TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Second isomorphism theorem for rings: $S/(S\cap I)\cong(S+I)/I$

Statement

Second isomorphism theorem for rings: $S/(S\cap I)\cong(S+I)/I$ .

If $S$ is a unital subring of $R$ and $I\mathrel{\trianglelefteq}R$ , then this is an isomorphism of unital rings.

Facts & Assumptions

Given: A unital subring $S\subseteq R$ and a two-sided ideal $I\mathrel{\trianglelefteq}R$ .

[L1]

$S+I$ is a subring, $I\mathrel{\trianglelefteq}S+I$ , and $S\cap I\mathrel{\trianglelefteq}S$ (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 $S$ ).

[L2]

The first ring isomorphism theorem identifies a ring modulo a kernel with its image (First isomorphism theorem for rings: $R/\ker f\cong\operatorname{im}f$ ).

[L3]

The canonical quotient map is a surjective ring homomorphism (The canonical projection $R\to R/I$ is a surjective ring homomorphism with kernel $I$ ).

Proof

technique · direct

1.1

Restrict the quotient map $S+I\to(S+I)/I$ to $\phi:S\to(S+I)/I$ , $\phi(s)=s+I$ .

L1L2L3givenconstruct

2.1

Its kernel is $S\cap I$ , and every $(s+i)+I$ equals $s+I$ , so its image is all of $(S+I)/I$ .

step 1.1L1L2L3givenalgebra

3.1

The kernel and image computation gives $S/(S\cap I)\cong(S+I)/I$ .

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 44 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

Second isomorphism theorem for rings: S/(S∩I)≅(S+I)/IS/(S\cap I)\cong(S+I)/IS/(S∩I)≅(S+I)/I