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Primes and G-primes in Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {Z}$$\end{document}-nearalgebras

Primes and G-primes in Z\documentclass[12pt]{minimal} \usepackage{amsmath}... Arab. J. Math. https://doi.org/10.1007/s40065-023-00426-z Arabian Journal of Mathematics Shalini Chandel · Ram Parkash Sharma Primes and G-primes in Z-nearalgebras Received: 30 January 2022 / Accepted: 2 March 2023 © The Author(s) 2023 Abstract Let N be a Z-nearalgebra; that is, a left nearring with identity satisfying k(nn ) = (kn)n = n(kn ) for all k ∈ Z, n, n ∈ N and G be a finite group acting on N . Then the skew group nearring N ∗ G of the group G over N is formed. If N is 3-prime (aNb = 0 implies a = 0or b = 0), then a nearring of quotients Q (N ) is constructed using semigroup ideals A (a multiplicative closed set A ⊆ N such that A N ⊆ A ⊇ NA )of i i i i i N and the maps f : A → N satisfying (na) f = n(af ), n ∈ N and a ∈ A . Through Q (N ), we discuss the i i i i i 0 relationships between invariant prime subnearrings (I -primes) of N ∗ G and G-invariant prime subnearrings (GI -primes) of N . Particularly we describe all the I -primes P of N ∗ G such that each P ∩ N ={0},a i i GI -prime of N . As an application, we settle Incomparability and Going Down Problem for N and N ∗ G in this situation. Mathematics Subject Classification 16Y30 · 16W22 1 Introduction Group theory plays an important role when combined with ring theory, semiring theory and various other algebraic structures in mathematics. The action of groups on these structures is one of the important areas of research which was initiated as an attempt to develop Galois theory for them after the inception of Galois theory, when Liouville reviewed Galois manuscript and published it in October–November 1846 issue of the Journal de Mathematiques Pures et Appliques after his death in 1832. In the case of rings, the theory was initially extended to division rings by Jacobson [2] in 1940. The conditions in the case of non-commutative rings were complex, so a fresh approach was made by starting with some simple questions about the relationships of the structure of a ring with identity to the structure of fixed subring R and skew group ring R ∗ G.For G-prime ideals of a ring R with a finite group action on it and prime ideals of R ∗ G, the “Incomparability” and “Going Down” problems were studied and settled by Passman [6], Lorenz and Passman [3], and Montgomery [4,5]. Then these problems were settled for prime kernel functors of group graded rings and their smash products by Parvathi and Sharma [9] which is the generalization of the work of M. Cohen and and Montgomery [1] “Group Graded Rings, Smash Products and Group Actions” and further Sharma et al. [8,10–14] studied group actions on different algebraic structures including semirings. But nearrings, which are the generalization of rings with only one distributive law and are best tools to study non-linear algebraic structures, remain untouched as far as the action of groups on them are concerned. In rings with finite group actions, ideals play an important S. Chandel · R. P. Sharma ( ) Department of Mathematics and Statistics, Himachal Pradesh University, Summer Hill, Shimla 171005, India E-mail: rp_math_hpu@yahoo.com S. Chandel E-mail: shalinichandel1992@gmail.com 123 Arab. J. Math. role in settling various questions regarding the relationships between a ring and its skew group ring; but the conditions are different for nearrings in the absence of one distributive law. Hence the generalization of the results provedin[3] could not be attained for nearrings as it is. For example, a nearring of quotients cannot be defined identical to the ring of quotients (c.f. [3], Sect. 2). This made us to explore the possibilities of different algebraic substructures in nearrings to establish relationships between nearrings and their skew group nearrings. We succeeded to use semigroup ideals to define a nearring of quotients, through which we could settle ‘Incomparability’ and ‘Going Down Problem’ for invariant subnearrings (which are, in fact, alike ideals in rings) of a 3-prime nearring. Let N be a left nearring with identity denoted by 1 (or by 1 when it has to be distinguished from the integer 1) and Z the ring of integers. Then N becomes Z-group; that is, the map Z × N → N satisfies (i) (k + k )n = k n + k n and (ii) (k k )n = k (k n) for all k , k ∈ Z and n ∈ N . Throughout this paper, N 1 2 1 2 1 2 1 2 1 2 is a Z-nearalgebra which will be frequently called a nearring; that is, a Z-group N which satisfies k(nn ) = (kn)n = n(kn ) for all k ∈ Z and n, n ∈ N . It is pertinent to note here that if N is a Z-nearalgebra then (i) N is zerosymmetric, (ii) (−1 )n =−n for all n ∈ N and (iii) N is abelian (n + n = n + n,for all n, n ∈ N ). For, (i) 0 .n = (0.1 )n = 1 (0.n) = 1 .0 = 0 , (ii) (−1 )n = (−1.1 )n =−1((1 )n)) =−1.n =−n N N N N N N N N N and (iii) n + n = (−1 )(−n) + (−1 )(−n ) = (−1 )((−n) + (−n )) =−((−n) + (−n )) = n + n.First, N N N we give an example of a Z-nearalgebra which shows that class of such nearrings arises in a natural way. Example Let G be any additive abelian group and W (G) ={ f : G → G | (±a ± a ... k times) f = ±(af ) ± (af )... k times} (here (af ) is the image of a under f ), be a set of weak homomorphism from G into G itself. Then (1) W (G) is a Z-group under the map f + f + ··· k − times, if k ≥ 0 kf = (− f ) + (− f ) + ···|k|− times, if k < 0. (2) The maps 0, i (identity) and −i (a −→ −a) are in W (G). (3) W (G) is an abelian zerosymmetric left nearring with identity i, satisfying (i) 0 ◦ f = 0 (ii) (−a) f =−(af ) (iii) (−i ) ◦ f =− f ,where a(− f ) =−(af ), and (iv) k( f ◦ f ) = (kf ) ◦ f = f ◦ (kf ) for any integer k, f, f ∈ W (G) and hence W (G) is a Z-nearalgebra. Proof (1) and (2) are obvious. (3) Let f, f , f ∈ W (G) and a ∈ G. Then obviously f + f ∈ W (G) and f ◦ f ∈ W (G), as G is abelain. Clearly f ◦ i = i ◦ f = f , f ◦ ( f + f ) = f ◦ f + f ◦ f ,and f ◦ 0 = 0 ◦ f = 0, proving (i). (ii) (−a) f + af = (−a + a) f = (0) f = 0giving that (−a) f =−(af ). Now (iii ) follows from (ii ).(iv)For k > 0, we have a k( f ◦ f ) = a f ◦ f + f ◦ f +· · · k times = (af ) f +· · · k times = (af + ··· k times) ◦ f = (a ( f + ··· k times)) ◦ f = a(kf ◦ f ) implying k( f ◦ f ) = (kf ) ◦ f and k( f ◦ f ) = f ◦ (kf ) follows from the left distributive law in W (G).For k < 0, we have k( f ◦ f ) = (−( f ◦ f )) + (−( f ◦ f )) + · ··|k|-times = (−i ◦ ( f ◦ f )) + (−i ◦ ( f ◦ f )) + · ··|k|-times =−i ◦ ( f ◦ f + f ◦ f + ···|k|-times) =−i ◦ (|k|( f ◦ f )) =−i ◦ ((|k| f ) ◦ f ) = (−i ◦ (|k| f )) ◦ f = (|k|(−i ◦ f )) ◦ f = (|k|(− f )) ◦ f = (kf ) ◦ f . Similarly, we get the second half of the equality for k < 0. Let G = {g = e, g ,..., g } be a finite multiplicative group acting on N . The action of g ∈ G on n ∈ N 1 2 n g e g g gg  g g g  g g g is denoted by n .So n = n, (n ) = n , (n + n ) = n + n and (nn ) = n n ,where n, n ∈ N and g, g ∈ G.Let N ∗ G = n g |n ∈ N , g ∈ G . Define addition in N ∗ G componentwise; that is, i i i i i =1 n g + n g = (n + n )g .Since N is abelian, it is easy to see that N ∗ G is an additive abelian i i i i i i i i i i group. Thus collecting the cofficients of common g , here afterwards, without loss of generality any element of N ∗G is taken of the type n g ,where n ∈ N and g ∈ G are distinct. The map f : (N ∗G)× N −→ N ∗G i i i i −1 defined by n g , n −→ (n n )g makes it a right N -module. Two elements of N ∗G are equal if and i i i i i i only if corresponding coefficients n s are equal. Thus {1.g |g ∈ G} forms an N -basis for N ∗G.Further N ∗G i i becomes an abelian (commutative with respect to addition) nearring having identity 1.e ( also simply denoted −1 by 1) with addition defined as above and multiplication: ng n h = (nn )(gh), g∈G h∈G g,h∈G where ng, n h are two elements in N ∗ G. In the absence of right distributive law, it is pertinent g∈G h∈G to note here that before multiplication of two elements of N ∗ G, it is necessary to write them in the form where all g s are distinct. The following definitions play important role in this paper. 123 Arab. J. Math. Definition 1.1 [7] A non-empty subnearring A of N is called right invariant (resp. left invariant) if AN ⊆ A (resp. NA ⊆ A). If A is both right invariant and left invariant, then it is said to be an invariant subnearring of N . Definition 1.2 (i) An invariant subnearring P of N is called invariant prime if for invariant subnearrings A, B of N , AB ⊆ P implies that A ⊆ P or B ⊆ P. (ii) An invariant subnearring A of N is called nilpotent if there exists a positive integer n such that A ={0}. (iii) A G-invariant, invariant subnearring P of N is called G−invariant prime if for all G-invariant, invariant subnearrings A, B of N , AB ⊆ P implies that A ⊆ P or B ⊆ P.Here G-invariant, invariant subnearring g G means an invariant subnearring P of N such that P = P = P . g∈G The following Lemma motivates us to investigate Incomparability and Going Down relation between invariant prime subnearrings of N ∗ G and G-invariant prime subnearrings of N . Here afterwards, an invariant prime subnearring will be called I -prime and G-invariant prime subnearring will be called GI -prime. Lemma 1.3 If P is an I −prime of N ∗ G, then P ∩ N is a G I -prime in N . Conversely, if A is a G I -prime of N , then there exists atleast one I -prime P of N ∗ G such that P ∩ N = A. g −1 Proof If P is an invariant subnearring of N ∗G,then P is G-invariant because for x ∈ P ∩ N , x = g xg ∈ P for all g ∈ G,as (N ∗ G)P ⊆ P ⊇ P(N ∗ G). Hence x ∈ P ∩ N for all g ∈ G.Also P ∩ N is an invariant subnearring of N,since N (P ∩ N ) ⊆ NP ⊆ (N ∗ G)P ⊆ P implies that N (P ∩ N ) ⊆ P ∩ N . Similarly P ∩ N ⊇ (P ∩ N )N . To prove P ∩ N is GI -prime, suppose that A and B are G-invariant, invariant subnearrings of N such that AB ⊆ P ∩ N.Then A ∗ G = a g |a ∈ A, g ∈ G is an invariant subnearring of N ∗ G. i i i i For, if a g , a g ∈ A ∗ G,then a g + a g = (a + a )g and a g n h = i i i i i i i i i i j j i i i i i i j i i i −1 (a n )(g h ) ∈ A ∗ G for all n g ∈ N ∗ G. Similarly A ∗ G ⊇ (N ∗ G)(A ∗ G) proving i i j j j i, j j j that A ∗ G is an invariant subnearring of N ∗ G. Moreover a g = (1.g )a ∈ (N ∗ G)A and i i i i i i b g = b (1.g ) ∈ B(N ∗ G),sowehave A ∗ G ⊆ (N ∗ G)A and B ∗ G ⊆ B(N ∗ G) implying that i i i i i i (A ∗ G)(B ∗ G) ⊆ (N ∗ G)AB(N ∗ G) ⊆ (N ∗ G)(P ∩ N )(N ∗ G) ⊆ (N ∗ G)P(N ∗ G) ⊆ P.Since P is an I -prime of N ∗ G,wehave A ∗ G ⊆ P or B ∗ G ⊆ P. Suppose A ∗ G ⊆ P.Since A ⊆ (A ∗ G) ∩ N yields that A = (A ∗ G) ∩ N ⊆ P ∩ N , therefore P ∩ N is a GI -prime. Conversely, assume that A is a GI -prime subnearring of N . As observed above A ∗ G is an invariant subnearring of N ∗ G and (A ∗ G) ∩ N = A.So = {P ⊆ N ∗ G | P ’s invariant subnearring of N ∗ G and P ∩ N = A} i i i is non-empty and partially ordered set with respect to inclusion of subsets. Let C = {P ∈ } be a chain in ,then P is an invariant subnearring of N ∗ G and P ∩ N = A,as P ∩ N = A for all i.This i i i i i implies that P is in . Thus by Zorn’s lemma we can choose P ∈ , a maximal element. To prove that P is I -prime, suppose I, J are invariant subnearrings of N ∗ G such that I ⊃ P, J ⊃ P with IJ ⊂ P.Then IJ ∩ N ⊂ P ∩ N.Since I ∩ N , J ∩ N are invariant subnearrings of N and (I ∩ N )(J ∩ N ) ⊆ IJ ∩ N,wehave (I ∩ N )(J ∩ N ) ⊂ P ∩ N = A.Soby GI -primeness of A, I ∩ N ⊂ A (suppose). Also A = P ∩ N ⊂ I ∩ N . This implies that I ∩ N = A. Thus I ∈  and by maximality of P, P = I implying that P is I -prime. The main results, in this paper, derived are as follows: Let N be 3-prime and G a finite group acting on N . Incomparability:If P is any I -prime of N ∗ G with P ∩ N ={0} and M any invariant subnearring of N ∗ G such that P  M then M ∩ N ={0}. Going Down Problem:For every GI -prime {0} = A of N , there exist I -primes P and P of N ∗ G with P ∩ N = A such that P  P and P ∩ N ={0}. The above results are generalizations or some equivalent forms of the results proved in [3]. 2 Nearring of quotients Here afterwards, the nearring N has assumed to be 3-prime. We recall that a non-empty multiplicative closed subset M of N is called right semigroup ideal (resp. left semigroup ideal) if MN ⊆ M (resp. NM ⊆ M). If M is both right semigroup ideal and left semigroup ideal, then it is said to be a semigroup ideal of N.Let be the collection of all semigroup ideals of N.Since N is 3-prime, A ∩ A and A A become non-zero i j i j semigroup ideals of N for all non-zero semigroup ideals A and A of N . Consider the set all functions i j f : A −→ N with (na) f = n(af ),where n ∈ N ,a ∈ A and A ∈ . Two functions f and f are i j 123 Arab. J. Math. equivalent if both agree on a common domain; that is, f ≡ f iff af = af on A ∩ A .Let f denotes the i j i j i j equivalence class of f and S = Q (N ) = f | f : A −→ N , A ∈  and (na) f = n(af ) .Then S becomes a left distributive abelian nearring with respect to addition and multiplication defined as f + f = f + f i j i j and f . f = f f in S,where f + f : A ∩ A −→ N and f f : A A −→ N . We call S, a nearring of i j i j i j i j i j j i quotients. We note that N is a subnearring of S.For,let n ∈ N,define n : N −→ N , the right multiplication by n.For all n , n ∈ N , n n n = n (n n) = n (n n ).Define amap ϕ : N −→ S by nϕ = n . r r r If n = n ,then nn = nn for all n ∈ N so n = n proving ϕ is well defined This is a nearring 1 2 1 2 1r 2r homomorphism as we have n(n + n ) = n(n + n ) = nn + nn = nn + nn = n(n + n ) and 1 2 r 1 2 1 2 1r 2r 1r 2r n(n n ) = n(n n ) = (nn )n = (nn )n = n(n n ).Since 1 ∈ N , ϕ is one-one. 1 2 r 1 2 1 2 1r 2r 1r 2r Lemma 2.1 Let S = Q (N ) be as above. Then (i) If f ∈ S and A f = {0} for some non-zero semigroup ideal A of N , then f = f ,where f is the zero 0 0 function. (ii) If f , f ,..., f ∈ S, then there exists a non-zero semigroup ideal A of N with A f ⊂ N for all i. 1 2 n i (iii) Assume that g be any automorphism of N and A,B be two non-zero semigroup ideals of N . Let f : A −→ B be a bijection which satisfies (nan ) f = n(af )n , for all n, n ∈ N, a ∈ A. Then f is a unit in S and conjugation by f induces an automorphism g on N . Proof Suppose that f : A −→ N and a ∈ A. Consider a : N −→ N defined by na = na ∈ N.Then f = r r a f is defined on N,asfor any n ∈ N we have n(a f ) = (na ) f = (na) f . Hence n(af ) = (na) f = n(af ) . r r r r (i) Let f ∈ S with Af = {0} for some non-zero semigroup ideal A of N.Since f vanishes on non-zero semigroup ideal A,thisimpliesthat f ∈ f ,but f ∈ f which further gives f = f . 0 0 (ii) This follows easily as N is 3-prime. (iii) Here f is a bijection, so there exists an inverse map f : B −→ A which is also a bijection and −1 −1 −1 g g  g  g nbn f = (n(af )(n ) ) f = nan = n(bf )n (1) −1 Since f and f are both in S,where ff is an identity on A and f f is an identity on B,so f = f . −1 −1 −1 For any n ∈ N , f n f is defined on B,soby (1) for all b ∈ B,wehave b( f n f ) = (bf )(n f ) = r r r −1 −1 g g g −1 g g −1 g −1 ((bf )(n ) ) f = (bn ) f f = bn = b(n ) . This implies that f n f = (n ) ,thatis, f n f = r r r r g −1 (n ) .Since N is a subnearring of S, this yields f n f = n for all n ∈ N . Lemma 2.2 (i) If N is 3-prime, then so is S. (ii) Any automorphism of N extends uniquely to an automorphism of S. ρ ρ (iii) Suppose that ρ is an automorphism of N and x , y be any non-zero elements of N satisfying xny = yn x for all n ∈ N , then there exists a unit f ∈ S such that x f = y and conjugation by f induces an automorphism ρ on N . Proof (i) Suppose that f , f = f in S such that f S f = f , then there exist A , A non-zero semigroup i j 0 i j 0 i j ideals of N on whom f and f are defined with A f = {0} and A f = {0}.Since N ⊂ S and f n f i j i i j j i r j is defined on A A ,sowehave (A A )( f n f ) = {0} for all n ∈ N , implying that (A (A f )N ) f = j i j i i r j j i i j {0}.Since A (A f )N is a semigroup ideal of N,for a (a f )n in A (A f )N and n, n in N , we have j i i j i i j i i n (a (a f )n) = (n a )(a f )n ∈ A (A f )N and (a (a f )n)n ∈ A (A f )N.Also A (A f )N = {0}, j i i j i i j i i j i i j i i j i i if not then A (A f ) ⊂ A (A f )N = {0}, implies that (A A ) f = {0}.But N is 3-prime, so A A = {0} j i i j i i j i i j i and thus by Lemma 2.1(i) f = f , which is a contradiction. Thus A (A f )N = {0} andbyLemma 2.1(i), i 0 j i i { } (A (A f )N ) f = 0 implies that f = f , again a contradiction. Hence f S f = f . j i i j j 0 i j 0 g g g g g (ii) Let g be an automorphism of N and f : A −→ N . Define a map f : A −→ N by a f = (af ) .Since −1 −1 −1 g g g g g g g g g f ∈ S, for all a ∈ A and n ∈ N,wehave (na ) f = (n a) f = ((n a) f ) = (n (af )) = g g g n(af ) = n(a f ). Define a map ϕ : S −→ S by f ϕ = f . This is an automorphism from S onto g g itself. For, f , f in S,if f ≡ f then f = f on some common domain implies that f = f on a i j i j i j i j g g common domain. Thus f ≡ f proving that ϕ is well defined. Reversing the steps, ϕ is easily seem to i j g g g be one-one. To prove this is a homomorphism, ( f + f )ϕ = f + f ϕ = f + f = f + f = i j i j i j i j g g g g g g g f + f = f ϕ + f ϕ and ( f . f )ϕ = f f ϕ = f f = f f = f . f = ( f )ϕ.( f )ϕ. To prove i j i j i j i j i j i j i j i j 123 Arab. J. Math. −1 −1 −1 g g g onto, if f ∈ S,take f ∈ S and we have f ϕ = f = f ∈ S. Therefore this gives rise to an automorphism of S extended by g. Now we show that if σ is an automorphism of S fixing N elementwise, then σ is an identity automorphism of S.For,let A be a non-zero semigroup ideal of N with A f ⊂ N , σ σ σ σ σ then a f = a f = a f = a f giving A f − f = {0}. So by Lemma 2.1(i), f = f for all f ∈ S. Now for uniqueness of extension, suppose that ϕ and ϕ are two automorphisms of S extended 1 2 −1 −1 −1 −1 −1 by g.Since n ϕ ϕ = (nϕ ) ϕ = (ng) g = n gg = n,weget ϕ ϕ is an identity on N 1 1 1 2 2 2 therefore ϕ = ϕ on S. 1 2 (iii) Set X = n xn |n , n ∈ N and Y = n yn |n , n ∈ N . Define maps f : X → Y 1 2 1 2 1 2 1 2 finite finite −1 ρ ρ and f : Y → X by (n xn ) f = n yn and (n yn ) f = n xn .Here f and f are well-defined. For, 1 2 1 1 2 1 2 2 −1 −1 −1 −1 −1 ρ ρ ρ ρ ρ take n xn = n xn .Since (xny) = y nx (given), then (xn y) n = y n xn = y n xn = 1 2 1 2 1 2 1 2 1 2 −1 −1 −1 −1 −1 −1 ρ ρ −1 ρ ρ ρ ρ  ρ (xn y) n which implies x (n y n − n y n ) = 0. Since N is 3-prime and x ={0}, 1 2 1 1 2 ρ ρ hence n yn = n yn . Similarly f can be shown to be well-defined. The functions are well defined on 2 1 2 the finite product of such elements as it is independent of the order on which action of ρ is applied. Since f −1 and f are in S,where ff = f on X and f f = f on Y , f = f is a unit in S.Also x f is defined on 1 1 r N,sofor all n ∈ N we have n(x f ) = (nx ) f = n(xf ) = ny = ny ,thatis, xf = y.Again forany n ∈ N , r r −1 ρ ρ f n f is defined on Y,sowehave (n yn )( f n f ) = ((n xn )n) f = (n yn )n = (n yn )n ,so r 1 2 r 1 1 2 1 2 r −1 f n f = n ;thatis, f n f = n . r r Now we are able to define inner automorphisms of N . Definition 2.3 An automorphism of N is said to be G-inner iff it is induced by conjugation of a unit in S; −1 that is, these automorphisms arise from those units f ∈ S with f N f = N.Since N is a subnearring of S, g −1 by Lemma 2.2(ii), for any g ∈ G , there exists f corresponding unit in S such that f = f f f for all inn g g g f ∈ S.The setofall G-inner automorphisms of N is denoted by G . inn Lemma 2.4 G is a normal subgroup of G. inn Proof For, if g, g ∈ G , then there exist f , f (both not equal to f ) units in S such that for all n ∈ N , inn g 0 −1  −1  −1 −1 g gg g −1 n = f n f and n = ( f n f ) = f  f n f f  = ( f f  ) n( f f  ). Thus the unit f f g g g g g g g g g g g g g g −1 gg g in S gives rise to G-inner automorphism gg .For g ∈ G and g ∈ G ,wehave n = f n f .So inn g g −1 −1 −1 −1 −1 −1 −1 −1 g −1 g gg g g g g g −1 g (n ) = ( f n f ) = ( f ) n( f ) = ( f ) n( f ) .Here f is a unit in S, thus g g g g g g g −1 ggg ∈ G . inn Finally, in this section, we prove result which is required for our work in next section. Lemma 2.5 Let  be a non-zero additive subgroup of N ∗G with N , N ⊆ and T = {g = e, g ,..., g } ⊂ 1 2 m G. Suppose ∩ (N ∗ T ) = {0} and  ∩ (N ∗ T ) = {0} for all T T(1) For each i = 1, 2,..., m, define ⎧ ⎫ ⎨ ⎬ A = n ∈ N |∃α = n g ∈ with coefficient of g as n i j j i ⎩ ⎭ j =1 then (i) Each A is a non-zero invariant subnearring of N . (ii) A = A , then for each i, there exist a natural bijection f : A −→ A which satisfies (nan ) f = 1 i i i −1 n(af )n for all n, n ∈ N and a ∈ A, where f is the identity function. i 1 (iii) The elements of  ∩ (N ∗ T ) are precisely of the form α = (af )g with a ∈ A. i i i =1 123 Arab. J. Math. { } { } Proof (i) Let A be as above. Then A = 0 ;for,if A = 0 , then we have elements of type i i i α = n g ∈  ∩ (N ∗ T ) with coefficient of g as 0, contradicting (1). Take n, n ∈ A , j j i i j =1, j =i m m so there exist α = n g and α = n g in  with coefficient of g as n and n respec- j j j i j =1 j =1 tively. Since α + α = (n + n )g ∈  with coefficient of g as n + n so by definition of A , j j i i j =1 we have n + n ∈ A . Similarly for n ∈ N,wehave n α = (n n )g ∈  with coefficient of i j j j =1 g as n n implying that n n ∈ A .Further,for any n ∈ N , there exists n ∈ N such that n = n , i i −1 −1 m g g j  i αn = (n n )g ∈  with coefficient of g as nn implying that nn ∈ A . Hence A is an j j i i i j =1 invariant subnearring of N . (ii) Note that for each a ∈ A , there exists a unique α = n g such that a = n . The existence is i i j j i i j =1 m m obvious. For uniqueness, suppose that α = n g and α = n g in T such that coefficients j j j j =1 j =1 j of g in both α and α are equal, then we have α − α ∈  ∩ (N ∗ T ) so α = α (using (1)). Hence the map f : A −→ A defined by af = a , a ∈ A is well defined, one-one and onto. Finally let α be as i i i i −1 m g above and let n, n ∈ N , a ∈ A.Since nαn = (nn n )g ∈  and g = e,thisimpliesthat j j 1 j =1 −1 −1 g g i i (nan ) f = na n = n(af )n . i i i (iii) From (ii), we have α = (af )g ∈  ∩ (N ∗ T ) and conversely it is also obvious that every element i i i =1 of  ∩ (N ∗ T ) is of this form. 3 Incomparability and going down problem In this section, we settle the Incomparability and Going Down Problem for N and N ∗ G. For any positive integer k,let kN = {kn | n ∈ N }.Then kN is a semigroup ideal of N.For,let n ∈ N , kn ∈ kN,then n (kn) = k(n n) and (kn)n = k(nn ) in kN as N is a Z-nearalgebra. For any ∈ Q, q > 0and qn ∈ qN , pn, if p ≥ 0 p p p define f : qN −→ N by (qn) f = Then f ∈ S.For,let n ∈ N , qn ∈ qN,then q q q | p|(−n), if p < 0. p(nn ) = n (pn), if p ≥ 0 p p p p we have (n (qn)) f = (q(n n)) f = So (n (qn)) f = n (qn) f ; q q q q | p|(−n n) = n (| p|n), if p < 0. p p that is, f ∈ S.Define Z = f | ∈ Q, q > 0 . q q Lemma 3.1 Let Z be as above. p p (i) For any ∈ Q, q > 0, we have f =− f ,if p < 0 and hence f −p =− f ,if p ≥ 0. | p| q q q p p 1 2 (ii) For any , ∈ Q, q > 0 and q > 0; we have 1 2 q q 1 2 p p (a) f 1 + f 2 = f p p ; 1 2 q q 1 2 q q 1 2 (b) f p . f p = f . 1 2 p p 1 2 q q 1 2 q q 1 2 p p p (c) Z is closed with respect to addition and multiplication of maps and f n = n f for all n ∈ N, f ∈ Z r r q q q and thus Z is contained in the centre of S. (iii) Z is a subnearring of S isomorphic to Q and hence is a field. Proof (i) For any qn ∈ qN,wehave (qn) f =| p|(−n) =| p|(n(−1 )) = (| p|n)(−1 ) =−(| p|n) = N N p p − (qn) f | p| implying (qn) f + f | p| = {0} for all qn ∈ qN . Hence f =− f | p| ,if p < 0. q q q q q (ii) (a) Firstly the maps on both sides of (a) are defined on qN,where q = lcm(q , q ).Let t = and 1 2 1 p p t = .Then f + f = f = f if p ≥ 0and p ≥ 0. In case, p and p are 1 2 p t +p t p p 2 1 2 1 2 1 1 2 2 1 2 q q + 1 2 q q q 1 2 negative; using (i), the sum is equal to − f + f = f = f . If one of | p | | p | | p |t +| p |t p t +p t 1 2 1 1 2 2 1 1 2 2 q q q q p or p is negative, then also we get the same result. 1 2 123 Arab. J. Math. p p (b) The product of f 1 and f 2 is defined on q q N and obviously this is contained in both q N and q N . 1 2 1 2 q q 1 2 p p Further on q q N , f . f = f irrespective of the sign of p and p as | p p |=| p || p |. 1 2 p p 1 2 1 2 1 2 1 2 1 2 q q 1 2 q q 1 2 (c) The closure properties in Z follow from (a) and (b).Let f p ∈ Z , n ∈ N , f p n is defined on q q (pa)n, if p ≥ 0 p p qN,sofor all qa ∈ qN we have (qa) f n = (qa) f n = = r r q q (| p|(−a))n, if p < 0 p(an), if p ≥ 0 p p p p = q(an) f = ((qa)n) f = (qa) n f , proving that f n = r r q q q q | p|(−(an)), if p < 0 n f . Therefore, conjugation by f induces an automorphism of S which is trivial on N . Hence Lemma p p 2.2(ii) implies that this automorphism is trivial on S;thatis, f f = f f for all f ∈ S, proving that q q Z is contained in the centre of S. (iii) First part follows from (ii) (a & b). For isomorphism between Q and Z,define amap from Q to Z by p p −→ f , ∈ Q, q > 0. Then obviously, this map is well defined and it follows from (ii) (a & b) that q q p p this is an onto homomorphism. For one-one, suppose that f = f .If p and p both are positive or 1 2 1 2 q q 1 2 p p 1 2 negative, then this equality gives = . So suppose that one of p or p is negative (say p is negative). 1 2 1 q q 1 2 p p Then − f = f or f = f + f = f ;thatis, (| p |t + p t ) n = 0for all n ∈ N.So | p | 2 | p | 2 | p |t +p t 0 1 1 2 2 1 1 1 1 2 2 q q q 2 q 2 q 1 1 q q p p 1 2 | p |t + p t = 0 implying −| p |t = p t or −| p | = p and finally we get = , proving 1 1 2 2 1 1 2 2 1 2 q q q q 1 2 1 2 the lemma. Let Z be as in Lemma 3.1, Z (S) centre of S and C centralizer of S in S ∗ G. Then in the absence of right distributive law, unlike rings Z (S) and C are not closed with respect to ‘+’. However, we have −1 Lemma 3.2 C ⊆ S ∗ G and the set { g = f g | g ∈ G } where f is the unit in S inducing the inn g inn g automorphism g on N , forms S-basis for S ∗ G . inn Proof Firstly, we prove that C ⊆ S ∗ G .Let α ∈ C and g ∈ Supp(α),then α = fg + ··· . Thus by inn Lemma 2.1(i & ii), there exists a ∈ N with aα ∈ N ∗ G.Since α ∈ C, n a α = αn a for all n ∈ N r r r r −1 and we have a(n a ( fg)) = a(( fg)n a ) implying a((n a f )g)) = a(( f (n a ) g) which further implies r r r r r r r r −1 −1 −1 −1 g g g g that an(af ) = (af )n a . Let b = af be a non-zero element of N,sowehave anb = bn a for all −1 n ∈ N . Hence Lemma 2.2(iii ) implies that g ∈ G ;thatis, g ∈ G showing that C ⊆ S ∗ G .Now inn inn inn −1 for each g ∈ G , choose f ∈ S inducing G-inner automorphism g on N and let  g = f g. All elements inn g g −1 −1 −1 g = f g form S-basis for S ∗ G . For, consider f f g = f e,where f ∈ S implies f f = f . g inn i g 0 i i g 0 Since f is a unit in S, hence each f = f . g i 0 −1 −1 −1 Note. Each  g is a unit in S ∗ G which acts trivially on N by conjugation. For,  g n  g = g f n f g = r g r g −1 −1 −1 −1 −1 g g g −1 g (n ) g = (n ) = n because, since f n f = (n ) and unit corresponding to g is f . r r r g r g r g Hence  g ∈ C.As C ⊆ S ∗ G , any element of C is of the form α = f g , with f ∈ S. Since each f g inn i i i i i centralizes S and g  is a unit in C,so f ∈ S ∩ C = Z (S).Sothat α = f g , f ∈ Z (S). From Lemma 3.1, i i i i i the field Z ⊆ Z (S),sowedefine C = α ∈ C | α = f g  with each f ∈ Z Z i i i −1 Obviously, the set { g = f g | g ∈ G }⊆ C . Now we prove a result analogous to Lorenz and Passman g inn Z ([3], Lemma 2.2). Lemma 3.3 Let C be as above. Then (i) C forms finite dimensional group algebra Z [G ] with Z-basis { g | g ∈ G }.Further S ∗ G Z inn inn inn S ⊗ C . Z Z 123 Arab. J. Math. (ii) If I is a G-invariant ideal of C ,then I SG = SG I is an invariant subnearring of S ∗ G. Moremore SG I is contained in S-direct summand of S ∗ G with SG I ∩ (S ∗ G ) = S I and SG I ∩ S ={0}. inn p p p p Proof (i) Obviously C is closed w.r.t. ‘+’. If f g , f g  ∈ C ,then f g  f g  = Z i i j j Z i i j j q q q q i i j j −1 −1 −1 −1 −1 p p p p p p p p f f g f f g = f f f f f f g g = f f  g g = f  g g . i g i j g j i g g j g g i j i j i j i j i j i j i i j i q q q q q q q q i i i j j j i j As seen in Lemma 3.2, the elements  g, g ∈ G are S-linearly independent and hence Z-linearly ndepen- inn dent. Let α = f g  ∈ C , f ∈ Z.Since Z is central in C ,weget S ∗ G  S ⊗ C under the i i Z i Z inn Z Z map g −→  g,for g ∈ G . inn (ii) Let I be a G-invariant ideal of C . Since by definition of I , IS = SI and the G-invariance of I implies Ig = −1 gI for all g ∈ G. Observe that SG I = ISG.For,let f g i ∈ SG I , f g i = f i g = j j j j j j j j j j j −1 i f g ∈ ISG.Let G be the transversal of G in G;thatis, g G = G. So any element j inn inn j j j g ∈G of SG I is of the form   fg f  g ∈ SG I.Let I be the Z-complement of I in C .Since z Z g ∈G g∈G inn C is a finite dimensional vector space over Z and I its subspace, by (i), S ∗ G = SI ⊕ SI ,sowe Z inn have S ∗ G =   (SIg ⊕ SI g ). Thus SG I is contained in S-direct summand of S ∗ G. SG I is an g ∈G g −1 j   k invariant subnearring of S ∗ G.For, f g i f h i = f f g h (i i ) ∈ SG I. j 1 j k 2k j k 2k j j k k j,k j k 1 j g −1 j  k Again let f h ∈ S ∗ G,thenwehave f g i f h = f f g h i and k k j 1 j k k j k k k j j k j,k j j −1 f h f g i = f f h g i ∈ SG I. Note that SG I ∩ (S ∗ G ) = SI . Finally if k k j 1 j k j k 1 j inn k j j k, j j c c I = C , then we can choose I contains unity f e. This implies that   SI g contains S and hence Z 1 g ∈G SG I ∩ S = {0}. The following correspondence between G-invariant ideals of C and invariant subnearrings of N ∗ G is going to play a central role in proving the main results of this paper. Definition 3.4 (i) If I is a G-invariant ideal of C , then define I = SG I ∩ (N ∗ G). (ii) For any invariant subnearring M of N ∗ G,define M ={α ∈ C | Aα ⊆ M for some non-zero semigroup ideal A of N }. Lemma 3.5 (i) I is an invariant subnearring of N ∗ G. (ii) M is a G-invariant ideal of C . Proof (i) By Lemma 3.3(ii),wehave SG I is an inavriant subnearring of S ∗ G which implies that I becomes invariant subnearring of N ∗ G. (ii) Let α ,α ∈ M and β ∈ C . Then there exist A and A two non-zero semigroup ideals of N with 1 2 Z 1 2 A α, A α ⊆ M and B a semigroup ideal of N such that Bβ ⊆ N ∗G.Since A ∩ A is a non-zero semigroup 1 2 1 2 ideal of N,for α + α ∈ C , (A ∩ A )(α + α ) ∈ M implies α + α ∈ M .For any a b ∈ A B 1 2 Z 1 2 1 2 1 2 1 1 and ba ∈ BA , (a b)(α β) = (a b )(α β) = a { b α β)}= (a α )(bβ) ⊆ M (N ∗ G) ⊆ M implies 1 1 1 1 1 r 1 1 r 1 1 1 d d d d α β ∈ M and similarly, βα ∈ M .For G-invariance, let α ∈ M , so by definition of M , there exists 1 1 1 g g g g −1 a non-zero semigroup ideal A of N with Aα ⊂ M, hence A α = (Aα) ⊂ M = g Mg ⊂ M because g g g d M is an invariant subnearring of N ∗ G,sowehave A α ⊂ M which implies α ∈ M for all g ∈ G. Definition 3.6 Any invariant subnearring M of N ∗ G is N -cancelable iff for any α ∈ N ∗ G and non-zero G-invariant semigroup ideal A of N such that Aα ⊂ M implies α ∈ M. The I -primes of N ∗ G which intersect trivially with N and G-invariant ideals of C are N -cancelable. Lemma 3.7 (i) Any I -prime P of N ∗ G satisfying P ∩ N ={0} is N -cancelable. (ii) For any G-invariant ideal I of C ,I becomes N -cancelable. u u u (iii) If I , I are G-invariant ideals of C ,then I I ⊂ (I I ) . 1 2 Z 1 2 1 2 123 Arab. J. Math. Proof (i) Let A be any non-zero G-invariant semigroup ideal of N such that Aα ⊂ P for some non-zero α ∈ N ∗ G.Then Aα ∩ N ⊂ P ∩ N = {0},but Aα ∩ N = n ∈ N |n = (an )g gives An ={0} for i i i all i.Since N is 3-prime and ANn ⊆ An ={0},wehave n ={0} for all i. By this we can say that there i i i is no non-zero α ∈ N ∗ G such that Aα ⊂ P. Therefore, P is N -cancelable. (ii) Assume A is any non-zero G-invariant semigroup ideal of N such that Aα ⊂ I where α ∈ N ∗ G.By definition of I ,wehave Aα ⊂ SG I . Since by Lemma 3.3(ii) we know that SG I is contained in S-direct summand of S ∗ G;solet α = α + α where α ∈ SG I and α ∈ K,where K is a left and right 1 2 1 2 S-submodule of S ∗ G.Since N ∗ G ⊂ S ∗ G, we see that A(α − α ) = A(α + α − α ) = Aα ∈ K 1 1 2 1 2 implying Aα ∈ SG I ∩ K ={0}.SobyLemma 2.1(i), α is zero. Hence α = α ∈ SG I ∩ (N ∗ G) = I . 