A NOTE ON MULTIPLICATION S−ACT 1. Introduction

A NOTE ON MULTIPLICATION S−ACT A. A. ESTAJI AND M. SHABANI

Abstract. In this article, we study some properties of multiplication S-acts and their prime subacts. we verify the conditions of ACC and DCC on subacts of multiplication act. we show that pure subacts of multiplication act are multiplication and strongly faithful multiplication acts are torsion-free. Also, we have seen that subact B of S-act A is uniform (essential) if and only if the ideal (B : A) of S is uniform (essential).

1. Introduction Through this paper S will denote a monoid with zero. A non empty set A is called right S-act and denoted by AS , if there exists an action (a, s) −→ as from A × S in to A such that i) a(st) = (as)t for a ∈ A and s, t ∈ S, ii) a1 = 1 for all a ∈ A. A subset B of A is subact of AS and written as B ≤ A, if bs ∈ B for all b ∈ B and s ∈ S. Thus the subacts of the S-act SS (resp. S S) are right (resp. left) ideal of S. An element θ ∈ AS is called fixed element of A if for all s ∈ S, θs = θ. All AS in this paper has an unique fixed element which is denoted by θ, such that for all a ∈ A and s ∈ S, θs = θ and a0 = θ; θ will be called the zero of A. If I is an ideal of S then the Rees factor of S modulo I will be denoted by SI ; we recall that the equivalence classes of SI are I (the zero of SI ) and every single element set {a} with a ∈ S \ I. From now on ”S−act” means ”right S−act”. An S-act A is called a multiplication S-act if for each subact B of A there exists an ideal I of S with B = AI. An S-act A is a multiplication S-act if and only if for each a ∈ A there exists an ideal I of S such that aS = AI. If B is a subact of an S-act A, the ideal {s ∈ S : As ⊆ B} will be denoted by (B : A). If B is a subact of multiplication S-act A, then B = A(B : A). we define Q = {s ∈ S; s is not unit} and it is clear that Q is unique maximal ideal of S. For every s-act A, we put T (A) = {a ∈ A : there exists q ∈ Q s.t a = aq} A ideal I of S is called prime if for a, b ∈ S, the inclusion aSb ⊆ I implies that either a ∈ I or b ∈ I. Equivalently, I is prime if and only if for any right ideal J and K of S, the set inclusion JK ⊆ I implies that J ⊆ I or K ⊆ I. Trivially, unique maximal ideal Q of S is prime ideal. An subact B of A is prime subact if for any a ∈ A and r ∈ S, the set inclusion aSr ⊆ B implies either a ∈ B or Ar ⊆ B. A S-act A itself is called prime if the subact (θ) of A is prime.(See [2]) 2000 Mathematics Subject Classification. 20M30. Key words and phrases. direct union, essential, finitely cogenerated, finite uniform dimention, multiplication S-act, prime subact, pure subact, torsion free, uniform. 1

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S is called duo, if every one sided ideal is two sided ideal, that is for every x ∈ S, xS = Sx. We refer the reader to [7], for other terminology and notations which are not given in this paper. In view of the connection between S-acts and modules, in this paper, we study multiplication S-act by using the techniques similar to those used in multiplication module. 2. Multiplication S-act In this section, we study of the relation between multiplication S-act and cyclic S-act. Also we show that for any commutative monoid S if A is a strongly faithful multiplication S-act and B ≤ A, then (1) There exists a unique ideal I of S such that B = AI. (2) B is essential subact of A if and only if (B : A) is essential ideal of S. (3) B is uniform subact of A if and only if (B : A) is uniform ideal of S. (4) A has finite uniform dimension if and only if SS too. (5) A is torsion-free S-act. We are now going to start with follow proposition. Proposition 2.1. Let S be a duo monoid and AS a multiplication S-act. Then (1) Q is unique maximal ideal of S, (2) If A ̸= AQ, then for every a ∈ A \ AQ, aS = A and A is cyclic S-act. Proof. (1) It is clear. (2) Suppose that a ∈ A \ AQ, then there exists ideal I of S such that aS = AI. Clearly I ̸⊆ Q and hence I = S, which implies aS = A.  Proposition 2.2. Let S be a commutative monoid and AS a multiplication S-act. If A = AQ, then A = T (A). Proof. Let a ∈ A, then there exists ideal I of S such that aS = AI. Hence aS = AQI = AIQ = aSQ = aQ, it follows that for some s ∈ S, a = as and a ∈ T (A). Thus A = T (A).  Example 2.3. Let N∞ = N ∪ {∞}, where for each n ∈ N, n < ∞. We define for each x, y ∈ N∞ , xy = min{x, y}, then N∞ is a monoid with zero and for each x ∈ N∞ , x∞ = x and x1 = 1, i.e., 1 is zero element and ∞ is fixed element. If I is an ideal of N∞ , then I = N∞ , I = N, or there exists n ∈ N such that I = {1, 2, . . . , n}. It is easy to check that NN ∞ is a multiplication N ∞ − act, T (N) = N and NQ = N. Thus Proposition 2.2, is satisfies. Proposition 2.4. Let S be a duo monoid . If AS is a cyclic S-act then AS is a multiplication S-act. A S-act AS is called strongly faithful if for s, t ∈ S the equality as = at for some θ ̸= a ∈ A implies that s = t. ( See [6]).

