TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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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.
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First isomorphism theorem for modules: $M / \ker f ≅ im f$

Statement

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

$M / \ker f ≅ im f,$

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 $\overset{ˉ}{f} : M / \ker f \to N$ with $\overset{ˉ}{f} (m + \ker f) = f (m)$ .

L1L2L3given

2.1

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

step 1.1L1L3given

2.2

The corestriction has trivial kernel: $\overset{ˉ}{f} (m + \ker f) = 0_{N}$ 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

Dependency tree · two levels

12 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

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

First isomorphism theorem for modules: M/ker⁡f≅im⁡fM/kerf≅imf