# Answer to Question #23463 in Abstract Algebra for jeremy

Question #23463

Let R be a subring of a right noetherian ring Q with a set S ⊆ R ∩ U(Q) such that every element q ∈ Q has the form rs^−1 for some r ∈ R and s ∈ S. Show that:the converse of " if Q is prime (resp. semiprime), then so is R " is true even without assuming Q to be right noetherian.

Expert's answer

Assume

*R*is prime, and (*as^−*1)*Q*(*bt^−*1)= 0 where*a*,*b**∈**R*and*s*,*t**∈**S*. Then 0 = (*as^−*1)(*sR*)*b*=*aRb*,so*a*= 0 or*b*= 0, thus ideal 0 is prime. The semiprime case issimilar, by setting*a*=*b*.
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