CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1L1givenalgebra

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].

1.2L1givenalgebra

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.

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

The nilradical and reduced rings

The reduced quotient by the nilradical

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources