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A domain finite over a polynomial ring has dimension at least the number of variables

Statement

Assume the Axiom of Choice.

Let $k$ be a field, let $d \geq 0$ , and let $A$ be an integral domain that is module-finite over the polynomial ring $k [z_{1}, \dots, z_{d}]$ via an injective $k$ -algebra map $k [z_{1}, \dots, z_{d}] ↪ A .$ Then $\dim A \geq d .$

Facts & Assumptions

Given: The Axiom of Choice, a field $k$ , an integer $d \geq 0$ , an integral domain $A$ , and an injective $k$ -algebra map $k [z_{1}, \dots, z_{d}] ↪ A$ making $A$ module-finite over $k [z_{1}, \dots, z_{d}]$ .

[L1]

Krull dimension is the supremum of the lengths of chains of prime ideals (Krull dimension of a nonzero ring).

[L2]

Integral extensions lift finite prime chains from the base (Integral extensions lift finite prime chains from the base).

Proof

technique · direct

1.1givenalgebra

Because $A$ is finitely generated as a module over $R : = k [z_{1}, \dots, z_{d}]$ , multiplication by any $a \in A$ is an $R$ -linear endomorphism of a finite $R$ -module. Cayley-Hamilton therefore gives a monic polynomial over $R$ satisfied by $a$ , so $A$ is integral over $R$ .

1.2givenalgebra

The polynomial ring $R$ has the prime chain $(0) \subset (z_{1}) \subset (z_{1}, z_{2}) \subset \dots \subset (z_{1}, \dots, z_{d}),$ of length $d$ : each quotient by $(z_{1}, \dots, z_{i})$ is again a polynomial ring over $k$ and hence an integral domain, so those ideals are prime.

2.1L1L2step 1.1step 1.2∎

Apply [L2] to the chain from step 1.2 and the integral inclusion $R \subseteq A$ from step 1.1. This gives a chain of $d + 1$ prime ideals in $A$ , so by [L1] the dimension of $A$ is at least $d$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Lemma (15.9) (standard reference, not scraped)