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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.

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.

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