【摘要】 设R为结合环。文献[3]证明了:设R是具有正则元的半质环,如果R满足条件:对于任意的x,y∈R,都存在一个与x,y有关的整数n=n(x,y)≥1,使得(xy)n+k=xn+kyn+k,k=0,1,2,则R为交换环。给出上述结果的一个简短证明,并将其推广,证明了定理:设R是具有正则元的半质环,如果R满足条件:对于任意的x,y∈R,都存在一个与x,y有关的整数n=n(x,y)≥1,使得(xy)n+k=yn+kxn+k,k=0,1,2,则R为交换环。更多还原

【Abstract】 Let R be an associative ring. The paper [1] gives the following result: Let R be a semi-prime ring with normal element, if R satisfies the condition: For Vx, y∈R, there is an integer n = n(x, y)≥1, related to x, y such that (xy)n+k = xn+kyn+k, k = 0, 1, 2, then R is commutative. An easier proof is presented of the above results. The results of the paper [1] are extened, the theorem is proved: Let R be a semi -prime ring with normal elements, if R satisfies the condition: For Vx, y∈R, there is an integer n = n(x, y)≥1, related to x, y, such that (xy)n+k = yn+kxn+k, k = 0, 1, 2, then R is commutative.更多还原