aa r X i v : . [ m a t h . R A ] F e b . NIL-REVERSIBLE RINGS
SANJIV SUBBA † AND TIKARAM SUBEDI †∗ Abstract.
This paper introduces and studies nil-reversible rings whereinwe call a ring R nil-reversible if the left and right annihilators of everynilpotent element of R are equal. Reversible rings (and hence reducedrings) form a proper subclass of nil-reversible rings and hence we pro-vide some conditions for a nil-reversible ring to be reduced. It turns outthat nil-reversible rings are abelian, 2-primal, weakly semicommutativeand nil-Armendariz. Further, we observe that the polynomial ring overa nil-reversible ring R is not necessarily nil-reversible in general, but itis nil-reversible if R is Armendariz additionally. Keywords:
Reversible ring, nil-reversible ring, abelian ring. INTRODUCTION
Throughout this paper, unless otherwise mentioned all rings consideredare associative with identity, R represents a ring and all modules are unital.For any a ∈ R , the notations r ( a ) and l ( a ) represent right and left anni-hilators of a respectively. We write Z ( R ) , P ( R ) , J ( R ) , N ( R ) and U ( R )respectively for the set of all central elements, the prime radical, the Ja-cobson radical, the set of all nilpotent elements, and the group of units of R . A ring or an ideal is said to be reduced if it contains no non-zeronilpotent elements. Reversible ring is defined in [1] and it generalizes reducedrings. R is said to be reversible ([1]) if for any w, h ∈ R , wh = 0 implies hw = 0. R is said to be central reversible ([20]) if wh = 0 implies hw ∈ Z ( R ), w, h ∈ R . R is semicommutative if wh = 0 implies wRh = 0, where w, h ∈ R .It is obvious that reversible rings are semicommutative rings. According to[3], a ring R is called weakly semicommutative if for any w, h ∈ R, wh = 0implies wrh ∈ N ( R ) for any r ∈ R . Clearly, weakly semicommutative ringsare generalizations of semicommutative rings. Following [19], R is calledunit-central whenever U ( R ) ⊆ Z ( R ).Following [20], R is called right ( left ) principally projective if the right(left) annihilator of an element of R is generated by an idempotent. A ring inwhich all idempotent elements are central is said to be abelian . R is directlyfinite if xy = 1 implies yx = 1, x, y ∈ R . It is well known that abelian ringsare directly finite. Following [7], R is said to be strongly regular if for any x in R , there is y in R satisfying x = yx . R is said to be 2 − primal if N ( R ) coincides with P ( R ). Rings in which N ( R ) is an ideal are said to be N I . Note that 2-primal rings are
N I . R is called a left ( right ) SF ([8]) if Mathematics Subject Classification.
Primary 16U80; Secondary 16S34, 16S36. all simple left (right) R -modules are flat. A left R -module M is said to be Y J − injective ([13]) if for every w ( = 0) in R , there is a positive integer m such that w m = 0 and every left R -homomorphism from Rw m to M extendsto one from R to M . Following [7], R is called left GP V -ring if every simpleleft R -module is Y J -injective. Following [9], a left R -module M is called W nil − injective if for any w ( = 0) ∈ N ( R ), there exists a positive integer m such that w m = 0 and any left R -homomorphism f : Rw m → M extendsto one from R to M . It is obvious that Y J -injective modules are left W -nilinjective.According to Rege and Chhawchharia ([10]), a ring R is called Ar-mendariz if f ( x ) = n P i =0 w i x i , g ( x ) = m P j =0 h j x j ∈ R [ x ] satisfy f ( x ) g ( x ) = 0,then w i h j = 0 for each i, j . In [6], R is called nil-Armendariz if f ( x ) = n P i =0 w i x i , g ( x ) = m P j =0 h j x j ∈ R [ x ] satisfy f ( x ) g ( x ) ∈ N ( R )[ x ], then w i h j ∈ N ( R ) for each i, j . According to [11], R is called central Armendariz iffor any f ( x ) = n P i =0 w i x i , g ( x ) = m P j =0 h j x j ∈ R [ x ] satisfy f ( x ) g ( x ) = 0, then w i h j ∈ Z ( R ) for each i, j .For an endomorphism α of R , R [ x ; α ] denotes the skew polynomial ringover R whose elements are polynomials n P i =0 w i x i , w i ∈ R , where addition isdefined as usual and the multiplication is defined by the rule xw = α ( w ) x for any w ∈ R . R [ x, x − ; α ] denotes the skew Laurent polynomial ring andits elements are finite sum of the form x − i wx j where w ∈ R and i and j arenon-negative integers. The skew power series ring is represented by R [[ x ; α ]]and its elements are of the form ∞ P i =0 w i x i , w i ∈ R and non-negative integer i .The skew Laurent power series ring is denoted by R [[ x, x − ; α ]] and it is thelocalization of R [[ x ; α ]] with respect to { x i } i ≥ . Thus, R [[ x ; α ]] is a subringof R [[ x, x − ; α ]]. If α becomes an automorphism of R , then the elementsof R [[ x, x − ; α ]] are of the form x p w p + x p +1 w p +1 + ... + w + w x + ... , w j ∈ R and integer p ≤ j ≥ p , and addition is defined as usualand the multiplication is defined by the relation xw = α ( w ) x for any w ∈ R . According to [18], R is called power-serieswise Armendariz if whenever f ( x ) = ∞ P i =0 w i x i , g ( x ) = ∞ P j =0 h j x j ∈ R [[ x ]] satisfy f ( x ) g ( x ) = 0, then w i h j = 0for all i, j . R is said to be skew power-serieswise Armendariz (or SPA)([16])if for any g ( x ) = ∞ P i =0 w i x i , h ( x ) = ∞ P j =0 h j x j ∈ R [[ x ; α ]] , g ( x ) h ( x ) = 0 if andonly if w i h j = 0 for all i, j .In this paper, we introduce a class of rings called nil-reversible ringswhich is a strict generalization of reversible rings. It is proved that each ofthe classes of abelian, 2-primal, weakly semicommutative and nil-Armendariz IL-REVERSIBLE RINGS 3 rings strictly contains the class of nil-reversible rings. It is also shown thatthe class of nil-reversible rings strictly contains the class of unit central rings.Moreover, the nil-reversibility property is shown to be preserved under var-ious ring extensions. 2.
Nil-reversible rings
Definition 2.1.
We call a ring R nil reversible if l ( a ) = r ( a ) for any a ∈ N ( R ). Remark 2.2.
By definition, it is clear that nil-reversible rings are closedunder subrings.It is evident that a reversible ring is nil-reversible. However, there existsa nil-reversible ring which is not reversible as given in the next example.
Example 2.3.
Consider the ring R = Z [ x, y ] /I , where xy = yx and I =
Nil-reversible rings are (1) abelian. (2) (3) weakly semicommutative.Proof.
Let R be a nil-reversible ring.(1) Let e be an idempotent element of R . For any x ∈ R , ex − exe ∈ N ( R ). Note that ( ex − exe ) e = 0. By hypothesis, e ( ex − exe ) = 0,so ex = exe . Again, xe − exe ∈ N ( R ) and e ( xe − exe ) = 0. Soby nil-revesibility of R , we have ( xe − exe ) e = 0, that is, xe = exe .Hence, ex = xe .(2) Note that P ( R ) ⊆ N ( R ). Suppose w ( = 0) ∈ N ( R ). Then there isa positive integer m ≥ w m = 0. Thus, Rw m − w = 0.This implies that wRw m − = 0 as R is nil-reversible. This yields( Rw ) m = 0, so w ∈ P ( R ).(3) Let a, b ∈ R be such that ab = 0. Then, ba ∈ N ( R ). For any r ∈ R, rbaba = 0. Since R is nil-reversible, barba = 0. Then,( arb ) = arbarbarb = 0, whence arb ∈ N ( R ). Hence, R is weaklysemicommutative. (cid:3) One may think whether an abelian or a 2-primal or a weakly semicom-mutative ring is nil-reversible. Hence, we provide the following examples.
Example 2.5. (1) Let Z be the ring of integers, and consider the ring R = (cid:26)(cid:20) x yz w (cid:21) : x ≡ w ( mod , y ≡ z ( mod , x, y, z, w ∈ Z (cid:27) . By SANJIV SUBBA AND TIKARAM SUBEDI [[12],Example 2.7], R is abelian. Now, (cid:20) (cid:21) ∈ N ( R ). We have, (cid:20) (cid:21) (cid:20) (cid:21) = (cid:20) (cid:21) . But (cid:20) (cid:21) (cid:20) (cid:21) = (cid:20) (cid:21) .So, R is not nil-reversible.(2) Take R = (cid:20) R R R (cid:21) , where R denotes the filed of real numbers.Then, R is 2 − primal . Observe that (cid:20) (cid:21) ∈ N ( R ), (cid:20) (cid:21) (cid:20) (cid:21) =0 , (cid:20) (cid:21) (cid:20) (cid:21) = 0. So, R is not nil-reversible.(3) We use the ring given in [[4],example 2.1]. Let F be a field. F
Let R be a nil-reversible ring. Then R is reduced if R satisfies any one of the following conditions: (1) R is right (left) principally projective . (2) R is semiprime . (3) R is left SF (4)
R is right SF (5)
Every simpler singular left R-module is Wnil-injective (6)
Every simpler singular right R-module is Wnil-injective (7)
Every maximal essential left ideal of R is W-nil injective (8)
Every maximal essential right ideal of R is Wnil-injectiveProof. (1) Let R be a right principally projective ring. Suppose w ∈ R such that w = 0. Then, w ∈ N ( R ). As R is right principallyprojective, r ( w ) = eR for some idempotent e ∈ R . This implies w = ew . As e ∈ r ( w ), we = 0. Since R is nil-reversible, ew = 0.Thus, w = 0 and so R is reduced. Similarly, we can prove that R isreduced if R is left principally projective.(2) Let R be a semiprime ring. Suppose w ∈ R such that w = 0.Then Rww = 0. Since R is nil-reversible, wRw = 0. By hypothesis, w = 0. Thus, R is reduced.(3) Let w ( = 0) ∈ R and w = 0. Then l ( w ) = R and l ( w ) is containedin some maximal left ideal H of R . As w ∈ l ( w ) ⊆ H and R isleft SF , there exists h ∈ H such that w = wh . So, w (1 − h ) = 0. IL-REVERSIBLE RINGS 5
Since R is nil-reversible, (1 − h ) w = 0, (1 − h ) ∈ l ( w ) ⊆ H . Then,1 = (1 − h ) + h ∈ H . This is a contradiction.(4) The proof is similar to (3).(5) Let w (0 =) be in R satisfying w = 0. Then, l ( w ) is contained insome maximal left ideal H . If possible, let us assume that H is notessential. Then there exists an idempotent e ∈ R with the property H = l ( e ). Since w ∈ H , we = 0. As R is abelian (by Proposition2.4 ), ew = 0, so e ∈ l ( w ) ⊆ H = l ( e ). Thus, e = 0 = e . Thisis a contradiction. Therefore, H is essential and R/H is simplesingular left R -module which is W nil-injective by hypothesis. Define f : Rw → R/H by f ( rw ) = r + H . This is well-defined as l ( W ) ⊆ H .By hypothesis, there exists a left R -homomorphism g : R → H whichextends f . Therefore, 1 + H = f ( w ) = g ( w ) = ws + H , for some s ∈ R . So, 1 − ws ∈ H . Now, w s = 0 implies ws ∈ r ( w ) = l ( w ) ⊆ H .So, ws ∈ H , 1 = 1 − ws + ws ∈ H . This is a contradiction.(6) The proof is same as in (5).(7) Let w ( = 0) ∈ R and w = 0. Then, l ( w ) ⊆ H for some maximalleft ideal H of R . As in (5), H is essential. Define f : Rw → H as f ( rw ) = rw . By hypothesis, there exists a left R -homomorphism g : R → H which extends f . Therefore, w = f ( w ) = g ( w ) = wg (1) = wh for some h ∈ H . As R is nil-reversible, 1 − h ∈ l ( w ) ⊆ H .So, 1 = (1 − h ) + h ∈ H , which is a contradiction.(8) Similar to (7). (cid:3) Corollary 2.7.
Nil-reversible left (right) SF ring is strongly regular.Proof. Follows from [14, Theorem 4.10] and the
Proposition 2.6 . (cid:3) Let P be an (R,R)-bimodule. The trivial extention of R by P is the ring T ( R, P ) = R L P , where the addition is usual and the multiplication isgiven by:( r , p )( r , p ) = ( r r , r p + p r ) , p i ∈ P, r i ∈ R and i = 1 ,
2. This ring isisomophic to the ring (cid:26)(cid:20) x p x (cid:21) : x ∈ R, p ∈ P (cid:27) , where the operations are usual matrix opera-tions. Proposition 2.8. If T ( R, R ) is nil-reversible then R is reversible.Proof. Let w, h ∈ R and wh = 0. Observe that (cid:20) w (cid:21) ∈ N ( T ( R, R ))and (cid:20) h h (cid:21) ∈ T ( R, R ) and (cid:20) w (cid:21) (cid:20) h h (cid:21) = (cid:20) wh (cid:21) = (cid:20) (cid:21) .Since T ( R, R ) is nil-reversible, (cid:20) (cid:21) = (cid:20) h h (cid:21) (cid:20) w (cid:21) = (cid:20) hw (cid:21) .This gives hw = 0. (cid:3) SANJIV SUBBA AND TIKARAM SUBEDI
Proposition 2.9. If R be a ring such that ( wh ) = 0 implies wh = 0 and ( hw ) = 0 implies hw = 0 , w ∈ N ( R ) , h ∈ R then R is nil-reversible.Proof. Let wh = 0 , w ∈ N ( R ) , h ∈ R . Then ( hw ) = 0. So by hypothesis, hw = 0. This implies r ( w ) ⊆ l ( w ). Similarly, l ( w ) ⊆ r ( w ) for w ∈ N ( R ).Thus, R is nil-reversible. (cid:3) Proposition 2.10. If R is unit-central then R is nil-reversible.Proof. Let w ∈ N ( R ) , h ∈ R . Now, (1 − w ) ∈ U ( R ) ⊆ Z ( R ). So, (1 − w )(1 − h ) = (1 − h )(1 − w ). This implies hw = wh . Therefore, hw = 0 if and onlyif wh = 0. Hence R is nil-reversible. (cid:3) But the converse does not hold.
Example 2.11.
Take R = T ( R , R ) and a = (cid:20) (cid:21) ∈ R , where R denotesthe field of real numbers. Then R is nil-reversible by [5, Proposition 1.6].Obviously, a ∈ U ( R ) but a / ∈ Z ( R ). Proposition 2.12.
Suppose R is a central reversible ring containing asthe only central nilpotent element then R is nil-reversible.Proof. Let a ∈ N ( R ) and b ∈ r ( a ). We have a m = 0 for some positive integer m . Since R is central reversible, ba ∈ Z ( R ). Now, ( ba ) m = b m a m = 0. Thus, ba ∈ N ( R ) ∩ Z ( R ) = { } . It gives ba = 0, and so r ( a ) ⊆ l ( a ). Similarly, l ( a ) ⊆ r ( a ). Hence r ( a ) = l ( a ). (cid:3) Let A be an algebra (not necessarily with identity) over a commutativering C . The Dorroh extension of A by C is the ring denoted by A L D C ,with the operations ( a , s )+( a , s ) = ( a + a , s + s ) and ( a , s )( a , s ) =( a a + s a + s a , s s ) for all a i ∈ A and s i ∈ C, i = 1 , Proposition 2.13.
Dorroh Extension of a nil-reversible ring R by Z is nil-reversible.Proof. Let ( a, ∈ N ( R L D Z ) and ( s, m ) ∈ l ( a, ,
0) = (( s + m ) a, s + m ) a = 0. Since R is nil-reversible, ( s + m ) ∈ r ( a ). So( a, s, m ) = (0 , l ( a, ⊆ r ( a, r ( a, ⊆ l ( a, . Hence, R L D Z is nil reversible. (cid:3) Proposition 2.14.
The subdirect product of arbitrary family of nil-reversiblerings is nil-reversible.Proof.
Let { I λ | λ ∈ ∆ } be a family of ideals of R such that ∩ λ ∈ ∆ I λ = 0 and R/I λ ’s are nil-reversible rings, where ∆ is some index set. Take a ∈ N ( R )and b ∈ R satisfying ab = 0. There is a positive integer m with the property a m = 0. Thus, a + I λ ∈ N ( R/I λ ). Since N ( R/I λ ) is nil-reversible and( a + I λ )( b + I λ ) = 0, ( b + I λ )( a + I λ ) = 0. This implies ba + I λ = 0. So ba ∈ I λ for all λ ∈ ∆, i.e., ba ∈ ∩ λ ∈ ∆ I λ = 0. Following similar argument, ba = 0 implies ab = 0. Therefore, the subdirect product of nil-reversiblerings is nil-reversible. (cid:3) IL-REVERSIBLE RINGS 7
Proposition 2.15.
