1. Introduction
Let
be a finite multiplicatively written commutative semigroup with identity
. By a sequence over
, we mean a finite unordered sequence of terms from
where repetition is allowed. For a sequence T over
we denote by
the product of its terms and we say that T is a product-one sequence if
. If
is a finite abelian group, the Davenport constant
of
is the smallest positive integer
such that every sequence T over
of length
has a nonempty product-one subsequence. The Davenport constant has mainly been studied for finite abelian groups but also in more general settings (we refer to [1] [2] [3] [4] [5] for work in the setting of abelian groups, to [6] [7] for work in case of non-abelian groups, and to [8] [9] [10] [11] [12] for work in commutative semigroups).
In the present paper we study the Erdös-Burgess constant
of
which is defined as the smallest positive integer
such that every sequence T over
of length
has a non-empty subsequence
whose product
is an idempotent of
. Clearly, if
happens to be a finite abelian group, then the unique idempotent of
is the identity
, whence
. The study of
for general semigroups is initiated by a question of Erdös and has found renewed attention in recent years (e.g., [13] [14] [15] [16] [17] ). For a commutative unitary ring R, let
be the multiplicative semigroup of the ring R, and
the group of units of R, noticing that the group
is a subsemigroup of the semigoup
. We state our main result.
Theorem 1.1. Let
be an integer, and let
be the ring of integers modulon. Then
where
is the number of primes occurring in the prime-power decomposition of n counted with multiplicity, and
is the number of distinct primes. Moreover, if n is a prime power or a product of pairwise distinct primes, then equality holds.
2. Notation
Let
be a finite multiplicatively written commutative semigroup with the binary operation *. An element
of
is said to be idempotent if
. Let
be the set of idempotents of
. We introduce sequences over semigroups and follow the notation and terminology of Grynkiewicz and others (cf. [4] , Chapter 10] or [6] [18] ). Sequences over
are considered as elements in the free abelian monoid
with basis
. In order to avoid confusion between the multiplication in
and multiplication in
, we denote multiplication in
by the boldsymbol
and we use brackets for all exponentiation in
. In particular, a sequence
has the form
(1)
where
are the terms of T, and
is the multiplicity of the term a in T. We call
the length of T. Moreover, if
and
, then
has length
has length
,
is a sequence of length 2. If
and
, then
. Any sequence
is called a subsequence of T if
for every element
, denoted
. In particular, if
, we call
a proper subsequence of T, and let
denote the resulting sequence by removing the terms of
from T.
Let T be a sequence as in (1). Then
・
is the product of all terms of T, and
・
is the set of subsequence products of T.
We say that T is
・ a product-one sequence if
,
・ an idempotent-product sequence if
,
・ product-one free if
,
・ idempotent-product free if
.
Let
be an integer. For any integer
, we denote
the congruence class of
modulo n. Any integer
is said to be idempotent modulo n if
, i.e.,
in
. A sequence T of integers is said to be idempotent-product free modulo n provided that T contains no nonempty subsequence
with
being idempotent modulo n. We remark that saying a sequence T of integers is idempotent-product free modulo n is equivalent to saying the sequence
is idempotent-product free in the multiplicative semigroup of the ring
.
3. Proof of Theorem 1.1
Lemma 3.1. Let
be a positive integer where
,
, and
are distinct primes. For any integer
, the congruence
holds if and only if
or
for every
.
Proof. Noted that
if and only if
divides
for all
, since
, it follows that
holds if and only if
divides
or
, i.e.,
or
for every
, completing the proof.
Proof of Theorem 1. 1. Say
(2)
where
are distinct primes and
for all
. It is observed that
(3)
and
(4)
taking a sequence V of integers of length
such that
(5)
and
(6)
Now we show that the sequence
is idempotent-product free modulo n, supposing to the contrary that
contains a nonempty subsequence W, say
, such that
is idempotent modulo n, where
is a subsequence of V and
It follows that
(7)
If
, then
is a nonempty subsequence of V. By (5) and (6), there exists some
such that
and
. By Lemma 3.1,
is not idempotent modulo n, a contradiction. Otherwise,
for some
, say
(8)
Since
, it follows from (7) that
. Combined with (8), we have that
and
. By Lemma 3.1, we conclude that
is not idempotent modulo n, a contradiction. This proves that the sequence
is idempotent-product free modulo n. Combined with (3) and (4), we have that
(9)
Now we assume that n is a prime power or a product of pairwise distinct primes, i.e., either
or
in (2). It remains to show the equality
holds. We distinguish two cases.
Case 1.
in (2), i.e.,
.
Taking an arbitrary sequence T of integers of length
, let
and
. By the Pigeonhole Principle, we see that either
or
. It follows either
, or
. By Lemma 3.1, the sequence T is not idempotent-product free modulo n, which implies that
. Combined with (9), we have that
Case 2.
in (2), i.e.,
.
Then
(10)
Taking an arbitrary sequence T of integers of length
, by the Chinese Remainder Theorem, for any term
of T we can take an integer
such that for each
,
(11)
Note that
and thus
. Since
, it follows that
, and so there exists a nonempty subsequence W of T such that
for each
. Combined with (11), we derive that
or
, where
. By Lemma 3.1, we conclude that
is idempotent modulo n. Combined with (10), we have that
. It follows from (9) that
, completing the proof.
We close this paper with the following conjecture.
Conjecture 3.2. Let
be an integer, and let
be the ring of integers modulo n. Then
.
Acknowledgements
This work is supported by NSFC (Grant No. 61303023, 11301381, 11501561).