Weak McCoy Ore Extensions

Transcription

1 Internatonal Mathematcal Forum, Vol. 6, 2, no. 2, Weak McCoy Ore Extenon R. Mohammad, A. Mouav and M. Zahr Department of Pure Mathematc, Faculty of Mathematcal Scence Tarbat Modare Unverty, P.O. Box: 45-34, Tehran, Iran Correpondng author: Abtract A rng R called nl-emcommutatve f for every a, b nl(r, ab = mple arb =. Nelen n [] prove that reverble rng are McCoy and gve an example of a emcommutatve rng whch not rght McCoy. At the ame tme, he alo how that emcommutatve rng do have a property cloe to the McCoy condton. Accordng to [3] and [5], a rng R ad to be rght weak McCoy f the equaton f(xg(x =, where f(x = m a x, g(x = n j= b jx j R[x] \ {}, mple that there ext R \ {} uch that a nl(r for all m. Weak McCoy rng are a common generalzaton of Mc- Coy rng and emcommutatve rng. For every rng R, the n-by-n upper trangular matrx rng T n (R weak McCoy. For each nlemcommutatve rng R, we prove that, f R α-compatble then R[x; α] weak McCoy and when R δ-compatble then R[x; δ] weak McCoy. Mathematc Subject Clafcaton: 6S36; 6E5. Keyword: Ore extenon, N l-emcommutatve rng, Weak McCoy rng. Introducton Throughout th artcle, R denote an aocatve rng wth unty, α : R R an endomorphm and δ : R R a dervaton of R, that, δ an addtve map uch that δ(ab = δ(ab + aδ(b, for each a, b R. We denote by R[x; α] the kew polynomal rng whoe element are the polynomal over R, the addton defned a uual, and the multplcaton ubject to the relaton xa = α(ax, for each a R. The dfferental polynomal rng denoted by R[x; δ] whoe element are the polynomal n r x R[x; δ], r R, where

2 76 R. Mohammad, A. Mouav and M. Zahr the addton defned a uual and the multplcaton by xb = bx + δ(b for any b R. The ubet of all nlpotent element of R denoted by nl(r. Recall that a rng R called reduced f a 2 = mple that a =, for all a R; R reverble f ab = mple ba =, for all a, b R; R ymmetrc f abc = mple acb =, for all a, b, c R; R emcommutatve (or ha IFP f ab = mple arb =, for all a, b R. Due to Brkenmeer et al. [2], a rng R called 2-prmal f P (R = nl(r, where P (R the prme radcal of R. Accordng to Nelen [], a rng R called a McCoy rng f the equaton f(xg(x =, where f(x, g(x R[x]\{}, mple that there ext, t R\{} uch that f(x = tg(x =. It well known that commutatve rng are alway McCoy rng [9, Theorem 2 ], but t not true for noncommutatve rng (ee [6]. In [], reverble rng are hown to be McCoy rng, but the convere doe not hold, for example, there ext a non-reverble (rght McCoy rng (ee []. Followng Z. Le, J. Chen and Z. Yng [7], the polynomal rng R[x] over R McCoy f and only f R McCoy. Nelen n [] prove that reverble rng are McCoy and gve an example of a emcommutatve rng whch not rght McCoy. At the ame tme, he alo how that emcommutatve rng do have a property cloe to the McCoy condton. In [5], L. Ouyang and H. Chen ntroduced the noton of weak McCoy rng, and nvetgated the properte of everal extenon of reverble weak McCoy rng. Accordng to [3] and [5], a rng R ad to be rght weak McCoy f the equaton f(xg(x =, where f(x = m a x, g(x = n j= b jx j R[x] \ {}, mple that there ext R \ {} uch that a nll(r for all m. Left weak McCoy rng are defned analogouly. If a rng both left and rght weak McCoy, t called a weak McCoy rng. Clearly, McCoy rng and weak Armendarz rng are weak McCoy. The followng mplcaton hold: reduced reverble emcommutatve nl emcommutatve 2-prmal commutatve McCoy weak McCoy In [], ome example of nl-emcommutatve rng whch are not emcommutatve provded. In th paper we extend the reult n [5] to the more general cae where R a nl- emcommutatve rng. We prove that, when R a weak α-rgd and nl-emcommutatve rng, then R[x; α] a weak McCoy rng. Notce that α-compatble rng are weak α-rgd. We alo how that, when R a δ-compatble and nl-emcommutatve rng, then R[x; δ] a weak McCoy rng.

