Ratio Mathematica Volume 47, 2023 Bi-amalgamated algebra with (n, p)-weakly clean like properties Aruldoss Antonysamy* Selvaraj Chelliah† Abstract Let f : A −→ B and g : A −→ C be two ring homomorphisms and let K and K′ be two ideals of B and C, respectively such that f−1(K) = g−1(K′). In this paper, we give a characterization for the bi-amalgamation of A with (B, C) along (K, K′) with respect to (f, g) (denoted by A ▷◁f,g (K, K′)) to be a (n, p)-weakly clean ring. Keywords: (n, p)-clean ring ; (n, p)- weakly clean ring; bi-amalgamation algebra along ideals. 2020 AMS subject classifications:16N40, 16U40, 16S99, 16U60. 1 *Periyar University, Salem, India; aruldossa529@gmail.com. †Periyar University, Salem, India; selvavlr@yahoo.com. 1Received on September 15, 2022. Accepted on December 15, 2022. Published online on January 10, 2023. DOI: 10.23755/rm.v41i0.937. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This paper is published under the CC-BY licence agreement. 324 A. Aruldoss and C. Selvaraj 1 Introduction Throughout this paper all rings are commutative with identity elements. Let A and B be two rings with unity, let K be an ideal of B and let f : A → B be a ring homomorphism. In D’Anna et al. [2009], the authors introduced and studied the new ring structure the following subring of A × B: A ▷◁f K := {(a, f(a) + k) | a ∈ A, k ∈ K} called the amalgamation of A with B along K with respect to f. This new ring structure construction is a generalization of the amalgamated duplication of a ring along an ideal. The amalgamated duplication of a ring along an ideal was intro- duced and studied in (D’Anna [2006], D’Anna and Fontana [2007]). In [D’Anna et al., 2009, Section 4], the authors studied the amalgamation can be in the frame of pullback constructions and also the basic properties of this construction (e.g., characterizations for A ▷◁f K to be a Noetherian ring, an integral domain, a reduced ring) and they characterized those distinguished pullbacks that can be expressed as an amalgamation. Let α : A −→ C, β : A −→ C and f : A −→ B be ring homomorphisms. In D’Anna et al. [2009], the authors studied amalgamated algebras within the frame of pullback α × β such that α = β ◦ f [D’Anna et al., 2009, Proposition 4.2 and 4.4]. In this motivation, the authors created the new constructions, called bi-amalgamated algebras which arise as pullbacks α × β such that the following diagram of ring homomorphims C D A B f g α β is commutative with α ◦ πB(α × β) = α ◦ f(A), where πB denotes the canonical projection of B×C over B. Namely, let f : A −→ B and g : A −→ C be two ring homomorphisms and let K and K′ be two ideals of B and C, respectively, such that f−1(K) = g−1(K′). The bi-amalgamation of A with (B, C) along (K, K′) with respect to (f, g) is the subring of B × C given by A ▷◁f,g (K, K′) := {(f(a) + k, g(a) + k′)|a ∈ A, (k, k′) ∈ K × K′} Following Kabbaj et al. [2013], the above definition was introduced by and studied by Kabbaj, Louartiti and Tamekkante. Following Nicholson [1977], an element a in a ring A is called a clean if a is a sum of a unit and an idempotent in A. A ring is clean if all its elements are clean. 