LemmaStatement: 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.

If $S$ is a subring and $I$ is an ideal of $R$ , then $S + I$ is a subring, $I$ is an ideal of $S + I$ , and $S \cap I$ is an ideal of $S$

Statement

If $S$ is a subring and $I$ is an ideal of $R$ , then $S + I$ is a subring, $I$ is an ideal of $S + I$ , and $S \cap I$ is an ideal of $S$ .

Facts & Assumptions

Given: A unital subring $S \subseteq R$ and a two-sided ideal $I ⊴ R$ .

[L1]

A subring contains the ambient identity and is closed under ring operations (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L2]

The subring criterion tests subtraction and multiplication (Subring criterion: $S \subseteq R$ is a subring if and only if $1_{R} \in S$ and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ; and an intersection of subrings is a subring).

[L3]

An ideal is an additive subgroup with two-sided absorption (Left, right and two-sided ideals).

[L4]

Intersections of ideals and the ideal criterion are valid (Ideal criteria and intersections of ideals).

Proof

technique · direct

1.1

$S + I$ contains $1_{R}$ and is subtraction-closed; expanding $(s + i) (t + j) = s t + s j + i t + i j$ proves multiplication closure.

L1L2L3L4givenalgebra

2.1

For $s + i \in S + I$ and $j \in I$ , both $(s + i) j$ and $j (s + i)$ lie in $I$ , while $S \cap I$ is subtraction-closed and absorbed by $S$ .

step 1.1L1L2L3L4givenalgebra

3.1

Hence $S + I$ is a unital subring, $I ⊴ S + I$ , and $S \cap I ⊴ S$ .

step 2.1∎

Depends on

Used by

Second isomorphism theorem for rings: S/(S∩ I)≅(S+I)/I Theorem

Dependency tree · two levels

11 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

Judson, Abstract Algebra: Theory and Applications, Ring Homomorphisms and Ideals (standard reference, not scraped)

If SS is a subring and II is an ideal of RR, then S+IS+I is a subring, II is an ideal of S+IS+I, and S∩IS∩I is an ideal of SS