3 Weak-McCoy Ore extenon 77 2 On weak Mccoy kew polynomal rng Accordng to Krempa [5], an endomorphm α of a rng R called rgd f aα(a = mple a = for a R. A rng R ad to be α-rgd f there ext a rgd endomorphm α of R. Note that any rgd endomorphm of a rng monomorphm and α-rgd rng are reduced. A rng R ad to be α-compatble f for each a, b R, ab = aα(b =. Moreover, R ad to be δ-compatble f for each a, b R, ab = aδ(b =. If R both α-compatble and δ-compatble, R ad to be (α, δ-compatble. For more detal ee [4]. Recently, Z. Le, J. Chen and Z. Yng proved that the polynomal rng over a McCoy rng alway McCoy (ee [7]. In th ecton, we extend the weak McCoy properte of rng to the kew polynomal rng. Lemma 2.. [4, Lemma 2.] Let R be an α-compatble rng. Then we have the followng: ( If ab =, then aα n (b = a n (ab = for any potve nteger n. ( If α k (ab = for ome potve nteger k, then ab =. ( If aα k (b = for ome potve nteger k, then ab =. Lemma 2.2. [] If R a nl emcommutatve rng, then nl(r an deal of R. Example 2.3. We refer to the argument n [4, Example.2]. Let R be an α-rgd rng for an endomorphm α of R and an α-dervaton δ of R. Let a b c R 3 := a b a, b, c R a be a ubrng of T 3 (R. The endomorphm α of R extended to the endomorphm α : R 3 R 3 defned by α((a j = (α(a j and the α-dervaton δ of R alo extended to δ : R 3 R 3 defned by δ((a j = (δ(a j. It hown that, ( R an (α, δ-compatble rng, ( R 3 not α-rgd. By [6, Propoton.2] and [6, Example.5], R 3 em-commutatve but not reverble. Lemma 2.4. Let R be an α-compatble and nl-emcommutatve rng. Then nl(r[x; α] nl(r[x; α]. Proof. Suppoe that f(x = a +a x+ +a n x n and a m =, =,,, n. Let k = n m +. Then nk ( (a + a x + + a n x n k = = k = a α v 2 (a 2 α v k (a k x,

4 78 R. Mohammad, A. Mouav and M. Zahr where v p (2 p k are potve nteger and a j {a, a,, a n } for j k. Then conder a α v 2 (a 2 α v k (ak. If the number of a n a α v 2 (a 2 α v k (ak greater then m, then we can wrte b (α v (a j b (α v 2 (a j2 b t (α v t (a jt b t a a α v 2 (a 2 α v k (ak, where j + j j t > m, l t, v l {v 2, v 3,, v k } and b q a product of ome element choong from the et {a, α v 2 (a 2,, α v k (ak } or equal to. By a j +j 2 + +j t =, and α-compatb lty of R, we get the followng mplcaton a j a j 2 a jt = a j a j 2 α v t (a jt = a j a j 2 (α v t (a jt = α v t (a j a j 2 a j t (α v t (a jt = a j a j 2 (α v t (a j t (α v t (a jt =. (α v (a j (α v 2 (a j2 (α v t (a jt =. Snce R nl-emcommutatve we have, b (α v (a j b (α v 2 (a j2 b t (α v t (a jt b t =. Next f the number of a n a α v 2 (a 2 α v k (ak greater than m, then a mlar dcuon yeld that a α v 2 (a 2 α v k (a k =. So k = a α v 2 (a 2 α v k (ak =. Therefore f(x a nlpotent element of R[x; α]. Accordng to [4], a rng R wth an endomorphm α weak α-rgd f, aα(a nl(r f and only f a nl(r. One can how that α-compatble rng and hence α-rgd rng are weak α-rgd. Accordng to [4], a rng R called weak α-kew Armendarz f whenever polynomal p = m a x, and q = n j= b jx j R[x; α] atfy pq =, then a α (b j a nlpotent element of R for each, j. Propoton 2.5. [4, Theorem 3.3] Let R be a weak α-rgd rng wth nl(r an deal of R. Then R a weak α-kew Armendarz rng. Theorem 2.6. Let R be a weak α-rgd rng. If R nl-emcommutatve, then R[x, α] a weak McCoy rng.

