ON A Topological Γ-Rings

In this study we introduce the concepts of topological Γ-ring as generalization of Topological ring. We also present and study the concepts of topological Γ-ring, Homomorphism of topological Γ-ring, and compact of topological Γ-ring. The results have confirmed that: if (G, +, . ,J) is a compact topological Γ-ring and ƒ : (G,+ ,. ,J)→( ,+',.', ) epimorphism topological Γ-ring Then is compact.


INTRODUCTION
The concepts of topological ring is one of the most important topics in topological algebra. The objective of the concept of topological Γ-ring is to generalize the definition of topological ring. The concept of Γ-ring was presented by Nobusawa 1964 in [1], and it was generalized by Barnes 1966 [2] as below: Let M and Γ be two additive abelian groups. Suppose that there is a mapping from M×Γ×M→M, the image of (a,α,b) is denoted by aαb, a,b M and α Γ, that satisfies the following for all a,b,c M, α,β Γ: i) (a+b)αc= aαc+ bαc a(α +β )c = aαc + aβc aα(b +c ) = aα b + aα c ii) ( aα b)βc = aα(bβc) Then M is called a Γ-ring., where every ring is a Γ-ring. M is said to be 2-torsion free if 2a = 0 implies a=0 for all a∈ M. Besides, M is called a prime Γ-ring if for all a,b M, aMΓMb = (0) implies either a=0 or b=0, and M is called a semiprime if aMΓMa=(0) with a∈M implies a=0. Note that every prime Γ-ring is obviously a semiprime [3], [4] , [5][6][7][8][9][10].

ISSN: 0067-2904
Tawfeeq Iraqi Journal of Science, 2022, Vol. 63, No. 3, pp: 1258-1264 In this paper we present the concept of topological Γ-ring as generalization of topological ring where we get every topological ring is topological Γ-ring. In general the converse is not true . In this paper we introduce and study the concepts of topological Γ-ring, norm, homomorphism of topological Γ-ring some type of homomorphism kernel of homomorphism, and natural mapping as well as we present the compact of topological Γ-rings. We also prove that: If :(G,+, . , ) → ( ́ is homomorphism of topological Γ-ring G onto topological Γring G 1 ,then 1)If is an ideal of topological Γ-ring of G then is an ideal of topological Γring . 2) If J is an ideal of topological Γ-ring of G 1 then (J) is an ideal of topological Γ-ring G. The results have confirmed that: If f:(G,+, . , )→ ( ,+',.', ) is isomorphism topological Γ-ring from G into compact topological Γ-ring then G is compact topological Γ-ring.

