CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Distinct primes have distinct closures, so the spectrum is T0

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring. Distinct prime ideals of $R$ have distinct closures in $Spec (R)$ . Equivalently, $Spec (R)$ is $T_{0}$ .

Facts & Assumptions

Given: A commutative ring $R$ , distinct prime ideals $p, q \subseteq R$ , and the Axiom of Choice.

[L1]

In a prime spectrum, $q$ is a specialisation of $p$ exactly when $p \subseteq q$ (Specialisation in a prime spectrum is reverse inclusion).

[A1]

A space is $T_{0}$ exactly when distinct points have distinct closures.

Proof

technique · direct

1.1

If $\overline{{p}} = \overline{{q}}$ , then each point lies in the closure of the other. Hence each is a specialisation of the other, so [L1] gives both $p \subseteq q$ and $q \subseteq p$ . Therefore $p = q$ , contrary to the hypothesis.

L1given

2.1

Thus distinct prime ideals have distinct closures. By [A1], this is exactly the $T_{0}$ property.

A1step 1.1

3.1

Therefore $Spec (R)$ is $T_{0}$ .

step 2.1∎

Depends on

Specialisation in a prime spectrum is reverse inclusion

Used by

A Zariski-closed subset is irreducible exactly when its radical defining ideal is prime, and then it has a unique generic point Theorem

Dependency tree · two levels

3 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, §14 (standard reference, not scraped)
The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)