Formal triangular matrix ring with nil clean index 4
aa r X i v : . [ m a t h . R A ] F e b Formal triangular matrix ring with nil clean index
Department of Mathematical Sciences, Tezpur University,Napaam, Tezpur-784028, Assam, India.Email: [email protected]
Jayanta Bhattacharyya
Department of Mathematical Sciences, Tezpur University,Napaam, Tezpur-784028, Assam, India.Email: [email protected]
Abstract:
For an element a ∈ R , let η ( a ) = { e ∈ R | e = e and a − e ∈ nil( R ) } . The nil clean index of R , denoted by Nin( R ), is defined asNin( R ) = sup {| η ( a ) | : a ∈ R } . In this article we have characterized formaltriangular matrix ring ( A M B ) with nil clean index 4. Key words:
Nil clean ring, nil clean index.2010
Mathematics Subject Classification:
Throughout this article R denotes a associative ring with unity. The set of nilpo-tents and set of idempotents are denoted by nil( R ) and idem( R ) respectively. Thecyclic group of order n is denoted by C n and | S | denotes the cardinality of the set S . For an element a ∈ R , if a − e ∈ nil( R ) for some e ∈ idem( R ), then a = e +( a − e )is said to be a nil clean expression of a in R and a is called a nil clean element[3, 2].The ring R is called nil clean if each of its elements is nil clean.For an element a ∈ R , let η ( a ) = { e ∈ R | e = e and a − e ∈ nil( R ) } . The nilclean index of R , denoted by Nin( R ), is defined as Nin( R ) = sup {| η ( a ) | : a ∈ R } [1].Characterization of arbitrary ring with nil clean indices 1, 2 and few sufficientcondition for a ring to be of nil clean index 3 is given in [1]. In this article we havecharacterized formal triangular matrix ring (cid:0) A M B (cid:1) with nil clean index 4, where A and B are rings and M is a A − B − bimodule. Following results about nil cleanindex will be used in this article. emma 1.1. ([ , Lemma . ]) Let R = (cid:18) A M B (cid:19) , where A and B are rings, A M B is a bimodule. Let Nin ( A ) = n and Nin ( B ) = m . Then (i) Nin ( R ) ≥ | M | . (ii) If ( M, +) ∼ = C p k , where p is a prime and k ≥ , then Nin ( R ) ≥ n +[ n )( | M |− , where [ n ) denotes the least integer greater than or equal to n . (iii) Either Nin ( R ) ≥ nm + | M | − or Nin ( R ) ≥ nm . Lemma 1.2. ([ , Lemma . ]) Let R = (cid:18) A M B (cid:19) , where A and B are rings, A M B is a bimodule with ( M, +) ∼ = C r . Then Nin ( R ) = 2 r Nin ( A ) Nin ( B ) . Theorem 1.3. ([ , Theorem . ]) Nin ( R ) = 2 if and only if R = (cid:18) A M B (cid:19) , where Nin ( A ) = Nin ( B ) = 1 and A M B is a bimodule with | M | = 2 . Theorem 1.4. ([ , Proposition . ]) If R = (cid:18) A M B (cid:19) , where Nin ( A ) = Nin ( B ) =1 and A M B is a bimodule with | M | = 3 then Nin ( R ) = 3 . Theorem 2.1.
Let R = (cid:18) A M B (cid:19) , where A and B are rings, A M B is a nontrivial bimodule. Then Nin( R ) = 4 if and only if one of the following holds: (1) ( M, +) ∼ = C and Nin( A ) Nin( B ) = 2 . (2) ( M, +) ∼ = C and Nin( A ) = Nin( B ) = 1 . (3) ( M, +) ∼ = C ⊕ C plus one of the following ( a ) Nin( A ) = Nin( B ) = 1 . ( b ) Nin( A ) = 1 , B = (cid:18) S W T (cid:19) , where Nin( S ) = Nin( T ) = 1 and | W | = 2 , and eM (1 B − f ) + (1 A − e ) M f = 0 for all e = e ∈ A and f ∈ η ( b ) , where b ∈ B with | η ( b ) | = 2 . ( c ) Nin( B ) = 1 , A = (cid:18) S W T (cid:19) , where Nin( S ) = Nin( T ) = 1 and | W | = 2 , and eM (1 B − f ) + (1 A − e ) M f = 0 for all e = e ∈ B and f ∈ η ( a ) , where a ∈ A with | η ( a ) | = 2 . roof : ( ⇐ ) If (1) holds then by Lemma 1.2 , we get Nin( R ) = 4.If (2) holds then Nin( R ) ≥ | M | = 4. Now, for any α = (cid:18) a x b (cid:19) ∈ R , η ( α ) = (cid:26)(cid:18) e w f (cid:19) ∈ R : e ∈ η ( a ) , f ∈ η ( b ) , w = ew + f w (cid:27) . Because | M | = 4 , | η ( a ) | ≤ | η ( b ) | ≤
1, it follows that | η ( α ) | ≤
4. HenceNin( R ) = 4.Let (3) ( a ) holds, then Nin( R ) ≥ | M | = 4. Now, for any α = (cid:18) a x b (cid:19) ∈ R , η ( α ) = (cid:26)(cid:18) e w f (cid:19) ∈ R : e ∈ η ( a ) , f ∈ η ( b ) , w = ew + f w (cid:27) . Because | M | = 4 , | η ( a ) | ≤ | η ( b ) | ≤
