A
ˆ
+B = {a + b


a ∈ A, b ∈ B, a 6= b}.
Let |A| = k and |B| = l. If k 6= l, then
|A
ˆ
+B| ≥ min(p, k + l − 2}.
Proof. Let |A| = k and |B| = l. We can assume that
1 ≤ l < k ≤ p.
If k + l − 2 > p, let l
0
= p − k + 2. Then
2 ≤ l
0
< l < k
and
k + l
0
− 2 = p.
Choose B
0
⊆ B such that |B
0
| = l
0
. If the Theorem holds for the sets A and B
0
,
then
|A
ˆ
+B| ≥ |A
ˆ
+B
0
| ≥ k + l
0
− 2 = p = min(p, |A| + |B| − 2).
Therefore, we can assume that
k + l − 2 ≤ p.
Let C = A
ˆ
+B. We must prove that
|C| ≥ k + l − 2.
Suppose that
|C| ≤ k + l − 3.
Choose m so that
m + |C| = k + l − 3.
We shall construct three polynomials f
0
, f
1
, and f in F [x, y] as follows: Let
f
0
(x, y) =
Y
c∈C
(x + y − c).
Then deg(f
0
) = |C| ≤ k + l − 3 and
f
0
(a, b) = 0 for all a ∈ A, b ∈ B, a 6= b.
Let
f
1
(x, y) = (x − y)f
0
(x, y).
3

Then deg(f
1
) = 1 + |C| ≤ k + l − 2 and
f
1
(a, b) = 0 for all a ∈ A, b ∈ B.
Multiplying f
1
by (x + y)
m
, we obtain the polynomial
f(x, y) = (x − y)(x + y)
m
Y
c∈C
(x + y − c)
of degree exactly k + l − 2 such that
f(a, b) = 0 for all a ∈ A, b ∈ B.
Then
f(x, y) =
X
i,j≥0
i+j≤k+l−2
f
i,j
x
i
y
j
= (x − y)(x + y)
k+l−3
+ lower order terms.
Since 1 ≤ l < k ≤ p and 1 ≤ k + l − 3 < p, it follows that the coefficient
f
k−1,l−1
of the monomial x
k−1
y
l−1
in f(x, y) is

k + l − 3
k − 2

−

k + l − 3
k − 1

=
(k − l)(k + l − 3)!
(k − 1)!(l − 1)!
6≡ 0 (mod p).
By Lemma 2, for every m ≥ k there exists a polynomial g
m
(x) of degree
at most k − 1 such that g
m
(a) = a
m
for all a ∈ A, and for every n ≥ l there
exists a polynomial h
n
(y) of degree at most l − 1 such that h
n
(b) = b
n
for all
b ∈ B. We use the polynomials g
m
(x) and h
n
(y) to construct a new polynomial
f
∗
(x, y) from f(x, y) as follows: If x
m
y
n
is a monomial in f(x, y) with m ≥ k,
then we replace x
m
y
n
with g
m
(x)y
n
. Since deg(f (x, y)) = k + l − 2, it follows
that if m ≥ k, then n ≤ l − 2, and so g
m
(x)y
n
is a sum of monomials x
i
y
j
with
i ≤ k − 1 and j ≤ l − 2. Similarly, if x
m
y
n
is a monomial in f (x, y) with n ≥ l,
then we replace x
m
y
n
with x
m
h
n
(y). If n ≥ l, then m ≤ k − 2, and so x
m
h
n
(y)
is a sum of monomials x
i
y
j
with i ≤ k − 2 and j ≤ l − 1. This determines a new
polynomial f
∗
(x, y) of degree at most k − 1 in x and l − 1 in y. The process of
constructing f
∗
(x, y) from f(x, y) does not alter the coefficient f
k−1,l−1
of the
term x
k−1
y
l−1
, since this monomial does not occur in any of the polynomials
g
m
(x)y
n
or x
m
h
n
(y). On the other hand,
f
∗
(a, b) = f (a, b) = 0
for all a ∈ A and b ∈ B. It follows immediately from Lemma 1 that the poly-
nomial f
∗
(x, y) is identically zero. This contradicts the fact that the coefficient
f
k−1,l−1
of x
k−1
y
l−1
in f
∗
(x, y) is nonzero, and completes the proof. 2
4

Theorem 2 (Dias da Silva-Hamidoune [3]) Let p be a prime number, and
let F = Z/pZ. Let A ⊆ F , and let |A| = k ≥ 2. Let 2
∧
A denote the set of all
sums of two distinct elements of A. Then
|2
∧
A| ≥ min(p, 2k − 3).
Proof. Let A ⊆ F , |A| ≥ 2. Choose a ∈ A, and let B = A \ {a}. Then
|B| = |A| − 1 and, by Theorem 1,
|2
∧
A| ≥ |A
ˆ
+B| ≥ min(p, |A| + |B| − 2) = min(p, 2|A| − 3).
This completes the proof of the Erd˝os-Heilbronn conjecture.2
Let k + l − 2 ≤ p, 1 ≤ l < k ≤ p. Let A = {0, 1, 2, . . . , k − 1} and B =
{0, 1, 2, . . . , l−1}. Then A
ˆ
+B = {1, 2, . . . , k+l−2} and 2
∧
A = {1, 2, . . . , 2k−3}.
This example shows that the lower bounds in Theorem 1 and Theorem 2 are
sharp.
3 Further applications of the method
The polynomial method is a powerful new technique to obtain results in additive
number theory. For example, it gives the following simple proof of the Cauchy-
Davenport theorem. Let A and B be subsets of Z/pZ, and let C = A + B. Let
|A| = k and |B| = l. We can assume that k + l − 1 ≤ p. If |C| ≤ k + l − 2, let
m = k + l − 2 − |C|, and consider the polynomial
f(x, y) = (x + y)
m
Y
c∈C
(x + y − c).
Then f(a, b) = 0 for all a ∈ A and b ∈ B. The polynomial has degree k + l − 2,
and the coefficient of the monomial x
k−1
y
l−1
is exactly

k + l − 2
k − 1

6≡ 0 (mod p).
The proof proceeds exactly as the proof of Theorem 1.
As a final example of the method, we state and prove the following new
result.
Theorem 3 Let A and B be nonempty subsets of F = Z/pZ, and let
C = {a + b


a ∈ A, b ∈ B, ab 6= 1}.
Let |A| = k and |B| = l. Then
|C| ≥ min(p, k + l − 3}.
5

Frequently asked questions

Q1. What contributions have the authors mentioned in the paper "Adding distinct congruence classes modulo a prime" ?

Erdős and Heilbronn conjectured 30 years ago that there are at least min ( p, 2k − 3 ) congruence classes that can be written as the sum of two distinct elements of A. Erdős has frequently mentioned this problem in his lectures and papers ( for example, Erdős-Graham [ 4, p. 95 ] ). The purpose of this paper is to give a simple proof of the Erdős-Heilbronn conjecture that uses only the most elementary properties of polynomials. 

Q2. What is the result of the polynomial method?

The results in this paper hold for addition in any field F , where p is equal to the characteristic of F if the characteristic is a prime, and p = ∞ if the characteristic is zero. 

Q3. What is the simplest proof of the Erds-Heilbronn conjecture?

The Cauchy-Davenport theorem states that if A and B are nonempty sets of congruence classes modulo a prime p, and if |A| = k and |B| = l, then the sumset A + B contains at least min(p, k + l − 1) congruence classes. 

Q4. What is the proof of the k linear equations?

This is a system of k linear equations in the k unknowns u0, u1, . . . , uk−1, and it has a solution if the determinant of the coefficients of the unknowns is nonzero. 

Q5. What is the inverse of the inverse of the inverse?

Fix b ∈ B. Thenu(x) = k−1∑ i=0 vi(b)xiis a polynomial of degree at most k−1 in x such that u(a) = 0 for all a ∈ A. Since u(x) has at least k distinct roots, it follows that u(x) is the zero polynomial, and so vi(b) = 0 for all b ∈ B. Since deg(vi(y)) ≤ l − 1 and |B| = l, it follows that vi(y) is the zero polynomial, and so fi,j = 0 for all i and j. 

Q6. What is the purpose of this paper?

Research supported in part by grants from the PSC-CUNY Research Award Program ‡Mathematical Institute of the Hungarian Academy of Sciences, Budapest, P.O.B. 127, H1364, Hungary. 

Q7. What is the inverse of the fk1yl1?

Since 1 ≤ l < k ≤ p and 1 ≤ k + l − 3 < p, it follows that the coefficient fk−1,l−1 of the monomial xk−1yl−1 in f(x, y) is(k + l − 3 k − 2 ) − ( k + l − 3 k − 1 ) = (k − l)(k + l − 3)! (k − 1)!(l − 1)! 

Q8. What is the inverse of the gm(x)?

By Lemma 2, for every m ≥ k there exists a polynomial gm(x) of degree at most k − 1 such that gm(a) = am for all a ∈ A, and for every n ≥ l there exists a polynomial hn(y) of degree at most l − 1 such that hn(b) = bn for all b ∈ B. The authors use the polynomials gm(x) and hn(y) to construct a new polynomial f∗(x, y) from f(x, y) as follows: