# Homomorphism of Commutative Rings, Prime Ideals, Maximal Ideals

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.

https://brainmass.com/math/ring-theory/homomorphism-of-commutative-rings-prime-ideals-maximal-ideals-101848

#### Solution Preview

Proof:

a) Let Q = phi^-1(P).

For any xy in Q, since phi is a homomorphism, then we have

phi(xy)=phi(x)phi(y) is in P.

Since P is a prime ideal of S, then phi(x) is in P or phi(y) is in P.

If phi(x) is in P, then x is in phi^-1(P)=Q; if phi(y) is in P, then y is in phi^-1(P)=Q

Thus from xy in Q, we can induce that x is in Q or y is in Q.

So if Q=phi^-1(P) is not equal to R, then Q must be a prime ideal ...

#### Solution Summary

Homomorphism of Commutative Rings, Prime Ideals, Maximal Ideals are investigated.