Prove that if m is a maximal ideal of s and phi is


Let phi:R->S be a homomorphism of commutative rings

a) Prove that if P is a prime ideal of S then either phi^-1(P)=R or phi^-1(P) is a prime ideal of R. Apply this to the special case when R is a subring of S and phi is the inclusion homomorphism to deduce that if P is a prime ideal of S then PR is either R or prime ideal in R

b) Prove that if M is a maximal ideal of S and phi is surjective then phi^-1(M) is maximal ideal of R. Give an example to show that this need not be the case if phi is not surjective.

Solution Preview :

Prepared by a verified Expert
Algebra: Prove that if m is a maximal ideal of s and phi is
Reference No:- TGS01373716

Now Priced at $20 (50% Discount)

Recommended (94%)

Rated (4.6/5)