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Proposition 2.5. Let S be a commutative monoid. If AS is a strongly faithful multiplication S-act then A ̸= AQ and AS is a cyclic S-act. Proof. Since AS is a strongly faithful S-act then T (A) ̸= A. Thus by Propositions 2.1 and 2.2, AS is a cyclic S-act.  Example 2.6. (1) By Example 2.3, it is clear that NN∗ is a multiplication S−act, but it is not strongly faithful. Also N ̸= aN∗ ; for any a ∈ N and NN∗ isn’t cyclic. Hence condition of strongly faithful is necessary in Proposition 2.5. (2) By Example 2.3, it is clear that N∗ N∗ is a multiplication cyclic S−act, but it is not strongly faithful. (3) Let S = 2N ∪ {0, 1} and A = 2N ∪ {0}. It is clear that A is strongly faithful S-act under ordinary multiplication, but it isn’t cyclic. Also it isn’t multiplication S−act. Hence condition of multiplication S−act is necessary in Proposition 2.5. (4) We define for each s, t ∈ N∞ , st = M ax{s, t}. Then N∞ is a multiplication N∞ -act and we have Q = N∞ \ {1}, QN∞ ̸= N∞ and T (N∞ ) ̸= N∞ . By above propositions we have Proposition 2.7. Let S be a commutative monoid and A a strongly faithful multiplication S-act. Then (1) If I and J are ideals of S such that AI ⊆ AJ then I ⊆ J. (2) For each subact B of A there exists a unique ideal I of S such that B = AI. (3) A ̸= AI for any proper ideal I of S. Corollary 2.8. Let I be an ideal in commutative monoid S. If A is a strongly faithful multiplication S-act then AI is so. Corollary 2.9. Let S be a commutative monoid and A ̸= (θ) be a strongly faithful ∪ S-act. If there exists {aλ }λ∈Λ ⊆ A and ideals Iλ of S such that A = λ∈Λ aλ S and for every λ ∈ Λ, aλ S = AIλ then A = aλ S for some λ ∈ Λ. Proof. Let for every λ ∈ Λ, A ̸= aλ S. Then for every λ ∈ Λ, Iλ ⊆ Q. It follows ∪ that A = λ∈Λ aλ S = AQ. Since AS is a strongly faithful S-act, by Proposition 2.2, A = T (A) = (θ), contradiction.  Proposition 2.10. Let S be a commutative monoid. Consider the following statements for an S-act A.

∩ ∩ (1) For every non-empty collection of ideals {Iλ }λ∈Λ of S, λ∈Λ (AIλ ) = A( λ∈Λ Iλ ). (2) For every subact B of A and ideal I of S such that B ⊂ AI there exists an ideal J of S with J ⊂ I and B ⊆ AJ.

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Then the statements (1) and (2) implies that A is a multiplication S-act and if A is a strongly faithful multiplication S-act then the statements (1) and (2) are true. Proof. Suppose that (1) and (2) hold. Let B be an subact of A. Let A = {I : I is an ideal of S and B ⊆ AI}. Clearly S ∈ A. Since the statement (1) is correct by Zorn’s Lemma A has a minimal member I (say). Then B ⊆ AI. Suppose that B ̸= AI. By the statement (2), there exists an ideal J of S with J ⊂ I and B ⊆ JA. In this case J ∈ A, contradicting the choice of I. Thus B = AI. It follows that A is a multiplication S-act. ∩ Now suppose that A is a multiplication S-act. It is clear that A( λ∈Λ Iλ ) ⊆ ∩ ∩ ∩ λ∈Λ (AIλ ). Let I = λ∈Λ Iλ and x ∈ λ∈Λ (AIλ ). By Proposition 2.5, there exists a ∈ A \ AQ such that A = aS. Then for every λ ∈ Λ, there exists aλ ∈ A and iλ ∈ Iλ such that x = aλ iλ and also there exists sλ ∈ S such that aλ = asλ , it follows that x = aλ iλ = asλ iλ . Choose λ, µ ∈ Λ. Since AS is a strongly faithful and asλ iλ = x = asµ iµ then sλ iλ = sµ iµ ∈ Iλ , it follows that sλ iλ ∈ I and x ∈ AI. ∩ This shows that λ∈Λ (AIλ ) ⊆ AI and the statement (1) is proved. Now let B be an subact of A and ideal I of S such that B ⊂ AI. There exists an ideal J of S such that B = AJ. Let K = I ∩ J. Clearly K ⊂ I and by the statement (1), B = AI ∩ AJ = A(I ∩ J) = AK ⊆ AI. This proves the statement (2).  An S-act A is called finitely ∩ cogenerated provided for every non-empty collection of subact ∩ {Aλ }λ∈Λ of A with λ∈Λ Aλ = (θ) there exists a finite subset Λ′ of Λ such that λ∈Λ′ Aλ = (θ). Corollary 2.11. Let A be a strongly faithful multiplication S-act. If S is a commutative monoid, then the following statements equivalent: (1) A is a finitely cogenerated. (2) S is a finitely cogenerated. Before stating the next result we introduce some notation. Let {Aλ }λ∈Λ be a non-empty collection of subacts of an S-act A. Throughout this note we use the following notation: ∩ ∪ ∪ ∩ Aˇλ = λ̸=µ∈Λ Aλ , A˘Λ = λ∈Λ Aλ , Aˆλ = λ̸=µ∈Λ Aλ , and A¯Λ = λ∈Λ Aλ . ˇ λ ∩ Aλ = (θ), in this case Also {Aλ }λ∈Λ is called independent if for each λ ∈ Λ, A ⨿ ˘ we denoted by AΛ = Aλ and it is called 0-direct union of {Aλ }λ∈Λ . λ∈Λ

Proposition 2.12. Let S be a commutative monoid. Let {Aλ }λ∈Λ is a non-empty ⨿ collection of subacts of an S-act A such that A = λ∈Λ Aλ . Consider the following statements: (1) For each λ ∈ Λ, Aλ is a multiplication S-act and (2) For each λ ∈ Λ, there exits an ideal Iλ of S such that AIλ = Aλ and Aˇλ Iλ = (θ).

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If A is a multiplication S-act then the statements (1) and (2) are true and if A is a strongly faithful S-act then the statements (1) and (2) implies that A is a multiplication S-act. Proof. First, suppose A is multiplication. Let B is an subact of Aλ0 , for some λ0 ∈ Λ. Then B is an subact of A, and so there exists an ideal Iλ0 of S such that ⨿ B = AIλ0 = Aλ0 Iλ0 Aˇλ Iλ0 . Clearly Aλ0 Iλ0 ∩ Aˇλ Iλ0 ⊆ Aλ0 ∩ Aˇλ = (θ), it follows that B = Aλ0 Iλ0 and Aˇλ = (θ). Thus the statements (1) and (2) are true. Now suppose that A is a θ−strongly faithful S-act and the statements (1) and (2) are true. If for each λ ∈ Λ, T (Aλ ) = Aλ , then A = T (A) = (θ), a contradiction. Therefore, we may suppose that there exists λ0 ∈ Λ such that T (Aλ0 ) ⊂ Aλ0 . Let a ∈ Aλ0 \ T (Aλ0 ), then by Propositions 2.1 and 2.2, Aλ0 = aS. It follows that AIλ0 = Aλ0 = aS. If Iλ0 is a proper ideal of S then Iλ0 ⊆ Q and Aλ0 = AIλ0 ⊆ AQ ⊆ Aλ0 , it follows that Aλ0 = Aλ0 Q and by Proposition 2.2, T (Aλ0 ) = Aλ0 , a contradiction. Thus S = Iλ0 and since Aˇλ0 = Aˇλ0 Iλ0 = (θ), so A = Aλ0 = aS. By Proposition 2.4, A is a multiplication S-act.  A proper subact B of AS is called essential provided C ∩ B ̸= (θ) for every subact C ̸= (θ) of A. An subact U of A is uniform if B ∩ C ̸= (θ); for all subacts B, C ̸= (θ) of U . An S-act A has finite uniform dimension if it does not contain an infinite independent collection of subact in A. An ideal I of S is essential or uniform provided is an essential or uniform subact of SS , respectively. Proposition 2.13. Let S be a commutative monoid and A be a strongly faithful multiplication S-act, then (1) subact B of A is essential if and only if there exists essential ideal E of S such that B = AE. (2) subact B of A is uniform if and only if ideal I = (B : A) of S is an essential. (3) A has finite uniform dimension if and only if SS too. Proof. (1) Suppose B subact A is an essential. Since A is multiplication, B = AE; for an ideal E of S. Let (0) ̸= I be an ideal of S and E ∩ I = (0). So B ∩ AI = AE ∩ AI = A(E ∩ I). By hypothesis, AI = (θ) and hence I = (0). Thus E is an essential. Conversely let B = AE for essential ideal E of S and AE ∩ C = B ∩ C = (θ) for subact (θ) ̸= C of A. Then there exists ideal (0) ̸= I of S such that C = AI. So A(E ∩ I) = AE ∩ AI = θ. Thus E ∩ I = (0), because A is strongly faithful. (2) Consider uniform subact B of A. Since A is multiplication, B = A(B : A). Let J and K be ideals of S and contained I = (B : A) such that K ∩ J = (0). Hence, by Proposition 2.1, BK ∩ BJ ⊆ AK ∩ AJ = A(K ∩ J) = (θ). Thus by hypothesis, BK = (θ) or BJ = (θ) and so J = (0) or K = (0), it follows that I is an uniform.

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Conversely suppose I = (B : A) is an essential of S. Let X, Y ̸= (θ) be subacts of B. So (X : A) and (Y : A) are contained (B : A), X = A(X : A) and Y = A(Y : A). If (X : A) = (0) or (Y : A) = (0) then X = (θ) or Y = (θ), we have contradiction. Thus (X : A) ̸= (0) and (Y : A) ̸= (0). We have X ∩ Y = A(X : A) ∩ A(Y : A) = A((X : A) ∩ (Y : A)). If X ∩ Y = (θ) then (X : A) ∩ (Y : A) = (0), we have contradiction. Thus X ∩ Y ̸= (θ) and B is an uniform subact of A. ⨿ (3) Suppose first that A has finite uniform dimension. Let λ∈Λ Iλ be a 0-direct union of non-empty collection {Iλ }λ∈Λ of nonzero ideals of S. By Proposition 2.1, for any λ′ ∈ Λ we have ∪ ∪ ∪ AIλ ) = (AIλ′ )∩[A( Iλ )] = A[Iλ′ ∩( Iλ )] = A(0) = (θ). (AIλ′ )∩( λ′ ̸=λ∈Λ

λ′ ̸=λ∈Λ

λ′ ̸=λ∈Λ

Since A has finite uniform dimension then Λ is finite, it follows that SS has finite uniform dimension. ⨿ Conversely, suppose that S has finite uniform dimension. λ∈Λ Bλ be a 0-direct union of non-empty collection {Bλ }λ∈Λ of nonzero subacts of A. For any λ ∈ Λ, there exists ideal Iλ of S such that Bλ = AIλ and for each λ′ ∈ Λ we have ∪ ∪ ∪ A[Iλ′ ∩ ( λ′ ̸=λ∈Λ Iλ )] = AIλ′ ∩ [A( λ′ ̸=λ∈Λ Iλ )] = (AIλ′ ) ∩ ( λ′ ̸=λ∈Λ AIλ ) ∪ = (Bλ′ ) ∩ ( λ′ ̸=λ∈Λ Bλ ) = (θ) ⨿ ∪ So by hypothesis, Iλ′ ∩ ( λ′ ̸=λ∈Λ Iλ ) = (0) and λ∈Λ Iλ is a 0-direct union of nonempty collection {Iλ }λ∈Λ of nonzero ideals of S. Hence Λ is finite, it follows that A has finite uniform dimension.



An subact B of A is pure subact if Bs = B ∩ As for each s ∈ S. if B is the pure subact of A, then for each ideal I of S BI = B ∩ AI. we know that subacts of multiplication S-act is not necessarily multiplication,but Proposition 2.14. Pure subacts of multiplication S-act are multiplication. Proof. Suppose A is multiplication S-act and B a pure subact of A. Let C be an subact of B and so A. Thus C = AI for an ideal I of S. But B is pure, so we have C = C ∩ B = AI ∩ B = BI.  Remark 2.15. A mapping f : AS → BS is called homomorphism of S-acts or Shomomorphism if f (as) = f (a)s for all a ∈ A and s ∈ S. An S-homomorphism f : A → B is called an S-isomorphism if f is bijective. In this case, we say that A and B are isomorphic and write A ∼ = B. Note that if f : A → B is an S-homomorphism then Imf = f (A) is a subact of B.(See [7]) Proposition 2.16. Let S be a monoid. Every S-homomorphism image of multiplication S-act is a multiplication S-act. Proof. Suppose A be a multiplication S-act and f : A → B is S-homomorphism for S-act B. Let B ′ = f (A) and b ∈ B ′ . Then b = f (a) for some a ∈ A. Since A is

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multiplication, there is an ideal I of S such that aS = AI. Thus B ′ I = f (A)I = f (AI) = f (aS) = f (a)S = bS. Therefore, B ′ is multiplication S-act.



