Armendariz ring with weakly semicommutativity
aa r X i v : . [ m a t h . R A ] J a n Armendariz ring with weakly semicommutativity
Sushma Singh and Om PrakashDepartment of MathematicsIndian Institute of Technology Patna, BihtaPatna, INDIA- 801 [email protected] & [email protected]
Abstract
In this article, we introduce the weak ideal-Armendariz ring which combines Armendarizring and weakly semicommutative properties of rings. In fact, it is a generalisation of anideal-Armendariz ring. We investigate some properties of weak ideal Armendariz rings andprove that R is a weak ideal-Armendariz ring if and only if R [ x ] is weak ideal-Armendarizring. Also, we generalise weak ideal-Armendariz as strongly nil-IFP and a number of prop-erties are discussed which distinguishes it from other existing structures. We prove that if I is a semicommutative ideal of a ring R and RI is a strongly nil-IFP, then R is stronglynil-IFP. Moreover, if R is 2-primal, then R [ x ] / < x n > is a strongly nil-IFP. Mathematical subject classification : 16N40, 16N60, 16U20, 16Y99.
Key Words : Armendariz ring, weak Armendariz ring, semicommutative ring, weakly semi-commutative ring, ideal-Armendariz ring, weak ideal-Armendariz ring, strongly nil-IFP.
1. INTRODUCTION
Throughout this article, R denotes an associative ring with identity, otherwise it is mentionedand R [ x ] is a polynomial ring over R with an indeterminate x . For any polynomial f ( x ), C f ( x ) denotes the set of all coefficients of f ( x ). M n ( R ) and U n ( R ) denote the n × n full matrix ringand upper triangular matrix ring over R respectively. D n ( R ) is the ring of n × n upper trian-gular matrices over R whose diagonal entries are equal and V n ( R ) is ring of all matrices ( a ij ) in D n ( R ) such that a pq = a ( p +1)( q +1) for p = 0 , , , . . . , n − q = 0 , , , . . . , n −
1. We use e ij todenote the matrix with ( i, j )th-entry 1 and elsewhere 0. Z n is the ring of residue classes moduloa positive integer n and GF ( p n ) denotes the Galois field of order p n . Here, N ∗ ( R ) , N ( R ) and J ( R ) represent the prime radical (lower nil radical), set of all nilpotent elements and Jacobsonradical of the ring R respectively. R + represents the additive abelian group ( R, +) and cardi-nality of a given set S is denoted by | S | .Recall that a ring R is reduced if it has no non-zero nilpotent element. It was Armendariz [ [3],Lemma 1] who initially observed that a reduced ring always satisfies the below condition: If f ( x ) and g ( x ) ∈ R [ x ] satisfies f ( x ) g ( x ) = 0 , then ab = 0 for each a ∈ C f ( x ) and b ∈ C g ( x ) .In 1997, Rege and Chhawchharia [19] had coined the term Armendariz for those rings whichare satisfying above condition. Clearly, subring of an Armendariz ring is an Armendariz. Aring R is said to be an abelian ring if every idempotent element is central. It is known thatArmendariz ring is an abelian ring [7].In 2006, Liu et al. generalised the Armendariz ring which is called weak Armendariz [16] anddefined as if two polynomials f ( x ) and g ( x ) ∈ R [ x ] such that f ( x ) g ( x ) = 0 implies ab ∈ N ( R )1or each a ∈ C f ( x ) and b ∈ C g ( x ) . A ring R is reversible if, for any a, b ∈ R, ab = 0 im-plies ba = 0 [6]. Later, by using the concept of reversible, D. W. Jung et. al [9] introducedquasi-reversible-over-prime-radical ( QRP R ). A ring R is said to be ( QRP R ) if ab = 0 implies ba ∈ N ∗ ( R ) for a, b ∈ R .A right (left) ideal I of a ring R is said to have the insertion-of-factor-property (IFP) if ab ∈ I implies aRb ⊆ I for a, b ∈ R . A ring R is said to be an IFP if the zero ideal of R has the IFPProperty [4]. IFP rings are also known as semicommutative rings [18]. In 1992, Birkenmeier et.al [5], are called a ring R is 2-primal if and only if N ∗ ( R ) = N ( R ). It can be easily seen thatsemicommutative rings are 2- primal. A ring R is called an N I ring if N ∗ ( R ) = N ( R ) [17].In 2012, Kwak [14], studied Armendariz ring with IFP and introduced ideal-Armendariz ring.He proved that a ring R is Armendariz and IFP if and only if f ( x ) g ( x ) = 0 implies aRb = 0,for each a ∈ C f ( x ) and b ∈ C g ( x ) .Later, in 2007, Liang [15] introduced the weakly semicommutative ring which satisfy for any a, b ∈ R , ab = 0 implies arb ∈ N ( R ) for each r ∈ R . By definition, it is clear that any semicom-mutative ring is weakly semicommutative ring but converse is not true. Also, for a reduced ring R , D n ( R ) is weakly semicommutative ring for n ≥ D n ( R ) is a weakly semicommutative but by Example 3 of [11], for n ≥
4, it is not an Armen-dariz ring. Moreover, by Example(4.8) of [2], if K is a field and R = K [ a, b ] / < a > , then R isan Armendariz ring. In this ring, ( ba ) a = 0 but ( ba ) b ( a ) is not nilpotent. Therefore, R is nota weakly semicommutative ring. Thus, Armendariz ring and weakly semicommutative ring donot imply each other.In light of above study, we introduced the concept of weak ideal-Armendariz ring andStrongly nil-IFP in Section 2 and Section 3 respectively. The below flow chart will also give anidea about relation of these two notions with other existing structures: Ideal − Armendariz W eak ideal − Armendariz IF PStrongly nil − IF P W eakly semicommutativeW eakArmendariz
2. WEAK IDEAL ARMENDARIZ RINGSDefinition 2.1.
Weak Ideal Armendariz Rings A ring R is said to be a weak ideal-Armendarizring if R is an Armendariz ring and weakly semicommutative. It is clear that every ideal-Armendariz ring is a weak ideal-Armendariz but converse is nottrue. Towards this, we have the following example:
Example 2.1.
