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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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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