CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $φ : R \to A$ is a ring homomorphism to a reduced commutative ring $A$ , then there is a unique ring homomorphism $\overline{φ} : R / N \to A$ with $φ = \overline{φ} \circ π$ , where $π : R \to R / N$ is the quotient map.

Facts & Assumptions

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

[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

1.1

Let $x + N \in R / N$ be nilpotent. Then $(x + N)^{m} = N$ for some $m \geq 1$ , so $x^{m} \in N$ . By [L1], some power of $x^{m}$ is zero, hence some power of $x$ is zero, so $x \in N$ . Therefore $x + N = 0 + N$ , and the only nilpotent element of $R / N$ is zero. Thus $R / N$ is reduced by [L1].

L1givenalgebra

1.2

Let $φ : R \to A$ with $A$ reduced. If $x \in N$ , then $x^{m} = 0$ for some $m \geq 1$ , so $φ (x)^{m} = 0$ . Reducedness of $A$ forces $φ (x) = 0$ , so $N \subseteq \ker φ$ . Therefore $\overline{φ} (x + N) : = φ (x)$ is well-defined, and it is unique because $π$ is surjective.

L1givenalgebra

2.1

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

step 1.1step 1.2∎

Depends on

The nilradical and reduced rings

Used by

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Dependency tree · two levels

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Sources

M. Hochster, Introduction to Commutative Algebra, Math 614 notes (2020) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §2 Ideals (standard reference, not scraped)