325 Bi-amalgamated algebra with (n, p)-weakly clean like properties Clean rings were initially developed by Nicholson [1977], as a natural class of rings which have the exchange property. This paper aims at studying the transfer of the notion of n-clean rings, (n, p)- weakly clean rings to bi-amalgamation of algebra along ideals. We denote by U(A), set of all unit elements of A. 2 (n, p)-clean ring We start with definition of (n, p)-clean rings. Definition 2.1. (Chen and Qua [2014]) An element a ∈ A is said to be (n, p)- clean if a = u1 + ... + un + x for some unit ui ∈ A(i = 1, ..., n) and xp = x, where x is called p-potent element. The ring A is said to be (n, p)-clean if all of its elements are (n, p)-clean. Remark 2.1. Every clean rings are (1, 2)-clean rings and n-clean rings are (n, 2)- clean rings. Following Chhiti [2018], the above definition is studied the authors in 2018. Now, we start with following example of (2, 2)-clean ring. Example 2.1. Let A := Z4, B := Z4 × Z4 and C := Z2 and let J := 0 × Z4 and J′ := Z2 are ideals of B and C respectively. Consider the map f : A −→ B is defined by f(a) = (a, 0) for all a ∈ A and the map g : A −→ C is defined by g(0) = g(2) = 0 and g(1) = g(3) = 1. Hence, A ▷◁f,g (J, J′) ={ ((0, 0), 0), ((0, 1), 0), ((0, 2), 0), ((0, 3), 0), ((2, 0), 0), ((2, 1), 0), ((2, 2), 0), ((2, 3), 0), ((1, 0), 1), ((1, 1), 1), ((1, 2), 1), ((1, 3), 1), ((3, 0), 1), ((3, 1), 1), ((3, 2), 1), ((3, 3), 1) } is a 2-clean ring. Therefore, by the remark 2.1 A ▷◁f,g (J, J′) is (2, 2)-clean ring. Proposition 2.1. The class of (n, p)-clean ring is closed under homomorphic im- ages. Proof. The proof is straightforward.2 Proposition 2.2. If A ▷◁f,g (K, K′) is a (n, p)-clean ring then f(A) + K and g(A) + K′ are (n, p)-clean rings. Proof. Clearly, by Proposition 2.1 (n, p)-clean ring is a (n, p)-clean ring. Thus, in view of Kabbaj et al. [2013] [Proposition 4.1], we have the following isomorphism of rings A▷◁ f,g(K,K′) 0×K′ ∼= f(A) + K and A▷◁ f,g(K,K′) K×0 ∼= g(A) + K′. Hence, f(A) + K and g(A) + K′ are (n, p)-clean rings.2 326 A. Aruldoss and C. Selvaraj Definition 2.2. A ring is called uniquely (n, p)-clean ring if each element in A can be written as unique way. Theorem 2.1. Assume that A is (n, p)-clean ring and f(A) + K K and g(A) + K′ K′ are uniquely n-clean rings. Then A ▷◁f,g (K, K′) is (n, p)-clean ring if and only if f(A) + K and g(A) + K′ are (n, p)-clean rings. Proof. If A ▷◁f,g (K, K′) is a (n, p)-clean ring, then so f(A) + K and g(A)+K′ by Proposition 2.1. Conversely, assume that f(A)+K and g(A)+K′ are (n, p)-clean rings. Since A is a (n, p)-clean ring, we can write a = u1+...+un+x for some unit ui ∈ A(i = 1, ..., n) and xp = x. On the other hand, since f(A)+K is a (n, p)-clean ring, f(a)+k = (f(x1)+k1)+...+(f(xn)+kn)+f(p)+k∗ with f(xi) + ki(i = 1, ..., n) and f(y) + k∗ are respectively units and (f(p) + k∗) p = f(p) + k∗ element of f(A) + K. It is clear that f(x1) = f(x1) + k1 (resp., f(u1)),...,f(xn) = f(xn) + kn (resp., f(un)) and f(p) = f(p) + k∗ (resp., f(x)), are respectively units and an p-potent element of f(A) + K K , and we have f(a) = f(u1)+ ...