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.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Specialisation in a prime spectrum is reverse inclusion

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $p, q \in Spec (R)$ . Then $q$ is a specialisation of $p$ if and only if $p \subseteq q .$

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

[L1]

A point $y$ is a specialisation of $x$ exactly when $y \in \overline{{x}}$ (Specialisations, generalisations, and generic points).

[L2]

technique · direct

1.1L1L2

By [L1] and [L2], $q$ is a specialisation of $p$ exactly when $q \in V (p)$ .

2.1step 1.1algebra

By the definition of $V (p)$ , the condition $q \in V (p)$ is exactly $p \subseteq q$ .

3.1step 1.1step 2.1∎

Therefore specialisation in $Spec (R)$ is reverse inclusion of prime ideals.

Depends on