Let R ⊂ 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 ∩ R.

Question

Let [latex]R \subset S[/latex] be a pair of rings such that [latex]S[/latex] is integral over [latex]R[/latex]. If [latex]P[/latex] is a prime ideal in [latex]R[/latex], then prove that there exists a prime ideal [latex]Q[/latex] in [latex]S[/latex] such that [latex]P=Q \cap R[/latex].

Accepted Answer

The idea is to analyze the situation locally. Let [latex]D[/latex] denote the complement of [latex]P[/latex] in [latex]R[/latex]. Then [latex]D[/latex] is a multiplicative monoid. Notice that [latex]D[/latex] is a multiplicative monoid in [latex]S[/latex] as well.

By Problem 217 we know that [latex]D^{-1} S[/latex] is integral over [latex]D^{-1} R[/latex] and also if [latex]M[/latex] is a prime ideal in [latex]D^{-1} S[/latex] then [latex]D^{-1} S / M[/latex] is integral over [latex]D^{-1} R / D^{-1} R \cap M[/latex]. Now, suppose [latex]M[/latex] is a maximal ideal in [latex]D^{-1} S[/latex]. (The existence of [latex]M[/latex] is guaranteed by the Zorn's lemma.) Since [latex]D^{-1} S / M[/latex] is a field, by Problem 213 we see that [latex]D^{-1} R / M \cap D^{-1} R[/latex] is a field, also. In other words, the ideal [latex]M_{0}:=M \cap D^{-1} R[/latex] is maximal in [latex]D^{-1} R[/latex]. Since [latex]D^{-1} R[/latex] has a unique maximal ideal (generated by [latex]P[/latex] ) we have [latex]M_{0} \cap R=P[/latex]. See Problem 219 .

Similarly, [latex]M \cap S[/latex] is a prime ideal in [latex]S[/latex] and furthermore [latex]M[/latex] does not intersect [latex]D[/latex]. Therefore, its restriction to [latex]R[/latex], namely [latex]M \cap R[/latex] does not intersect [latex]D[/latex] neither. Thus it is contained in [latex]P=R-D[/latex]. Combining this with the fact that [latex]M_{0} \cap R=P[/latex], we see that [latex]P=M[/latex].

Question 1.220: Let R ⊂ S be a pair of rings such that S is integral over R.......

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