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CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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Passing to the reduced quotient does not change the prime spectrum

Statement

Let R be a commutative ring, let N=Nil(R), and let π:RR/N be the quotient map. Then contraction along π induces an inclusion-preserving bijection Spec(R/N)Spec(R). If JR/N and I=π1(J), this bijection identifies V(J) with V(I).

Facts & Assumptions

Given: A commutative ring R, its nilradical N=Nil(R), and the quotient map π:RR/N.

[L1]

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).

[L2]

Every element of N is nilpotent (The nilradical and reduced rings).

Proof

technique · direct
1.1

Let pSpec(R). If xN, then xm=0 for some m1, so xmp. Because p is prime, this forces xp. Thus every prime ideal of R contains N.

L2givenalgebra
2.1

Applying [L1] to the quotient map π and using step 1.1, one gets an inclusion-preserving bijection from Spec(R/N) onto all of Spec(R). Moreover, for an ideal JR/N with pullback I, a prime q of R/N contains J exactly when its contraction π1(q) contains I. So the same bijection carries V(J) onto V(I).

L1step 1.1
3.1

Therefore passing from R to its reduced quotient does not change the prime spectrum or the vanishing sets attached to quotient ideals.

step 2.1

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