EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 2, 2024, 1206-1212 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Prime Radical of Nearrings Which is Kurosh-Amitsur Kilaru J. Lakshminarayana1,∗, V.B.V.N. Prasad1, Srinivasa Rao Ravi2, A.V. Ramakrishna3 1 Department of Engineering Mathematics Koneru Lakshmaiah Education Foundation, Vaddeswaram-522502 Guntur (Dist.), Andhra Pradesh, India 2 Department of Mathematics, University College of Sciences, Acharya Nagarjuna University, Nagarjuna Nagar-522510 Guntur (Dist.), Andhra Pradesh, India 3 Department of Mathematics, R.V.R and J.C College of Engineering Chowdavaram-522019 Guntur (Dist.), Andhra Pradesh, India Abstract. A prime radical of near-rings is introduced by defining a new class of prime modules of near-rings. It is a generalization of the Prime radical of rings. Properties of the radical are studied. It is established that this radical is a Kurosh-Amitsur radical of near-rings. 2020 Mathematics Subject Classifications: 16Y30 Key Words and Phrases: Near-ring, N -group, primeN -groups of type 2, Prime radical of type 2 1. Introduction N is a near-ring and all near-rings are zero-symmetric. One may look for more defi- nitions and results of near-rings in Pliz [4]. An additive group H is a right N -group if there is a mapping (h, x) → hx of H ×N into H such that: (i) h(xy) = (hx)y; (ii) h(x+ y) = hx+ hy for all h ∈ H,x, y ∈ N . If K is a right ideal of N then K is a right N -group under the multiplication in N . Also the quotient group N/K is a right N -group under the operation (x +K)y = xy +K for all x, y ∈ N . A subgroup (normal subgroup) C of the right N -group H is a right N -subgroup (ideal) of H if cx ∈ C for all c ∈ C, x ∈ N . ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i2.5032 Email addresses: 2002511005@kluniversity.in (K. J. Lakshminarayana), vbvnprasad@kluniversity.in (V.B.V. N. Prasad), dr rsrao@yahoo.com (S. Rao Ravi), amathi7@gmail.com (A. V. Ramakrishna) https://www.ejpam.com 1206 © 2024 EJPAM All rights reserved. K. J. Lakshminarayana et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1206-1212 1207 An element h0 ∈ H is a distributive element if h0(x+ y) = h0x+ h0y for all x, y ∈ N . Since only right N -groups are considered, hereon we call a right N -group just an N -group and a right N -subgroup just an N -subgroup. Unlike in rings the prime radical of near-rings is not a Kurosh-Amitsur radical [1]. In 1990, near-ringers could introduce a Kurosh-Amitsur prime radical of near-rings [2], called the equiprime radical. A characterization of the prime radical of near-rings was given in [5] using right modules of near-rings. With this motivation right representation of radicals of right near-ring was presented in [7] and a prime radical for near-rings, the right prime radical of type 1, was defined and studied in [6] which is a non-ideal hereditary Kurosh- Amitsur radical. This is the second known Kurosh-Amitsur prime radical of near-rings. In this paper, using right modules, another prime radical is introduced for near-rings which is a Kurosh-Amitsur radical. 