Let F be a field and A = F [ a, b, c ] be a free algebra of polynomials with constant erms zero in noncommuting indeterminates a, b, c over F . Then A is a ring without identity.Consider an ideal I of F + A generated by cc, ac, crc for all r ∈ A . Let R = ( F + A ) /I . FromExample 14 of [7], R is an Armendariz but it is not a semicommutative ring because ac ∈ I but abc / ∈ I . We denote a + I by a . In order to show R is a weakly semicommutative ring, we havethe following cases: Case (1) ac ∈ I , Case (2) cc ∈ I , Case (3) crc ∈ I . Case (1) If ac ∈ I for a, c ∈ A , then ( asc ) ∈ I for each s ∈ A . Consider, atc = a ( α + t ′ ) c ,where t = α + t ′ ∈ F + A . Then ( atc ) ∈ I . Since, r acr ∈ I , therefore ( r arcr ) ∈ I for any r , r , r ∈ A . Similarly, we can prove that ( r atcr ) ∈ I for any r , r , ∈ A , and t ∈ F + A .Again, we have ( r acr ) ∈ I for any r , r ∈ F + A and this implies ( r arcr ) ∈ I for any r ∈ A as well as ( r atcr ) ∈ I for t ∈ F + A .Similarly, rest cases can be easily checked. Remark 2.1. (1)
Every reduced ring is a weak ideal-Armendariz ring. (2)
Subring of a weak ideal-Armendariz ring is a weak ideal-Armendariz ring. (3) If R is a reduced ring, then D ( R ) is a weak ideal-Armendariz ring. (4) For any ring R and n ≥ , U n ( R ) and M n ( R ) are not Armendariz by [11] and [19]respectively. Hence, they are not weak ideal-Armendariz rings. (5) By Example of [11], for any ring A , D n ( A ) for n ≥ is not an Armendariz ring. Hence,it is not a weak ideal Armendariz ring. But, when A is a reduced ring, then D n ( A ) is aweakly semicommutative ring. (6) It is nevertheless to say that polynomial ring over weakly semicommutative ring need notbe weakly semicommutative. Therefore, there is a huge scope to extend and study thepolynomial rings over weakly semicommutative rings with the Armendariz condition.
Proposition 2.1.
A ring R is weak ideal-Armendariz ring if and only if R [ x ] is weak ideal-Armendariz ring.Proof. Since, R is Armendariz ring if and only if R [ x ] is Armendariz ring [1]. In order toprove the result, we show that R is weakly semicommutative ring if and only if R [ x ] is weaklysemicommutative. For this, let f ( x ) and g ( x ) ∈ R [ x ] such that f ( x ) g ( x ) = 0. Since R isArmendariz, therefore ab = 0 for each a ∈ C f ( x ) and b ∈ C g ( x ) . Since R is weakly semicommu-tative ring, therefore arb ∈ N ( R ) for each r ∈ R . Hence, f ( x ) R [ x ] g ( x ) ⊆ N ( R )[ x ]. By Lemma(2.6) of [2], we have N ( R )[ x ] ⊆ N ( R [ x ]), hence f ( x ) R [ x ] g ( x ) ⊆ N ( R [ x ]). Thus, R [ x ] is weakideal-Armendariz ring. Also, R being subring of R [ x ], converse is true. Proposition 2.2.
Let N be a nil ideal of a ring R . If | N | = p , where p is a prime, then N isa commutative Armendariz ring without identity such that N = 0 .Proof. If | N | = p , then N is nilpotent by Hungerford [Proposition 2.13 of [8]], since J ( N ) = N ∩ J ( R ).By Corollary (1 . .
11) of [13], if R is nilpotent ring of order p n , then exp( R ) ≤ n + 1. Hence N = 0.Suppose N + is cyclic. To prove N is Armendariz, let f ( x ) , g ( x ) ∈ N [ x ] such that f ( x ) g ( x ) =0. Then f ( x ) , g ( x ) can be written as f ( x ) = af ( x ) and g ( x ) = ag ( x ), where f ( x ) , g ( x ) ∈ Z p [ x ]. But, we know that Z p is Armendariz by Proposition (2.1) of [19]. Therefore, γδ = 0,for each γ ∈ C f ( x ) and δ ∈ C g ( x ) . Hence N is an Armendariz ring.If N + is noncyclic. Then by Theorem 2.3.3 of [13], there is a basis { a, b } for N such that3 har a = char b = p and one of the following conditions holds:( i ) a = b = ab = ba = 0 ( ii ) a = b, a = 0.To prove N is an Armendariz ring, let f ( x ) , g ( x ) ∈ N [ x ] such that f ( x ) g ( x ) = 0. If N satisfiesthe case ( i ) a = b = ab = ba = 0, then N = { , a, a, . . . , ( p − a, b, b, . . . , ( p − b, a + b, a +2 b, . . . , a + ( p − b, a + b, a + 2 b, . . . , a + ( p − b, . . . , ( p − a + b, ( p − a + 2 b, . . . , ( p − a + ( p − b } . In this case, γδ = 0 for each γ ∈ C f ( x ) and δ ∈ C g ( x ) , because product of anytwo elements of N is zero.When N satisfies the condition (ii) a = b , a = 0, then N = { , a, a, . . . , ( p − a, a , a , . . . , ( p − a , a + a , a + 2 a , . . . , a + ( p − a , a + a , a + 2 a , . . . , a + ( p − a , ( p − a + a , ( p − a + 2 a , . . . , ( p − a + ( p − a } . In this case, only possibility for the product oftwo non-zero elements of N is zero in which one of elements in the product is from the set { a , a , . . . , ( p − a } . Here, we can write f ( x ) = a m f ( x ) , g ( x ) = a n g ( x ) and f ( x ) g ( x ) = 0when m + n ≥
3. Thus, f ( x ) g ( x ) = 0 implies γδ = 0, for each γ ∈ C f ( x ) and δ ∈ C g ( x ) . Proposition 2.3.
Let K be an ideal of a ring R such that every element of R \ K is regularand K = 0 . Then R is a weak ideal-Armendariz ring.Proof. Since each element of R \ K is regular and K = 0, therefore by Proposition (3.7) of [12], R is an Armendariz ring. To prove R is weakly semicommutative ring, let ab = 0 in R . Then ab + K = K . Since R/K is a domain, so a ∈ K or b ∈ K . Also, arb ∈ K for any r ∈ R and K = 0, therefore ( arb ) = 0. Hence, R is a weakly semicommutative ring. Thus, R is a weakideal-Armendariz ring.Given a ring R and a bimodule R M R , the trivial extension of R by M is the ring T ( R, M )with the usual addition and multiplication defined by( r , m )( r , m ) = ( r r , r m + m r ) . This is isomorphic to the ring of all matrices of the form (cid:18) r m r (cid:19) with usual addition andmultiplication of matrices, where r ∈ R and m ∈ M . Proposition 2.4.
For a ring R , the following are equivalent: (1) R is a reduced ring. (2) R [ x ] / < x n > is a weak ideal-Armendariz ring, where < x n > is the ideal generated by x n for any positive integer n . (3) The trivial extension T ( R, R ) is a weak ideal-Armendariz ring.Proof. (1) ⇒ (2): By Theorem 5 of [1], R [ x ] / < x n > is an Armendariz ring. Also, reducedring is weakly semicommutative ring, therefore R [ x ] / < x n > is a weak ideal-Armendariz ring.(2) ⇒ (3) and (3) ⇒ (1) are obvious. Proposition 2.5.
Let I be a reduced ideal of a ring R . If R/I is a weak ideal-Armendariz ring,then R is a weak ideal-Armendariz ring.Proof. Let ab = 0 for a, b ∈ R . Since I is reduced ideal of R , so bIa = 0. Again, ab = 0 implies ab ∈ I . Since R/I is a weakly semicommutative, so ( aRb + I ) ⊆ N ( R/I ). Therefore, ( arb ) n ∈ I for any r ∈ R . Moreover, ( arb ) n I ( arb ) n I ⊆ I and this implies ( arb ) n − ( arb ) I ( arb )( arb ) n − I =0. This shows that (( arb ) n I ) = 0. Since I is reduced, so ( arb ) n I = 0 and ( arb ) n ∈ ( arb ) n I implies ( arb ) n = 0. Therefore, ( arb ) ∈ N ( R ) for any r ∈ R . Hence, R is a weakly semi-commutative ring. Also by Theorem 11 of [7], R is an Armendariz ring. Thus, R is a weakideal-Armendariz ring. 4 roposition 2.6. Finite direct product of weak ideal-Armendariz rings is weak ideal-Armendariz.Proof.
Let R , R , . . . , R n be Armendariz rings and R = Q ni =0 R i . Let f ( x ) = a + a x + a x + · · · + a n x n and g ( x ) = b + b x + b x + · · · + b m x m ∈ R [ x ] such that f ( x ) g ( x ) = 0, where a i = ( a i , a i , . . . , a in ) and b i = ( b i , b i , . . . , b im ). Define f p ( x ) = a p + a p x + · · · + a np x n , g p ( x ) = b p + b p x + · · · + b mp x m . By f ( x ) g ( x ) = 0, we have a b = 0, a b + a b = 0, a b + a b + a b = 0 , . . . , a n b m = 0 thisimplies a b = 0 , a b = 0 , . . . a n b n = 0 a b + a b = 0 , . . . , a n b n + a n b n = 0 a n b m = 0 , a n b m = 0 , . . . , a nn b nm = 0 . Therefore, f p ( x ) g p ( x ) = 0 in R p [ x ], for 1 ≤ p ≤ n . Since each R p is an Armendariz ring, a ip b jp = 0 for each i, j and 1 ≤ p ≤ n and hence a i b j = 0 for each 0 ≤ i ≤ n and 0 ≤ j ≤ m .Thus, R is a weak ideal-Armendariz ring.It is known that N ( Q ni =0 R i ) = Q ni =0 N ( R i ).Let ( a , a , . . . , a n ) , ( b , b , . . . , b n ) ∈ N ( Q ni =0 R i ) such that ( a , a , . . . , a n )( b , b , . . . , b n ) = 0.Then a i b j = 0 for each 1 ≤ i, j ≤ n . Since R i is a weakly semicommutative ring, therefore a i R i b j ⊆ N ( R i ) for each i, j . So, there exists a maximal positive integer k among all pos-itive integers (all nilpotency) such that a i R i b i ⊆ N ( R i ) for each i = 1 , , . . . , n . Therefore,( a , a , . . . , a n ) Q ni =0 R i ( b , b , . . . , b n ) ∈ Q ni =0 N ( R i ) = N ( Q ni =0 R i ).Let R be a ring and let S − R = { u − a | u ∈ S, a ∈ R } with S a multiplicative closed subsetof the ring R consisting of central regular elements. Then S − R is a ring. Proposition 2.7.
Let S be multiplicative closed subset of a ring R consisting of central regularelements. Then R is a weak ideal-Armendariz if and only if S − R is a weak ideal-Armendariz.Proof. Since R is an Armendariz ring if and only if S − R is an Armendariz ring. Also, byProposition (3 .
1) of [15], S − R is a weakly semicommutative ring. Hence, S − R is a weakideal-Armendariz ring. It is known that subring of a weak ideal-Armendariz ring is weak ideal-Armendariz, therefore converse is true. Corollary 2.1.
For any ring R , following conditions are equivalent: (1) R [ x ] is weak ideal-Armendariz ring. (2) R [ x, x − ] is weak ideal-Armendariz ring.Proof. (1) ⇔ (2) Clearly, S = { , x, x , . . . } is a multiplicative set in R [ x ] and R [ x, x − ] = S − R [ x ]. Rest follows from Proposition 2.7.For a semiprime ring R , note that R is reduced if and only if N ∗ ( R ) = N ( R ) if and onlyif R is semicommutative ring if and only if R is weakly semicommutative ring. Since, for asemiprime ring R , { } = N ∗ ( R ) = N ( R ). A ring R is said to be a right Ore ring if for a, b ∈ R with b is a regular, there exist a , b ∈ R with b is regular such that ab = ba . It is knownthat a ring R is a right Ore ring if and only if the classical right quotient ring Q ( R ) of R exists.5 roposition 2.8. Let R be a semiprime right Goldie ring. Then following statements areequivalent: (1) R is an ideal-Armendariz ring. (2) R is an Armendariz ring. (3) R is reduced ring. (4) R is semicommutative ring. (5) R is weakly semicommutative ring. (6) Q ( R ) is an ideal Armendariz. (7) Q ( R ) is reduced ring. (8) Q ( R ) is semicommutative ring. (9) Q ( R ) is weakly semicommutative ring. (10) R is weak ideal-Armendariz ring. (11) Q ( R ) is weak ideal-Armendariz ring.Proof. Implication from (1) to (9) are obvious.(9) ⇒ (10): Q ( R ) is weakly semicommutative so is R (being subring). Moreover, Q ( R ) issemicommutative, therefore, R is an Armendariz by Corollary 13 of [ [7]].(10) ⇒ (11): Since R is a semiprime right Goldie ring. Therefore, Armedariz ring and weaklysemicommutative both are equivalent and from Theorem 12 of [7], R is an Armendariz if andonly if Q ( R ) is an Armendariz ring.(11) ⇒ (1): It is obvious.For an algebra R over commutative ring S , the Dorroh extension of R by S is an abeliangroup D = R ⊕ S with multiplication defined by ( r , s )( r , s ) = ( r r + s r + s r , s s ),where r , r ∈ R and s , s ∈ S . Theorem 2.1.