1, it follows that | η ( α ) | ≤
4. HenceNin( R ) = 4.Suppose (3)( c ) holds, then clearly Nin( R ) ≥ | M | = 4 . Let α = (cid:18) a w b (cid:19) ∈ R .We show that | η ( α ) | ≤ R ) = 4 holds. Since Nin( B ) = 1, we canassume that η ( b ) = { f } . Then as above we have η ( α ) = (cid:26)(cid:18) e z f (cid:19) ∈ R : e ∈ η ( a ) , z = ez + zf (cid:27) If | η ( a ) | ≤
1, then | η ( α ) | ≤ | η ( a ) | . | M | ≤
4. So we can assume that | η ( a ) | = 2.Write η ( a ) = { e , e } . Thus η ( α ) = T S T , where T i = (cid:26)(cid:18) e i z f (cid:19) ∈ R : (1 A − e i ) z = zf (cid:27) ( i = 1 , . Since η (1 A − a ) = { A − e , A − e } , the assumption (3)( c ) shows that { z ∈ M :(1 A − e i ) z = zf } is a proper subgroup of ( M, +); so | T i | ≤ i = 1 ,
2. Hence | η ( α ) | ≤ | T | + | T | ≤ ⇒ ) Suppose Nin( R ) = 4. Then 2 ≤ | M | ≤ Nin( R ) = 4. If | M | = 2 thenNin( A ) Nin( B ) = 2 by Lemma 1.2 , so (1) holds.Suppose | M | = 3, then we have by Lemma 1.1 , Nin( A ) + | M | ≤ Nin( R ), showingNin( A ) ≤
2, similarly Nin( B ) ≤
2. But Nin( A ) = 2 = Nin( B ) will give Nin( R ) ≥ Lemma 1.1 and Nin( A ) = Nin( B ) = 1 will give Nin( R ) = 3 by Theorem1.4 . Hence the only possibility is Nin( A ) Nin( B ) = 2, so without loss of generalitywe assume that Nin( A ) = 2 and Nin( B ) = 1. Write M = { , x, x } . Now by Theorem 1.3 , we have A = (cid:18) T N S (cid:19) , where T & S are rings, T N S is bimodule ith Nin( T ) = Nin( S ) = 1 and | N | = 2. Note that for e ∈ idem( A ) , ex ∈ { , x } ,for if ex = 2 x , we have 2 x = ex = e ( ex ) = e (2 x ) = e ( x + x ) = ex + ex = 2 x + 2 x =4 x = x which is not true.Now Let a = (cid:18) T
00 0 (cid:19) ∈ A such that a = (cid:18) T
00 0 (cid:19) + (cid:18) (cid:19) = (cid:18) T y (cid:19) + (cid:18) − y (cid:19) .Let us denote, e = (cid:18) T
00 0 (cid:19) , e = (cid:18) T y (cid:19) , n = (cid:18) (cid:19) and n = (cid:18) − y (cid:19) , where e , e ∈ idem( A ) & n , n ∈ nil( R ). Now we havefollowing cases: Case I:
Let e x = e x = 0, then we have an element β = (cid:18) (1 A − a ) 00 0 (cid:19) ∈ R such that β = (cid:18) (1 A − e ) z (cid:19) + (cid:18) − n − x (cid:19) ∀ z ∈ M = (cid:18) (1 A − e ) z (cid:19) + (cid:18) − n − x (cid:19) ∀ z ∈ M are six nil clean expressions for β, which implies | η ( β ) | ≥
6, that is Nin( R ) ≥ Case II:
Let e x = e x = x , then we have an element α = (cid:18) a
00 0 (cid:19) ∈ R such that α = (cid:18) e z (cid:19) + (cid:18) n − x (cid:19) ∀ z ∈ M = (cid:18) e z (cid:19) + (cid:18) n − x (cid:19) ∀ z ∈ M are six nil clean expressions for α, which implies | η ( α ) | ≥
6, that is Nin( R ) ≥ Case III:
Let e x = x and e x = 0, then we have ( e − e ) x = x .Let j = e − e , then clearly j ∈ nil( A ) and we have jx = x ⇒ (1 A − j ) x = 0 ⇒ x = 0 , ( as (1 − j ) ∈ U( A )).Which is not possible. ase IV: Let e x = 0 and e x = x , as in case III, we get a contradiction.Hence if M ∼ = C , Nin( R ) is never 3.Suppose | M | = 4. If ( M, +) ∼ = C , then Nin( A ) Nin( B ) = 1 byLemma 1.2, So (2) holds. Let ( M, +) ∼ = C ⊕ C . Since Nin( R ) = 4, by Lemma1.1 , we have Nin( A ) Nin( B ) ≤
2. If Nin( A ) Nin( B ) = 1 then (3)( a ) holds. IfNin( A ) Nin( B ) = 2, without loss of generality we can assume Nin( A ) = 2 andNin( B ) = 1. So by Theorem 1.3 , we have A = (cid:18) S W T (cid:19) where Nin( S ) =Nin( T ) = 1, and | W | = 2. To complete the proof suppose in contrary that eM (1 B − f ) + (1 A − e ) M f = 0 for some f = f ∈ B and e ∈ η ( a ), where a ∈ A with | η ( a ) | = 2 . Then ew = wf for all w ∈ M . It is easy to check that η ( a ) = { e, e + j } where j = (cid:18) w (cid:19) ∈ A with 0 = w ∈ W . Thus, for γ := (cid:18) A − e f (cid:19) , η ( γ ) ⊇ (cid:26)(cid:18) A − e f (cid:19) , (cid:18) A − ( e + j ) 00 f (cid:19) , (cid:18) A − e w f (cid:19) : w ∈ M (cid:27) So | η ( γ ) | ≥
5, a contradiction. Hence (3)( c ) holds, similarly (3)( b ) can be proved ✷ References [1] Basnet, D. Kr. and Bhattacharyya, J. Nil clean index of rings Internationa,
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