+f(un)+f(p) = f(x1)+ ...+f(xn)+f(x). Thus, f(u1) = f(x1),..., f(un) = f(xn) and f(p) = f(x) since f(A) + K K is an uniquely (n, p)-clean ring. Consider k∗1, ..., k ∗ n, k ∗ l ∈ K such that f(x1) = f(u1)+k ∗ 1,..., f(xn) = f(un)+k ∗ n and f(x) = f(p) + k∗l and also since g(A) + K ′ is a (n, p)-clean ring, g(a) + k′ = (g(x′1) + k ′ 1) + ... + (g(x ′ n) + k ′ n) + g(p ′) + k′∗ with g(x′i) + k ′ i(i = 1, ..., n) and g(p′) + k′∗ are respectively units and p-potent element of g(A) + K′. It is clear that g(x′1) = g(x ′ 1) + k ′ 1 (resp., g(u1)),...,g(x′n) = g(x′n) + k′n (resp., g(un)) and g(p′) = g(p′) + k′∗ (resp., g(x)), are respectively units and an p-potent element of g(A) + K′ K′ , and we have g(a) = g(u1) + ... + g(un) + g(p) = g(x′1) + ... + g(x′n) + g(p ′). Thus, g(u1) = g(x′1),..., g(un) = g(x′n) and g(x) = g(p′) since g(A) + K′ K′ is an uniquely (n, p)-clean ring. Consider k′∗1 , ..., k ′∗ n , k ′∗ l ∈ K such that g(x′1) = g(u1) + k ′∗ 1 ,..., g(x ′ n) = g(un) + k ′∗ n and g(y ′) = g(p) + k′∗l . We have (f(a)+k, g(a)+k′) = {(f(u1)+k∗1 +k1)+...+(f(un)+k∗n +kn)+(f(p)+k∗l + k∗), (g(u1)+k ′∗ 1 +k ′ 1)+...+(g(un)+k ′∗ n +k ′ n)+(g(p ′)+k′∗l +k ′∗)} = (f(u1)+k∗1+ k1, g(u1)+k ′∗ 1 +k ′ 1)+...+(f(un)+k ∗ n+kn, g(un)+k ′∗ n +k ′ n)+(f(p)+k ∗ l +k ∗, g(p′)+ k′∗l +k ′∗). It is clear that (f(p)+k∗l +k ∗, g(p′)+k′∗l +k ′∗) is an p-potent in A ▷◁f,g (K, K′). Hence, we have only to prove that (f(u1) + k∗1 + k1, g(u1) + k ′∗ 1 + k ′ 1) is invertible in A ▷◁f,g (K, K′). Since f(u1) + k∗1 + k1 is invertible in f(A) + K, there exists an element f(β) + k0 such that (f(u1) + k∗1 + k1)(f(β) + k0) = 1. Thus f(u1)f(β) = 1. Then f(β) = f(u −1 1 ). So there exists k ∗ 0 ∈ K such that f(β) = f(u−11 ) + k ∗ 0. Similarly, g(u1) + k ′∗ 1 + k ′ 1 is invertible in g(A) + K ′. Hence, (f(u1)+k∗1 +k1, g(u1)+k ′∗ 1 +k ′ 1)(f(u −1 1 )+k ∗ 0 +k0, g(u −1 1 )+k ′∗ 0 +k ′ 0) = 327 Bi-amalgamated algebra with (n, p)-weakly clean like properties (f(u1) + k ∗ 1 + k1, g(u1) + k ′∗ 1 + k ′ 1)(f(β) + k0, g(γ) + k ′ 0) = (1, 1). Therefore, (f(u1) + k ∗ 1 + k1, g(u1) + k ′∗ 1 + k ′ 1) is invertible in A ▷◁ f,g (K, K′). Similarly, each term are invertible. This completes the proof.2 Proposition 2.3. Let f : A → B and g : A → C be two surjective ring ho- momorphisms, let K and K′ be two ideals of B and C respectively such that f−1(K) = g−1(K′) = I0 and let A is a (n, p)-clean ring and A/I0 is an uniquely (n, p)-clean ring. Then A ▷◁f,g (K, K′) is a (n, p)-clean ring. Proof. It is clear that B and C are (n, p)-clean rings. So, since f(A)+K = B and g(A) + K′ = C, we conclude that A ▷◁f,g (K, K′) is a (n, p)-clean ring by Theorem 2.1.2 Let A be a commutative ring with identity and let M be a unitary A-module. The idealization of M in A(or trivial extension of A by M) is the commutative ring A ∝ M = {(a, m)|a ∈ R, m ∈ M} under the usual addition and the multiplica- tion defined as (a1m1)(a2m2) = (a1a2, a1m2 +a2m1) for all (a1, m1), (a2, m2) ∈ A ∝ M. Theorem 2.2. Consider n and p two positive integers (p ≥ 2). Let f : A −→ C, g : A −→ C be two ring homomorphisms. Assume that A is (n, p)-clean ring. Let K be an ideal of B such that f(u) + k is invertible (in B) for each u ∈ U(A) and k ∈ K and K′ be an ideal of C such that g(u) + k′ is invertible (in C) for each u ∈ U(A) and k′ ∈ K′. Then A ▷◁f,g (K, K′) is a (n, p)-clean ring if and only if f(A) + K and g(A) + K′ are (n, p)-clean ring. Proof. In light of Proposition 2.1, homomorphic image of (n, p)-clean ring is (n, p)-clean ring. Thus, in view of Kabbaj et al. [2013] [Proposition 4.1], we have the following isomorphism of rings A▷◁ f,g(K,K′) 0×K′ ∼= f(A) + K and A▷◁ f,g(K,K′) K×0 ∼= g(A) + K′. Hence, f(A) + K and g(A) + K′ are (n, p)-rings. Conversely, we assume that A is (n, p)-clean and K be an ideal of B such that f(u)+k is invertible (in B) and K′ be an ideal of C such that g(u) + k′ is invertible (in C). Then there exist v1 ∈ B such that (f(u1 + k)v1) = 1 and there exist v2 ∈ C such that (g(u1 + k ′)v2) = 1. Hence, (f(u1) + k)(f(u −1 1 ) − v1f(u −1 1 )k) = f(u1)f(u −1 1 ) + kf(u−11 ) − (f(u1) + k)v1f(u −1 1 )k) = 1 + kf(u −1 1 ) − f(u −1 1 )k = 1 and (g(u1) + k′)(g(u−11 )−v2g(u −1 1 )k ′) = g(u1)g(u −1 1 )+k ′g(u−11 )−(g(u1)+k′)v2g(u −1 1 )k ′) = 1 + k′g(u−11 ) − g(u −1 1 )k ′ = 1. Thus, (f(u1) + k, g(u1) + k′) is invertible in A ▷◁f,g (K, K′). Hence, (f(a) + k, g(a) + k′) = (f(u1 + u2 + ... + un + x) + k, g(u1 + u2 + ... + un + x) + k ′) = (f(u1) + k, g(u1) + k ′) + (f(u2), g(u2)) + ... + (f(un), g(un)) + (f(x), g(x)), where (f(u1) + k, g(u1) + k′) ∈ U(A ∝ M), (f(ui), g(ui) ∈ U(A ∝ M)(i = 2, 3, ..., n) and (f(x), g(x)p = (f(x), g(x)). Consequently, A ▷◁f,g (K, K′) is (n, p)-clean ring.2 328 A. Aruldoss and C. Selvaraj 3 (n, p)-weakly clean ring Now we introduce the new class of ring. Definition 3.1. A ring A is called (n, p)-weakly clean ring if a = u1 +...+un +x or a = u1 + ... + un − x for some unit ui ∈ A(i = 1, ..., n) and xp = x, where x is p-potent element. If the above representation is unique, we say that A is uniquely (n, p)-weakly clean ring. Note that (n, p)-clean ring is weakly (n, p)-clean ring. In this section we study the tranfer of (n, p)-weakly clean ring property to the ring A ▷◁f,g (K, K′) is defined above. We establishes necessary and sufficient conditions for A ▷◁f,g (K, K′) to be (n, p)-weakly clean. Proposition 3.1. The class of (n, p)-weakly clean is closed under homomorphic images. Proof. The proof is straightforward.2 Proposition 3.2. A ring A is called uniquely (n, p)-weakly clean ring if the rep- resentation of a (n, p)-weakly clean element in a unique way. Our first main result gives a necessary and sufficient conditions for A ▷◁f,g (K, K′) to be (n, p)-weakly clean ring. Theorem 3.1. Let f : A → B and g : A → C be two ring homomorphisms, let K and K′ be two ideals of B and C respectively such that f−1(K) = g−1(K′) = I0. Assume that the following conditions hold: a) A is a (n, p)-weakly clean ring and A/I0 is an uniquely (n, p)-weakly clean ring. b)f(A) + K and g(A) + K′ are (n, p)-weakly clean rings and atmost one of them is not a (n, p)-clean ring. Then A ▷◁f,g (K, K′) is a (n, p)-weakly clean ring. Proof. Without loss of generality, we assume that f(A) + K is a n-weakly clean ring and g(A) + K′ is a n-clean ring. Let a ∈ A. Then a can be written as a = u1 + ... + un + x or a = u1 + ... + un − x for some ui ∈ A(i = 1, ..., n) and p-potent x ∈ A. Since A is (n, p)-weakly clean ring and since f(A) + K is a (n, p)-weakly clean ring, f(a)+k = (f(x1)+k1)+...