ExampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Assuming the Axiom of Choice, the Chinese-remainder map Z/6Z -> Z/2Z direct-sum Z/3Z is an isomorphism by local tests

Example

Assume the Axiom of Choice.

Consider the $Z$ -module homomorphism

ϕ : Z / 6 Z ⟶ Z / 2 Z \oplus Z / 3 Z, [a]_{6} ⟼ ([a]_{2}, [a]_{3}) .

This map is an isomorphism because every prime localisation of $ϕ$ is an isomorphism.

Facts & Assumptions

Given: The $Z$ -module homomorphism $ϕ : Z / 6 Z \to Z / 2 Z \oplus Z / 3 Z$ , $[a]_{6} \mapsto ([a]_{2}, [a]_{3})$ .

[L1]

A module homomorphism is an isomorphism exactly when all of its prime localisations are isomorphisms (Assuming the Axiom of Choice, local criteria for zero modules and for injective, surjective, and bijective maps).

[L2]

The arithmetic of $Z / n Z$ is the usual modular arithmetic (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Verification

technique · direct

1.1

At the prime $(2)$ , the localisation of $Z / 6 Z$ is $Z / 2 Z$ , while the localisation of $Z / 2 Z \oplus Z / 3 Z$ is $Z / 2 Z \oplus 0$ , because $3$ becomes a unit. The localised map is therefore the identity on the surviving $Z / 2 Z$ summand.

L2algebra

1.2

At the prime $(3)$ , the same computation gives the identity $Z / 3 Z \to 0 \oplus Z / 3 Z$ .

L2algebra

1.3

At every other prime $(q)$ , both $Z / 6 Z$ and $Z / 2 Z \oplus Z / 3 Z$ localise to $0$ because both $2$ and $3$ become units.

L2algebra

2.1

Steps 1.1, 1.2, and 1.3 show that every prime localisation of $ϕ$ is an isomorphism, so [L1] gives that $ϕ$ is an isomorphism.

L1step 1.1step 1.2step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Corollary 5.17 (standard reference, not scraped)