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Dołączył: 17 Gru 2010
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| Temat postu: 1 of _81 with Rings |
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1 with the exchange of rings
Positive integer households, there is a [, d] a '[, d By the Lemma 2, b, d] = 0, Yan aCC (R). To NCC (R). Under license D (R) volume c (R). Under license D (R) c (R). In (1) to 1 + on behalf of: (1 +)[,]=( 1 +)'[,()]( 1 + there (1 +),]=( 1 +). R A ■, ()] a (1 +).=( 1 +) 'I,] (1 + (2) take = (;. b 1) a (0 ¨ 0】 I1 a JlJ generation (2), because, m, q is not divisible by 3 at the same time, it does not satisfy (2). by the Lemma 1, D (R) R, Nil ideals. and Ⅳ C), column D (R) C (R). by (1) and Lemma 3 was :'[,]=¨ k,] = 2 h ,()]=[,)] workers stare - [,] eleven ([,]) a '(b ,'()]) =. r'+r'+'[, from the bow of a ,()]= 0 I Li 32.r '[, a plant'()]= 0 2 ,-()] by the Lemma 2 a 2, ()] 10 (3) where F ()=-() in (1) to F () on behalf of the y have a [, F ()] a. r '[, G ()] (1) where () a (F () ) for any positive integer, there -,]=+'[,()] generation z 3 from the bow reason I have 4 -~--! Ix,] = sx 'a I [, plant' ()] from 3 have 3x ¨,] = 3x3 (m4 2 [a ()] 0 dumb broadcast from Seoul Yan Normal School # 2 years of l999,] = 2 ,,'()] pan 2 [, G ()] = 2 [, (y)] = 0 by (5) to know 3. [, F ()] +2 x, F ()]=。',[link widoczny dla zalogowanych]。' H, G ,()]+ 2x.G ()] from (3) and (6) was, F ,()]=。, G, ()] from (4) know X3 (1) know that pour, F, ()] a 3k (,()])=(-+, H ()])== 0 by Lemma 2, F (y) is a H ,()]= 0 () a H, () ∈ c (R), ie Y a (()+,()) ∈ c (R) while () + H () ∈ z (), by [3] known as the commutative ring R. same reason. Theorem 2 l R is a ring, if for any yER, a (f) a fZ ( ,) and not divisible by 3 while the non-negative integer, m.g meet: one by one'()]= 0, V.27 ∈ R then R is commutative ring. reference text indigo 1Kez] an, TP.Anoteonec, mmutativltyofsemiprimePI-ri Aln ~ r.J.Math.1953, 75:864 ~ 87lTHEC0MMUTATIVITY0FRINGSWITHlTianShurong (NavalAermmutlea [Engineeringlnstltute.Y ~ mi) Thispaperprovedacommutativltytheoremofassociativeringswith1whichsatisfiedani-dentity.Keywords: Associativeringswith1; Center ~ Commutativity
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