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

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Krull's height theorem

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $R$ be a Noetherian commutative ring, let $I = (x_{1}, \dots, x_{n})$ be an ideal generated by $n \geq 1$ elements, and let $p$ be a prime ideal minimal over $I$ . Then $ht (p) \leq n$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian commutative ring $R$ , an $n$ -generated ideal $I = (x_{1}, \dots, x_{n})$ with $n \geq 1$ , and a prime ideal $p$ minimal over $I$ .

[L1]

The case $n = 1$ is Krull's principal ideal theorem (Krull's principal ideal theorem).

[L2]

Under the assumed Axiom of Choice, for $n \geq 2$ and a positive-length chain, one may replace the first generator by an element $b$ from $p_{1}$ of a same-length chain so that $p$ is minimal over $(b, x_{2}, \dots, x_{n})$ (Choose the first generator's minimal prime inside the target prime).

[L3]

Under the same Choice assumption, after quotienting by that chosen $b$ , the image $p / (b)$ is minimal over $n - 1$ generators and a chain ending at $p$ loses one step (Quotienting by the first minimal prime reduces the remaining height count).

Proof

technique · induction on the number of generators

1.1L1basegiven

If $n = 1$ , [L1] gives $ht (p) \leq 1$ .

1.2L2ihgiven

Assume $n \geq 2$ and that the theorem is known for $(n - 1)$ -generated ideals. Suppose for contradiction that $ht (p) \geq n + 1$ . Then there exists a strict prime chain $p = p_{d} ⊋ p_{d - 1} ⊋ \dots ⊋ p_{0}$ with $d \geq n + 1 \geq 3$ . Under the assumed AC, [L2] replaces $x_{1}$ by a suitable element $b \in p_{1}$ of a same-length chain for which $p$ is minimal over $(b, x_{2}, \dots, x_{n})$ .

2.1L3step 1.2ih

By [L3], the quotient prime $p / (b)$ is minimal over the $(n - 1)$ -generated ideal generated by the images of $x_{2}, \dots, x_{n}$ , and the above chain descends to a strict chain of length $d - 1 \geq n$ ending at $p / (b)$ in $R / (b)$ . This contradicts the induction hypothesis for $(n - 1)$ generators.

3.1step 1.1step 2.1discharge-induction∎

Therefore the assumption $ht (p) \geq n + 1$ is impossible, and $ht (p) \leq n$ .

Krull's height theorem

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources