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Specialisation in a prime spectrum is reverse inclusion
Statement
Assume the Axiom of Choice.
Let be a commutative ring and let . Then is a specialisation of if and only if
Facts & Assumptions
Given: A commutative ring , prime ideals , and the Axiom of Choice.
A point is a specialisation of exactly when (Specialisations, generalisations, and generic points).
The closure of is (The closure of a prime is its vanishing set).
Proof
By [L1] and [L2], is a specialisation of exactly when .
By the definition of , the condition is exactly .
Therefore specialisation in is reverse inclusion of prime ideals.
Depends on
Used by
- Distinct primes have distinct closures, so the spectrum is T0 Corollary
- A generic point and a distinct specialization cannot be separated in the Zariski topology Example
- A local PID gives a two-point spectrum with one generic point and one closed point Example
- The spectrum of the integers has one generic point, closed points (p), and basic opens D(n) Example
- A quasi-compact immersion with nonclosed image has a boundary specialization Lemma
- Proper closed subsets of a curve are finite Lemma
- Proper normal curve rational function map Lemma
- The support of any module is closed under specialisation Lemma
- Valuation lifts detect universal closedness Lemma
- Finite-dimensional projective space is proper over every base Theorem
Dependency tree · two levels
6 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)