Let J be a reduced ideal of a ring R such that R/J isnil-reversible. Then R is nil-reversible.Proof. Let w ∈ N ( R ) and h ∈ R be such that wh = 0. This implies¯ w ¯ h = 0. As ¯ w ∈ N ( R/J ) and
R/J is nil-reversible, we have ¯ h ¯ w = 0. Thus, hw ∈ J ∩ N ( R ). Since J is reduced, we have hw = 0. As above, hw = 0yields wh = 0. Thus, R is nil-reversible. (cid:3) In Proposition 2.15 , reducedness of J is necessary. As shown by thefollowing example that in deficiency of this condition the Proposition 2.15 is unlikely to be guaranteed.
Example 2.16.
Consider the ring R = (cid:20) F F F (cid:21) , where F is a field. Let e ij be matrix units with 1 at theentry ( i, j ) and zero elsewhere and J = e ij R . Then J = (cid:20) F (cid:21) . Onecan easily verify that J is not a reduced ideal . Now, consider the quotientring R/J = (cid:26)(cid:20) x y (cid:21) + J : x, y ∈ F (cid:27) . Clearly, R/J is reduced and so nil-reversible. Observe that e = 0, e ∈ N ( R ). e e = 0. This implies e ∈ r ( e ). But e e = e = 0. Thus, r ( e ) = l ( e ).A homomorphic image of a nil-reversible ring may not be nil-reversible. Example 2.17.
Let R = D [ x ], where D is a division ring and J = < xy > ,where xy = yx . As R is a domain, R is nil-reversible. Clearly yx ∈ N ( R/J )and x ( yx ) = xyx = 0. But, ( yx ) x = yx = 0. This implies R/J is notnil-reversible.
Proposition 2.18. If R is a nil-reversible ring and J an ideal consisting ofnilpotent elements of bounded index ≤ m in R , then R/J is nil-reversible.Proof.
Let a ∈ N ( R/J ) , b ∈ R/J such that ab = 0. Then ab ∈ J and a ∈ N ( R/J ). This implies a ∈ N ( R ). So ( ab ) m = 0 = a ( ba ) m − b for somepositive integer m . Since R is nil-reversible, ( ba ) m = 0 implying ba ∈ J and r ( a ) ⊆ l ( a ). Proceed similarly to get l ( a ) ⊆ r ( b ). Therefore R/J isnil-reversible. (cid:3) Polynomial and power series extensions of nil-reversiblerings
The polynomial ring over a reduced ring is reduced. However, the polyyno-mial ring over a reversible ring may not be semicommutative by [5, Example2.1]. Based on this observation, it is quite natural to question whether thepolynomial ring over a nil-reversible ring is nil-reversible. But the followingexample gives the answer in negative.
SANJIV SUBBA AND TIKARAM SUBEDI
Example 3.1.
Take Z as the field of integers modulo 2 and let Ω = Z [ w , w , w , h , h , h , d ] be the free algebra of polynomials with zero con-stant terms in non-commuting indeterminates w , w , w , h , h , h and d over Z . Take an ideal J of the ring Z + Ω generated by w h , w h + w h , w h + w h + w h , w h + w h , w h , w ph , w ph , h w , h w + h w , h w + h w + h w , h w + h w , h w , h pw , h pw , ( w + w + w ) p ( h + h + h ) , ( h + h + h ) p ( w + w + w ) and p p p p where p, p , p , p , p ∈ Ω. Take R = Z +Ω J . Then we have R [ x ] ∼ = ( Z +Ω)[ x ] J [ x ] .Now, by [5, Example 2.1], R is reversible and hence nil-reversible. Now,( w + w x + w x ) ∈ N ( R [ x ]) and ( h + h x + h x ) ∈ R [ x ] and d ( h + h x + h x )( w + w x + w x ) ∈ J [ x ] which means d ( h + h x + h x ) ∈ l ( w + w x + w x ). But ( w + w x + w x ) d ( h + h x + h x ) / ∈ J [ x ], since w dh + w dh / ∈ J . This implies d ( h + h x + h x ) / ∈ r ( w + w x + w x ).Thus, R [ x ] is not nil-reversible.Given a nil-reversible ring R , we investigate some conditions under which R [ x ] also becomes nil-reversible. Proposition 3.2. If R is an Armendariz ring, then R is nil-reversible ifand only if R [ x ] is nil-reversible.Proof. Suppose R is nil-reversible. Let g ( x ) = m P i =0 w i x i ∈ N ( R [ x ]), h ( x ) = n P j =0 h j x j ∈ R [ x ] be such that g ( x ) h ( x ) = 0. By [6, Corollary 5.2], N ( R [ x ]) = N ( R )[ x ]. So w i ∈ N ( R ) for all i . Since R is Armendariz, w i h j = 0 for all i, j . By nil-reversibility, h j w i = 0 for all i, j . So, h ( x ) g ( x ) = 0. Followingthe similar argument, h ( x ) g ( x ) = 0 implies g ( x ) h ( x ) = 0. Therefore, R [ x ]is nil-reversible. The proof of the converse is trivial. (cid:3) Proposition 3.3.
Let ∆ denotes a multiplicatively closed subset of R con-sisting of central non-zero divisors. Then R is nil-reversible if and only if ∆ − R is nil-reversible.Proof. Suppose R is nil-reversible. Let s − a ∈ N (∆ − R ) . Then ( s − a ) n =0 for some positive integer n . This implies a n = 0, so a ∈ N ( R ). Forany ( s − b ) ∈ ∆ − R having the property that ( s − b )( s − a ) = 0, we have( ss ) − ba = 0, ba = 0. Since R is nil-reversible, ab = 0, so s − s − ab = 0which further yields ( s − a )( s − b ) = 0. So, l ( s − a ) ⊆ r ( s − a ). Similarly, r ( s − a ) ⊆ l ( s − a ). Hence ∆ − R is nil-reversible.The converse part is trivial. (cid:3) Corollary 3.4. R [ x ] is nil-reversible if and only if R [ x, x − ] is nil-reversible.Proof. Suppose R [ x ] is nil-reversible. Take ∆ = { , x, x , x , ... } ⊆ R [ x ].Clearly, ∆ is multiplicatively closed subset consisting of central non-zero IL-REVERSIBLE RINGS 9 divisors of R [ x ] and R [ x, x − ] = ∆ − R [ x ]. So R [ x, x − ] is nil-reversible.The proof of the converse is trivial. (cid:3) Corollary 3.5.
For an Armendariz ring R , the following are equivalent; (1) R is nil-reversible. (2) R [ x ] is nil-reversible. (3) R [ x, x − ] is nil-reversible. Proposition 3.6.
For a right principally projective ring R and a naturalnumber m ≥ , R is nil-reversible if and only if R [ x ]
Let R be a nil-reversible ring. If f , f , ..., f n ∈ R [ x ] besuch that C f f ...f n ⊆ N ( R ) , then C f C f ...C f n ⊆ N ( R ) , where C f representsthe set of coefficients of f .Proof. From the
Proposition 2.4 , N ( R ) is an ideal of R . Clearly, R/N ( R )is reduced, and so R/N ( R ) is Armendariz. Let f ( x ) = m P i =0 w i x i ∈ R [ x ]. De-fine ζ : R [ x ] → RN ( R ) [ x ] via ζ ( f ( x )) = ¯ f ( x ), where ¯ f ( x ) = m P i =0 ( w i + N ( R )) x i .Clearly, ζ is a ring homomorphism. As C f f ...f n ⊆ N ( R ), ¯ f ¯ f ... ¯ f n = f f ...f n = 0. So, by [2, Proposition 1], ¯ p ¯ p ... ¯ p n = 0 , p i ∈ C f i , i =1 , , ..., n . Now, it is easy to verify that C f C f ...C f n ⊆ N ( R ). (cid:3) Corollary 3.8. If R is nil-reversible ring, then N ( R [ x ]) ⊆ N ( R )[ x ] . Proposition 3.9. If R is nil-reversible ring, then N ( R [ x ]) = N ( R )[ x ] .Proof. By Corollary 3.8 , N ( R [ x ]) ⊆ N ( R )[ x ]. Let f ( x ) = n P i =0 w i x i ∈ N ( R )[ x ]. For each i , there exists a positive integer s i with w s i i = 0. Take l = s + s + ... + s n + 1. Now, ( f ( x )) l = ( w + w x + ... + w n x n ) l = nl P p =0 P i + ... + i l = p w i w i ...w i l ! x p , where w i , w i , ..., w i l ∈ { w , ..., w n } . If thenumber of w ’s in the product w i ...wi l is more than s , then we can ex-press w i ...wi l as h w j h w j h ...h t − w j t h t , where t is a non-negative in-teger and 1 ≤ j , j , ..., j t , s ≤ j + j + ... + j t and for each i, ≤ i ≤ t , h i is a product of some elements choosen from { w , ..., w n } or equalto 1. Here, we have w j + ... + j l = 0 , w j ( w j ...w j l h ) = 0. Since R isnil-reversible, w j ...w j l h w j = 0 = w j ( w j ...w j l h w j ) h . This implies( w j ...w j l h w j ) h w j = 0. Continuing this way, we get h w j ...w j t h t = 0. Thus, w i ...w i l = 0. If the number of w i ’s in w i ...w i l is more than s i ’s, thenproceeding as above, we get w i ...w i l = 0. Hence P i + i + ... + i l = p w i w i ...w i l =0, ( f ( x )) l = 0, N ( R )[ x ] ⊆ N ( R [ x ]). Therefore, N ( R )[ x ] = N ( R [ x ]). (cid:3) Proposition 3.10.
Nil-reversible rings are nil-Armendariz.Proof.
Let R be a nil-reversible ring and f ( x ) = n P i =0 w i x i , g ( x ) = m P j =0 h j x j ∈ R [ x ] and f ( x ) g ( x ) ∈ N ( R )[ x ]. Then f ( x ) g ( x ) = m + n P s =0 P i + j = s w i h j ! x s , P i + j = s w i h j ∈ N ( R ). Now, we use induction to show that w i h j ∈ N ( R ) foreach i, j . For s = 0, P i + j =0 w i h j = w h ∈ N ( R ) which implies h w ∈ N ( R ).For s = 1, w h + w h ∈ N ( R ). R is 2 − primal (by Proposition 2.4 ), w h w + w h w ∈ N ( R ). As h w ∈ N ( R ), w h w ∈ N ( R ). Thus,( w h ) ∈ N ( R ) and so w h ∈ N ( R ), h w ∈ N ( R ). It immediately impliesthat w h ∈ N ( R ) and so h w ∈ N ( R ). Assume w i h j ∈ N ( R ) for i + j = t ,2 ≤ t ≤ m + n −
1. Now, P i + j = t +1 w i h j = w i h j l + w i h j l − + ... + w i l h j o ∈ N ( R ), where 0 ≤ l, i k ∈ { , , , ..., n } , j k ∈ { , , ..., m } , k ∈ { , , ..., l } and i ≤ i ≤ ... ≤ i l , j ≤ j ≤ .... ≤ j l . If there is only one termin P i + j = t +1 w i h j , then we are through. Suppose P i + j = t +1 w i h j contains morethan one term. Then we can take either j l > i or i > j l . Firstly, take j l > i . Now, P i + j = t +1 w i h j = w i h j l + w i h j l − + ... + w i l h j ∈ N ( R ).This implies that w i h j l w i + w i h j l − w i + ... + w i l h j w i ∈ N ( R ). But j l − + i ≤ t, j l − + i ≤ t, ..., j + i ≤ t . Thus, by induction hypothesis h j − w i , ..., h j w i ∈ N ( R ). Therefore, w i h j l w i ∈ N ( R ), w i h j l ∈ N ( R ).Proceeding similarly, we get that w i h j ∈ N ( R ) for each i, j satisfying i + j = t + 1. Similar approach can be applied if i > j l . Hence R isnil-Armendariz. (cid:3) Proposition 3.11. If R is nil-reversible, then R[x] is nil-Armendariz.Proof. Let R be nil-reversible. By Proposition 3.10 , R is nil-Armendarizand N ( R [ x ]) = N ( R )[ x ] ( Proposition 3.9 ). Therefore, by [6, Theorem5.3], R [ x ] is nil-Armendariz. (cid:3) Lemma 3.12.