Let R be an algebra over a commutative domain S and D be a Dorroh extensionof R by S . Then R is a weak ideal Armendariz if and only if D is a weak ideal Armendarizring.Proof. Let D be a weak ideal-Armendariz ring. Since D is an extension of R , therefore R is aweak ideal-Armendariz ring.To prove the converse, let R be a weak ideal-Armendariz ring. Take ( r , s )( r , s ) = 0. Then r r + s r + s r = 0 and s s = 0. Case (1) If s = 0, then r r + s r = 0. This implies r ( r + s ) = 0, and hence by defi-nition of weakly semicommutative ring r R ( r + s ) ⊆ N ( R ). Therefore, ( r , a, b )( r , s ) =( r a + br , r , s ) = ( r ( a + b ) , r , s ) = ( r ( a + b ) r + r ( a + b ) s ,
0) = ( r ( a + b )( r + s ) , ⊆ N ( R ). Thus, ( r , s ) D ( r , s ) ⊆ N ( D ). Case (2) If s = 0, then r r + s r = 0, so ( r + s ) r = 0. Therefore, ( r + s ) Rr ⊆ N ( R ).Also, ( r , s )( a, b )( r ,
0) = ( r a + s a + r b, s b )( r ,
0) = ( r ar + s ar + r br + s br ,
0) = ( r ( a + b ) r + s ( a + b ) r ,
0) = (( r + s )( a + b ) r , ⊆ N ( R ). This implies ( r , s )( a, b )( r , s ) ⊆ N ( D ).Hence, D is a weakly semicommutative ring. By Theorem (3.4) of [14], D is an Armendarizring. Thus, D is a weak ideal-Armendariz ring.6 ring R is abelian if every idempotent element is central. In 1998, Anderson and Camilloproved that Armendariz rings are abelian [1], but weakly semicommutative ring is not abelian,therefore, weak ideal-Armendariz ring is not abelian. In this regards we have the following: Example 2.2.
Let D be a domain and R , R be two rings such that R = (cid:18) D D (cid:19) , R = (cid:18) D D (cid:19) . Let = A = (cid:18) a b (cid:19) , = B = (cid:18) a b (cid:19) ∈ R such that AB = 0 .Then a a = 0 and a b = 0 . It is only possible when a = 0 . This implies ACB ∈ N ( R ) foreach C ∈ R . Hence, R is a weakly semicommutative ring. Since, e e = e and e e = 0 ,whereas e = e , therefore R is a non-abelian ring.Similarly, we can prove that (cid:18) D D (cid:19) is a non-abelian weakly semicommutative ring.Also, from Example (2.14) of [14], R and R both are Armendariz rings. Thus, R and R both are non-abelian weak ideal-Armendariz rings.
3. STRONGLY NIL-IFP
In this section we have introduced the concept of strongly nil-IFP which is the generalisationof weak ideal-Armendariz ring. Towards this, we have the following definition:
A ring R is said to be strongly nil-IFP if for any f ( x ) , g ( x ) ∈ R [ x ] such that f ( x ) g ( x ) = 0 ,then arb ∈ N ( R ) , for all a ∈ C f ( x ) , b ∈ C g ( x ) and each r ∈ R . By definition, it is clear that weak ideal-Armendariz ring is strongly nil-IFP but converse isnot true.
Example 3.1.
Suppose R = Z [ x, y ] / < x , x y , y > , where Z is a Galois field of order , Z [ x, y ] is the polynomial ring with two indeterminates x and y over Z and ( x , x y , y ) isthe ideal of Z [ x, y ] generated by x , x y , y . Let R [ t ] be the polynomial ring over R with anindeterminate t . We consider f ( t ) = ¯ x + ¯ yt , g ( t ) = 3¯ x + 4¯ x ¯ yt + 3¯ y t ∈ R [ t ] . Then f ( t ) g ( t ) = 0 but ¯ x. x ¯ y = ¯0 . Therefore, R is not an Armendariz ring, hence it is not a weak ideal-Armendarizring. But R is a strongly nil-IFP, because for any f ( t ) , g ( t ) ∈ R [ t ] if f ( t ) g ( t ) = 0 , then arb ∈ N ( R ) for each r ∈ R , a ∈ C f ( t ) and b ∈ C g ( t ) . Example 3.2. ( [7], Example (2)) Let Z be the field of integers modulo 2 and A = Z [ a , a , a , b , b , b , c ] be the free algebra of polynomials with zero constant terms in noncommuting indeterminates a , a , a , b , b , b , c over Z . Here, A is a ring without identity. Consider an ideal I of Z + A ,generated by a b , a b + a b , a b + a b + a b , a b + a b , a b , a rb , a rb ( a + a + a ) r ( b + b + b ) , r r r r where r, r , r , r , r ∈ A . Then clearly, A ⊆ I . Now, let R = ( Z + A ) /I . By Example (2)of [7] R is semicommutative ring. Therefore, by Proposition 3.2, R is a strongly nil-IFP. If wetake f ( t ) = a + a t + a t , g ( t ) = b + b t + b t ∈ R [ t ] , then f ( t ) g ( t ) = 0 but a b = 0 . So, R is not an Armendariz ring. Thus, R is not a weak ideal Armendariz ring. roposition 3.1. (1) The classes of strongly nil-IFP is closed under subrings. (2)
The classes of strongly nil-IFP is closed under direct sums.Proof. (1) It is obvious that subring of strongly nil-IFP is strongly nil-IFP.(2) Suppose that R α is strongly nil-IFP for each α ∈ Γ and let R = L α ∈ Γ R α . If f ( x ) g ( x ) =0 ∈ R [ x ], then f α ( x ) g α ( x ) = 0, where f α ( x ) , g α ( x ) ∈ R α ( x ) for each α ∈ Γ. Since each R α is strongly nil-IFP, a α r α b α ∈ N ( R α ) for each a α ∈ C f α and b α ∈ C g α and for all r α ∈ R α .We know that N ( R ) = L α ∈ Γ N ( R α ), therefore R is a strongly nil-IFP. Proposition 3.2.
Every semicommutative ring is strongly nil-IFP.Proof.
It is known that semicommutative rings are weak Armendariz. So, for any f ( x ) and g ( x ) ∈ R [ x ] such that f ( x ) g ( x ) = 0 implies ab ∈ N ( R ) for each a ∈ C f ( x ) and b ∈ C g ( x ) . Since R is a semicommutative ring, therefore ab ∈ N ( R ) implies arb ∈ N ( R ) for each r ∈ R . Hence, R is a strongly nil-IFP.Converse of above is not true. See the below given example: Example 3.3.
