Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-27
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The reduced quotient by the nilradical

Statement

Let R be a commutative ring and let N=Nil⁡(R). Then R/N is reduced. Moreover, if φ:R→A is a ring homomorphism to a reduced commutative ring A, then there is a unique ring homomorphism φ‾:R/N→A with φ=φ‾∘π, where π:R→R/N is the quotient map.

Facts & Assumptions

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

[L1]

The nilradical is the ideal of nilpotent elements, and a ring is reduced exactly when its nilradical is zero (The nilradical and reduced rings).

Proof

technique · direct
1.1L1givenalgebra

Let x+N∈R/N be nilpotent. Then (x+N)m=N for some m≥1, so xm∈N. By [L1], some power of xm is zero, hence some power of x is zero, so x∈N. Therefore x+N=0+N, and the only nilpotent element of R/N is zero. Thus R/N is reduced by [L1].

1.2L1givenalgebra

Let φ:R→A with A reduced. If x∈N, then xm=0 for some m≥1, so φ(x)m=0. Reducedness of A forces φ(x)=0, so N⊆ker⁡φ. Therefore φ‾(x+N):=φ(x) is well-defined, and it is unique because π is surjective.

2.1step 1.1step 1.2∎

The quotient by the nilradical is reduced and is universal among maps from R to reduced rings.

Depends on

Used by

Dependency tree · two levels

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Sources