2. Prime N-groups of type 2 Let H be an N -group. The annihilator of H in N will be denoted by An(H) := {x ∈ N | hx = 0 for all h ∈ H}. The largest ideal of N contained in An(H), if it exists, will be denoted by (H : 0). Definition 1. Let N be a near-ring and H be an N -group. H is a prime N -group of type 2 if : (i) HN ̸= {0}; (ii) for each 0 ̸= h ∈ H, hN has a distributive element h0(̸= 0); (iii) for each 0 ̸= h ∈ H, An(hN) = An (H). If the near-ring N is a ring then from the conditions (i) and (iii) of the prime N -group of type 2 it follows that it is a prime module [3]. It is clear that a non-zeroN -subgroup of a primeN -group of type 2 is also a primeN -group of type 2. Example 1. We give an example of a prime N -group of type 2. Let H := {0, a, b, c} be the additive non-cyclic group of order 4. Consider the near-ring M0(H) of mappings of H into H fixing 0. We claim that the M0(H)-group M0(H) is a prime M0(H)-group of type 2. It is clear that the distributive elements of the near-ring M0(H) are precisely the endomorphisms of group H and are also the distributive elements of the M0(H)-group M0(H). (i) we have M0(H)M0(H) ̸= {0}; (ii) let 0 ̸= f ∈ M0(H). We suppose without loss of generality that f(a) ̸= 0 and f(a) = x, x ∈ {a, b, c}. We choose g ∈ M0(H) such that g(a) = a = g(c), g(b) = 0. Now fg is an endomorphism of H and hence a distributive element in fM0(H); (iii) let 0 ̸= f ∈ M0(H). It is clear that An(fM0(H)) = {0} = An(M0(H)). Therefore M0(H) is a prime M0(H)-group of type 2. K. J. Lakshminarayana et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1206-1212 1208 We give an example of a prime N -group of type 1 [6] which is not of type 2. Definition 2. Let H be an N -group with HN ̸= {0}. Then H is a prime N -group of type 1 if: (i) every non-zero N -subgroup of H has a non-zero distributive element; (ii) hNx = {0}, 0 ̸= h ∈ H,x ∈ N implies Hx = {0}. Example 2. Let H be a cyclic group of order p, where p is a prime number greater than 2. Clearly M0(H) is a M0(H)-group. Since H has exactly two subgroups, {0} and M0(H) are the only M0(H)-subgroups of M0(H). Therefore M0(H) is a prime M0(H)-group of type 1. Note that any non-zero endomorphism of H is an automorphism of H. Choose a non-zero function f ∈ M0(H) such that the image of f is not equal to H. We have no g ∈ M0(H) such that fg is an automorphism of H, that is, a non-zero endomorphism of H, that is, a non-zero distributive element of M0(H). This shows that fM0(H) has no non-zero distributive element. Therefore M0(H) is not a prime M0(H)-group of type 2. Now we study some properties of prime N -groups of type 2. Proposition 1. Let H be a prime N -group of type 2. Then (H : 0)N exists. Proof. Suppose that H is a prime N -group of type 2. Now H has a distributive element 0 ̸= h0. It is clear that {0} ̸= h0N is an N -subgroup of H. We have (h0x)0 = h0(x0) = h00 = 0 for all x ∈ N , that is, (h0N)0 = {0}. So H0 = {0}. Let K, L be ideals of N contained in An (H). We have (h0x)(k + l) = h0(x((k + l) − xk + xk) = h0(x((k + l) − xk) + h0(xk) = 0 + 0 = 0 for all x ∈ N, k ∈ K, l ∈ L. Therefore (h0N)(K + L) = {0}. Since {0} ≠ h0N is an N -subgroup of H, H(K + L) = {0}. Hence there is a largest ideal of N contained in An (H), that is, (H : 0)N exists. Proposition 2. Let H be a prime N -group of type 2 and K be an ideal of N and HK = {0}. Then h0N is a prime N/K-group of type 2 for any distributive element 0 ̸= h0 ∈ H. Moreover (h0N : 0)N/K = (H : 0)N/K. Proof. K is an ideal of N and H is a prime N -group of type 2 and HK = {0}. Let 0 ̸= h0 ∈ H be a distributive element. Clearly h0N = {h0x | x ∈ N} is a subgroup of (H,+) and is an N -subgroup of H. Let h0x ∈ h0N, x, y, z ∈ N . Define (h0x)(y +N) := (h0x)y. This operation is well-defined. For this suppose that let y+K = z+K. Now −z+y ∈ K. We have (h0x)y = (h0x)[z+(−z+y)]−(h0xz)+(h0xz) = h0[(x(z+(−z+y))−xz]+(h0xz) = 0+(h0xz) = (h0x)z. Therefore the above operation is well defined. It can be easily verified that with this operation h0N is anN/K-group. Moreover h0N is a primeN -group of type 2 beingN -subgroup ofH. From this it follows that h0N is a primeN/K-group of type 2. Let M be the largest ideal of N contained in An (H), that is, (H : 0)N = M . Clearly K ⊆ M . Since HM = {0}, (h0N)M = {0}. So (h0N)(M/K) = {0}, that is, M/K ⊆ (h0N : 0)N/K , that is, (H : 0)N/K ⊆ (h0N : 0)N/K . Similarly if (h0N : 0)N/K = T/K, then the ideal T of N is contained in (H : 0)N and hence (h0N : 0)N/K ⊆ (H : 0)N/K. Therefore (h0N : 0)N/K = (H : 0)N/K. K. J. Lakshminarayana et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1206-1212 1209 Proposition 3. Let H be a prime N/K-group of type 2, K an ideal of N . Then H is a prime N -group of type 2 and (H : 0)N/K = (H : 0)N/K . Proof. Suppose that H is a prime N/K-group of type 2, K is an ideal of N . For h ∈ H,x ∈ N define hx := h(x+K). Note thatHK = {0}. ClearlyH is anN -group under the above operation. It is an easy observation that this operation also satisfies all the three conditions of a prime N -group of type 2. So H is a prime N -group of type 2. Let M/K be the largest ideal of N/K contained in An N/K(H), that is, (H : 0)N/K = M/K. Since H(M/K) = {0}, we have HM = {0}. So (H : 0)N/K ⊆ (H : 0)N/K. Let T be the largest ideal of N contained in An N (H), that is (H : 0)N = T . Now K ⊆ T and H(T/K) = {0} as HT = {0}. So (H : 0)N/K ⊆ (H : 0)N/K . Therefore (H : 0)N/K = (H : 0)N/K . 3. The right prime radical of type 2 N denotes the class of zero-symmetric near-rings. An ideal-mapping onN is a mapping R from N into itself such that R(N) is an ideal of N for all N ∈ N . An ideal-mapping is a Hoehnke radical or H-radical if: (i) t(R(N)) ⊆ R(t(N)) for all homomorphisms t of N in N ; (ii) R(N/R(N)) = {0} for all N in N . For a class of near-rings M ⊆ N and for N in N , we have (N)M := ∩(J | J is an ideal of N and N/J ∈ M). Corresponding to any class of near-rings M ⊆ N , we have an ideal-mapping R defined by R(N) := (N)M and it is well known that this ideal-mapping R is a H-radical. Definition 3. An ideal P of a near-ring N is a right prime ideal of type 2 if P = (H : 0) for a prime N -group H of type 2. It can be observed that a right prime ideal of N of type 2 is a prime ideal of N [4]. If N is a ring then a right prime ideal of N of type 2 is a prime ideal of the ring N . Definition 4. A near-ring is a right prime near-ring of type 2 if the zero ideal of N is a right prime ideal of type 2. Definition 5. The right prime radical of N of type 2 is the intersection of all right prime ideals of N of type 2 and it will be denoted by P2(r)(N). Now it will be proved that P2(r) is a H-radical. Theorem 1. P2(r) is a H-radical. Proof. Let A be the H-radical determined by the class of right prime near-ring of type 2. Let Q be a right prime ideal of N of type 2. We get a prime N -group H of type 2 such that Q = (H : 0). By Proposition 2, there is a prime N/Q-group B of type 2 such that (B : 0)N/Q = (H : 0)N/Q = Q/Q = {0}. So N/Q is right prime near-ring of type 2. Therefore P2(r)(N) ⊇ A(N). On the other hand suppose that P is an ideal of N and N/P is right prime near-ring of type 2. We get a prime N/P -group A of type K. J. Lakshminarayana et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1206-1212 1210 2 such that (A : 0)N/P = {0}. By Proposition 3, A is a prime N -group of type 2 and (A : 0)N/P = (A : 0)N/P = {0}. So (A : 0)N = P . Therefore P2(r)(N) ⊆ A(N). Hence P2(r)(N) = A(N). Since A is a H-radical, P2(r) is also a H-radical. Theorem 2. Let H be a prime N -group of type 2. If K is an ideal of N and K ̸⊆ (H : 0) then H is a prime K-group of type 2 and (H : 0)K ⊇ K ∩ (H : 0)N . Proof. Suppose that H is a prime N -group of type 2 and K is an ideal of N and HK ̸= {0}. We claim that H is a prime K-group of type 2. Under restriction, clearly H is a K-group. Let 0 ̸= h ∈ H. Assume that hK = {0}. Now (hR)K = h(RK) ⊆ hK = {0} and hence AK = {0}, where A is the N -subgroup of H generated by hK. So HK = {0}, a contradiction. Therefore hK ̸= {0}. Let 0 ̸= hk ∈ hK, k ∈ K. We have that (hk)N contains a non-zero distributive element (hk)x, x ∈ N . Clearly (hk)x = h(kx) ∈ hK as required. Finally we prove that AnK(hK) = AnK(H). As seen above, we have 0 ̸= hk ∈ hK, k ∈ K. Note that AnN ((hk)N) = AnN (H). Since (hk)N = h(kN) ⊆ hK ⊆ H we have AnN (H) ⊆ AnN (hK) ⊆ AnN ((hk)N). Therefore, AnN (H) = AnN (hK) = AnN ((hk)N). So AnK(H) = K∩ AnN (H) = K∩ AnN (hK) = AnK(hK). Now it is also clear that K ∩ (H : 0)N ⊆ (H : 0)K . A H-radical R is complete if K ⊆ R(N) for all ideals K of N for which R(K) = K. Theorem 3. The H-radical P2(r) is complete. Proof. Let K be an ideal N and P2(r)(K) = K. Suppose that K ̸⊆ P2(r)(N). So there is a prime N -group H of type 2 such that K ̸⊆ (H : 0)N . By Theorem 2, H is a prime K-group of type 2. This contradicts P2(r)(K) = K. Therefore K ⊆ P2(r)(N). Hence P2(r) is complete. Theorem 4. Let H be prime K-group of type 2 and K be an ideal of N . Then there is a K-subgroup C of H which is a prime N -group of type 2 and (C : 0)N ∩K ⊆ (H : 0)K . Proof. Suppose that K is an ideal of N and H is a prime K-group of type 2. We have a distributive element h0 ∈ H. So h0(y + z) = h0y + h0z for all y, z ∈ K. Clearly h0K := {h0k | k ∈ K} is a non-zero K-subgroup of H. The claim now is h0K is an N -group. For this, define (h0k)x := h0(kx) for all h0k ∈ K,x ∈ N , where k ∈ K. To show that this operation is well defined, suppose that h0y = h0z, y, z ∈ K. Let