TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-12

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Affine algebraic sets and reduced affine k-algebras at the object level

Statement

Let $k$ be an algebraically closed field.

If $X \subseteq A_{k}^{n}$ is an affine algebraic set, then its coordinate ring $k [X]$ is a reduced affine $k$ -algebra.
If $A$ is a reduced affine $k$ -algebra, then there exist $n \geq 0$ and an affine algebraic set $X \subseteq A_{k}^{n}$ such that $A ≅ k [X]$ as $k$ -algebras.

Thus affine algebraic sets correspond to reduced affine $k$ -algebras at the object level, with the morphism half deferred to the next page.

Facts & Assumptions

Given: An algebraically closed field $k$ .

[L1]

For an affine algebraic set $X \subseteq A_{k}^{n}$ , $k [X] = k [x_{1}, \dots, x_{n}] / I (X)$ (The coordinate ring of an affine algebraic set).

[L2]

A reduced affine $k$ -algebra is a reduced commutative $k$ -algebra of finite type over $k$ (A reduced affine k-algebra).

[L3]

A commutative $k$ -algebra is of finite type exactly when it is isomorphic to a quotient $k [x_{1}, \dots, x_{n}] / I$ for some $n$ and some ideal $I$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L4]

Over an algebraically closed field, algebraic sets correspond exactly to radical ideals (Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals).

Proof

technique · direct

1.1L1L3given

Let $X \subseteq A_{k}^{n}$ be an affine algebraic set. By [L1], the ring $k [X]$ is a quotient of the polynomial ring $k [x_{1}, \dots, x_{n}]$ , so [L3] shows that $k [X]$ is of finite type over $k$ .

1.2L2L3givenalgebra

Let $A$ be a reduced affine $k$ -algebra. By [L2] and [L3], there exist $n \geq 0$ and an ideal $I \subseteq k [x_{1}, \dots, x_{n}]$ with $A ≅ k [x_{1}, \dots, x_{n}] / I$ . Because $A$ is reduced, if $f^{m} \in I$ then the class of $f$ in $A$ is nilpotent, hence zero; therefore $f \in I$ . So $I$ is a radical ideal.

2.1L2L4step 1.1algebra

To see that $k [X]$ is reduced, let $\overline{f} \in k [X]$ satisfy ${\overline{f}}^{m} = 0$ for some $m \geq 1$ . Choosing a representative $f$ in the polynomial ring, this means $f^{m} \in I (X)$ . Since $I (X)$ is radical by [L4], we have $f \in I (X)$ , so $\overline{f} = 0$ . Thus $k [X]$ is reduced. Together with step 1.1 and [L2], this proves part (1).

2.2L1L4step 1.2

By [L4], the radical ideal $I$ is the vanishing ideal of the algebraic set $X : = V (I) \subseteq A_{k}^{n}$ . Then [L1] gives $k [X] = k [x_{1}, \dots, x_{n}] / I (X) = k [x_{1}, \dots, x_{n}] / I$ , so $A ≅ k [X]$ .

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 establish the stated object-level correspondence between affine algebraic sets and reduced affine $k$ -algebras.

Affine algebraic sets and reduced affine k-algebras at the object level

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources