Answer to Question #23463 in Abstract Algebra for jeremy

Question

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.

Answer 1

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.