Question 1.221: Suppose P is a maximal ideal of R, and S is ring that contai......
Suppose P is a maximal ideal of R, and S is ring that contains R. If there exists a finite number of elements s_{1}, \ldots, s_{n} \in S that generate S as a ring over R (that is to say, the coefficients of the polynomial expressions in s_{1}, \ldots, s_{n} are all from R ), then prove that there are only finitely many maximal ideals Q in S such that Q \cap R=P.
By Problem 220 we know that S has a prime ideal Q such that Q \cap R=P. We claim that there exists a maximal ideal M in S such that M \cap R=P. If Q is maximal, then there is nothing to do, so we assume that Q is contained in a maximal ideal M. Clearly M \cap R contains P. If this is a proper containment, since P is maximal, then M \cap R has to contain a unit, which is absurd. Therefore, M \cap R=P.
Next, we claim that there only finitely many maximal ideals M of S such that M \cap R= P. Notice that, since M \cap R=P, the canonical injection R \hookrightarrow S induces an injection R / P \hookrightarrow S / M. Therefore, S / M is a field extension of R / P. We analyze the generators of S / M as a field over R / P. It is clear that the images of s_{i} ‘s generate S / M as field with coefficients from R / P. Since s_{i} ‘s are integral over R, we have polynomials f_{i}(x) \in R[x] such that f_{i}\left(s_{i}\right)=0. Reducing the coefficients of f_{i} ‘s modulo P, we see that the minimal polynomial over R / P of s_{i} ‘s are divisors of f_{i}(x) \bmod P. We conclude that the field S / M is obtained from R / P by adjoining some roots of f_{i}(x) \bmod P. Since f_{i}(x) has finitely many roots, there are finitely many different field extensions of the form S / M over R / P. This proves that there are only finitely many different maximal ideals of S that restricts to the same maximal ideal P in R.