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).
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