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

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 of R.
Now we consider the special case that R is a subring of S and phi is an inclusion.
Then if P is a prime ideal of S, we want to show that PR is either R or a prime ideal of R.
We note that R is a commutative ring, then each element in PR has the form pr, where p
is in P and r is in R.
From the above proof, since phi is an inclusion, then phi^-1(P) = P. Then P=R or P is a prime
ideal of R. On the other hand, P is an ideal, then for any r in R, Pr=P. So PR=P.
Therefore, PR is either R or a prime ideal of R.
b) Let N=phi^-1(M). We want to show that N is the maximal ideal of R. If not, we can find
an ideal N' in R, such that N is properly contained in N' and N' is properly contained in R.
It is easy to verify that phi(N') is an ideal in S and phi(N') contains M. Since M is a maximal
ideal, then phi(N')=M or phi(N')=S.
If phi(N') = M, then N'=phi^-1(M)=N, this is a contradiction.
If phi(N') = S, since phi is surjective, then N'=phi^-1(S)=R, this is also a contradiction.
Therefore, N must be a maximal ideal of R.
If phi is not surjective, the conclusion may not hold. Here is an example.
S=Z is the integer ring. R=2Z. phi(x)=2x. P=5Z is a maximal ideal in S, but phi^-1(P)=20Z,
which is not a maximal ideal in R=2Z.