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Every Zariski-closed subset has a unique radical defining ideal

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $Z \subseteq Spec (R)$ be Zariski-closed. Then $I (Z) : = ⋂_{p \in Z} p$ is a radical ideal, $Z = V (I (Z))$ , and if $Z = V (J)$ for an ideal $J$ , then $I (Z) = \sqrt{J}$ . In particular, $Z$ has a unique radical defining ideal.

Facts & Assumptions

Given: A commutative ring $R$ , a Zariski-closed subset $Z \subseteq Spec (R)$ , and the Axiom of Choice.

[L1]

For every ideal $J$ , its radical is the intersection of the prime ideals containing $J$ (The radical of an ideal is the intersection of the prime ideals containing it).

[L2]

Two vanishing sets are equal exactly when the radicals of their defining ideals are equal (Vanishing sets detect radicals).

[A1]

Because $Z$ is Zariski-closed, there exists an ideal $J ⊴ R$ with $Z = V (J)$ .

Proof

technique · direct

1.1L1A1

Choose $J$ with $Z = V (J)$ as in [A1]. Then $I (Z) = ⋂_{p \in V (J)} p = \sqrt{J}$ by [L1]. In particular, $I (Z)$ is radical.

2.1L2step 1.1A1

Since step 1.1 gives $I (Z) = \sqrt{J}$ , fact [L2] yields $V (I (Z)) = V (\sqrt{J}) = V (J) = Z$ .

2.2L2step 1.1A1

If also $Z = V (K)$ for an ideal $K$ , then $V (K) = V (J)$ , so [L2] gives $\sqrt{K} = \sqrt{J} = I (Z)$ . Thus any radical ideal defining $Z$ equals $I (Z)$ .

3.1step 2.1step 2.2∎

Therefore every Zariski-closed subset has the unique radical defining ideal $I (Z)$ .

Every Zariski-closed subset has a unique radical defining ideal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources