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A semi- prime ring commutativity theorem

Nilpotent element . Then by Proposition 6. For any n. b ∈ R, are n Therefore R is a commutative ring 【 . Similarly . Can prove Theorem 2 Let R be a half grommet , if R satisfies ( ) : for any n, 6 ∈ R, there exists a word > K and contains () ( or ( ) ) the word Wx (.) and a Be Wx (,) divides the Integer Polynomial Pi (.) makes aXb-gx (a, 6) ∈ z (R). Where K is a given positive integer , Wx (-r,) and gx (, y) can be with X = (n, 6) and change. So R must be for the exchange of rings. Corollary If the ring R satisfies the condition (K) ( or ( )). Then (1) R 's commutator ideal is locally nilpotent ; (2) R for all nilpotent R, Boer made under the nil radical . Prove that B (R) is Baer ring R, under the nil radical . The R / B (R) is a satisfying (K) ( or ( wind )) semi- grommet . Therefore, by Theorem 1 (or Theorem 2), R / B (R) is a commutative ring. So for any n. b ∈ R, both n6 a ∈ B (R). To the commutator ideal of R c (R) B (R) as B (R) is locally nilpotent , so c (R) is no different. And because the half grommet free exchange of non-zero nilpotent element . Therefore, all nilpotent R in health i (R) in the . F 24 Fujian Normal University Natural Science) in June 1995 as B (R) is a nil ring R ideal, and therefore all nilpotent R made B (R). Note the general ring R , the conditions ( ) ( or ( ) ) is a commutative ring R is not a sufficient condition . Indeed . f000] Let F be a field, so that R = {a00IIn, b, ∈ Fl to F, full matrix ring of order 3 FJ of : f- ring, then on the 【 60J any ∈ R, either can be ((Or (yx)) divisible Integer Polynomial f (x,) and a positive integer K, are . but not a commutative ring R .