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A maximal ideal of an affine algebra has finite residue field over the base field

Statement

Let $k$ be a field, let $A$ be a finite-type $k$ -algebra, and let $m$ be a maximal ideal of $A$ . Then the residue field $A / m$ is a finite extension of $k$ .

Facts & Assumptions

Given: A field $k$ , a finite-type $k$ -algebra $A$ , and a maximal ideal $m \subset A$ .

[L1]

Finite type means generated by finitely many algebra elements (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

A quotient by a maximal ideal is a field ( $R / M$ is a field if and only if $M$ is a maximal ideal).

[L3]

A field finitely generated as a $k$ -algebra is finite over $k$ (A field finitely generated as a k-algebra is a finite extension of k).

Proof

technique · direct

1.1L1givenchoose

By [L1], choose generators $a_{1}, \dots, a_{n}$ of $A$ as a $k$ -algebra. Their images in $A / m$ generate the quotient as a $k$ -algebra, so $A / m$ is again a finite-type $k$ -algebra.

2.1L2L3step 1.1∎

By [L2], the quotient $A / m$ is a field. Applying [L3] to this finite-type field over $k$ shows that $A / m$ is finite over $k$ .

Depends on

Dependency tree · 2 levels

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Corollary 13.2