+(f(xn)+kn)+(f(p)+k∗) or f(a)+k = (f(x1)+k1)+ ...+(f(xn)+kn)−(f(p)+k∗) with f(xi)+ki(i = 1, ..., n) and f(p) + k∗ are respectively units and p-potent element of f(A) + K. Therefore, f(a) = f(u1) + ... + f(un) + f(x) or f(a) = f(u1) + ... + f(un) − f(x). Then, in f(A) + K/K we have: f(a) = f(u1) + ... + f(un) + f(x) 329 Bi-amalgamated algebra with (n, p)-weakly clean like properties or f(a) = f(u1) + ... + f(un) − f(x). It is clear that f(x1) = f(x1) + k1 (resp., f(u1)),...,f(xn) = f(xn) + kn (resp., f(un)) and f(p) = f(p) + k∗ (resp., f(x)), are respectively units and p-potent element of f(A) + K K , and we have f(a) = f(u1) + ... + f(un) + f(x) = f(x1) + ... + f(xn) + f(p) or f(a) = f(u1)+...+f(un)−f(x) = f(x1)+...+f(xn)−f(p). Thus, f(u1) = f(x1),..., f(un) = f(xn) and f(x) = f(p) since f(A) + K K is an uniquely (n, p)-weakly clean ring. Consider k∗1, ..., k ∗ n, k ∗ l ∈ K such that f(x1) = f(u1) + k ∗ 1,..., f(xn) = f(un) + k ∗ n and f(p) = f(x) + k ∗ l . Hence, f(a) + j = (f(u1) + k ∗ 1 + k1) + ... + (f(un)+k ∗ n+kn)+(f(x)+k ∗ l +k ∗) or f(a)+j = (f(u1)+k∗1 +k1)+...+(f(un)+ k∗n + kn) − (f(x) + k∗l + k ∗). Thus, using the same technique of the previous part g(a) + k′ = (g(u1) + k ′∗ 1 + k ′ 1) + ... + (g(un) + k ′∗ n + k ′ n) + (g(p) + k ′∗ l + k ′∗) or g(a) + k′ = (g(u1) + k ′∗ 1 + k ′ 1) + ... + (g(un) + k ′∗ n + k ′ n) − (g(p) + k′∗l + k ′∗) since g(A)+K′/K′ ∼= A/I0 is an uniquely n-weakly clean ring. This implies that (f(a)+k, g(a)+k′) = {(f(u1)+k∗1 +k1)+...+(f(un)+k∗n +kn)+(f(x)+k∗l + k∗), (g(u1)+k ′∗ 1 +k ′ 1)+...+(g(un)+k ′∗ n +k ′ n)+(g(x)+k ′∗ l +k ′∗)} = (f(u1)+k∗1 + k1, g(u1)+k ′∗ 1 +k ′ 1)+...+(f(un)+k ∗ n+kn, g(un)+k ′∗ n +k ′ n)+(f(x)+k ∗ l +k ∗, g(x)+ k′∗l + k ′∗). Now, the same argument follows from Theorem 2.1 in the remaining case, f(a) + j = (f(u1) + k∗1 + k1) + ... + (f(un) + k ∗ n + kn) − (f(x) + k∗l + k ∗). Let g(a)+k′ = (g(u1)+k′∗1 +k ′ 1)+...+(g(un)+k ′∗ n +k ′ n)−(g(x)+k′∗l +k ′∗). We have (f(a)+k, g(a)+k′) = {(f(u1)+k∗1 +k1)+...+(f(un)+k∗n +kn)+(f(x)+ k∗l + k ∗), (g(u1) + k ′∗ 1 + k ′ 1) + ... + (g(un) + k ′∗ n + k ′ n) + (g(x) + k ′∗ l + k ′∗)} = (f(u1) + k ∗ 1 + k1, g(u1) + k ′∗ 1 + k ′ 1) + ... + (f(un) + k ∗ n + kn, g(un) + k ′∗ n + k′n) − (f(x) + k∗l + k ∗, g(x) + k′∗l + k ′∗). In all cases, (f(a) + k, g(a) + k′) is a (n, p)-weakly clean elements of A ▷◁f,g (J, J′). This completes the proof. 2 4 Conclusions Through the above we have studied the characterization for the bi-amalgamation of A with (B, C) along (K, K′) with respect to (f, g) to be a (n, p- clean ring along with an example. Further, we have studied the necessary and sufficient conditions for A ▷◁f,g (K, K′) to be a (n, p)-weakly clean ring. 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