Abstract
LetR be a ring with a subringA such that a power of every element ofR lies inA. The following results are proved: IfR has no nonzero nil right ideals, neither doesA; if moreoverR is prime,A is also prime. IfR is semiprime Goldie, so isA. IfA has no nonzero nilpotent elements, then the nilpotent elements ofR form an ideal. Finally ifR has no nil right ideals andA is Goldie, thenR is Goldie.
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