5 Weak-McCoy Ore extenon 79 Proof. Let F (y = m f y, G(y = n j= g jy j R[x, α][y] \ {}, and uppoe that F (yg(y = where f = p = a x and g j = q j t= b jt x t be n R[x; α]. A n the proof of D.D. Anderon and V. Camllo n [], let k = m degf + n j= degg j, where the degree a polynomal n x and the degree of the zero polynomal taken to be. Then F (x k = m f x k, G(x k = n j= g jx jk R[x; α] and the et of coeffcent of the f (repectvely g j equal the et of coeffcent of F (x k (repectvely G(x k. Snce F (yg(y = and x commute wth element of R, F (x k G(x k =. By Propoton 2.5, we can ee that a α (b jt nl(r for all, j, and t. Thu a and b jt for ome, j, and t, by [4, Propoton 2.3] we have a b jt nl(r. Set r = b jt and u = a uch that both of them are nonzero element of R. We have ug j = u q j t= b jt x t = q j t= ub jt x t where ub jt nl(r; o from Lemma 2.4, ug j nl(r[x; α]. Then R[x; α] left weak McCoy rng. Next we prove that f r nl(r[x; α] for all. We have f r = ( p = a x r = p = a α (rx. Snce a r nl(r, by [4, Propoton 2.3] we have α (a r nl(r. Thu from Lemma 2.4, f r nl(r[x; α] and R[x; α] rght weak McCoy. Corollary 2.7. If R an α-compatble and reverble rng, then R[x; α] a weak McCoy rng. Example 2.8. The convere of Theorem 2.4 not true n general. Let R be a reduced rng and conder the rng of 4 4 upper trangular matrx rng T 4 (R, whch a weak McCoy rng. Let α = d. So T 4 (R α-compatble. Now we how that T 4 (R[x] weak McCoy. Take F (y = m f y, G(y = n j= g jy j T 4 (R[x][y] \ {}, wth f = p = a x and g j = q t= b j t x. Snce nl(t 4 (R = a 2 a 3 a 4 a 23 a 24 a 34 [nl(t 4 (R] 4 =. Note that f r = u = a j R,, j 4, then and that rg j nl(t 4 (R[x; α] and f u nl(t 4 (R[x; α]. We have (f u 4 = and (rg j 4 =. So T 4 (R[x; α] weak McCoy. Notce that T 4 (R not nlemcommutatve. Th becaue

6 8 R. Mohammad, A. Mouav and M. Zahr but =. =, Propoton 2.9. Every weak Armendarz rng a weak McCoy rng. Proof. Let f(x = m a x, g(x = n j= b jx j R[x] \ {}, and uppoe that f(xg(x =. Snce R weak Armendarz, a b j nl(r for all and j. There ext a k, b t R \ {} where k m, t n, becaue f(x and g(x are nonzero. Set r = b t and = a k ; thu a r nl(r and b j nl(r. Therefore R weak McCoy. N l-emcommutatve rng are weak McCoy by the followng reult: Theorem 2.. Let R be a nl-emcommutatve rng. Then R and R[x] are weak McCoy rng. Proof. By [], nl-emcommutatve rng are weak Armendarz. Thu, by Propoton 2.9, nl-emcommutatve rng are weak McCoy. Therefore, by Theorem 2.4, wth α = d, we deduce that R[x] weak McCoy. 3 Weak-McCoy dfferental polynomal rng Thee are the man reult of the paper. Let δ : R R be a dervaton on R, that, δ an addtve map uch that δ(ab = δ(ab + aδ(b, for a, b R. In th ecton we are concerned wth the dfferental polynomal rng R[x; δ]. Accordng to the rule xb = bx + δ(b for any b R, an nductve argument can be made to calculate an expreon for x j a, for any potve nteger j and a R. One can how wth a routne computaton that : x j a = ( j j δ j (ax. Lemma 3.. Let R be a δ-compatble rng and nl(r be an deal of R. Then we have the followng: ( If ab nl(r, then aδ m (b nl(r and δ n (ab nl(r for any po-

