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Noether normalisation yields module finiteness over a polynomial subring

Statement

Let $k$ be a field and let $A$ be a nonzero finite-type $k$ -algebra. Then there exist algebraically independent elements $z_{1}, \dots, z_{d} \in A$ such that $A$ is a module-finite algebra over the polynomial ring $k [z_{1}, \dots, z_{d}]$ .

Facts & Assumptions

Given: A field $k$ and a nonzero finite-type $k$ -algebra $A$ .

[L1]

Noether normalisation provides algebraically independent $z_{1}, \dots, z_{d} \in A$ such that $A$ is integral over $k [z_{1}, \dots, z_{d}]$ (Induction produces a polynomial subalgebra over which the affine algebra is integral).

[L2]

A subalgebra generated by finitely many integral elements over a base ring is module-finite over that base ring (A subalgebra generated by finitely many integral elements is module-finite).

Proof

technique · direct

1.1

By [L1], choose algebraically independent elements $z_{1}, \dots, z_{d} \in A$ such that $A$ is integral over $R : = k [z_{1}, \dots, z_{d}]$ .

L1choose

2.1

Because $A$ is of finite type over $k$ , choose generators $a_{1}, \dots, a_{m} \in A$ with $A = k [a_{1}, \dots, a_{m}]$ . Since $R$ contains the image of $k$ and lies in $A$ , we also have $A = R [a_{1}, \dots, a_{m}]$ . Each $a_{i}$ is integral over $R$ by step 1.1, so [L2] implies that $A$ is module-finite over $R$ .

L2step 1.1given

3.1

Hence $A$ is module-finite over the polynomial ring $k [z_{1}, \dots, z_{d}]$ .

step 2.1∎

Depends on

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Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Lemma (15.1) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Theorem 8.1 (standard reference, not scraped)