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Finite type algebras over fields have bounded dimension

PDF p. 4: Finite valuative dimension

Statement

Assume the Axiom of Choice. If a commutative $k$ -algebra $A$ is generated by $n$ elements over a field $k$ , then $A$ is noetherian and $\dim A \leq n$ . An algebra is generated by those elements when the evaluation map from the polynomial ring in $n$ variables is surjective.

Facts & Assumptions

Given: The Axiom of Choice, a field $k$ , and a surjection $k [t_{1}, \dots, t_{n}] \to A$ .

Goal: Prove noetherianity and the dimension bound.

[F1]

A finite type algebra over a noetherian ring is noetherian. Every algebra of finite type over a Noetherian ring is a Noetherian ring

[F2]

Under Choice, adjoining one polynomial variable to a finite-dimensional noetherian ring increases its Krull dimension by one. A Noetherian polynomial ring has dimension one larger The Axiom of Choice

[F3]

Dimension does not increase under quotients. Dimension decreases under quotients

Proof

1.1F1F2givenalgebra

A field is noetherian and has dimension zero: its only ideals are zero and the whole field, and its only prime is zero. Apply [F1] successively to its finite-variable polynomial rings. Induction using [F2] gives $\dim k [t_{1}, \dots, t_{n}] = n$ ; at each step the ring is noetherian and finite-dimensional, as required by [F2].

2.1F1F3givenstep 1.1∎

The surjection identifies $A$ with a quotient of that polynomial ring. By [F3] its dimension is at most $n$ , and [F1] gives its noetherianity because it is a finite type $k$ -algebra.

Depends on

Used by · 1

Finite valuative dimension for noetherian and finite type schemes

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