Construction of example is same as Example 3.2. Here, consider the ideal I of Z + A which is generated by a b , a b + a b , a b + a b + a b , a b + a b , a b , ( a + a + a ) r ( b + b + b ) , r r r r where r, r , r , r , r ∈ A . Then, A ⊆ I . Now, let R = ( Z + A ) /I . Since, a b = 0 and a b b = 0 , therefore R is not a semicommutative ring. Also, whenever, f ( t ) g ( t ) ∈ R [ t ] suchthat f ( t ) g ( t ) = 0 , then aRb ⊆ N ( R ) for any and each a ∈ C f ( t ) and b ∈ C g ( t ) . Thus, R is astrongly nil-IFP. Proposition 3.3.
Every strongly nil-IFP ring is a weakly semicommutative ring.Proof.
It is obvious.Below given two examples show that weakly semicommutative ring/ Armendariz ring is notstrongly nil-IFP.
Example 3.4.
Let K be a field and K [ a , a , b , b ] be the free algebra with noncommuting in-determinates a , a , b , b over K . Let I be an ideal of K [ a , a , b , b ] generated by a b , a b + a b , a b , ( r a r b r ) , ( r a r b r ) , r ( a + a ) r ( b + b ) r ,where r , r , r , r , r , r , r , r , r ∈ K [ a , a , b , b ] . Next, let R = K [ a , a , b , b ] /I . Toprove R is not a strongly nil IFP, consider f ( x ) = a + a x and g ( x ) = b + b x . Then f ( x ) g ( x ) = a b + ( a b + a b ) x + a b x = 0 but a a b / ∈ N ( R ) . Therefore, R is not astrongly nil IFP. By construction of ideal, it is clear that R is a weakly semicommutative ring. Example 3.5.
Let K be a field and R = K [ a, b ] / < a > . Then R is an Armendariz ring byExample (4.8) of [2]. Therefore R is a weak Armendariz ring. Here, if we take f ( t ) = ba + bat , g ( t ) = a + at , then f ( t ) g ( t ) = 0 but ( ba ) b ( a ) is not nilpotent. Thus, R is not a strongly nil-IFP. roposition 3.4. If R is Armendariz and QRP R , then R is strongly nil-IFP.Proof. Let f ( x ) = P mi =0 a i x i and g ( x ) = P nj =0 b j x j ∈ R [ x ] such that f ( x ) g ( x ) = 0. Then a i b j = 0 for each i, j . Also, R is quasi-reversible-over-prime-radical, therefore b j a i ∈ N ∗ ( R ) andthis implies b j a i R ⊆ N ∗ ( R ) and b j a i R ⊆ N ( R ). Hence, a i Rb j ⊆ N ( R ) for each i, j . Thus, R isa strongly nil-IFP. Proposition 3.5.
Strongly nil-IFP ring is weak Armendariz ring.Proof.
Let f ( x ) = P mi =0 a i x i , g ( x ) = P nj =0 b j x j ∈ R [ x ] such that f ( x ) g ( x ) = 0. Then a i Rb j ⊆ N ( R ) for each i, j . If R has unity, then a i b j ∈ N ( R ) for each i, j . If ring has no unity, then a i ( b j a i ) b j ∈ N ( R ) for each i, j . This implies ( a i b j ) ∈ N ( R ) and hence a i b j ∈ N ( R ) for each i, j . Thus, R is a weak Armendariz ring. Remark 3.1. (1)
By Example 3.5, we can say that converse of above is not true. (2)
Every reversible ring is strongly nil-IFP but converse is not true.
Example 3.6.
Let Z be the field of integers modulo and A = Z [ a , a , a , b , b , b ] be the freealgebra of polynomials with zero constant term in noncommuting indeterminates a , a , a , b , b , b over Z . Here, A is a ring without identity. Consider an ideal of Z + A , say I , generated by a b , a b + a b , a b + a b + a b , a b + a b , a b and s s s s s s where s , s , s , s , s , s ∈ A . Let R = Z + AI . Here, R is a strongly nil-IFP. Since, a b = 0 but b a = 0 , therefore R isnot a reversible ring. Lemma 3.1. If R is strongly nil-IFP with no non-zero nil ideals, then R is reversible ring.Proof. To prove R is reversible ring, let ab = 0. Then axbx = 0 and this implies aRb ⊆ N ( R ).Therefore, Rba ⊆ N ( R ), hence Rba is nil one sided ideal of R . Also, Rba = { } , so ba = 0.Hence, R is reversible ring. Proposition 3.6.
Let R be a 2-primal ring. If f ( x ) R [ x ] g ( x ) ∈ N ∗ ( R )[ x ] , then aRb ⊆ N ∗ ( R ) for each a ∈ C f ( x ) and b ∈ C g ( x ) .Proof. Let f ( x ) , g ( x ) ∈ R [ x ] such that f ( x ) R [ x ] g ( x ) ∈ N ∗ ( R )[ x ]. Since R/N ∗ ( R ) is reduced,therefore R/N ∗ ( R ) is an Armendariz ring and hence aRb ⊆ N ∗ ( R ) for each a ∈ C f ( x ) and b ∈ C g ( x ) . Proposition 3.7. (1)
Every 2-primal ring is strongly nil-IFP. (2)
Every NI ring is strongly nil-IFP.Proof. (1) Let f ( x ) = P mi =0 a i x i and g ( x ) = P nj =0 b j x j ∈ R [ x ] such that f ( x ) g ( x ) = 0. Then f ( x ) g ( x ) = 0 in ( R/N ∗ ( R ))[ x ]. Since, R/N ∗ ( R ) is reduced ring, therefore R/N ∗ ( R ) isan Armendariz ring. This implies a i b j + N ∗ ( R ) = N ∗ ( R ), a i b j ∈ N ∗ ( R ) for each i, j andhence b j a i ∈ N ∗ ( R ), so b j a i R ∈ N ∗ ( R ). This implies a i Rb j ∈ N ∗ ( R ). Thus, R is a stronglynil-IFP.(2) It is obvious. Proposition 3.8.
Let R be a ring of bonded index of nilpotency . Then the following conditionsare equivalent: (1) R is strongly nil-IFP. R is NI.Proof. (1) ⇒ (2) Let a, b ∈ N ( R ). Then a = 0 , b = 0. Now, ( a + b ) = a + ab + ba + b = ab + ba and ( a + b ) = abab + baba = 0 and so ( a + b ) = 0. Thus, a + b ∈ N ( R ).Again, ab = 0 implies Rab = 0 and hence
RaRb ⊆ N ( R ). In particular, a = 0, then RaRa ⊆ N ( R ) and hence Ra ⊆ N ( R ). Thus, ar, ra ∈ N ( R ).(2) ⇒ (1) It is obvious by Proposition 3.7. Proposition 3.9.
For a ring R , let R/I be strongly nil IFP. If I is a semicommutative idealof R , then R is a strongly nil IFP.Proof. Let f ( x ) = P mi =0 a i x i and g ( x ) = P nj =0 b i x j ∈ R [ x ] be such that f ( x ) g ( x ) = 0. Thisimplies m + n X k =0 ( X i + j = k a i b j ) x k = 0 . (3.1)Hence, we have the following equations X i + j = l a i b j = 0 , l = 0 , , , . . . , m + n. (3.2)Also, f ( x ) g ( x ) = 0 implies that f ( x ) g ( x ) = 0 ∈ ( R/I )[ x ]. Since R/I is strongly nil IFP, so a i R b j ⊆ N ( R/I ) for each i, j for 0 ≤ i ≤ m and 0 ≤ j ≤ n . This implies, there exists apositive integer n ij such that ( a i rb j ) n ij ∈ I for each i, j and all r ∈ R . Now, we use principle ofinduction on i + j to prove a i rb j ∈ N ( R ).If i + j = 0, we have a b = 0. Also from above, there exists a positive integer p suchthat ( a rb ) p ∈ I . This implies ( a rb ) p a , b ( a rb ) p ∈ I . Therefore, ( a rb ) p a b ( a rb ) p = 0.Since, rb ( a rb ) p , ( a rb ) p a r ∈ I , so ( a rb ) p a rb ( a rb ) p b ( a rb ) p = 0 and again ( a rb ) p a rb ( a rb ) p ( a rb ) p a rb ( a rb ) p = 0, hence ( a rb ) p +2 = 0.Thus, a Rb ⊆ N ( R ).Let the result is true for all positive integers less than l , i.e. a i Rb j ⊆ N ( R ), when i + j < l .Let there exists a positive integer q such that ( a rb l ) q ∈ I for a fixed r ∈ R . Also, by assumption,we have a Rb l − ⊆ N ( R ), so b l − a ∈ N ( R ). Then there exists s such that ( b l − a ) s = 0. Hence,we have (( a b l − )( a rb l ) q +1 a )( b l − a ) s ( b l − ( a rb l ) q +1 ) = 0 . This implies (( a b l − )( a rb l ) q +1 a )( b l − a )( b l − a ) s − ( b l − ( a rb l ) q +1 ) = 0 . Since rb l ( a rb l ) q a ∈ I and I is semicommutative, then(( a b l − )( a rb l ) q +1 a )( b l − a )( rb l ( a rb l ) q a )( b l − a ) s − ( b l − ( a rb l ) q +1 ) = 0 . This implies, ((( a b l − )( a rb l ) q +1 ) a )( b l − a )( b l − a ) s − ( b l − ( a rb l ) q +1 ) = 0 . Again, by using semicommutativity of I ,((( a b l − )( a rb l ) q +1 ) a )( b l − a ) rb l ( a rb l ) q a ( b l − a ) s − ( b l − ( a rb l ) q +1 ) = 0 . a b l − )( a rb l ) q +1 ) s +2 = 0 and hence (( a b l − )( a rb l ) q +1 ) ∈ N ( I ).Similarly, we can show ( a i b l − i )( a rb l ) q +1 ∈ N ( I ), for 2 ≤ i ≤ l . By equation (3.2), we have a b l + a b l − + a b l − + · · · + a l b = 0 . (3.3)Multiplying in equation (3.3) by ( a rb l ) q +1 from right, we get a b l ( a rb l ) q +1 + a b l − ( a rb l ) q +1 + a b l − ( a rb l ) q +1 + · · · + a l b ( a rb l ) q +1 = 0 . Since N ( I ) is an ideal and ( a i b l − i )( a rb l ) q +1 ∈ N ( I ), for 2 ≤ i ≤ l , therefore a b l ( a rb l ) q +1 ∈ N ( I ). Also ( a rb l ) q ∈ I , so ( a rb l ) q a b l ( a rb l ) q +1 ∈ N ( I ). This implies, ( b l ( a rb l ) q +1 )( a rb l ) q a ) ∈ N ( I ). Since, (( a rb l ) q a r ) , ( rb l ( a rb l ) q ) ∈ I , therefore (( a rb l ) q a r )( b l ( a rb l ) q +1 )(( a rb l ) q a )( rb l ( a rb l ) q ) ∈ N ( I ) and hence ( a rb l ) q +3 ∈ N ( I ). Hence, ( a rb l ) ∈ N ( R ) andthus a Rb l ⊆ N ( R ).Let ( a rb l − ) t ∈ I , therefore by above argument ( a i b l − i )( a rb l − ) t +1 ∈ N ( I ), for 2 ≤ i ≤ l .Since a b l ∈ N ( R ), then ( a b l ) u = 0. Therefore, ( a rb l − ) t +1 ( a b l ) u ( a rb l − ) t +1 = 0. Thisimplies, ( a rb l − ) t +1 ( a b l )( a b l ) u − ( a rb l − ) t +1 = 0 . Again, I is semicommutative, we have( a rb l − ) t +1 ( a b l )( a rb l − ) t +1 ( a b l ) u − ( a rb l − ) t +1 = 0 . ( a rb l − ) t +1 ( a b l )( a rb l − ) t +1 ( a b l )( a b l ) u − ( a rb l − ) t +1 = 0 . Continuing this process, we get (( a rb l − ) t +1 ( a b l )) u ( a rb l − ) t +1 = 0 and so (( a rb l − ) t +1 ( a b l )) u +1 = 0, therefore (( a b l )( a rb l − ) t +1 ) u +2 = 0. Thus, ( a b l )( a rb l − ) t +1 ∈ N ( I ).Multiplying in equation (3.3) by ( a rb l − ) t +1 from right side, we get ( a b l − )( a rb l − ) t +1 ∈ N ( I ). Again, by above analogy, we get ( a rb l − ) t +3 ∈ N ( I ) and hence ( a rb l − ) ∈ N ( R ).Thus, a Rb l − ⊆ N ( R ). Similarly, we can show that a Rb l − , a Rb l − , . . . , a l Rb ∈ N ( R ).Therefore, by induction we have a i Rb j ⊆ N ( R ) for each i, j . Proposition 3.10. If I is a nilpotent ideal of a ring R and R/I is strongly nil-IFP, then R isa strongly nil-IFP.Proof. Let f ( x ) and g ( x ) ∈ R [ x ] such that f ( x ) g ( x ) = 0. Then f ( x ) g ( x ) = 0, therefore( arb ) n ∈ I for each a ∈ C f ( x ) , b ∈ C g ( x ) and for all r ∈ R , since R/I is strongly nil-IFP. Thisimplies (( arb ) n ) m = 0, because I is nilpotent ideal with nilpotency m . Thus, R is a stronglynil-IFP. Theorem 3.1.