Let S be a semigroup. An element s ∈ S is called left (resp. right )cancellable if sr = st (resp. rs = ts) implies that r = t, for r, t ∈ S. If s is left and right cancellable then is called cancellable. An S-act A is called torsion-free provided that for any x, y ∈ A and for any right cancellable element c ∈ S, the equality xc = yc implies x = y. Proposition 2.17. If S is a commutative monoid then every strongly faithful multiplication S-acts is torsion-free. Proof. Suppose A is a Strongly faithful multiplication S-act which is not torsionfree. Hence there exists x, y ∈ A and right cancellable element c ∈ S such that xc = yc and x ̸= y. Also there exists two ideals I and J of S such that xS = AI and yS = AJ. So AIc = xSc = xcS = ycS = ySc = AJc. Hence Ic = Jc. But c is right cancellable, so I = J. Thus there exists s, s′ ∈ S such that x = ys and y = xs′ , this is x1 = xs′ s and s′ s is unit. Hence x = y and we have contradiction. Thus A is torsion-free.  Proposition 2.18. Let S be a commutative monoid such that all its elements are right cancellable. If (θ) ̸= A is a torsion-free multiplication S-act then there exists an ideal I of S such that A ∼ = I. Proof. Consider θ ̸= a ∈ A. There exists ideal J of S such that aS = AJ. Clearly J ̸= (0). Let 0 ̸= r ∈ J. Define ϕ : A −→ aS by ϕ(x) = xr; (x ∈ A). Since A is torsion-free, it follows that ϕ is S−monomorphism. Also aS ∼ = S, thus A is S-isomorphism to an ideal I of S.



Proposition 2.19. Let S be a commutative monoid and (θ) ̸= A a strongly faithful S-act. Then A is multiplication if and only if there exists proper ideal I of S such that A ∼ = SI and SI is multiplication. Proof. suppose that A is multiplication. There exists a ∈ A\AQ such that A = aS. Let I = ((θ) : aS) and define ϕ : A −→ SI as follows ϕ(x) = {s}, if s ̸∈ I and ϕ(x) = {I}, if s ̸∈ I, where s is any element of S such that x = as, for each x ∈ A. Obviously A ∼ = S and S is multiplication. I

I

 3. prime subact In this section, we are continuing our study of the prime and minimal prime subact. We show that for any commutative monoid S if A is a multiplication S-act, then B is prime subact of A if and only if (B : A) is prime ideal of S and If S satisfies ACC (DCC) on prime ideals, then A satisfies ACC (DCC) on prime subacts. Also for any commutative monoid S if A is a strongly faithful multiplication S-act, then

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(1) If A is satisfied ACC (DCC) on prime subacts, then S satisfies ACC (DCC) on prime ideals. (2) If B is a minimal prime subact of A then (B : A) is a minimal prime ideal of S. (3) If every prime ideal of S is finitely generated then A contains only a finitely many minimal prime subact. (4) If every minimal prime subact of A is finitely generated, then S contains only a finite number of minimal prime ideal. But we must begin at the beginning, with the basic proposition. Proposition 3.1. Let S be a duo monoid, B ≤ A and A a multiplication S-act, then the following statements are equivalent. (1) B is prime subact of A. (2) (B : A) is prime ideal of S. (3) There exists prime ideal P of S such that B = AP and for every ideal I of S, AI ⊆ B implies that I ⊆ P . Proof. (1) ⇒ (2) Let a, b ∈ S and aSb ⊆ (B : A). Then AaSb ⊆ B. If a ̸∈ (B : A), then Aa ̸⊆ B. Hence there exists a′ ∈ A such that a′ a ̸∈ B. Since B is a prime subact of A, then Ab ⊆ B, i.e., b ∈ (B : A). Thus (B : A) is prime ideal of S. (2) ⇒ (3) It is routine to check that Zorn’s Lemma applies to the family of F = {P : B = AP and P is an ideal of S } with respect to ordinary inclusion. Thus, we can pick a P to be maximal element of F. Let a, b ∈ S and aSb ⊆ P . Then AaSb ⊆ AP = B and it follows that aSb ⊆ (B : A). By hypothesis a ∈ (B : A) or b ∈ (B : A) and P ⊆ (B : A) implies that P = (B : A), i.e., a ∈ P or b ∈ P , Thus P is prime ideal of S. (3) ⇒ (1) Let prime ideal P of S such that B = AP and for every ideal I of S, AI ⊆ B implies that I ⊆ P . It is clear that P = (B : A). Let r ∈ S and a ∈ A such that aSr ⊆ B. Since A is a multiplication S-act, then there exists an ideal I of S such that aS = AI, it follows that Ir ⊆ (B : A) = P . We also note that (B : A) is prime ideal of S, then r ∈ (B : A) or I ⊆ (B : A), i.e., Ar ⊆ B or a ∈ aS = AI ⊆ B and we are trough.  Remark 3.2. If P is a prime ideal of S then, by induction, we can prove that I1 ∩ I2 ∩ ... ∩ Ik ⊆ P implies that Ij ⊆ P for some 1 ≤ j ≤ k. Corollary 3.3. Let S be a duo monoid, A a multiplication S-act and B1 , B2 , ..., Bk be subacts of A. Let P be a prime subact of A. Then the following statements are equivalent. (1) Bj ⊆ P for some 1 ≤ j ≤ k.

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∩k i=1

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Bi ⊆ P .

Lemma 3.4. Suppose S is a commutative monoid. If A be a prime S-act and (θ : A) = (0), then for every a, b ∈ S, ab = 0 implies that a = 0 or b = 0. Proposition 3.5. Let P be a prime ideal of a commutative monoid S and A a strongly faithful multiplication S-act. Let s ∈ S, a ∈ A satisfy as ∈ AP . Then s ∈ P or a ∈ AP . Proof. If A = (θ), it is clear. Let A be a non-zero multiplication S-act. Since A is a strongly faithful S-act, there exists θ ̸= b ∈ A such that A = bS by Proposition 2.5. Now suppose s ̸∈ P . Hence, there exists p ∈ P and s1 , s2 ∈ S such that a = bs1 and as = bs2 p, it follows that bs1 s = bs2 p. So s1 s = s2 p ∈ P and hence s1 ∈ P , because P is a prime ideal. Thus a = bs1 ∈ AP , as required.  Let S be a commutative monoid and A a strongly faithful multiplication S-act. If P is a prime ideal of S then AP is a prime subact of A. Note that A ̸= AP since A ̸= AQ. Let A be an S-act and B be a subact of A. Then, the radical of B denoted by A−rad(B) or r(B) is defined to be intersection of all prime subacts of A containing B. If B is not contained in any prime subact of A, then A − rad(B) = A. If I is an ideal of the√monoid S, then A − radical I (Considered as a subact of the S) √ of √ is denoted by I. Also for any k ∈ N, I n = I, because P is prime ideal of S if and only if for any ideals I and J of S, the set inclusion IJ ⊆ P implies that I ⊆ P or J ⊆ P .(See [2]) Proposition 3.6. Let S be a monoid and B a proper subact of a strongly faithful √ multiplication S-act A. Then A − rad(B) = A I, where I = (B : A). Proof. Let P denotes the collection of all prime ideals P of S such that I ⊆ P . If √ ∩ ∩ J = I then J = p∈P P and hence, by Proposition 2.10, AJ = p∈P (AP ). Let P ∈ P. A ̸= AP , B = AI ⊆ AP and by Proposition 3.1, AP is prime subact of A. Thus A − rad(B) ⊆ AP . Therefore A − rad(B) ⊆ AJ. Conversely, suppose K is a prime subact of A containing B. By Proposition 3.1, √ there is a prime ideal P of S such that I ⊆ P and K = AP . Hence, I ⊆ P and √ √ so A I ⊆ AP = K. Thus A I ⊆ A − rad(B).  Proposition 3.7. Let A be multiplication S-act. (1) If S satisfies ACC (DCC) on prime ideals, then A satisfies ACC (DCC) on prime subacts. (2) If A is an strongly faithful S-act and it satisfies ACC (DCC) on prime subacts, then S satisfies ACC (DCC) on prime ideals. Proof. (1) Assume B1 ⊆ B2 ⊆ . . . is a chain of prime subact of A. By Proposition 3.1, (B1 : A) ⊆ (B2 : A) ⊆ . . . is a chain of prime ideal of S. By hypothesis

