Question 1.220: Let R ⊂ S be a pair of rings such that S is integral over R.......
Let R \subset S be a pair of rings such that S is integral over R. If P is a prime ideal in R, then prove that there exists a prime ideal Q in S such that P=Q \cap R.
The idea is to analyze the situation locally. Let D denote the complement of P in R. Then D is a multiplicative monoid. Notice that D is a multiplicative monoid in S as well.
By Problem 217 we know that D^{-1} S is integral over D^{-1} R and also if M is a prime ideal in D^{-1} S then D^{-1} S / M is integral over D^{-1} R / D^{-1} R \cap M. Now, suppose M is a maximal ideal in D^{-1} S. (The existence of M is guaranteed by the Zorn’s lemma.) Since D^{-1} S / M is a field, by Problem 213 we see that D^{-1} R / M \cap D^{-1} R is a field, also. In other words, the ideal M_{0}:=M \cap D^{-1} R is maximal in D^{-1} R. Since D^{-1} R has a unique maximal ideal (generated by P ) we have M_{0} \cap R=P. See Problem 219 .
Similarly, M \cap S is a prime ideal in S and furthermore M does not intersect D. Therefore, its restriction to R, namely M \cap R does not intersect D neither. Thus it is contained in P=R-D. Combining this with the fact that M_{0} \cap R=P, we see that P=M.