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Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals

Statement

Assume the Axiom of Choice.

Let $k$ be an algebraically closed field.

The assignments $X ⟼ I (X), J ⟼ V (J)$ induce mutually inverse inclusion-reversing correspondences between affine algebraic sets $X \subseteq A_{k}^{n}$ and radical ideals $J \subseteq k [x_{1}, \dots, x_{n}]$ .
Under this correspondence, nonempty irreducible affine algebraic sets correspond exactly to prime ideals.

Facts & Assumptions

Given: An algebraically closed field $k$ and the Axiom of Choice.

[L1]

For every subset $X \subseteq k^{n}$ and ideal $J \subseteq k [x_{1}, \dots, x_{n}]$ , $X \subseteq V (J) ⟺ J \subseteq I (X),$ so in particular $X \subseteq V (I (X))$ and $J \subseteq I (V (J))$ (Vanishing ideals and zero loci form a Galois connection).

[L2]

For every ideal $J \subseteq k [x_{1}, \dots, x_{n}]$ , $I (V (J)) = \sqrt{J}$ (Strong Nullstellensatz: I(V(I)) equals the radical of I).

[L3]

For ideals $I, J$ one has $V (I) \cup V (J) = V (I J)$ (Zero loci in affine space are the closed sets of the classical Zariski topology).

Proof

technique · direct

1.1L1given

Let $X \subseteq k^{n}$ be affine algebraic, say $X = V (S)$ for some set $S \subseteq k [x_{1}, \dots, x_{n}]$ . Then $X \subseteq V (I (X))$ by [L1]. The reverse inclusion holds because every element of $S$ vanishes on $X$ , so $S \subseteq I (X)$ and therefore $V (I (X)) \subseteq V (S) = X$ by [L1]. Hence $X = V (I (X))$ .

1.2L1L3givenalgebra

Conversely, assume $X$ is nonempty and irreducible, and let $f g \in I (X)$ . Then every point of $X$ annihilates $f g$ , so $X \subseteq V (f g) = V (f) \cup V (g)$ by [L3]. Irreducibility gives $X \subseteq V (f)$ or $X \subseteq V (g)$ , and then [L1] yields $f \in I (X)$ or $g \in I (X)$ . Also $1 \notin I (X)$ because $X \neq \emptyset$ . Therefore $I (X)$ is prime.

2.1L2step 1.1

Let $J \subseteq k [x_{1}, \dots, x_{n}]$ be an ideal. Then [L2] gives $I (V (J)) = \sqrt{J}$ . Therefore $J$ is radical exactly when $I (V (J)) = J$ . Together with step 1.1, this makes $X \mapsto I (X)$ and $J \mapsto V (J)$ inverse correspondences between algebraic sets and radical ideals.

2.2L1L2L3choosealgebra

Let $X$ be a nonempty algebraic set. If $I (X)$ is prime and $X = Y \cup Z$ with algebraic subsets $Y = V (J)$ and $Z = V (K)$ , then [L3] gives $X = V (J K)$ . Applying step 1.1 and then [L2], $I (X) = I (V (J K)) = \sqrt{J K} .$ Hence $J K \subseteq I (X)$ . If neither $J$ nor $K$ were contained in $I (X)$ , we could choose $f \in J ∖ I (X)$ and $g \in K ∖ I (X)$ , and then $f g \in J K \subseteq I (X)$ would contradict primality of $I (X)$ . So $J \subseteq I (X)$ or $K \subseteq I (X)$ , which by [L1] means $X \subseteq Y$ or $X \subseteq Z$ . Since $Y, Z \subseteq X$ , one has $X = Y$ or $X = Z$ . Thus $X$ is irreducible.

3.1step 2.1step 2.2step 1.2∎

Steps 2.2 and 1.2 identify the irreducible algebraic sets precisely with the prime ideals under the correspondence from step 2.1.

Depends on

Used by

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Sources

J. S. Milne, Algebraic Geometry, Corollary 2.17 and Summary 2.28 (standard reference, not scraped)
Donu Arapura, Notes on Basic Algebraic Geometry, §1.5-1.6 (standard reference, not scraped)