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there exists k ∈ N such that for every i ≥ k, (Bi : A) = (Bk : A). It follows that Bi = A(Bi : A) = A(Bk : A) = Bk . Thus A satisfies ACC. (2) Assume P1 ⊆ P2 ⊆ . . . is a chain of prime ideals of S. First Suppose that for i ∈ N, A ̸= APi . By Proposition 3.5, AP1 ⊆ AP2 ⊆ . . . is a chain of prime subact of A. By hypothesis there exists k ∈ N such that for every i ≥ k, APk = APi . Let i ≥ k, p ∈ Pi , and one may take some a ∈ A \ APi . So ap ∈ APi = APk . By Proposition 3.5, p ∈ Pk , it follows that Pk = Pi . Therefore we may suppose that there exists k ∈ N, A ̸= APk then for every i ≥ k, APk = APi , it follows that for every i ≥ k, Pk = Pi .  A prime ideal (prime subact) P in S (AS ) is called a minimal prime ideal (prime sub act) of ideal (subact) I if I ⊆ P and there is no prime ideal (prime subact) P ′ such that I ⊆ P ′ ⊂ P . Let M in(I) denote the set of minimal prime ideals (prime subacts) of I in S (AS ), and every element of M in((0)) (M in((θ))) is called minimal prime ideal (prime subact). Proposition 3.8. Let S be a commutative monoid and A a multiplication S-act and B ≤ A. (1) If (B : A) is a minimal prime ideal over (θ : A), then B is a minimal prime subact of A. (2) If A is an strongly faithful S-act and B is a minimal prime subact of A then (B : A) is a minimal prime ideal of S. Proof. (1) Assume B ′ is a prime subact of A such that B ′ ⊆ B. By Proposition 3.1, (B ′ : A) ⊆ (B : A) is a chain of prime ideal of S. By hypothesis (B ′ : A) = (B : A), it follows that B ′ = A(B ′ : A) = A(B : A) = B. Thus B is a minimal prime subact of A. (2) By Proposition 3.1, (B : A) is a prime ideal of S. Assume P is a prime ideal of S such that P ⊆ (B : A). Hence AP ⊆ A(B : A) and by our hypothesis AP = A(B : A). Let a ∈ A \ B. So for any q ∈ (B : A), aq ∈ A(B : A) = AP and there exists a′ ∈ A and p ∈ P such that a′ p = aq. But A is cyclic and there exists m ∈ A such that A = mS. Hence there exists s, s′ ∈ S such that a = ms and a′ = ms′ , it follows that ms′ p = msq. Since A is an strongly faithful S-act, then sq = s′ p ∈ P , it follows that q ∈ P or s ∈ P , because P is a prime ideal of a commutative monoid S. If s ∈ P , then a = ms ∈ AP = B which is a contradiction. So q ∈ P and P = (B : A). This means that (B : A) is a minimal prime ideal of S.  Proposition 3.9. Let I be a proper ideal of commutative monoid S. If each minimal prime ideal of S over I is finitely generated then M in(I) is finite set. Proof. Consider the set

A NOTE MULTIPLICATION S−ACT

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S = {P1 P2 ...Pn ; n ∈ N and Pi ∈ M in(I), for each 1 ≤ i ≤ n} and set ∆ = {K; K is an ideal of S and q ̸⊆ K, for each q ∈ S} which is the non-empty set, since I ∈ ∆. (∆, ⊆) is the partial ordered set. Suppose ∪ {Kλ }λ∈Λ is the chain of ∆ in which Λ ̸= ∅ and set K = λ∈Λ Kλ . It is clear that K is an ideal of S. Also, if for q ∈ S, we have had q ⊆ K, then q = P1 P2 ...Pn ∪n is finitely generated i.e., q = i=1 xi S for which x1 , x2 , ..., xn ∈ S and n ∈ N. But q ⊆ K implies that x1 , x2 , ..., xn ∈ K. Thus there exists λ ∈ Λ such that x1 , x2 , ..., xn ∈ Kλ and so q ⊆ Kλ , contradiction. Hence, for each q ∈ S, q ̸⊆ K and K ∈ ∆ is the upper band of this chain. By Zorhn’s lemma ∆ has maximal element such as q ∗ . Now if a ̸∈ q ∗ and b ̸∈ q ∗ for a, b ∈ S, then q ∗ ⊆ q ∗ ∪ aS and q ∗ ⊆ q ∗ ∪ bS. Maximality of q ∗ implies that q ∗ ∪ aS, q ∗ ∪ bS ̸∈ ∆. So there exists q1 and q2 in S such that q1 ⊆ q ∗ ∪ aS and q2 ⊆ q ∗ ∪ bS. Thus q1 q2 ⊆ q ∗ ∪ aSb. If aSb ⊆ q ∗ then q1 q2 ⊆ q ∗ which is contradiction, since q1 q2 ∈ S. Therefore aSb ̸⊆ q ∗ and q ∗ is a prime ideal of S contained I. So there exists a minimal prime ideal p∗ ⊆ q ∗ . But p∗ ∈ S, contradictory with q ∗ ∈ ∆. Above contradicts show that there exists q = p1 p2 ...pm ∈ S such that q ⊆ I. Now for each p ∈ M in(I) we have q ⊆ I ⊆ p and p1 p2 ...pm ⊆ p. It is clear that pj ⊆ p for some 1 ≤ j ≤ m. Thus pj = p, since p is minimal. Hence M in(I) = {p1 , p2 , ..., pm } and is finite.



Proposition 3.10. Let S be a commutative monoid and A a strongly faithful multiplication S-act. If every prime ideal of S is finitely generated then A contains only a finitely many minimal prime subact. Proof. Assume {Bλ }λ∈Λ is the family of minimal prime subacts of A. Set Iλ = (Bλ : A) for λ ∈ Λ. By Proposition 3.8, each Iλ is a minimal prime ideal of S. On the other hand, S contains only a finite number of minimal prime ideal as {I1 , I2 , . . . In }, by Proposition 3.9. Now suppose I is the minimal prime ideal of S. So I = Ii , for some 1 ≤ i ≤ n and B = AI = AIi . Thus {AI1 , AI2 , . . . , AIn } is the finite family of minimal prime subact of A.



Proposition 3.11. Let S be a commutative monoid and A a strongly faithful multiplication S-act such that every minimal prime subact of A is finitely generated, then S contains only a finite number of minimal prime ideal. Proof. Suppose I and J are two distinct minimal prime ideal. Clearly, A is cyclic and so A ̸= AI ̸= AJ. Also AI and AJ are prime subacts of A. Assume that B1 ⊆ AI and B2 ⊆ AJ are two minimal prime subacts of A. Hence B1 = A(B1 : A) and B2 = A(B2 : A), where (B1 : A) and (B2 : A) are minimal prime ideals of S,

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(by Proposition 3.8). On the other hand, Proposition 3.1, implies that (B1 : A) ⊆ I and (B2 : A) ⊆ J, since A is cyclic. Therefore (B1 : A) = I and (B2 : A) = J. This says that AI and AJ are two distinct minimal prime subacts of A. Now if S contains infinite many minimal prime ideals, then A must have infinitely many minimal prime subacts which is contradiction.



Example 3.12. Let X be an infinite discrete space and let S(X) denote the commutative monoid with zero of real continuous function on X (multiply function by multiplying their values). If χp is characteristic function on {p}, for every p ∈ X, then the principal ideal (χp ) is minimal prime ideal of S(X), it follows that S(X)−act S(X) contains infinitely many minimal prime subact. Also if I is minimal prime subact of S(X), then there exits p ∈ X such that I = (χp ) and is finitely generated. It is evident that S(X)−act S(X) is multiplication. Also, if p, q, r ∈ X are distinct points, then χp χq = χp χr = 0, but χq ̸= χr , i.e., S(X)−act S(X) is not strongly faithful. Hence condition of strongly faithful S−act is necessary in Proposition 3.11. References [1] M. Behboodi and H. Koohy, On minimal prime submodules, Far East J. Math. Sci., 6, No. 1 (2002), 83-88. [2] Javed Ahsan and Liu Zhongkui, Prime and semiprime acts over monoids with zero, Math. J., Ibaraki Vol 33, 2001. [3] Z. A. EL-Bast and P. F. Smith, Multiplication modules, Comm. Algebra 16(1988), 755-779. [4] A. A. Estaji And M. Shabani, Multiplication S-acts, 38 th Annual Iranian Mathematics Conference 3-6 September 2007 , university of ZanJan, Zanjan, Iran, 19-21. [5] L. Gillman, M. Jerison, Rings of Continuous Functions, Grad. Texts Math. 43, SpringerVerlag, Berlin-Heidelberg-New York, 1976. [6] Jone M. Howie, Fundamentals of semigroup theory, Oxford Uuniversity press Inc. New York, 1995. [7] Mati Klip, Ulrich Knauer, and Alexander V. Mikhalev, Monoids, Acts and Categories With Applications to Wreath Products and Graphs, Walter de Gruyter. Berlin. New York 2000. [8] H. Koohy, On Finiteness of Multiplication Modules, Acta Math. Hungar., 118 (1-2) (2008), 1-7. 1

Ali Akbar Estaji, Department of Mathematics, Tarbiat Moallem University 0f

Sabzevar, PO Box 397, Sabzevar, Iran. E-mail address: aa− [email protected]