LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-24 (gpt-6-sol)

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Quotienting by the first minimal prime reduces the remaining height count

Statement

Assume the Axiom of Choice (The Axiom of Choice). In the situation of Choose the first generator's minimal prime inside the target prime, with a chain of length $d \geq 1$ , choose $b$ so that $p$ is minimal over $(b, x_{2}, \dots, x_{n})$ , and let $p = p_{d} ⊋ p_{d - 1} ⊋ \dots ⊋ p_{0}$ be a strict prime chain with $b \in p_{1}$ . Then in $R / (b)$ the prime $p / (b)$ is minimal over the ideal generated by the images of $x_{2}, \dots, x_{n}$ , and the quotient chain $p_{d} / (b) ⊋ p_{d - 1} / (b) ⊋ \dots ⊋ p_{1} / (b)$ is strict of length $d - 1$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , an ideal $(x_{1}, \dots, x_{n})$ with $n \geq 2$ , a prime $p$ minimal over it, an element $b$ as in Choose the first generator's minimal prime inside the target prime, and a strict chain $p = p_{d} ⊋ p_{d - 1} ⊋ \dots ⊋ p_{0}$ of length $d \geq 1$ with $b \in p_{1}$ .

[L1]

Under the assumed Axiom of Choice, the chosen element $b$ makes $p$ minimal over $(b, x_{2}, \dots, x_{n})$ (Choose the first generator's minimal prime inside the target prime).

[L2]

Prime ideals of a quotient correspond to prime ideals upstairs containing the quotient ideal, with strict inclusions preserved (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

Proof

technique · direct

1.1L1given

By [L1], every prime ideal of $R$ containing $(b, x_{2}, \dots, x_{n})$ and lying inside $p$ is equal to $p$ . Hence $p / (b)$ is minimal over the images of $x_{2}, \dots, x_{n}$ in $R / (b)$ .

1.2L2given

Because $b \in p_{i}$ for every $1 \leq i \leq d$ , [L2] sends the given chain to $p_{d} / (b) ⊋ p_{d - 1} / (b) ⊋ \dots ⊋ p_{1} / (b) .$ The inclusions stay strict, because equality between consecutive quotient primes would contract back to equality between the corresponding primes of $R$ .

2.1step 1.1step 1.2∎

Thus quotienting by $(b)$ leaves a prime minimal over $n - 1$ generators and reduces this chosen chain by exactly one step.

Depends on

Used by

Krull's height theorem Theorem

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, §21 (standard reference, not scraped)
The Stacks Project, Section 10.60: Dimension (standard reference, not scraped)