Abstract
In a recent paper, Kaplansky [2] proved the following theorem: Let R be a
ring with centre Z, and such that xn(x) ∈ Z for every x∈ R. If R,
in addition, is semi-simple then it is also commutativeThe existence of non-commutative rings in which every element is nilpotent
rules out the possibility of extending this result to all rings. One might
hope, however, that if R is such that xn(x) ∈ Z for all x ∈ R and
the nilpotent elements of R are reasonably “well-behaved,” then Kaplansky's
theorem should be true without the restriction of semi-simplicity.
Publisher
Canadian Mathematical Society
Cited by
20 articles.
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