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Every Zariski-closed subset has a unique radical defining ideal
Statement
Assume the Axiom of Choice.
Let be a commutative ring and let be Zariski-closed. Then is a radical ideal, , and if for an ideal , then . In particular, has a unique radical defining ideal.
Facts & Assumptions
Given: A commutative ring , a Zariski-closed subset , and the Axiom of Choice.
For every ideal , its radical is the intersection of the prime ideals containing (The radical of an ideal is the intersection of the prime ideals containing it).
Two vanishing sets are equal exactly when the radicals of their defining ideals are equal (Vanishing sets detect radicals).
Because is Zariski-closed, there exists an ideal with .
Proof
Choose with as in [A1]. Then by [L1]. In particular, is radical.
Since step 1.1 gives , fact [L2] yields .
If also for an ideal , then , so [L2] gives . Thus any radical ideal defining equals .
Therefore every Zariski-closed subset has the unique radical defining ideal .
Depends on
Used by
- In a finite-type algebra over a field, closed points are dense in every closed subset of the spectrum Corollary
- A clopen decomposition of the spectrum comes from a nontrivial idempotent Lemma
- The closure of a prime is its vanishing set Lemma
- A Zariski-closed subset is irreducible exactly when its radical defining ideal is prime, and then it has a unique generic point Theorem
- The spectrum of a Noetherian ring is a Noetherian topological space Theorem
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(a) (standard reference, not scraped)
- The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)