Topological Γ-Rings:
Basic concepts related to the talked topic are given as follows with few important properties has been presented in the next section. Definition (2-1): A topology on a Γ-ring G is a topological Γ-ring, which is denoted by ( , +, . , ), iff: -m is continuous from G G into G where G is given topology and G G the Cartesian product topology determined by . Example ( 2-2) : Let G be any Γ-ring and be discrete topology on G then (G,+, . , ) is topology Γ-ring, which is called discrete topological Γ-ring Example (2-3) : Let G be any Γ-ring and be indiscrete topology on G then (G,+, . , ) is topology Γ-ring,which is called indiscrete topological Γ-ring Example (2-4) : Let R be the set of all real numbers and Q be the set of all rational numbers then R is Q-ring if is usual topology on R ,and then (R,+, . , ) is topological Q-ring, which is called usual topological Γ-ring. By Definition (2.1), G=R and Γ= Q, also the operations +, • and -are continuous from R × R into R. Since every ring is Γ-ring, then we can get the following lemma Lemma (2-5): Every topological ring is topological Γ-ring The next example shows that the converse of Lemma (2-5) is not true in general. Example Hence the definition 2-1 (3) holds, and the proof is finished Let f be a partial function from the carrier of S to R, we recall that f is said to be uniformly continuous on X if and only if the following conditions are satisfied (i) X ⊆ dom f, and (ii) for every r such that 0 < r there exists s such that 0 < s and for all x 1 , x 2 such that x 1 , x 2 ∈ X and ‖ ‖ < s holds |f(x 1 )f(x 2 ) | < r.) from G into R + { }. Theorem (2-9): Let N be a norm on a Γ-ring G. for all p,h G, then| p | ≤N(p-h ) ,and hence N is a uniformly continuous function .
The Cartesian product of topological Γ-rings is a topological Γ-ring . Proof: `Let ( , +, . , ) and ( , +, . , ) are two topological spaces and ℬ be a basis of and be a basis for then {B C B∈ℬ and C∈ } is a basis for the product topology (G×G1 ,+,., ) Let (x,y) be any point of G×G1and be a neighborhood of (x,y) Since E={X×Y: X∈ and Y∈ } is a basis for there exists a member X×Y of E such that (x,y)∈X×Y⊆ … Since X is -open and ℬ is a basis for there exists some B∈ℬ such that x∈B⊆X. Similarly there exists some C∈ suc t at y∈C⊆Y t follows t at x y ∈ B C ⊆ X Y …..(2) Hence from (1) and (2) , we get (x,y) ∈ B C⊆ T is i plies t at ℋ is basis for 1
́ ́ ∈ f and r̀ ́ . ́ ∈ f f is onto , then r ∈G s.t. , r̀ f r and x ∈ . s. t. f x since, r ∈ R , x ∈ and since I is an ideal of a ring x . .r ∈ and r . .x ∈ f(x . . r) ∈ f(I) and f(r . . x) ∈ f(I) f x ́ f r ∈ f and f r ́ ́ f x ∈ f ́ ́ r̀∈ f and r ̀ ́ ́ ∈ f Therefore, f(I) is an ideal of topological Γ-ring . 2)By using the same technic of (1) we get the require result.  Then , x , y ∈ , s .t. , f x and b f y Hence , M 1 is a commutative topological Γ-ring 2) Suppose that M be a topological Γ-ring with identity Let a ∈M 1 since f is onto therefore )=ƒ(E).hence ğ is onto. Hence g is isomorphism Γ-ring. Now, since ƒ is continuous therefore g is continuous. Let U be an open set in G/ker(f) Since g(E+ker(f))=ƒ(E) and f is continuous, so it g -1 is continuous g -1 : → er and ƒ is continuous then g -1 is continuous . Hence the topological Γ-ring er is isomorphic topological Γ-ring to . Theorem   (G i , + , . ) is a proper subrings of G, for all ∈ } is a cover topological Γ-ring of (G,+ , ., ). That means G= ⋃ ∈ . Therefore f(G)=G = ⋃ ∈ f(G i),where f( i ) ∈ Since G is compact ,it follows that there is a finite subcover ℋ ⊆ Such that ⊆ ℋ Since f is isomorphism we get G= ℋ .where ℋ is a finite subcover of G. Theorem (4-5): Every closed subset of a compact topological Γ-ring space is compact. Proof: Let be a closed subset of the compact topological Γ-ring (G,+,. , ). Let ={ G i ∈ : (G i , + , . ) is a proper subrings of G, for all ∈ } be a cover topological Γring of . Since B is closed then G-B is open, and = (G-B)is an open cover of G. Since G is compact topological ring, it has a finite subcover, containing only finitely many members G 1 ,…, n of and may contain G-B. Since G=(G-B)⋃ it follows that B⊆ ⋃ and has a finite subcover. Theorem (4-6): If (G, +, . , ) is a compact topological Γ-ring, and ƒ : (G,+ ,. , )→( ,+',.', ) epimorphism topological Γ-ring , then is compact. Proof: Assume that ={ ∈ : ( , +' , .' ) is a proper subrings of , for all ∈ } is a cover topological Γ-ring of that is G = ⋃ ∈ Since f is epimorphism topological Γ-ring this implies that G= ⋃ ∈ ,where ∈ Since G is compact topological Γ-ring ,this gives G=⋃ Hence ƒ is epimorphism topological Γ-ring =⋃ . Thus is compact topological Γ-ring Acknowledgements: The author wishes to express his gratitude to the reviewers for their insightful suggestions that improved the paper's presentation and for volunteering their time to assist in its publication. Financial support: This study did not receive any financial support from any organisation. Authors' declaration: -Conflicts of Interest: None.
-Ethical approval: None of the writers conducted any experiments with human subjects or animals for this publication.