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Passing to the reduced quotient does not change the prime spectrum
Statement
Let be a commutative ring, let , and let be the quotient map. Then contraction along induces an inclusion-preserving bijection . If and , this bijection identifies with .
Facts & Assumptions
Given: A commutative ring , its nilradical , and the quotient map .
Prime ideals of a quotient correspond exactly to prime ideals of the original ring containing the kernel (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).
Every element of is nilpotent (The nilradical and reduced rings).
Proof
Let . If , then for some , so . Because is prime, this forces . Thus every prime ideal of contains .
Applying [L1] to the quotient map and using step 1.1, one gets an inclusion-preserving bijection from onto all of . Moreover, for an ideal with pullback , a prime of contains exactly when its contraction contains . So the same bijection carries onto .
Therefore passing from to its reduced quotient does not change the prime spectrum or the vanishing sets attached to quotient ideals.
Depends on
Used by
Dependency tree · two levels
5 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 The spectrum of a ring (standard reference, not scraped)
- The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)