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

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A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring

Statement

A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring.

If $I\mathrel{\trianglelefteq}R$ , $f:R\to S$ is a ring homomorphism, and $I\subseteq\ker f$ , there is a unique ring homomorphism $\bar f:R/I\to S$ such that $\bar f(r+I)=f(r)$ .

Facts & Assumptions

Given: A two-sided ideal $I\mathrel{\trianglelefteq}R$ and a ring homomorphism $f:R\to S$ with $I\subseteq\ker f$ .

[L1]

The additive group of a ring is abelian (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L2]

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

[L3]

Every subgroup of an abelian group is normal (Every subgroup of an abelian group is normal).

[L4]

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

[L5]

A group homomorphism killing a normal subgroup factors uniquely through its quotient (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L6]

A ring homomorphism preserves addition, multiplication, and identity (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

Proof

technique · constructive

1.1

By [L1]--[L3], $I$ is normal in the additive group of $R$ ; applying [L5] to the additive homomorphism underlying $f$ defines $\bar f(r+I)=f(r)$ and proves representative independence.

L1L2L3L5L6givenconstruct

2.1

Since [L4] gives $(r+I)(s+I)=rs+I$ , one has $\bar f((r+I)(s+I))=f(rs)=f(r)f(s)$ , and $\bar f(1+I)=f(1)=1$ ; thus $\bar f$ is a ring homomorphism.

step 1.1L4L6givenalgebra

3.1

The factor identity $f=\bar f\circ\pi$ holds by step 1.1, and any ring-homomorphic factor is additive, so the uniqueness in [L5] proves its uniqueness as a ring factor.

step 1.1step 2.1L5L6discharge-construct∎

Depends on

Used by

First isomorphism theorem for rings: R/ker f congimf Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 34 results over 18 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

Ernst, An Inquiry-Based Approach to Abstract Algebra, Ideals and Quotient Rings (standard reference, not scraped)