TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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.

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ .

The empty sum is included and equals $0$ .

Given: A commutative ring $R$ and a subset $S \subseteq R$ .

[L1]

$(S)$ is the intersection of ideals containing $S$ (The ideal generated by a subset and principal ideals).

[L2]

The ideal criterion uses subtraction and absorption (Ideal criteria and intersections of ideals).

[L3]

Multiplication in a commutative ring commutes (Commutative ring).

technique · direct

1.1L1L2L3givenalgebra

Let $J$ be the finite sums $\sum r_{i} s_{i}$ ; it contains $S$ and is closed under subtraction and multiplication by arbitrary ring elements.

2.1step 1.1L1L2L3givenalgebra

Thus $J$ is an ideal containing $S$ , while every ideal containing $S$ contains each such finite sum.

3.1step 2.1∎

Hence $J = (S)$ ; taking $S = {a}$ gives $(a) = R a$ .

Depends on