2 2 1 u u u u u (iii) By definition of I , I I ⊂ (SG I )(SG I ) ⊆ SG(I I ) which implies that I I ⊂ SG(I I )∩(N ∗G) = 1 2 1 2 1 2 1 2 1 2 (I I ) . 1 2 Now we have ud Lemma 3.8 (i) For any G-invariant ideal I of C ,I = I . du (ii) If M is an invariant subnearring of N ∗ G, then M ⊆ M . Proof (i) If α ∈ I , then there exists a non-zero semigroup ideal A of N with Aα ⊂ N ∗G ⊂ SG I ∩ (N ∗G) = u ud ud ud I ,so α ∈ I and we have I ⊆ I .Conversely, if α ∈ I , then there exists a non-zero semigroup ideal A of N with Aα ⊂ I = SG I ∩ (N ∗ G) ⊂ SG I.Since Aα ⊂ S ∗ G , by Lemma 3.3 (ii), we inn have Aα ⊂ SG I ∩ (S ∗ G ) = SI.Let I be the Z-complement for I in C .Then α = α + α where inn Z 1 2 c c α ∈ I ,α ∈ I .Sowehave A(α − α ) = A(α + α − α ) = Aα ∈ SI ∩ SI ={0} implying α ={0} 1 2 1 1 2 1 2 2 ud (using Lemma 2.1(i)). Therefore α = α ∈ I which further gives I ⊆ I . du (ii) Let α ∈ M. We will show that α ∈ M , by induction on |Supp(α)|. Firstly, if Supp(α) ={0},then the case is trivial. So suppose α is non-zero and the result is true for all elements β ∈ M of smaller support size. Choose T ⊂ Supp(α), minimal with respect to property M ∩ (N ∗ G} ={0}.If g ∈ T , −1 −1 −1 −1 then Supp(αg ) = Supp (α) g ⊇ Tg .Since T has the minimal property, so does Tg and −1 −1 −1 du e = gg ∈ Tg hence it is sufficient to show that αg ∈ M . Thus without loss of generality, we −1 −1 can replace α by αg and T by Tg .If T = {e, g ,..., g }; then by Lemma 2.5, there exist invariant 2 m g−1 subnearrings A = A , A ,..., A of N satisfying (nan ) f = n(af )n for all a ∈ A, n, n ∈ N and 1 2 m i i af = a f for all i. Thus by Lemma 2.1(iii), there exist units f , f ,..., f ∈ S such that conjugation by i 1 2 m −1 −1 m f induces automorphism g on N and hence g ∈ G ;thatis, g ∈ G .Let β = f g ∈ S ∗G, inn i inn i i i i i i =1 where f g is a unit in S ∗ G which acts trivially on S and β ∈ C .Since af = a f for all a ∈ A,by i Z i i i Lemma 2.5(iii), we have aβ = (a f )g ∈ M for all a ∈ A;thatis, Aβ ⊂ M which implies β ∈ M i i i =1 (by definition of M ). Let t = tr (α) be the identity coefficient of α and a ∈ A,define γ = aα − aβt ∈ M, where Supp(β) = T ⊂ Supp(α).Thenwehave Supp(γ ) = Supp(aα −aβt ) = aSupp(α)−aSupp(βt ). Since e ∈ T and f = f for e ∈ Supp(β) implies that Supp(βt )  1. Thus |Supp(γ )| < |Supp(α)|, 1 0 du d so by induction γ ∈ M . By Lemma 3.5(ii), M is a G−invariant ideal of C ,soonusing Lemma d d du 3.3(ii), SG M becomes invariant subnearring of S ∗ G. Hence a.g βt ∈ SG M ∩ (N ∗ G) = M du g which further implies that aα = γ + aβt ∈ M for all a ∈ A.Then B = A becomes non-zero g∈G g du G-invariant, invariant subnearring of N,where Bα = A α ∈ M . Since by Lemma 3.7(ii) g∈G du du M is N -cancelable, we get α ∈ M . The following theorem makes the task of proving the main results easier. Theorem 3.9 Let N ∗ G be a skew group nearring of finite group G acting on 3-prime nearring N and let C = Z [G ] be as above. Z inn d du (i) If P is an I -prime of N ∗ G with P ∩ N ={0},then P is a G-prime ideal of C and P = P . u u ud (ii) If I is a G-prime ideal of C ,then I is an I -prime of N ∗ G with I ∩ N ={0} and I = I . du Proof (i) We first show that P = P for any I -prime P of N ∗ G with P ∩ N ={0}. By Lemma 3.8(ii), du du d k we have atleast P ⊆ P . Conversely, let α ∈ P = P SG ∩ (N ∗ G),so α = δ f g ,where i i i =1 i d d δ ∈ P , f ∈ S, g ∈ G. By definition of P , there exist non-zero semigroup ideals D of N with i i i δ D = D δ ⊆ P for each i. Then D = D becomes non-zero G-invariant semigroup i i i i g∈G i =1 i 123 Arab. J. Math. ideal of N such that Dδ ⊆ P for all i. Again there exist non-zero semigroup ideals B of N such that i i B f ⊆ N ∗ G for all i.If B = B , then it becomes non-zero G-invariant semigroup ideal i g∈G i =1 of N with B f ⊆ N ∗ G. Therefore (DB)α ⊆ (Dδ )(B f )g ⊆ P(N ∗ G) ⊆ P.Since P ∩ N ={0} i i i i i and by Lemma 3.7(i) P is N −cancelable, hence α ∈ P.Let I and I be G- invariant ideals of C such 1 2 Z d u u u du u that I I ⊆ P . By Lemma 3.7(iii), I I ⊆ (I I ) ⊆ P = P,soby I -primeness of P, I ⊆ P or 1 2 1 2 1 2 1 u ud d ud d d I ⊆ P. Using Lemma 3.8(i), I = I ⊆ P or I = I ⊆ P , implying that P is G-prime. 1 2 2 1 2 (ii) Let I be a G-prime ideal of C . Here C is a finite dimensional algebra over Z and a finite dimensional Z Z algebra over field is an artinian ring so satisfies the descending chain condition on ideals and also every prime ideal is a maximal ideal. Thus I is a G-maximal ideal of C . From Lemma 3.3(ii), SG I ∩ N ⊂ SG I ∩ S ={0},sowehave I ∩ N ={SG I ∩ (N ∗ G)}∩ N ={0}∩ (N ∗ G) ={0}.Let M be any invariant u d ud subnearring of N ∗ G properly containing I .Thenwehave M ⊃ I = I (by Lemma 3.8(i)) implying d d d d that I ⊂ M ⊆ C .Since I is G-maximal and M is G- invariant, so either M = I or M = C . Suppose Z Z d du u d that M = I,then M ⊆ M = I (using Lemma 3.8(ii)) which is a contradiction. Thus M = C ,sowe have M ∩ N ={0}. For, we know that f e ∈ C so there exists a non-zero semigroup ideal M of N with 1 Z 1 M ( f e) ⊂ M which gives M ⊂ M implying that {0} = M ∩ N ⊂ M ∩ N . From this we conclude that 1 1 1 1 u u I is I -prime. For, suppose M , M ⊇ I ,where M and M are two invariant subnearrings of N ∗ G.But 1 2 1 2 by above observation, M ∩ N , M ∩ N ={0} which gives that {0} = (M ∩ N )(M ∩ N ) ⊆ M M ∩ N . 1 2 1 2 1 2 u u Since I ∩ N ={0}, it follows that M M ⊂ I . 1 2 The incomparability, we stated in the beginning now follows easily. Corollary 3.10 If P is an I -prime of N ∗ G with P ∩ N ={0} and M any invariant subnearring of N ∗ G such that M  Pthen M ∩ N ={0}. u d Proof By Theorem 3.9(i), any I -prime P of N ∗ G with P ∩ N ={0} is of the form I ,where I = P is a G-prime ideal of C . Thus if M  P = I and now M ∩ N ={0} by Theorem 3.9(ii). Theorem 3.11 Let N be a 3-prime nearring. (i) Any I -prime P of N ∗ G is minimal if and only if P ∩ N ={0}. (ii) There are finitely many such minimal primes, say P , P ,..., P where k  |G |  |G|. 1 2 k inn (iii) J = P ∩ P ∩ ··· ∩ P is the unique largest nilpotent invariant subnearring of N ∗ G such that J = 1 2 k u |G | inn (rad(C )) ,where rad(C ) is the jacobson radical of C and J ={0}. Z Z Z Proof C is an artinian ring so has finitely many ideals. Let I , I ,..., I be all G-prime ideals of C . Z 1 2 k Z Then as in the case of rings [3, Lemma 2.7], it follows that k  dim (Z [G ]) =|G |  |G| and Z inn inn I ∩ I ∩ ··· ∩ I = rad(C ), the Jacobson radical of C .Let P = I , so by Theorem 3.9(ii), P , P ,..., P 1 2 k Z Z i 1 2 k are I -primes of N ∗ G with P ∩ N ={0}.Set J = P ∩ P ∩ ··· ∩ P ⊂ P = I for all i,sowehave i 1 2 k i d ud d du u J ⊂ I = I ;thatis, J ⊂ I ∩ I ∩· · · ∩ I = rad(C ). Thus by Lemma 3.8(ii), J ⊂ J ⊂ (rad(C )) . i 1 2 k Z Z We know that rad(C ) is nilpotent(Jacobson radical of an artinian ring is nilpotent), thus (rad(C )) and J Z Z are nilpotent. On other hand, since each P certainly contain all nilpotent invariant subnearrings of N ∗G, thus J u u contains all nilpotent invariant subnearrings of N ∗G, so must contain (rad(C )) . Therefore J = (rad(C )) , Z Z u k |G | inn where J is the largest nilpotent invariant subnearring of N ∗ G and also {(rad(C )) } = f e = J . Z 0 Finally let P be any I -prime of N ∗ G so we have P ⊂ P = J ⊂ P,as J is nilpotent and hence by I - i i i i primeness of P, P ⊂ P for some i. Thus the minimal primes of N ∗G are the members of set {P , P ,..., P }. i 1 2 k d d Observe that P ⊃ P implies that I = P ⊃ P = I . But since I is G-maximal, we must have i = j.This i j i j j i j shows that P , P ,..., P are only minimal primes of N ∗ G. 1 2 k As an easy application of above result, we now prove Going Down Problem. Proof of Going Down Problem. The existence of an I -prime such that P ∩ N = A follows from Lemma 1.3.Since A ={0}, by above result, P is not minimal prime. So P must properly contain some minimal prime P ;thatis, P ⊃ P for some i such that P ∩ N ={0}. i i Corollary 3.12 Let J = P ∩ P ∩ ··· ∩ P be as above. 1 2 k (i) There exist a sequence J = J , J ,... of N -cancelable invariant subnearrings of N ∗ G such that J J ⊂ 1 2 i J and J ={0} for some k  |G |. i +1 k 1 inn 123 Arab. J. Math. (ii) Let α ,α ,...,α be elements of J and let A be a non-zero semigroup ideal of N . Then there exists a 1 2 m non-zero G-invariant semigroup ideal B of N such that Bα ⊂ AJ for all i. Proof Set I = rad(C ).Then I is nilpotent and J = I (by Theorem 3.11(iii)). u u u (i) Set J = I ,so I becomes invariant subnearring of N ∗ G and N -cancelable (by definition of I and i i i u i u i +1 u k by Lemma 3.7(iii)) where JJ = I (I ) ⊆ (I ) = J .Since I is nilpotent then I ={0} for i i +1 k  |G | so we have J = SG I ∩ (N ∗ G) = f e. 1 inn k 0 (ii) Let γ = fgβ ∈ S ∗ G with f ∈ S, g ∈ G,β ∈ I . We firstly show that there exists a non-zero G-invariant semigroup ideal D of N such that Dγ ⊂ AJ . We can choose non-zero semigroup ideals D and D of 1 2 g g N with D f ⊂ N and D β ⊂ (N ∗ G) ∩ SG I = I .Set D = D and D = D then 1 2 g∈G g∈G 1 2 1 2 both become G-invariant semigroup ideals of N with D f ⊂ N and D β ⊂ I .Alsowecan finda 1 2 G-invariant semigroup ideal D of N with D ⊂ A,where D = A . Setting D = D D D then, 3 3 3 3 2 1 g∈G for all d d d ∈ D = D D D ,wehave (d d d )( fgβ) = (d d ) (d β ) fg = d ((d β )(d f )g) ⊆ 3 2 1 3 2 1 3 2 1 3 2r 1r 3 2 2 u u D (I (N ∗ G)) ⊆ AJ . Finally if α ,α ,...,α be elements of J = I , then α ∈ SG I for all i and 3 1 2 m i hence α can be written as γ where γ = f g β ; f ∈ S, g ∈ G,β ∈ I . By above we can find i ij ij ij ij ij ij ij ij non-zero G-invariant semigroup ideal D of N with D γ ⊂ JA for each i.Set B = D ,thenit ij ij ij ij i, j becomes non-zero G-invariant semigroup ideal of N . For, if not then D ⊂ D ={0} implying ij ij i, j i, j that D ={0} for some i and j, a contradiction and also each D is a G-invariant semigroup ideal of N , ij ij so is their intersection. Here B ⊂ D for all i, j implies that Bα ⊂ AJ for all i. ij i Acknowledgements The authors are thankful to learned referee for his perspicacious comments that resulted in the present form of the paper. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Funding No funding available for publication. Declarations Conflict of interest The authors have not disclosed any conflict of interests. 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Algebra. 17(7), 1533–1562 (1989) 10. Sharma, R.P.; Gupta, A.: Characterization of G-Prime Fuzzy ideals in a ring–An alternate approach. Comm. Algebra. 28(10), 1981–1987 (2000) 11. Sharma, R.P.; Banota, R.: Fuzzy ideals of group graded rings and their smash products. Comm. Algebra. 30(6), 3025–3036 (2002) 12. Sharma, R.P.; Samriti, Sharma: Group action on fuzzy ideals. Comm. Algebra. 26(12), 4207–4220 (1998) 13. Sharma, R.P.; Samriti, S.: On the range of a G-prime fuzzy ideals. Comm. Algebra. 27(6), 2913–2916 (1999) 14. Sharma, R.P.; Tilak Raj, S.: G-prime ideals in semirings and their skew group semirings. Comm. Algebra. 34, 4459–4465 (2006) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Arabian Journal of Mathematics Springer Journals