Let R be a ring and n be a positive integer. If R is 2-primal, then R [ x ] / < x n > is a strongly nil-IFP.Proof. Result is obvious for n = 1 by Proposition 3.7. For n ≥
2, we use the technique ofthe proof of [ [1], Theorem 5]. Here, we denote x by w and hence R [ x ] / < x n > = R [ w ] = R + Rw + Rw + · · · + Rw n − , where w commutes with the elements of R and w n = 0. Let f ( t ) = a ( w ) + a ( w ) t + a ( w ) t + · · · + a r ( w ) t r , g ( t ) = b ( w ) + b ( w ) t + b ( w ) t + · · · + b s ( w ) t s ∈ R [ w ][ t ] with f ( t ) g ( t ) = 0. In order to prove a i ( w ) R [ w ] b j ( w ) ∈ N ( R [ w ]), we can write a i ( w ) = a (0) i + a (1) i w + a (2) i w + · · · + a ( n − i w n − and b j ( w ) = b (0) j + b (1) j w + b (2) j w + · · · + b ( n − j w n − .On the other hand, we can write f ( t ) = f ( t ) + f ( t ) w + f ( t ) w + · · · + f n − ( t ) w n − and g ( t ) = g ( t ) + g ( t ) w + g ( t ) w + · · · + g n − ( t ) w n − , where f i ( t ) = a ( i ) o + a ( i )1 t + a ( i )2 t + · · · + a ( i ) r t r and g j ( t ) = b ( j ) o + b ( j )1 t + b ( j )2 t + · · · + b ( j ) s t s . Now, it is sufficient to proof that a ( i ) p w i R [ w ] b ( j ) q w j ∈ ( R [ w ]) for all i, j, p, q . From f ( t ) g ( t ) = 0, we have f i ( t ) g j ( t ) = 0 for i + j = n and this implies a ( i ) p w i R [ w ] b ( j ) q w j = 0 ∈ N ( R [ w ]). If i + j < n , then f ( t ) g ( t ) = 0, f ( t ) g ( t )+ f ( t ) g ( t ) = 0, . . . , f ( t ) g n − ( t ) + f ( t ) g n − ( t ) + f ( t ) g n − ( t ) + · · · + f n − ( t ) g ( t ) = 0. Since R is 2-primal, therefore R [ t ] is 2-primal and hence R [ t ] is strongly nil-IFP. Therefore, f i ( t ) R [ t ] g j ( t ) ∈ N ( R [ t ]) = N ( R )[ t ].By Proposition 3.6, we have a ( i ) p Rb ( j ) q ∈ N ( R ). Thus, a ( i ) p w i R [ w ] b ( j ) q w j ∈ N ( R [ w ]). Proposition 3.11.
For a ring R and n ≥ , the following are equivalent: (1) R is strongly nil-IFP. (2) U n ( R ) is strongly nil-IFP. (3) D n ( R ) is strongly nil-IFP. (4) V n ( R ) is strongly nil-IFP. (5) T ( R, R ) is strongly nil-IFP.Proof. It is sufficient to show (1) ⇒ (2). Let f ( x ) = A + A x + A x + · · · + A r x r and g ( x ) = B + B x + B x + · · · + B s x s ∈ U n ( R )[ x ] such that f ( x ) g ( x ) = 0, where A ′ i s, Bj ′ s are n × n upper triangular matrices over R as below: A i = a i a i . . . a i n a i . . . a inn ... ... . . . ...0 0 . . . a inn , B j = b j b j . . . b j n b j . . . b inn ... ... . . . ...0 0 . . . b jnn ∈ U n ( R ).Then, from f ( x ) g ( x ) = 0, we have ( P ri =0 a ipp x i ) ( P sj =0 b jpp x j ) = 0 ∈ R [ x ], for p = 1 , . . . n .Since R is strongly nil-IFP, therefore a ipp rb jpp ∈ N ( R ), for each p , for each i , j and for all r ∈ R .This implies ( a ipp rb jpp ) m ijp = 0 for each p and i , j and for all r ∈ R . Let m ij = m ij m ij ...m ijn .Therefore, ( A i CB j ) m ij = ∗ . . . ∗ . . . ∗ ... ... . . . ...0 0 . . . and (( A i CB j ) m ij ) n = 0, for each C ∈ U n ( R ) andfor each i, j . Hence, A i CB j ∈ N ( U n ( R )) for each i, j , where 0 ≤ i ≤ s , 0 ≤ j ≤ t and for all C ∈ U n ( R ). Thus, U n ( R ) is strongly nil-IFP. Proposition 3.12.
Let R be a strongly nil-IFP. If N ( R )[ x ] = N ( R [ x ]) , then R [ x ] is stronglynil-IFP.Proof. Let p ( y ) = f ( x ) + f ( x ) y + · · · + f m ( x ) y m , q ( y ) = g + g ( x ) y + · · · + g n y n ∈ R [ x ][ y ] suchthat p ( y ) q ( y ) = 0, where f i ( x ) , g j ( x ) ∈ R [ x ]. Write f i ( x ) = a i + a i x + · · · + a iu i x u i , g j ( x ) = b j + b j x + · · · + b jv j x v j , for each 0 ≤ i ≤ m and 0 ≤ j ≤ n , where a i , a i , . . . , a iu i , b j , b j , . . . , b jv j ∈ R . Choose a positive integer k such that k > deg ( f ( x )) + deg ( f ( x )) + · · · + deg ( f m ( x )) + deg ( g ( x )) + deg ( g ( x )) + · · · + deg ( g n ( x )). Since p ( y ) q ( y ) = 0 ∈ R [ x ][ y ], we get f ( x ) g ( x ) = 0 f ( x ) g ( x ) + f ( x ) g ( x ) = 0 . . . . . . . . . . . .. . . . . . . . . . . .. . . . . . . . . . . .f m ( x ) g n ( x ) = 0 12ow, put p ( x k ) = f ( x ) = f ( x ) + f ( x ) x k + f x k + · · · + f m ( x ) x mk ; q ( x k ) = g ( x ) = g ( x ) + g ( x ) x k + g x k + · · · + g n x nk . ( ∗∗ )Then, we have f ( x ) g ( x ) = f ( x ) g ( x ) + ( f ( x ) g ( x ) + f ( x ) g ( x )) x k + · · · f m ( x ) g n ( x ) x ( n + k ) . Therefore, by ( ∗∗ ), we have f ( x ) g ( x ) = 0 in R [ x ]. On the other hand, we have f ( x ) g ( x ) =( a + a x + · · · + a u x u + a x k + a x k +1 + · · · + a u x k + u + · · · + a m + a m x mk +1 + · · · + a mu m x mk + u m )( b + b x + b v x v + b x k + b x k +1 + · · · + b v x k + v + · · · + b n x nk + b n x nk +1 + · · · + b nv n x nk + v n ) = 0 in R [ x ]. Since R is strongly nil-IFP, so a ic Rb jd ∈ N ( R ) , forall 0 ≤ i ≤ m , 0 ≤ j ≤ n , c ∈ { , , . . . , u i } and d ∈ { , , . . . , v j } . Therefore, f i ( x ) Rg j ( x ) ∈ N ( R )[ x ] = N ( R [ x ]) for all 0 ≤ i ≤ m and 0 ≤ j ≤ n . Hence, R [ x ] is a strongly nil-IFP. Corollary 3.1.
Let R be a ring. If N ( R )[ x ] = N ( R [ x ]) , then the following conditions areequivalent: (1) R is strongly nil-IFP. (2) R [ x ] is strongly nil-IFP. (3) R [ x, x − ] is strongly nil-IFP. Proposition 3.13. If R is an Armendariz and QRP R , then R [ x ] is strongly nil-IFP.Proof. By Proposition (3.4), R is a strongly nil-IFP. In Armendariz ring, N ( R )[ x ] = N ( R [ x ]).It follows by Proposition 3.12, R [ x ] is strongly nil-IFP. Proposition 3.14.
Let S be a multiplicative closed subset of a ring R consisting of centralregular elements. Then R is strongly nil-IFP ring if and only if S − R is strongly nil-IFP.Proof. Let R be a strongly nil-IFP. Let P ( x ) , Q ( x ) ∈ S − R [ x ] for P ( x ) = u − p ( x ) and Q ( x ) = v − q ( x ) such that P ( x ) Q ( x ) = 0, where u, v ∈ S . This implies p ( x ) q ( x ) = 0. Since, R isstrongly nil-IFP, so arb ∈ N ( R ), for each a ∈ C p ( x ) , b ∈ C q ( x ) and for all r ∈ R . Therefore, u − aS − Rv − b ⊆ N ( S − R ) for each a ∈ C P ( x ) , b ∈ C Q ( x ) . Hence, S − R is a strongly nil-IFP. Proposition 3.15.
Let R be a finite subdirect sum of strongly nil-IFP rings. Then R is stronglynil-IFP.Proof. Let I l , where 1 ≤ l ≤ k be ideals of R such that R/I l be strongly nil-IFP and T kl =1 I l = 0.Suppose f ( x ) = Σ mi =0 a i x i and g ( x ) = Σ nj =0 b j x j ∈ R [ x ] such that f ( x ) g ( x ) = 0. This implies f ( x ) g ( x ) = 0. Then there exist s ijl ∈ N such that ( a i Rb j ) s ijl ⊆ I l . Set s ij = s ij s ij s ij . . . s ijk ,then ( a i Rb j ) s ij ⊆ I l for any l . This shows that ( a i rb j ) s ij = 0 for each i, j and for all r ∈ R. Hence, R is strongly nil-IFP. Theorem 3.2.
Let R be an algebra over commutative domain S and D be Dorroh extension of R by S . Then R is strongly nil-IFP if and only if D is strongly nil-IFP. roof. Let D be a strongly nil-IFP. Since D is a trivial extension of R by S . Therefore, R isstrongly nil-IFP.Conversely, let R be strongly nil-IFP. Let f ( x ) = P mi =0 ( a i , b i ) x i = ( f ( x ) , f ( x )) and g ( x ) = P ( c j , d j ) x j = ( g ( x ) , g ( x )) ∈ D [ x ] such that f ( x ) g ( x ) = 0 where f ( x ) = P mi =0 a i x i , f ( x ) = P mi =0 b i x i , g ( x ) = P nj =0 c j x j and g ( x ) = P nj =0 d j x j . From f ( x ) g ( x ) = 0, we have f ( x ) g ( x ) + f ( x ) g ( x ) + f ( x ) g ( x ) = 0 , f ( x ) g ( x ) = 0 . (3.4)Since S is a domain, therefore either f ( x ) = 0 or g ( x ) = 0. Case 1. If f ( x ) = 0, then from equation (3.4), we have f ( x ) g ( x ) + f ( x ) g ( x ) = 0 and thisimplies f ( x )( g ( x ) + g ( x )) = 0. Since R is strongly nil-IFP, therefore a i R ( c j + d j ) ⊆ N ( R ).Hence ( a i , r, s )( c j , d j ) = ( a i ( r + s ) c j + a i ( r + s ) d j , ∈ N ( R ) for each i, j and any ( r, s ) ∈ D .Thus, D strongly nil-IFP. Case 2. If g ( x ) = 0, then from equation (3.4), we have ( f ( x ) + f ( x )) g ( x ) = 0. Since R isstrongly nil-IFP, therefore ( a i + b i ) Rc j ⊆ N ( R ) . Hence ( a i , b i )( r, s )( c j ,
0) = ( a i + b i ( r + s ) c j , ∈ N ( R ), for each i, j and any ( r, s ) ∈ D . Thus, D is a strongly nil-IFP.Given Example 2.2, R is a strongly nil IFP but not an abelian. The following example showsthat abelian ring need not be strongly nil IFP. Example 3.7. ( [10], Example (12)) Let K be a field and A = [ a , a , b , b ] be the free algebragenerated by noncommuting indeterminates a , a , b , b . Let I be an ideal of A generated by a b , a b + a b , a b and R = A/I . Let f ( x ) = a + a x, g ( x ) = b + b x such that f ( x ) g ( x ) = 0 .But a b b is not nilpotent. Therefore, R is not strongly nil-IFP. By Example 12 of [10] R isan abelian ring. Acknowledgement
The authors are thankful to Indian Institute of Technology Patna for providing financial supportand research facilities.