TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03

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First isomorphism theorem for modules: $M/\ker f\cong\operatorname{im}f$

Statement

For every module homomorphism $f:M\to N$ , there is a module isomorphism

$M/\ker f\ \cong\ \operatorname{im}f,$

given by $m+\ker f\mapsto f(m)$ .

Facts & Assumptions

Given: A module homomorphism $f:M\to N$ .

[L1]

Its kernel and image are submodules, and a module homomorphism is injective exactly when its kernel is trivial (Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel).

[L2]

A homomorphism vanishing on a submodule factors uniquely through the quotient module (A module homomorphism vanishing on $N$ factors uniquely through $M/N$ ).

[L3]

Module isomorphisms are precisely bijective module homomorphisms; kernel and image have their displayed definitions (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

Since $\ker f$ is a submodule and $f$ vanishes on it, [L2] gives a module homomorphism $\bar f:M/\ker f\to N$ with $\bar f(m+\ker f)=f(m)$ .

L1L2L3given

2.1

Every value of $\bar f$ lies in $\operatorname{im}f$ , and every $f(m)$ is the value of $\bar f$ at $m+\ker f$ ; hence its corestriction $\bar f:M/\ker f\to\operatorname{im}f$ is a surjective module homomorphism.

step 1.1L1L3given

2.2

The corestriction has trivial kernel: $\bar f(m+\ker f)=0_N$ means $f(m)=0_N$ , hence $m\in\ker f$ and $m+\ker f=0_{M/\ker f}$ .

step 1.1L1L3given

3.1

The corestriction is injective by [L1], so it is bijective.

step 2.1step 2.2L1

4.1

By [L3], this bijective module homomorphism is the claimed module isomorphism.

step 3.1L3∎

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: 38 results over 10 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

McGerty, Algebra II: Rings and Modules, Section 3 (standard reference, not scraped)

First isomorphism theorem for modules: M/ker⁡f≅im⁡fM/\ker f\cong\operatorname{im}fM/kerf≅imf