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.

Quotienting by the first minimal prime reduces the remaining height count

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources