Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Frac⁡(Z) is canonically isomorphic to Q

Example

The field of fractions of Z is canonically isomorphic to Q by ab⟼ab,a,b∈Z, b≠0.

Facts & Assumptions

Given: The standard inclusion Z↪Q.

[F1]

The rationals are the equivalence classes written a/b for integers a,b with b≠0 (The rationals as equivalence classes of pairs of integers).

[F2]

The rationals form a field (The rationals form a field).

[F3]

An injective ring map from a domain into a field extends uniquely to its field of fractions (Every injective ring map from a domain into a field factors uniquely through its field of fractions).

[F4]

The integers form a commutative ring (The integers form a commutative ring).

[F5]

A product of two nonzero integers is nonzero (The integers have no zero divisors; multiplicative cancellation).

[F6]

The natural numbers embed injectively in the integers, while 0 and 1 are distinct natural numbers (The naturals embed in the integers, The natural numbers N (von Neumann)).

[F7]

An integral domain is a commutative ring with 1≠0 and no zero divisors (Zero divisor, and integral domain: a commutative ring with 1≠0 and no zero divisors).

Verification

technique · direct
1.1

Facts [F4] and [F5] give the commutative-ring and no-zero-divisor clauses for Z, while [F6] gives 1≠0; hence [F7] makes Z an integral domain. The map a↦a/1 from Z to Q is injective by the defining equivalence relation in [F1]. Since Q is a field by [F2], [F3] extends it uniquely to an injective homomorphism Frac⁡(Z)→Q with the displayed formula.

F1F2F3F4F5F6F7
2.1

Every rational is a/b with b≠0 by [F1], so the homomorphism is surjective and hence an isomorphism. It fixes each integer, which makes it canonical over Z.

F1step 1.1∎

Depends on

Used by

Dependency tree · two levels

39 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