LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-09-05

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The closure of a prime is its vanishing set

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $p \in Spec (R)$ . Then $\overline{{p}} = V (p) .$

Facts & Assumptions

Given: A commutative ring $R$ , a prime ideal $p \subseteq R$ , and the Axiom of Choice.

[L1]

Every Zariski-closed subset has a unique radical defining ideal (Every Zariski-closed subset has a unique radical defining ideal).

[L2]

$V (I)$ is the set of prime ideals containing $I$ (The prime spectrum and vanishing sets).

Proof

technique · direct

1.1

Since $p \supseteq p$ , the point $p$ lies in $V (p)$ by [L2]. Because $V (p)$ is closed, the closure $\overline{{p}}$ is contained in $V (p)$ .

L2given

1.2

Let $Z$ be a closed subset containing $p$ . By [L1], write $Z = V (I)$ for its radical defining ideal $I$ . Since $p \in Z$ , fact [L2] gives $I \subseteq p$ . Therefore every prime ideal containing $p$ also contains $I$ , so $V (p) \subseteq V (I) = Z$ .

L1L2

2.1

Step 1.2 shows that every closed set containing $p$ also contains $V (p)$ . Hence $V (p)$ is the smallest closed set containing $p$ , that is, $\overline{{p}} = V (p)$ .

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 14.4(b) (standard reference, not scraped)
The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)