Let S be an integral extension of R. Suppose P1 ⊆ · · · ⊆ Pn is a sequence of prime ideals of R and suppose that there exists a sequence of prime ideals Q1 ⊆ · · · ⊆ Qm of S such that Qi ∩ R = Pi for i = 1, . . . , m. Here, 1 ≤ m

Question

Let [latex]S[/latex] be an integral extension of [latex]R[/latex]. Suppose [latex]P_{1} \subseteq \cdots \subseteq P_{n}[/latex] is a sequence of prime ideals of [latex]R[/latex] and suppose that there exists a sequence of prime ideals [latex]Q_{1} \subseteq \cdots \subseteq Q_{m}[/latex] of [latex]S[/latex] such that [latex]Q_{i} \cap R=P_{i}[/latex] for [latex]i=1, \ldots, m[/latex]. Here, [latex]1 \leq m

Accepted Answer

This is a simple application of the Problem 220: We look at the integral extension [latex]S / Q_{m}[/latex] of [latex]R / P_{m}[/latex]. Since the image of [latex]P_{m+1}[/latex] in [latex]R / P_{m}[/latex] is prime, there exists a prime ideal [latex]Q_{m+1}[/latex] in [latex]S[/latex], whose image in [latex]S / Q_{m}[/latex] lies over that of [latex]P_{m+1}[/latex]. It is clear that [latex]Q_{m+1}[/latex] contains [latex]Q_{m}[/latex].

Question 1.222: Let S be an integral extension of R. Suppose P1 ⊆ · · · ⊆ Pn......

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