7 Weak-McCoy Ore extenon 8 tve nteger m, n. ( If abc nl(r, then abδ (c nl(r and aδ t (bc nl(r where, t are potve nteger. ( If abc nl(r, then aδ t (bδ (c nl(r for any potve nteger, t. Proof. ( Snce ab nl(r, there ext ome potve nteger k uch that (ab k = ab ab ab =. From δ-compatblty we have ab ab δ(ab =, then ab ab δ(ab + ab ab aδ(b =. nce ab ab aδ(b = and recent equalty, ab ab δ(ab =. Now ab ab δ 2 (ab =, then ab ab δ 2 (ab + 2ab ab δ(aδ(b + ab ab aδ 2 (b =. So ab ab δ 2 (ab =. Th proce yeld that ab ab δ n (ab =. Now from ab ab δ(abab = we have ab ab δ(abab =, and ab ab δ 2 (abab = mple that ab ab δ 2 (abab =. Therefore δ n (ab δ n (ab δ n (ab =. Thu δ n (ab nl(r. Smlar dcuon yeld that aδ n (b nl(r. ( It enough to how that abδ(c nl(r and aδ(bc nl(r. By (, we have abc nl(r mple that abδ(c nl(r, and aδ(bc = abδ(c + aδ(bc nl(r. Snce nl(r an deal of R, aδ(bc = aδ(bc abδ(c nl(r. ( It a conequence of ( and (. Whenever R a δ-compatble rng and nl(r an deal of R, from Lemma 3., we deduce that δ k (nl(r nl(r, for any potve nteger k. Lemma 3.2. Let R be a δ-compatble rng and nl(r be an deal of R. Then we have the followng: ( If ab =, then aδ m (b = and δ n (ab = for any potve nteger m, n. ( If abc =, then abδ (c = and aδ t (bc = where, t are potve nteger. ( If abc =, then aδ t (bδ (c = for any potve nteger, t. Proof. The proof mlar to that of Lemma 3.. Lemma 3.3. Let R be a δ- compatble nl-emcommutatve rng. Then nl(r[x; δ] nl(r[x; δ]. Proof. Suppoe that f(x = a +a x+ +a n x n and a m =, =,,, n. Let k = n m +. Then from ( a x 2 = ( a x a + ( a x a x + + ( a x a n x n =

8 82 R. Mohammad, A. Mouav and M. Zahr a (x a + a (x a x+ + a (x a n x n = a j= ( j δ j (a x j + ( ( ( ( a δ j (a x j x + + a δ j (a x j x n = j j j= j= ( ( ( ( ( a δ (a + a δ (a x+ + a δ (a x + = = ( ( ( ( +a n a x n + a δ (a + a δ (a x + + a n a x n x = ( ( ( ( + + a δ (a n + a δ (a n x + + a n a n x n x n. Thu we wll have ( 2 a x = + + = ( a δ (a + ( +t=k = ( = ( a δ (a + ( ( a δ (a t x k + + a n a n x 2n, ( a δ (a x where n and t n. It eay to check that the coeffcent of each term n ( n a x k can be wrtten n the form of ( q a p δ 2 (a 2 δ k (ak, where p (2 p k are potve nteger and a j {a, a,, a n } for j k. Conder a α 2 (a 2 α k (ak. If the number of a n a α 2 (a 2 α k (ak greater than m, then we can wrte a α 2 (a 2 α k (ak a b (α (a j b (α 2 (a j2 b v (α v (a jv b v, where j + j j v > m, l t, l { 2, 3,, k } and b q a product of ome element choong from the et {a, α 2 (a 2,, α k (ak } or equal to. But a j +j 2 + +j v =, nce R nl- emcommutatve and δ-compatble, by Lemma 3.2 we have a j a j 2 a jv = ; j j 2 j v {}}{{}}{{}}{ a a a a a a a a a = ; j j 2 {}}{{}}{ δ (a δ (a δ (a δ 2 (a δ 2 (a δ 2 (a j v {}}{ δ v (a δ v (a δ v (a = (δ (a j (δ 2 (a j2 (δ v (a jv =.