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Arab. J. Math. https://doi.org/10.1007/s40065-023-00426-z Arabian Journal of Mathematics Shalini Chandel · Ram Parkash Sharma Primes and G-primes in Z-nearalgebras Received: 30 January 2022 / Accepted: 2 March 2023 © The Author(s) 2023 Abstract Let N be a Z-nearalgebra; that is, a left nearring with identity satisfying k(nn ) = (kn)n = n(kn ) for all k ∈ Z, n, n ∈ N and G be a finite group acting on N . Then the skew group nearring N ∗ G of the group G over N is formed. If N is 3-prime (aNb = 0 implies a = 0or b = 0), then a nearring of quotients Q (N ) is constructed using semigroup ideals A (a multiplicative closed set A ⊆ N such that A N ⊆ A ⊇ NA )of i i i i i N and the maps f : A → N satisfying (na) f = n(af ), n ∈ N and a ∈ A . Through Q (N ), we discuss the i i i i i 0 relationships between invariant prime subnearrings (I -primes) of N ∗ G and G-invariant prime subnearrings (GI -primes) of N . Particularly we describe all the I -primes P of N ∗ G such that each P ∩ N ={0},a i i GI -prime of N . As an application, we settle Incomparability and Going Down Problem for N and N ∗ G in this situation. Mathematics Subject Classification 16Y30 · 16W22 1 Introduction Group theory plays an important role when combined with ring theory, semiring theory and various other algebraic structures in mathematics. The action of groups on these structures is one of the important areas of research which was initiated as an attempt to develop Galois theory for them after the inception of Galois theory, when Liouville reviewed Galois manuscript and published it in October–November 1846 issue of the Journal de Mathematiques Pures et Appliques after his death in 1832. In the case of rings, the theory was initially extended to division rings by Jacobson [2] in 1940. The conditions in the case of non-commutative rings were complex, so a fresh approach was made by starting with some simple questions about the relationships of the structure of a ring with identity to the structure of fixed subring R and skew group ring R ∗ G.For G-prime ideals of a ring R with a finite group action on it and prime ideals of R ∗ G, the “Incomparability” and “Going Down” problems were studied and settled by Passman [6], Lorenz and Passman [3], and Montgomery [4,5]. Then these problems were settled for prime kernel functors of group graded rings and their smash products by Parvathi and Sharma [9] which is the generalization of the work of M. Cohen and and Montgomery [1] “Group Graded Rings, Smash Products and Group Actions” and further Sharma et al. [8,10–14] studied group actions on different algebraic structures including semirings. But nearrings, which are the generalization of rings with only one distributive law and are best tools to study non-linear algebraic structures, remain untouched as far as the action of groups on them are concerned. In rings with finite group actions, ideals play an important S. Chandel · R. P. Sharma ( ) Department of Mathematics and Statistics, Himachal Pradesh University, Summer Hill, Shimla 171005, India E-mail: rp_math_hpu@yahoo.com S. Chandel E-mail: shalinichandel1992@gmail.com 123 Arab. J. Math. role in settling various questions regarding the relationships between a ring and its skew group ring; but the conditions are different for nearrings in the absence of one distributive law. Hence the generalization of the results provedin[3] could not be attained for nearrings as it is. For example, a nearring of quotients cannot be defined identical to the ring of quotients (c.f. [3], Sect. 2). This made us to explore the possibilities of different algebraic substructures in nearrings to establish relationships between nearrings and their skew group nearrings. We succeeded to use semigroup ideals to define a nearring of quotients, through which we could settle ‘Incomparability’ and ‘Going Down Problem’ for invariant subnearrings (which are, in fact, alike ideals in rings) of a 3-prime nearring. Let N be a left nearring with identity denoted by 1 (or by 1 when it has to be distinguished from the integer 1) and Z the ring of integers. Then N becomes Z-group; that is, the map Z × N → N satisfies (i) (k + k )n = k n + k n and (ii) (k k )n = k (k n) for all k , k ∈ Z and n ∈ N . Throughout this paper, N 1 2 1 2 1 2 1 2 1 2 is a Z-nearalgebra which will be frequently called a nearring; that is, a Z-group N which satisfies k(nn ) = (kn)n = n(kn ) for all k ∈ Z and n, n ∈ N . It is pertinent to note here that if N is a Z-nearalgebra then (i) N is zerosymmetric, (ii) (−1 )n =−n for all n ∈ N and (iii) N is abelian (n + n = n + n,for all n, n ∈ N ). For, (i) 0 .n = (0.1 )n = 1 (0.n) = 1 .0 = 0 , (ii) (−1 )n = (−1.1 )n =−1((1 )n)) =−1.n =−n N N N N N N N N N and (iii) n + n = (−1 )(−n) + (−1 )(−n ) = (−1 )((−n) + (−n )) =−((−n) + (−n )) = n + n.First, N N N we give an example of a Z-nearalgebra which shows that class of such nearrings arises in a natural way. Example Let G be any additive abelian group and W (G) ={ f : G → G | (±a ± a ... k times) f = ±(af ) ± (af )... k times} (here (af ) is the image of a under f ), be a set of weak homomorphism from G into G itself. Then (1) W (G) is a Z-group under the map f + f + ··· k − times, if k ≥ 0 kf = (− f ) + (− f ) + ···|k|− times, if k < 0. (2) The maps 0, i (identity) and −i (a −→ −a) are in W (G). (3) W (G) is an abelian zerosymmetric left nearring with identity i, satisfying (i) 0 ◦ f = 0 (ii) (−a) f =−(af ) (iii) (−i ) ◦ f =− f ,where a(− f ) =−(af ), and (iv) k( f ◦ f ) = (kf ) ◦ f = f ◦ (kf ) for any integer k, f, f ∈ W (G) and hence W (G) is a Z-nearalgebra. Proof (1) and (2) are obvious. (3) Let f, f , f ∈ W (G) and a ∈ G. Then obviously f + f ∈ W (G) and f ◦ f ∈ W (G), as G is abelain. Clearly f ◦ i = i ◦ f = f , f ◦ ( f + f ) = f ◦ f + f ◦ f ,and f ◦ 0 = 0 ◦ f = 0, proving (i). (ii) (−a) f + af = (−a + a) f = (0) f = 0giving that (−a) f =−(af ). Now (iii ) follows from (ii ).(iv)For k > 0, we have a k( f ◦ f ) = a f ◦ f + f ◦ f +· · · k times = (af ) f +· · · k times = (af + ··· k times) ◦ f = (a ( f + ··· k times)) ◦ f = a(kf ◦ f ) implying k( f ◦ f ) = (kf ) ◦ f and k( f ◦ f ) = f ◦ (kf ) follows from the left distributive law in W (G).For k < 0, we have k( f ◦ f ) = (−( f ◦ f )) + (−( f ◦ f )) + · ··|k|-times = (−i ◦ ( f ◦ f )) + (−i ◦ ( f ◦ f )) + · ··|k|-times =−i ◦ ( f ◦ f + f ◦ f + ···|k|-times) =−i ◦ (|k|( f ◦ f )) =−i ◦ ((|k| f ) ◦ f ) = (−i ◦ (|k| f )) ◦ f = (|k|(−i ◦ f )) ◦ f = (|k|(− f )) ◦ f = (kf ) ◦ f . Similarly, we get the second half of the equality for k < 0. Let G = {g = e, g ,..., g } be a finite multiplicative group acting on N . The action of g ∈ G on n ∈ N 1 2 n g e g g gg  g g g  g g g is denoted by n .So n = n, (n ) = n , (n + n ) = n + n and (nn ) = n n ,where n, n ∈ N and g, g ∈ G.Let N ∗ G = n g |n ∈ N , g ∈ G . Define addition in N ∗ G componentwise; that is, i i i i i =1 n g + n g = (n + n )g .Since N is abelian, it is easy to see that N ∗ G is an additive abelian i i i i i i i i i i group. Thus collecting the cofficients of common g , here afterwards, without loss of generality any element of N ∗G is taken of the type n g ,where n ∈ N and g ∈ G are distinct. The map f : (N ∗G)× N −→ N ∗G i i i i −1 defined by n g , n −→ (n n )g makes it a right N -module. Two elements of N ∗G are equal if and i i i i i i only if corresponding coefficients n s are equal. Thus {1.g |g ∈ G} forms an N -basis for N ∗G.Further N ∗G i i becomes an abelian (commutative with respect to addition) nearring having identity 1.e ( also simply denoted −1 by 1) with addition defined as above and multiplication: ng n h = (nn )(gh), g∈G h∈G g,h∈G where ng, n h are two elements in N ∗ G. In the absence of right distributive law, it is pertinent g∈G h∈G to note here that before multiplication of two elements of N ∗ G, it is necessary to write them in the form where all g s are distinct. The following definitions play important role in this paper. 123 Arab. J. Math. Definition 1.1 [7] A non-empty subnearring A of N is called right invariant (resp. left invariant) if AN ⊆ A (resp. NA ⊆ A). If A is both right invariant and left invariant, then it is said to be an invariant subnearring of N . Definition 1.2 (i) An invariant subnearring P of N is called invariant prime if for invariant subnearrings A, B of N , AB ⊆ P implies that A ⊆ P or B ⊆ P. (ii) An invariant subnearring A of N is called nilpotent if there exists a positive integer n such that A ={0}. (iii) A G-invariant, invariant subnearring P of N is called G−invariant prime if for all G-invariant, invariant subnearrings A, B of N , AB ⊆ P implies that A ⊆ P or B ⊆ P.Here G-invariant, invariant subnearring g G means an invariant subnearring P of N such that P = P = P . g∈G The following Lemma motivates us to investigate Incomparability and Going Down relation between invariant prime subnearrings of N ∗ G and G-invariant prime subnearrings of N . Here afterwards, an invariant prime subnearring will be called I -prime and G-invariant prime subnearring will be called GI -prime. Lemma 1.3 If P is an I −prime of N ∗ G, then P ∩ N is a G I -prime in N . Conversely, if A is a G I -prime of N , then there exists atleast one I -prime P of N ∗ G such that P ∩ N = A. g −1 Proof If P is an invariant subnearring of N ∗G,then P is G-invariant because for x ∈ P ∩ N , x = g xg ∈ P for all g ∈ G,as (N ∗ G)P ⊆ P ⊇ P(N ∗ G). Hence x ∈ P ∩ N for all g ∈ G.Also P ∩ N is an invariant subnearring of N,since N (P ∩ N ) ⊆ NP ⊆ (N ∗ G)P ⊆ P implies that N (P ∩ N ) ⊆ P ∩ N . Similarly P ∩ N ⊇ (P ∩ N )N . To prove P ∩ N is GI -prime, suppose that A and B are G-invariant, invariant subnearrings of N such that AB ⊆ P ∩ N.Then A ∗ G = a g |a ∈ A, g ∈ G is an invariant subnearring of N ∗ G. i i i i For, if a g , a g ∈ A ∗ G,then a g + a g = (a + a )g and a g n h = i i i i i i i i i i j j i i i i i i j i i i −1 (a n )(g h ) ∈ A ∗ G for all n g ∈ N ∗ G. Similarly A ∗ G ⊇ (N ∗ G)(A ∗ G) proving i i j j j i, j j j that A ∗ G is an invariant subnearring of N ∗ G. Moreover a g = (1.g )a ∈ (N ∗ G)A and i i i i i i b g = b (1.g ) ∈ B(N ∗ G),sowehave A ∗ G ⊆ (N ∗ G)A and B ∗ G ⊆ B(N ∗ G) implying that i i i i i i (A ∗ G)(B ∗ G) ⊆ (N ∗ G)AB(N ∗ G) ⊆ (N ∗ G)(P ∩ N )(N ∗ G) ⊆ (N ∗ G)P(N ∗ G) ⊆ P.Since P is an I -prime of N ∗ G,wehave A ∗ G ⊆ P or B ∗ G ⊆ P. Suppose A ∗ G ⊆ P.Since A ⊆ (A ∗ G) ∩ N yields that A = (A ∗ G) ∩ N ⊆ P ∩ N , therefore P ∩ N is a GI -prime. Conversely, assume that A is a GI -prime subnearring of N . As observed above A ∗ G is an invariant subnearring of N ∗ G and (A ∗ G) ∩ N = A.So = {P ⊆ N ∗ G | P ’s invariant subnearring of N ∗ G and P ∩ N = A} i i i is non-empty and partially ordered set with respect to inclusion of subsets. Let C = {P ∈ } be a chain in ,then P is an invariant subnearring of N ∗ G and P ∩ N = A,as P ∩ N = A for all i.This i i i i i implies that P is in . Thus by Zorn’s lemma we can choose P ∈ , a maximal element. To prove that P is I -prime, suppose I, J are invariant subnearrings of N ∗ G such that I ⊃ P, J ⊃ P with IJ ⊂ P.Then IJ ∩ N ⊂ P ∩ N.Since I ∩ N , J ∩ N are invariant subnearrings of N and (I ∩ N )(J ∩ N ) ⊆ IJ ∩ N,wehave (I ∩ N )(J ∩ N ) ⊂ P ∩ N = A.Soby GI -primeness of A, I ∩ N ⊂ A (suppose). Also A = P ∩ N ⊂ I ∩ N . This implies that I ∩ N = A. Thus I ∈  and by maximality of P, P = I implying that P is I -prime. The main results, in this paper, derived are as follows: Let N be 3-prime and G a finite group acting on N . Incomparability:If P is any I -prime of N ∗ G with P ∩ N ={0} and M any invariant subnearring of N ∗ G such that P  M then M ∩ N ={0}. Going Down Problem:For every GI -prime {0} = A of N , there exist I -primes P and P of N ∗ G with P ∩ N = A such that P  P and P ∩ N ={0}. The above results are generalizations or some equivalent forms of the results proved in [3]. 2 Nearring of quotients Here afterwards, the nearring N has assumed to be 3-prime. We recall that a non-empty multiplicative closed subset M of N is called right semigroup ideal (resp. left semigroup ideal) if MN ⊆ M (resp. NM ⊆ M). If M is both right semigroup ideal and left semigroup ideal, then it is said to be a semigroup ideal of N.Let be the collection of all semigroup ideals of N.Since N is 3-prime, A ∩ A and A A become non-zero i j i j semigroup ideals of N for all non-zero semigroup ideals A and A of N . Consider the set all functions i j f : A −→ N with (na) f = n(af ),where n ∈ N ,a ∈ A and A ∈ . Two functions f and f are i j 123 Arab. J. Math. equivalent if both agree on a common domain; that is, f ≡ f iff af = af on A ∩ A .Let f denotes the i j i j i j equivalence class of f and S = Q (N ) = f | f : A −→ N , A ∈  and (na) f = n(af ) .Then S becomes a left distributive abelian nearring with respect to addition and multiplication defined as f + f = f + f i j i j and f . f = f f in S,where f + f : A ∩ A −→ N and f f : A A −→ N . We call S, a nearring of i j i j i j i j i j j i quotients. We note that N is a subnearring of S.For,let n ∈ N,define n : N −→ N , the right multiplication by n.For all n , n ∈ N , n n n = n (n n) = n (n n ).Define amap ϕ : N −→ S by nϕ = n . r r r If n = n ,then nn = nn for all n ∈ N so n = n proving ϕ is well defined This is a nearring 1 2 1 2 1r 2r homomorphism as we have n(n + n ) = n(n + n ) = nn + nn = nn + nn = n(n + n ) and 1 2 r 1 2 1 2 1r 2r 1r 2r n(n n ) = n(n n ) = (nn )n = (nn )n = n(n n ).Since 1 ∈ N , ϕ is one-one. 1 2 r 1 2 1 2 1r 2r 1r 2r Lemma 2.1 Let S = Q (N ) be as above. Then (i) If f ∈ S and A f = {0} for some non-zero semigroup ideal A of N , then f = f ,where f is the zero 0 0 function. (ii) If f , f ,..., f ∈ S, then there exists a non-zero semigroup ideal A of N with A f ⊂ N for all i. 1 2 n i (iii) Assume that g be any automorphism of N and A,B be two non-zero semigroup ideals of N . Let f : A −→ B be a bijection which satisfies (nan ) f = n(af )n , for all n, n ∈ N, a ∈ A. Then f is a unit in S and conjugation by f induces an automorphism g on N . Proof Suppose that f : A −→ N and a ∈ A. Consider a : N −→ N defined by na = na ∈ N.Then f = r r a f is defined on N,asfor any n ∈ N we have n(a f ) = (na ) f = (na) f . Hence n(af ) = (na) f = n(af ) . r r r r (i) Let f ∈ S with Af = {0} for some non-zero semigroup ideal A of N.Since f vanishes on non-zero semigroup ideal A,thisimpliesthat f ∈ f ,but f ∈ f which further gives f = f . 0 0 (ii) This follows easily as N is 3-prime. (iii) Here f is a bijection, so there exists an inverse map f : B −→ A which is also a bijection and −1 −1 −1 g g  g  g nbn f = (n(af )(n ) ) f = nan = n(bf )n (1) −1 Since f and f are both in S,where ff is an identity on A and f f is an identity on B,so f = f . −1 −1 −1 For any n ∈ N , f n f is defined on B,soby (1) for all b ∈ B,wehave b( f n f ) = (bf )(n f ) = r r r −1 −1 g g g −1 g g −1 g −1 ((bf )(n ) ) f = (bn ) f f = bn = b(n ) . This implies that f n f = (n ) ,thatis, f n f = r r r r g −1 (n ) .Since N is a subnearring of S, this yields f n f = n for all n ∈ N . Lemma 2.2 (i) If N is 3-prime, then so is S. (ii) Any automorphism of N extends uniquely to an automorphism of S. ρ ρ (iii) Suppose that ρ is an automorphism of N and x , y be any non-zero elements of N satisfying xny = yn x for all n ∈ N , then there exists a unit f ∈ S such that x f = y and conjugation by f induces an automorphism ρ on N . Proof (i) Suppose that f , f = f in S such that f S f = f , then there exist A , A non-zero semigroup i j 0 i j 0 i j ideals of N on whom f and f are defined with A f = {0} and A f = {0}.Since N ⊂ S and f n f i j i i j j i r j is defined on A A ,sowehave (A A )( f n f ) = {0} for all n ∈ N , implying that (A (A f )N ) f = j i j i i r j j i i j {0}.Since A (A f )N is a semigroup ideal of N,for a (a f )n in A (A f )N and n, n in N , we have j i i j i i j i i n (a (a f )n) = (n a )(a f )n ∈ A (A f )N and (a (a f )n)n ∈ A (A f )N.Also A (A f )N = {0}, j i i j i i j i i j i i j i i j i i if not then A (A f ) ⊂ A (A f )N = {0}, implies that (A A ) f = {0}.But N is 3-prime, so A A = {0} j i i j i i j i i j i and thus by Lemma 2.1(i) f = f , which is a contradiction. Thus A (A f )N = {0} andbyLemma 2.1(i), i 0 j i i { } (A (A f )N ) f = 0 implies that f = f , again a contradiction. Hence f S f = f . j i i j j 0 i j 0 g g g g g (ii) Let g be an automorphism of N and f : A −→ N . Define a map f : A −→ N by a f = (af ) .Since −1 −1 −1 g g g g g g g g g f ∈ S, for all a ∈ A and n ∈ N,wehave (na ) f = (n a) f = ((n a) f ) = (n (af )) = g g g n(af ) = n(a f ). Define a map ϕ : S −→ S by f ϕ = f . This is an automorphism from S onto g g itself. For, f , f in S,if f ≡ f then f = f on some common domain implies that f = f on a i j i j i j i j g g common domain. Thus f ≡ f proving that ϕ is well defined. Reversing the steps, ϕ is easily seem to i j g g g be one-one. To prove this is a homomorphism, ( f + f )ϕ = f + f ϕ = f + f = f + f = i j i j i j i j g g g g g g g f + f = f ϕ + f ϕ and ( f . f )ϕ = f f ϕ = f f = f f = f . f = ( f )ϕ.( f )ϕ. To prove i j i j i j i j i j i j i j i j 123 Arab. J. Math. −1 −1 −1 g g g onto, if f ∈ S,take f ∈ S and we have f ϕ = f = f ∈ S. Therefore this gives rise to an automorphism of S extended by g. Now we show that if σ is an automorphism of S fixing N elementwise, then σ is an identity automorphism of S.For,let A be a non-zero semigroup ideal of N with A f ⊂ N , σ σ σ σ σ then a f = a f = a f = a f giving A f − f = {0}. So by Lemma 2.1(i), f = f for all f ∈ S. Now for uniqueness of extension, suppose that ϕ and ϕ are two automorphisms of S extended 1 2 −1 −1 −1 −1 −1 by g.Since n ϕ ϕ = (nϕ ) ϕ = (ng) g = n gg = n,weget ϕ ϕ is an identity on N 1 1 1 2 2 2 therefore ϕ = ϕ on S. 1 2 (iii) Set X = n xn |n , n ∈ N and Y = n yn |n , n ∈ N . Define maps f : X → Y 1 2 1 2 1 2 1 2 finite finite −1 ρ ρ and f : Y → X by (n xn ) f = n yn and (n yn ) f = n xn .Here f and f are well-defined. For, 1 2 1 1 2 1 2 2 −1 −1 −1 −1 −1 ρ ρ ρ ρ ρ take n xn = n xn .Since (xny) = y nx (given), then (xn y) n = y n xn = y n xn = 1 2 1 2 1 2 1 2 1 2 −1 −1 −1 −1 −1 −1 ρ ρ −1 ρ ρ ρ ρ  ρ (xn y) n which implies x (n y n − n y n ) = 0. Since N is 3-prime and x ={0}, 1 2 1 1 2 ρ ρ hence n yn = n yn . Similarly f can be shown to be well-defined. The functions are well defined on 2 1 2 the finite product of such elements as it is independent of the order on which action of ρ is applied. Since f −1 and f are in S,where ff = f on X and f f = f on Y , f = f is a unit in S.Also x f is defined on 1 1 r N,sofor all n ∈ N we have n(x f ) = (nx ) f = n(xf ) = ny = ny ,thatis, xf = y.Again forany n ∈ N , r r −1 ρ ρ f n f is defined on Y,sowehave (n yn )( f n f ) = ((n xn )n) f = (n yn )n = (n yn )n ,so r 1 2 r 1 1 2 1 2 r −1 f n f = n ;thatis, f n f = n . r r Now we are able to define inner automorphisms of N . Definition 2.3 An automorphism of N is said to be G-inner iff it is induced by conjugation of a unit in S; −1 that is, these automorphisms arise from those units f ∈ S with f N f = N.Since N is a subnearring of S, g −1 by Lemma 2.2(ii), for any g ∈ G , there exists f corresponding unit in S such that f = f f f for all inn g g g f ∈ S.The setofall G-inner automorphisms of N is denoted by G . inn Lemma 2.4 G is a normal subgroup of G. inn Proof For, if g, g ∈ G , then there exist f , f (both not equal to f ) units in S such that for all n ∈ N , inn g 0 −1  −1  −1 −1 g gg g −1 n = f n f and n = ( f n f ) = f  f n f f  = ( f f  ) n( f f  ). Thus the unit f f g g g g g g g g g g g g g g −1 gg g in S gives rise to G-inner automorphism gg .For g ∈ G and g ∈ G ,wehave n = f n f .So inn g g −1 −1 −1 −1 −1 −1 −1 −1 g −1 g gg g g g g g −1 g (n ) = ( f n f ) = ( f ) n( f ) = ( f ) n( f ) .Here f is a unit in S, thus g g g g g g g −1 ggg ∈ G . inn Finally, in this section, we prove result which is required for our work in next section. Lemma 2.5 Let  be a non-zero additive subgroup of N ∗G with N , N ⊆ and T = {g = e, g ,..., g } ⊂ 1 2 m G. Suppose ∩ (N ∗ T ) = {0} and  ∩ (N ∗ T ) = {0} for all T T(1) For each i = 1, 2,..., m, define ⎧ ⎫ ⎨ ⎬ A = n ∈ N |∃α = n g ∈ with coefficient of g as n i j j i ⎩ ⎭ j =1 then (i) Each A is a non-zero invariant subnearring of N . (ii) A = A , then for each i, there exist a natural bijection f : A −→ A which satisfies (nan ) f = 1 i i i −1 n(af )n for all n, n ∈ N and a ∈ A, where f is the identity function. i 1 (iii) The elements of  ∩ (N ∗ T ) are precisely of the form α = (af )g with a ∈ A. i i i =1 123 Arab. J. Math. { } { } Proof (i) Let A be as above. Then A = 0 ;for,if A = 0 , then we have elements of type i i i α = n g ∈  ∩ (N ∗ T ) with coefficient of g as 0, contradicting (1). Take n, n ∈ A , j j i i j =1, j =i m m so there exist α = n g and α = n g in  with coefficient of g as n and n respec- j j j i j =1 j =1 tively. Since α + α = (n + n )g ∈  with coefficient of g as n + n so by definition of A , j j i i j =1 we have n + n ∈ A . Similarly for n ∈ N,wehave n α = (n n )g ∈  with coefficient of i j j j =1 g as n n implying that n n ∈ A .Further,for any n ∈ N , there exists n ∈ N such that n = n , i i −1 −1 m g g j  i αn = (n n )g ∈  with coefficient of g as nn implying that nn ∈ A . Hence A is an j j i i i j =1 invariant subnearring of N . (ii) Note that for each a ∈ A , there exists a unique α = n g such that a = n . The existence is i i j j i i j =1 m m obvious. For uniqueness, suppose that α = n g and α = n g in T such that coefficients j j j j =1 j =1 j of g in both α and α are equal, then we have α − α ∈  ∩ (N ∗ T ) so α = α (using (1)). Hence the map f : A −→ A defined by af = a , a ∈ A is well defined, one-one and onto. Finally let α be as i i i i −1 m g above and let n, n ∈ N , a ∈ A.Since nαn = (nn n )g ∈  and g = e,thisimpliesthat j j 1 j =1 −1 −1 g g i i (nan ) f = na n = n(af )n . i i i (iii) From (ii), we have α = (af )g ∈  ∩ (N ∗ T ) and conversely it is also obvious that every element i i i =1 of  ∩ (N ∗ T ) is of this form. 3 Incomparability and going down problem In this section, we settle the Incomparability and Going Down Problem for N and N ∗ G. For any positive integer k,let kN = {kn | n ∈ N }.Then kN is a semigroup ideal of N.For,let n ∈ N , kn ∈ kN,then n (kn) = k(n n) and (kn)n = k(nn ) in kN as N is a Z-nearalgebra. For any ∈ Q, q > 0and qn ∈ qN , pn, if p ≥ 0 p p p define f : qN −→ N by (qn) f = Then f ∈ S.For,let n ∈ N , qn ∈ qN,then q q q | p|(−n), if p < 0. p(nn ) = n (pn), if p ≥ 0 p p p p we have (n (qn)) f = (q(n n)) f = So (n (qn)) f = n (qn) f ; q q q q | p|(−n n) = n (| p|n), if p < 0. p p that is, f ∈ S.Define Z = f | ∈ Q, q > 0 . q q Lemma 3.1 Let Z be as above. p p (i) For any ∈ Q, q > 0, we have f =− f ,if p < 0 and hence f −p =− f ,if p ≥ 0. | p| q q q p p 1 2 (ii) For any , ∈ Q, q > 0 and q > 0; we have 1 2 q q 1 2 p p (a) f 1 + f 2 = f p p ; 1 2 q q 1 2 q q 1 2 (b) f p . f p = f . 1 2 p p 1 2 q q 1 2 q q 1 2 p p p (c) Z is closed with respect to addition and multiplication of maps and f n = n f for all n ∈ N, f ∈ Z r r q q q and thus Z is contained in the centre of S. (iii) Z is a subnearring of S isomorphic to Q and hence is a field. Proof (i) For any qn ∈ qN,wehave (qn) f =| p|(−n) =| p|(n(−1 )) = (| p|n)(−1 ) =−(| p|n) = N N p p − (qn) f | p| implying (qn) f + f | p| = {0} for all qn ∈ qN . Hence f =− f | p| ,if p < 0. q q q q q (ii) (a) Firstly the maps on both sides of (a) are defined on qN,where q = lcm(q , q ).Let t = and 1 2 1 p p t = .Then f + f = f = f if p ≥ 0and p ≥ 0. In case, p and p are 1 2 p t +p t p p 2 1 2 1 2 1 1 2 2 1 2 q q + 1 2 q q q 1 2 negative; using (i), the sum is equal to − f + f = f = f . If one of | p | | p | | p |t +| p |t p t +p t 1 2 1 1 2 2 1 1 2 2 q q q q p or p is negative, then also we get the same result. 1 2 123 Arab. J. Math. p p (b) The product of f 1 and f 2 is defined on q q N and obviously this is contained in both q N and q N . 1 2 1 2 q q 1 2 p p Further on q q N , f . f = f irrespective of the sign of p and p as | p p |=| p || p |. 1 2 p p 1 2 1 2 1 2 1 2 1 2 q q 1 2 q q 1 2 (c) The closure properties in Z follow from (a) and (b).Let f p ∈ Z , n ∈ N , f p n is defined on q q (pa)n, if p ≥ 0 p p qN,sofor all qa ∈ qN we have (qa) f n = (qa) f n = = r r q q (| p|(−a))n, if p < 0 p(an), if p ≥ 0 p p p p = q(an) f = ((qa)n) f = (qa) n f , proving that f n = r r q q q q | p|(−(an)), if p < 0 n f . Therefore, conjugation by f induces an automorphism of S which is trivial on N . Hence Lemma p p 2.2(ii) implies that this automorphism is trivial on S;thatis, f f = f f for all f ∈ S, proving that q q Z is contained in the centre of S. (iii) First part follows from (ii) (a & b). For isomorphism between Q and Z,define amap from Q to Z by p p −→ f , ∈ Q, q > 0. Then obviously, this map is well defined and it follows from (ii) (a & b) that q q p p this is an onto homomorphism. For one-one, suppose that f = f .If p and p both are positive or 1 2 1 2 q q 1 2 p p 1 2 negative, then this equality gives = . So suppose that one of p or p is negative (say p is negative). 1 2 1 q q 1 2 p p Then − f = f or f = f + f = f ;thatis, (| p |t + p t ) n = 0for all n ∈ N.So | p | 2 | p | 2 | p |t +p t 0 1 1 2 2 1 1 1 1 2 2 q q q 2 q 2 q 1 1 q q p p 1 2 | p |t + p t = 0 implying −| p |t = p t or −| p | = p and finally we get = , proving 1 1 2 2 1 1 2 2 1 2 q q q q 1 2 1 2 the lemma. Let Z be as in Lemma 3.1, Z (S) centre of S and C centralizer of S in S ∗ G. Then in the absence of right distributive law, unlike rings Z (S) and C are not closed with respect to ‘+’. However, we have −1 Lemma 3.2 C ⊆ S ∗ G and the set { g = f g | g ∈ G } where f is the unit in S inducing the inn g inn g automorphism g on N , forms S-basis for S ∗ G . inn Proof Firstly, we prove that C ⊆ S ∗ G .Let α ∈ C and g ∈ Supp(α),then α = fg + ··· . Thus by inn Lemma 2.1(i & ii), there exists a ∈ N with aα ∈ N ∗ G.Since α ∈ C, n a α = αn a for all n ∈ N r r r r −1 and we have a(n a ( fg)) = a(( fg)n a ) implying a((n a f )g)) = a(( f (n a ) g) which further implies r r r r r r r r −1 −1 −1 −1 g g g g that an(af ) = (af )n a . Let b = af be a non-zero element of N,sowehave anb = bn a for all −1 n ∈ N . Hence Lemma 2.2(iii ) implies that g ∈ G ;thatis, g ∈ G showing that C ⊆ S ∗ G .Now inn inn inn −1 for each g ∈ G , choose f ∈ S inducing G-inner automorphism g on N and let  g = f g. All elements inn g g −1 −1 −1 g = f g form S-basis for S ∗ G . For, consider f f g = f e,where f ∈ S implies f f = f . g inn i g 0 i i g 0 Since f is a unit in S, hence each f = f . g i 0 −1 −1 −1 Note. Each  g is a unit in S ∗ G which acts trivially on N by conjugation. For,  g n  g = g f n f g = r g r g −1 −1 −1 −1 −1 g g g −1 g (n ) g = (n ) = n because, since f n f = (n ) and unit corresponding to g is f . r r r g r g r g Hence  g ∈ C.As C ⊆ S ∗ G , any element of C is of the form α = f g , with f ∈ S. Since each f g inn i i i i i centralizes S and g  is a unit in C,so f ∈ S ∩ C = Z (S).Sothat α = f g , f ∈ Z (S). From Lemma 3.1, i i i i i the field Z ⊆ Z (S),sowedefine C = α ∈ C | α = f g  with each f ∈ Z Z i i i −1 Obviously, the set { g = f g | g ∈ G }⊆ C . Now we prove a result analogous to Lorenz and Passman g inn Z ([3], Lemma 2.2). Lemma 3.3 Let C be as above. Then (i) C forms finite dimensional group algebra Z [G ] with Z-basis { g | g ∈ G }.Further S ∗ G Z inn inn inn S ⊗ C . Z Z 123 Arab. J. Math. (ii) If I is a G-invariant ideal of C ,then I SG = SG I is an invariant subnearring of S ∗ G. Moremore SG I is contained in S-direct summand of S ∗ G with SG I ∩ (S ∗ G ) = S I and SG I ∩ S ={0}. inn p p p p Proof (i) Obviously C is closed w.r.t. ‘+’. If f g , f g  ∈ C ,then f g  f g  = Z i i j j Z i i j j q q q q i i j j −1 −1 −1 −1 −1 p p p p p p p p f f g f f g = f f f f f f g g = f f  g g = f  g g . i g i j g j i g g j g g i j i j i j i j i j i j i i j i q q q q q q q q i i i j j j i j As seen in Lemma 3.2, the elements  g, g ∈ G are S-linearly independent and hence Z-linearly ndepen- inn dent. Let α = f g  ∈ C , f ∈ Z.Since Z is central in C ,weget S ∗ G  S ⊗ C under the i i Z i Z inn Z Z map g −→  g,for g ∈ G . inn (ii) Let I be a G-invariant ideal of C . Since by definition of I , IS = SI and the G-invariance of I implies Ig = −1 gI for all g ∈ G. Observe that SG I = ISG.For,let f g i ∈ SG I , f g i = f i g = j j j j j j j j j j j −1 i f g ∈ ISG.Let G be the transversal of G in G;thatis, g G = G. So any element j inn inn j j j g ∈G of SG I is of the form   fg f  g ∈ SG I.Let I be the Z-complement of I in C .Since z Z g ∈G g∈G inn C is a finite dimensional vector space over Z and I its subspace, by (i), S ∗ G = SI ⊕ SI ,sowe Z inn have S ∗ G =   (SIg ⊕ SI g ). Thus SG I is contained in S-direct summand of S ∗ G. SG I is an g ∈G g −1 j   k invariant subnearring of S ∗ G.For, f g i f h i = f f g h (i i ) ∈ SG I. j 1 j k 2k j k 2k j j k k j,k j k 1 j g −1 j  k Again let f h ∈ S ∗ G,thenwehave f g i f h = f f g h i and k k j 1 j k k j k k k j j k j,k j j −1 f h f g i = f f h g i ∈ SG I. Note that SG I ∩ (S ∗ G ) = SI . Finally if k k j 1 j k j k 1 j inn k j j k, j j c c I = C , then we can choose I contains unity f e. This implies that   SI g contains S and hence Z 1 g ∈G SG I ∩ S = {0}. The following correspondence between G-invariant ideals of C and invariant subnearrings of N ∗ G is going to play a central role in proving the main results of this paper. Definition 3.4 (i) If I is a G-invariant ideal of C , then define I = SG I ∩ (N ∗ G). (ii) For any invariant subnearring M of N ∗ G,define M ={α ∈ C | Aα ⊆ M for some non-zero semigroup ideal A of N }. Lemma 3.5 (i) I is an invariant subnearring of N ∗ G. (ii) M is a G-invariant ideal of C . Proof (i) By Lemma 3.3(ii),wehave SG I is an inavriant subnearring of S ∗ G which implies that I becomes invariant subnearring of N ∗ G. (ii) Let α ,α ∈ M and β ∈ C . Then there exist A and A two non-zero semigroup ideals of N with 1 2 Z 1 2 A α, A α ⊆ M and B a semigroup ideal of N such that Bβ ⊆ N ∗G.Since A ∩ A is a non-zero semigroup 1 2 1 2 ideal of N,for α + α ∈ C , (A ∩ A )(α + α ) ∈ M implies α + α ∈ M .For any a b ∈ A B 1 2 Z 1 2 1 2 1 2 1 1 and ba ∈ BA , (a b)(α β) = (a b )(α β) = a { b α β)}= (a α )(bβ) ⊆ M (N ∗ G) ⊆ M implies 1 1 1 1 1 r 1 1 r 1 1 1 d d d d α β ∈ M and similarly, βα ∈ M .For G-invariance, let α ∈ M , so by definition of M , there exists 1 1 1 g g g g −1 a non-zero semigroup ideal A of N with Aα ⊂ M, hence A α = (Aα) ⊂ M = g Mg ⊂ M because g g g d M is an invariant subnearring of N ∗ G,sowehave A α ⊂ M which implies α ∈ M for all g ∈ G. Definition 3.6 Any invariant subnearring M of N ∗ G is N -cancelable iff for any α ∈ N ∗ G and non-zero G-invariant semigroup ideal A of N such that Aα ⊂ M implies α ∈ M. The I -primes of N ∗ G which intersect trivially with N and G-invariant ideals of C are N -cancelable. Lemma 3.7 (i) Any I -prime P of N ∗ G satisfying P ∩ N ={0} is N -cancelable. (ii) For any G-invariant ideal I of C ,I becomes N -cancelable. u u u (iii) If I , I are G-invariant ideals of C ,then I I ⊂ (I I ) . 1 2 Z 1 2 1 2 123 Arab. J. Math. Proof (i) Let A be any non-zero G-invariant semigroup ideal of N such that Aα ⊂ P for some non-zero α ∈ N ∗ G.Then Aα ∩ N ⊂ P ∩ N = {0},but Aα ∩ N = n ∈ N |n = (an )g gives An ={0} for i i i all i.Since N is 3-prime and ANn ⊆ An ={0},wehave n ={0} for all i. By this we can say that there i i i is no non-zero α ∈ N ∗ G such that Aα ⊂ P. Therefore, P is N -cancelable. (ii) Assume A is any non-zero G-invariant semigroup ideal of N such that Aα ⊂ I where α ∈ N ∗ G.By definition of I ,wehave Aα ⊂ SG I . Since by Lemma 3.3(ii) we know that SG I is contained in S-direct summand of S ∗ G;solet α = α + α where α ∈ SG I and α ∈ K,where K is a left and right 1 2 1 2 S-submodule of S ∗ G.Since N ∗ G ⊂ S ∗ G, we see that A(α − α ) = A(α + α − α ) = Aα ∈ K 1 1 2 1 2 implying Aα ∈ SG I ∩ K ={0}.SobyLemma 2.1(i), α is zero. Hence α = α ∈ SG I ∩ (N ∗ G) = I . 2 2 1 u u u u u (iii) By definition of I , I I ⊂ (SG I )(SG I ) ⊆ SG(I I ) which implies that I I ⊂ SG(I I )∩(N ∗G) = 1 2 1 2 1 2 1 2 1 2 (I I ) . 1 2 Now we have ud Lemma 3.8 (i) For any G-invariant ideal I of C ,I = I . du (ii) If M is an invariant subnearring of N ∗ G, then M ⊆ M . Proof (i) If α ∈ I , then there exists a non-zero semigroup ideal A of N with Aα ⊂ N ∗G ⊂ SG I ∩ (N ∗G) = u ud ud ud I ,so α ∈ I and we have I ⊆ I .Conversely, if α ∈ I , then there exists a non-zero semigroup ideal A of N with Aα ⊂ I = SG I ∩ (N ∗ G) ⊂ SG I.Since Aα ⊂ S ∗ G , by Lemma 3.3 (ii), we inn have Aα ⊂ SG I ∩ (S ∗ G ) = SI.Let I be the Z-complement for I in C .Then α = α + α where inn Z 1 2 c c α ∈ I ,α ∈ I .Sowehave A(α − α ) = A(α + α − α ) = Aα ∈ SI ∩ SI ={0} implying α ={0} 1 2 1 1 2 1 2 2 ud (using Lemma 2.1(i)). Therefore α = α ∈ I which further gives I ⊆ I . du (ii) Let α ∈ M. We will show that α ∈ M , by induction on |Supp(α)|. Firstly, if Supp(α) ={0},then the case is trivial. So suppose α is non-zero and the result is true for all elements β ∈ M of smaller support size. Choose T ⊂ Supp(α), minimal with respect to property M ∩ (N ∗ G} ={0}.If g ∈ T , −1 −1 −1 −1 then Supp(αg ) = Supp (α) g ⊇ Tg .Since T has the minimal property, so does Tg and −1 −1 −1 du e = gg ∈ Tg hence it is sufficient to show that αg ∈ M . Thus without loss of generality, we −1 −1 can replace α by αg and T by Tg .If T = {e, g ,..., g }; then by Lemma 2.5, there exist invariant 2 m g−1 subnearrings A = A , A ,..., A of N satisfying (nan ) f = n(af )n for all a ∈ A, n, n ∈ N and 1 2 m i i af = a f for all i. Thus by Lemma 2.1(iii), there exist units f , f ,..., f ∈ S such that conjugation by i 1 2 m −1 −1 m f induces automorphism g on N and hence g ∈ G ;thatis, g ∈ G .Let β = f g ∈ S ∗G, inn i inn i i i i i i =1 where f g is a unit in S ∗ G which acts trivially on S and β ∈ C .Since af = a f for all a ∈ A,by i Z i i i Lemma 2.5(iii), we have aβ = (a f )g ∈ M for all a ∈ A;thatis, Aβ ⊂ M which implies β ∈ M i i i =1 (by definition of M ). Let t = tr (α) be the identity coefficient of α and a ∈ A,define γ = aα − aβt ∈ M, where Supp(β) = T ⊂ Supp(α).Thenwehave Supp(γ ) = Supp(aα −aβt ) = aSupp(α)−aSupp(βt ). Since e ∈ T and f = f for e ∈ Supp(β) implies that Supp(βt )  1. Thus |Supp(γ )| < |Supp(α)|, 1 0 du d so by induction γ ∈ M . By Lemma 3.5(ii), M is a G−invariant ideal of C ,soonusing Lemma d d du 3.3(ii), SG M becomes invariant subnearring of S ∗ G. Hence a.g βt ∈ SG M ∩ (N ∗ G) = M du g which further implies that aα = γ + aβt ∈ M for all a ∈ A.Then B = A becomes non-zero g∈G g du G-invariant, invariant subnearring of N,where Bα = A α ∈ M . Since by Lemma 3.7(ii) g∈G du du M is N -cancelable, we get α ∈ M . The following theorem makes the task of proving the main results easier. Theorem 3.9 Let N ∗ G be a skew group nearring of finite group G acting on 3-prime nearring N and let C = Z [G ] be as above. Z inn d du (i) If P is an I -prime of N ∗ G with P ∩ N ={0},then P is a G-prime ideal of C and P = P . u u ud (ii) If I is a G-prime ideal of C ,then I is an I -prime of N ∗ G with I ∩ N ={0} and I = I . du Proof (i) We first show that P = P for any I -prime P of N ∗ G with P ∩ N ={0}. By Lemma 3.8(ii), du du d k we have atleast P ⊆ P . Conversely, let α ∈ P = P SG ∩ (N ∗ G),so α = δ f g ,where i i i =1 i d d δ ∈ P , f ∈ S, g ∈ G. By definition of P , there exist non-zero semigroup ideals D of N with i i i δ D = D δ ⊆ P for each i. Then D = D becomes non-zero G-invariant semigroup i i i i g∈G i =1 i 123 Arab. J. Math. ideal of N such that Dδ ⊆ P for all i. Again there exist non-zero semigroup ideals B of N such that i i B f ⊆ N ∗ G for all i.If B = B , then it becomes non-zero G-invariant semigroup ideal i g∈G i =1 of N with B f ⊆ N ∗ G. Therefore (DB)α ⊆ (Dδ )(B f )g ⊆ P(N ∗ G) ⊆ P.Since P ∩ N ={0} i i i i i and by Lemma 3.7(i) P is N −cancelable, hence α ∈ P.Let I and I be G- invariant ideals of C such 1 2 Z d u u u du u that I I ⊆ P . By Lemma 3.7(iii), I I ⊆ (I I ) ⊆ P = P,soby I -primeness of P, I ⊆ P or 1 2 1 2 1 2 1 u ud d ud d d I ⊆ P. Using Lemma 3.8(i), I = I ⊆ P or I = I ⊆ P , implying that P is G-prime. 1 2 2 1 2 (ii) Let I be a G-prime ideal of C . Here C is a finite dimensional algebra over Z and a finite dimensional Z Z algebra over field is an artinian ring so satisfies the descending chain condition on ideals and also every prime ideal is a maximal ideal. Thus I is a G-maximal ideal of C . From Lemma 3.3(ii), SG I ∩ N ⊂ SG I ∩ S ={0},sowehave I ∩ N ={SG I ∩ (N ∗ G)}∩ N ={0}∩ (N ∗ G) ={0}.Let M be any invariant u d ud subnearring of N ∗ G properly containing I .Thenwehave M ⊃ I = I (by Lemma 3.8(i)) implying d d d d that I ⊂ M ⊆ C .Since I is G-maximal and M is G- invariant, so either M = I or M = C . Suppose Z Z d du u d that M = I,then M ⊆ M = I (using Lemma 3.8(ii)) which is a contradiction. Thus M = C ,sowe have M ∩ N ={0}. For, we know that f e ∈ C so there exists a non-zero semigroup ideal M of N with 1 Z 1 M ( f e) ⊂ M which gives M ⊂ M implying that {0} = M ∩ N ⊂ M ∩ N . From this we conclude that 1 1 1 1 u u I is I -prime. For, suppose M , M ⊇ I ,where M and M are two invariant subnearrings of N ∗ G.But 1 2 1 2 by above observation, M ∩ N , M ∩ N ={0} which gives that {0} = (M ∩ N )(M ∩ N ) ⊆ M M ∩ N . 1 2 1 2 1 2 u u Since I ∩ N ={0}, it follows that M M ⊂ I . 1 2 The incomparability, we stated in the beginning now follows easily. Corollary 3.10 If P is an I -prime of N ∗ G with P ∩ N ={0} and M any invariant subnearring of N ∗ G such that M  Pthen M ∩ N ={0}. u d Proof By Theorem 3.9(i), any I -prime P of N ∗ G with P ∩ N ={0} is of the form I ,where I = P is a G-prime ideal of C . Thus if M  P = I and now M ∩ N ={0} by Theorem 3.9(ii). Theorem 3.11 Let N be a 3-prime nearring. (i) Any I -prime P of N ∗ G is minimal if and only if P ∩ N ={0}. (ii) There are finitely many such minimal primes, say P , P ,..., P where k  |G |  |G|. 1 2 k inn (iii) J = P ∩ P ∩ ··· ∩ P is the unique largest nilpotent invariant subnearring of N ∗ G such that J = 1 2 k u |G | inn (rad(C )) ,where rad(C ) is the jacobson radical of C and J ={0}. Z Z Z Proof C is an artinian ring so has finitely many ideals. Let I , I ,..., I be all G-prime ideals of C . Z 1 2 k Z Then as in the case of rings [3, Lemma 2.7], it follows that k  dim (Z [G ]) =|G |  |G| and Z inn inn I ∩ I ∩ ··· ∩ I = rad(C ), the Jacobson radical of C .Let P = I , so by Theorem 3.9(ii), P , P ,..., P 1 2 k Z Z i 1 2 k are I -primes of N ∗ G with P ∩ N ={0}.Set J = P ∩ P ∩ ··· ∩ P ⊂ P = I for all i,sowehave i 1 2 k i d ud d du u J ⊂ I = I ;thatis, J ⊂ I ∩ I ∩· · · ∩ I = rad(C ). Thus by Lemma 3.8(ii), J ⊂ J ⊂ (rad(C )) . i 1 2 k Z Z We know that rad(C ) is nilpotent(Jacobson radical of an artinian ring is nilpotent), thus (rad(C )) and J Z Z are nilpotent. On other hand, since each P certainly contain all nilpotent invariant subnearrings of N ∗G, thus J u u contains all nilpotent invariant subnearrings of N ∗G, so must contain (rad(C )) . Therefore J = (rad(C )) , Z Z u k |G | inn where J is the largest nilpotent invariant subnearring of N ∗ G and also {(rad(C )) } = f e = J . Z 0 Finally let P be any I -prime of N ∗ G so we have P ⊂ P = J ⊂ P,as J is nilpotent and hence by I - i i i i primeness of P, P ⊂ P for some i. Thus the minimal primes of N ∗G are the members of set {P , P ,..., P }. i 1 2 k d d Observe that P ⊃ P implies that I = P ⊃ P = I . But since I is G-maximal, we must have i = j.This i j i j j i j shows that P , P ,..., P are only minimal primes of N ∗ G. 1 2 k As an easy application of above result, we now prove Going Down Problem. Proof of Going Down Problem. The existence of an I -prime such that P ∩ N = A follows from Lemma 1.3.Since A ={0}, by above result, P is not minimal prime. So P must properly contain some minimal prime P ;thatis, P ⊃ P for some i such that P ∩ N ={0}. i i Corollary 3.12 Let J = P ∩ P ∩ ··· ∩ P be as above. 1 2 k (i) There exist a sequence J = J , J ,... of N -cancelable invariant subnearrings of N ∗ G such that J J ⊂ 1 2 i J and J ={0} for some k  |G |. i +1 k 1 inn 123 Arab. J. Math. (ii) Let α ,α ,...,α be elements of J and let A be a non-zero semigroup ideal of N . Then there exists a 1 2 m non-zero G-invariant semigroup ideal B of N such that Bα ⊂ AJ for all i. Proof Set I = rad(C ).Then I is nilpotent and J = I (by Theorem 3.11(iii)). u u u (i) Set J = I ,so I becomes invariant subnearring of N ∗ G and N -cancelable (by definition of I and i i i u i u i +1 u k by Lemma 3.7(iii)) where JJ = I (I ) ⊆ (I ) = J .Since I is nilpotent then I ={0} for i i +1 k  |G | so we have J = SG I ∩ (N ∗ G) = f e. 1 inn k 0 (ii) Let γ = fgβ ∈ S ∗ G with f ∈ S, g ∈ G,β ∈ I . We firstly show that there exists a non-zero G-invariant semigroup ideal D of N such that Dγ ⊂ AJ . We can choose non-zero semigroup ideals D and D of 1 2 g g N with D f ⊂ N and D β ⊂ (N ∗ G) ∩ SG I = I .Set D = D and D = D then 1 2 g∈G g∈G 1 2 1 2 both become G-invariant semigroup ideals of N with D f ⊂ N and D β ⊂ I .Alsowecan finda 1 2 G-invariant semigroup ideal D of N with D ⊂ A,where D = A . Setting D = D D D then, 3 3 3 3 2 1 g∈G for all d d d ∈ D = D D D ,wehave (d d d )( fgβ) = (d d ) (d β ) fg = d ((d β )(d f )g) ⊆ 3 2 1 3 2 1 3 2 1 3 2r 1r 3 2 2 u u D (I (N ∗ G)) ⊆ AJ . Finally if α ,α ,...,α be elements of J = I , then α ∈ SG I for all i and 3 1 2 m i hence α can be written as γ where γ = f g β ; f ∈ S, g ∈ G,β ∈ I . By above we can find i ij ij ij ij ij ij ij ij non-zero G-invariant semigroup ideal D of N with D γ ⊂ JA for each i.Set B = D ,thenit ij ij ij ij i, j becomes non-zero G-invariant semigroup ideal of N . For, if not then D ⊂ D ={0} implying ij ij i, j i, j that D ={0} for some i and j, a contradiction and also each D is a G-invariant semigroup ideal of N , ij ij so is their intersection. Here B ⊂ D for all i, j implies that Bα ⊂ AJ for all i. ij i Acknowledgements The authors are thankful to learned referee for his perspicacious comments that resulted in the present form of the paper. 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Journal

Arabian Journal of Mathematics – Springer Journals

Published: Dec 1, 2023

Keywords: 16Y30; 16W22

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