LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Quotienting by the first minimal prime reduces the remaining height count

Statement

In the situation of Choose the first generator's minimal prime inside the target prime, 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: 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 $p = p_{d} ⊋ p_{d - 1} ⊋ \dots ⊋ p_{0}$ with $b \in p_{1}$ .

[L1]

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

1.1

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 $R / (b)$ .

L1given

1.2

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 $R$ .

L2given

2.1

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

step 1.1step 1.2∎

Depends on

Used by

Krull's height theorem Theorem

Dependency tree · two levels

10 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)