9 Weak-McCoy Ore extenon 83 Snce nl(r nl(r then by nl emcommutatvty of R we get b (δ (a j b (δ 2 (a j2 b v (δ v (a jv b v =. Next f the number of a n a α 2 (a 2 α k (ak greater than m, then mlar dcuon yeld that a α 2 (a 2 α k (ak =. So ( q p a δ 2 (a 2 δ k (ak =. Therefore f(x a nlpotent element of R[x; δ]. Lemma 3.4. Let R be a δ- compatble and nl-emcommutatve rng. Suppoe f(x = a + a + + a m, g(x = b + b + + b n R[x; δ] \ {} are uch that f(xg(x =. Then there ext r, u R \ {} uch that a r nl(r and ub j nl(r for all m, j n. Proof. We note that f(xg(x = ( m ( ( m ( m ( m a x b j x j = a x b + a x b x+ + a x b n x n j= m ( ( m ( ( m ( = a δ (b + a δ (b x+ + a δ (b x = = ( m ( ( m ( + +a n b x m + a δ (b + a δ (b x + + a n b x m = ( m ( ( m ( x+ + a δ (b n + a δ (b n x + + a m b n x m x n. +t=k = = Thu m ( ( m ( m ( = a δ (b + a δ (b + a δ (b x = ( ( m ( + + a δ (b t x k + + a m b n x m+n, where m and t n. Then we have the followng equaton: a m b n = ; a m b n + a m b n + ( m am δ(b m n = ; a m 2 b n + ( m am δ(b m 2 n + ( m am δ 2 (b m 2 n + a m b n + ( m am δ(b m n + a m b n 2 = ;.

10 84 R. Mohammad, A. Mouav and M. Zahr +t=k ( m ( = a δ (b t =. From equaton (, we get a m b n nl(r and o b n a m nl(r. Now we how that a b n nl(r for all m. Multplyng equaton (2 on the left by b n yeld ( m b n a m b n = b n a m b n b n a m δ(b n nl(r m nce nl(r an deal of R. Then a m b n nl(r. Now multplyng equaton (3 on the left by b n yeld ( ( m m b n a m 2 b n = b n a m δ(b n b n a m δ 2 (b n b n a m b n m 2 m 2 ( m b n a m δ(b n b n a m b n 2 nl(r. m So a m 2 b n. Contnung th proce yeld that a b n nl(r for m. By Lemma 3., we have a δ (b n nl(r for m and potve nteger. Thu we obtan (a + a + + a m x m (b + b + + b n x n = ( ( m ( a δ (b t x k nl(r[x; δ], m+n k= +t=k = where m and t n. Therefore m+n = a m b n nl(r ( ( m m+n 2 = a m b n 2 + a m b n + a m δ(b n nl(r (2 m. k = m +t=k = ( a δ (b t nl(r. (3 From equaton (, we get b n a m nl(r. Multplyng equaton (2 on the left by b n yeld ( m b n a m b n = b n m+n 2 b n 2 b n a m δ(b n nl(r. m Thu a m b n nl(r and o b n a m nl(r. Contnung th proce yeld that a b n nl(r for m. Then ung nducton on n, we obtan a b j nl(r for m and j n. Snce f(x, g(x R[x; δ]\{},

11 Weak-McCoy Ore extenon 85 wthout lo of generalty, we can aume that a m, b n. Take r = b n and u = a n, then a r, ub j nl(r, for all m and j n. Theorem 3.5. Let R be a δ-compatble rng. If R nl-emcommutatve, then R[x, δ] a weak McCoy rng. Proof. Let F (y = m f y, G(y = n j= g jy j R[x, δ][y] \ {}, and uppoe that F (yg(y = where f = p = a x and g j = q j t= b jt x t be n R[x; δ]. A n the proof of D.D. Anderon and V. Camllo n [], let k = m degf + n j= degg j, where the degree a polynomal n x and the degree of the zero polynomal taken to be. Then F (x k = m f x k, G(x k = n j= g jx jk R[x; δ] and the et of coeffcent of the f (repectvely g j equal the et of coeffcent of F (x k (repectvely G(x k. Snce F (yg(y = and x commute wth element of R, F (x k G(x k =. By Lemma 3.4, there are r, u R, uch that a r, ub jt nl(r for all, j, and t, a. We have ug j = u q j t= b jt x t = q j t= ub jt x t. Snce R δ-compatble, ub jt nl(r; o from Lemma 3.3, ug j nl(r[x; δ]. Then R[x; δ] left weak McCoy rng. Next we prove that f r nl(r[x; δ] for all. We have f r = ( p p ( a x r = = + ( p =t = ( t ( p a δ (r + = ( a δ t (r x t + + a p rx p. a δ (r x + Snce a r nl(r, by Lemma 3., a δ v (r nl(r for all potve nteger v. Thu ( p =t a t δ t (r nl(r for t =,,..., p. Then by Lemma 3.3, f r nl(r[x; δ]. Therefore R[x; δ] weak McCoy. ACKNOWLEDGEMENTS. The author would lke to thank the referee for valuable comment and uggeton. Reference [] D.D. Anderon and V. Camllo, Armendarz rng and Gauan rng, Comm. Algebra, 26(7 ( [2] G.F. Brkenmeer, H.E. Heatherly, E.K. Lee, Completely prme deal and aocated radcal, n: S.K. Jan, S.T. Rzv (Ed., Rng Theory,

12 86 R. Mohammad, A. Mouav and M. Zahr Granvlle, OH, 992, World ScentGc, Sngapore and Rver Edge, (993, pp MR 96e:625. [3] Sh. Ghalandarzadeh, M. Khoramdel, On weak McCoy rng, Tha Journal of Mathematc, Vol.6( [4] E. Hahem and A. Mouav, Polynomal extenon of qua-baer rng, Acta. Math. Hungar, 5( [5] J. Krempa, Some example of reduced rng, Algebra Colloq. 3(996, [6] N. Km, Y. Lee, Extenon of reverble rng, J. Pure Appl. Algebra, 85(23, [7] Z. Le, J. Chen and Z. Yng, A queton on McCoy rng, Bull Autral. Math. Soc, 76(27, [8] Z. Lu and R.Y. Zhao, On weak Armendarz rng, Comm. Algebra, 34 ( [9] N.H. McCoy, Remark on dvor of zero, Amer. Math. Monthly 49 ( [] R. Mohammad, A. Mouav and M.Zahr On nl-semcommutatve rng,, Preprnt. [] P.P. Nelen, Sem-commutatvty and the McCoy condton, J.Algebra 298 ( [2] A.R. Nar-Ifahan and A. Mouav, On Ore extenon of qua-baer rng, J. of Algebra and It Applcaton, Vol. 7, No. 2 (28-4. [3] A.R. Nar-Ifahan and A. Mouav, Ore extenon of kew Armendarz rng, Comm. n Algebra, Vol.36 (28-5. [4] L. Ouyang, Extenon of generalzed α-rgd rng, Internatonal Electronc Journal of Algebra, Vol.3, ( [5] L. Ouyang and Huanyn-Chen, Extenon of weak McCoy rng, Preprnt. [6] Z. Yng, J. Chen and Z. Le, Extenon of McCoy rng, Northeat. Math. J. 24 (28, no., Receved: Aprl, 2