x ∈ N . Now (h0(yx) − h0(zx))k = (h0(yx))k − (h0(zx))k = h0((yx)k) − h0((zx)k) = h0(y(xk))− h0(z(xk)) = (h0y)(xk) = (h0z)(xk) = ((h0y)− (h0z))(xk) = 0(xk) = 0 for all k ∈ K. Therefore h0(yx) = h0(zx) and that the operation is well defined. It can be easily verified that h0K is an N -group. We see now that h0K is a prime N -group of type 2. Let 0 ̸= h0t ∈ h0K, t ∈ K. Since H is a prime K-group of type 2, we get a distributive element 0 ̸= h0y ∈ (h0t)K (⊆ (h0t)N), y ∈ K. Also, for a, b ∈ N , [(h0y)(a+ b)− ((h0y)a+ (h0y)b)]k = (h0y)(ak+ bk)− ((h0y)ak + (h0y)bk) = ((h0y)ak + (h0y)bk) − ((h0y))ak + (h0y))bk) = 0 for all k ∈ K. Therefore (h0y)(a+ b) = (h0y)a+ (h0y)b and that h0y is distributive over N as required. REFERENCES 1211 We see now that AnN ((h0t)N) = AnN (h0K). Obviously AnN (h0K) ⊆ AnN ((h0t)N) as (h0t)N ⊆ h0K. Since H is a prime K-group of type 2, AnK((h0t)K) = AnK(H). We have ((h0t)K)K = (h0t)KK ⊆ (h0t)K ⊆ (h0t)N . Let x ∈ AnN ((h0t)N). Now ((h0t)K)Kx = {0}. So Kx ⊆ AnK((h0t)K) = AnK(H) and hence h0Kx = {0}, that is x ∈ AnN (h0K). Therefore AnN ((h0t)N) ⊆ AnN (h0K). This gives the required AnN ((h0t)N) = AnN (h0K). Hence C := h0K is a prime N -group of type 2. Finally, let T := (C : 0)N . Now T ∩ K is an ideal of K. Also T ∩ K ⊆ AnK(h0K) = AnK(H). Therefore T ∩K ⊆ (H : 0)K , that is, (C : 0)N ∩K ⊆ (H : 0)K . A H-radical R is idempotent if R(N) = R(R(N)) for all N . Theorem 5. The H-radical P2(r) is idempotent. Proof. Let K be an ideal of N . We claim that P2(r)(K) ⊇ K ∩ P2(r)(N). Let P be a right prime ideal of K of type 2. There is K-group H of type 2 with P = (H : 0). By Theorem 4, there is a K-subgroup C of the K-group H which is a prime N -group of type 2 and (C : 0)N ∩ K ⊆ (H : 0)K . Moreover Q := (C : 0)N is a right prime ideal of N type 2 and P ⊇ K ∩ Q. Therefore P2(r)(K) ⊇ K ∩ P2(r)(N). Now take K = P2(r)(N). This gives P2(r)(N)∩P2(r)(N) ⊆ P2(r)(P2(r)(N)), that is, P2(r)(N) ⊆ P2(r)(P2(r)(N)). The other inclusion is obvious and hence P2(r)(N) = P2(r)(P2(r)(N)). So the H-radical P2(r) is idempotent. A H-radical R which is idempotent and complete is a Kurosh-Amitsur radical or KA-radical. From Theorems 3 and 5 we have: Theorem 6. The H-radical P2(r) is a KA-radical. References [1] J. Daunsr. Prime modules. Tartu Rikkl. Ul. Toitmetised., 764:23–29, 1987. [2] N. J. Groenewald G. L. Booth and S. Veldsman. A Kurosh-Amitsur prime radical for near-rings. Comm. Algebra, 18(9):3111–3122, 1990. [3] K. Kaarli and T. Kriis. Prime ideals ofnear-rings. Reine Angew.Math., 298:156–181, 1978. [4] G. Pilz. Near-Rings. North-Holland Mathematical Studies, Amsterdam, 1983. [5] K. Naga Koteswara Rao R. Srinivasa Rao and K. Siva Prasad. A module theoretic characterization of the prime radical of near-rings. Beitr.Algebra Geom., 59(1):51–60, 2018. [6] K. Siva Prasad R. Srinivasa Rao, K. Naga Koteswara Rao and K. Jaya Lakshmi Narayana. A non-ideal - hereditary Kurosh-Amitsur prime radical for right near- rings. Afrika Matematika., 32:1333–1339, 2021. REFERENCES 1212 [7] R. Srinivasa Rao and S. Veldsman. Right representations of right near-ring radicals. Afrika Matematika., 30(1-2):37–52, 2019.