Let R be a ring such that ( wh ) = 0 implies wh = 0 and ( hw ) = 0 implies hw = 0 , w ∈ R, h ∈ N ( R ) . Then, we have the following; (1) If xyx = 0 or xy = 0 then xy = 0 for any x ∈ R, y ∈ N ( R ) . (2) If y x = 0 then yx = 0 for any x ∈ R, y ∈ N ( R ) .Proof. By Proposition 2.9 , R is nil-reversible. Let x ∈ R and y ∈ N ( R ).(1) Suppose xyx = 0. This implies ( xy ) = 0. By hypothesis xy = 0.If xy = 0, then yxy = 0 ( by nil-reversibility of R ). So ( xy ) = 0.This implies xy = 0. IL-REVERSIBLE RINGS 11 (2) By similar computation as above, y x = 0 yields yx = 0. (cid:3) Proposition 3.13. If R is semicommutative in which ( wh ) = 0 implies wh = 0 and ( hw ) = 0 implies hw = 0 for w ∈ R, h ∈ N ( R ) , then both R [ x ] and R [[ x ]] are nil-reversible.Proof. By Proposition 2.9 , R is nil-reversible. Recall that nil-reversiblerings are closed under subrings ( by Remark R [[ x ]] is nil-reversible. Let g ( x ) = ∞ P i =0 w i x i ∈ R [[ x ]] , h ( x ) = ∞ P j =0 h j x j ∈ N ( R [[ x ]]) and g ( x ) h ( x ) = 0. Since R is nil-reversible, N ( R ) is an ideal by Proposition 2.4(2) . By [15, Proposition 1], N ( R [[ x ]]) ⊆ N ( R )[[ x ]]. So h j ∈ N ( R ) for each j . Since g ( x ) h ( x ) = 0, we get(3.a) ∞ X k =0 X i + j = k w i h j x k = 0Now, we claim that w i h j = 0 for all i, j . We use induction on i + j to proveour claim. Observe that w h = 0. This shows that the claim holds true for i + j = 0. Now, suppose that the claim holds true for i + j ≤ m −
1. Fromthe equation 3.a, we get(3.b) m X i =0 w i h m − i = 0Multiply the equation 3.b by h from the right. Then, we obtain ( w h m + w h m − + ... + w m h ) h = 0 which leads to w m h = 0 as R is semicommu-tative. So w m h = 0 by the Lemma 3.12 . Then the equation 3.b becomes(3.c) w h m + w h m − + ... + w m − h = 0Multiplying the equation 3.c by h , h , ..., h m on the right, we get w m − h , ...,w h m − , w h m = 0. Following similar argument, we get w i h j = 0 for i + j = m . Thus, by induction we get w i h j = 0 for all i, j . As R is nil-reversible h j w i = 0 for all i, j . Hence h ( x ) g ( x ) = 0. Similarly, one can showthat h ( x ) g ( x ) = 0 implies g ( x ) h ( x ) = 0. Thus, R [[ x ]] is nil-reversible. (cid:3) Proposition 3.14.
Let R be a power serieswise Armendariz ring. Then thefollowing are equivalent: (1) R is nil-reversible. (2) R [ x ] is nil-reversible. (3) R [[ x ]] is nil-reversible.Proof. It is sufficient to show that R [[ x ]] is nil-reversible if R is nil-reversible.By [15, Lemma 2], N ( R [[ x ]]) ⊆ N ( R )[[ x ]]. Let g ( x ) = ∞ P i =0 w i x i ∈ N ( R [[ x ]]) , h ( x ) = ∞ P j =0 h j x j ∈ R [[ x ]] and g ( x ) h ( x ) = 0. By hypothesis w i h j = 0. Since R is nil-reversible and N ( R [[ x ]]) ⊆ N ( R )[[ x ]], h j w i = 0 for all i, j . So, h ( x ) g ( x ) = 0.Proceeding as above, we can show that if h ( x ) g ( x ) = 0 then g ( x ) h ( x ) = 0.Therefore R [[ x ]] is nil-reversible. (cid:3) Proposition 3.15.
Let R be an SPA ring and α an automorphism of R .Then the following are equivalent: (1) R is nil-reversible. (2) R [ x ; α ] is nil-reversible. (3) R [ x, x − ; α ] is nil-reversible. (4) R [[ x ; α ]] is nil-reversible. (5) R [[ x, x − ; α ]] is nil-reversible.Proof. It is enough to show that (1) = ⇒ (5). Suppose that R is nil-reversible. Let g ( x ) = x s w s + x s +1 w s +1 + ... + w + w x + ... ∈ N ( R [[ x, x − ; α ]]) , h ( x ) = x t h t + x t +1 h t +1 + ... + h + h x + ... ∈ R [[ x, x − ; α ]] and g ( x ) h ( x ) = 0, where s and t are integers ≤
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A generalization of reversible rings ,Iran. J. Sci. Technol. Trans. A Sci., (1)(2014), 43-48. † Department of Mathematics, NIT Meghalaya, Shillong 793003, India
Email address : [email protected] † Department of Mathematics, NIT Meghalaya, Shillong 793003, India
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