Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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Frac⁡(D) is a field and d↦d/1 embeds the integral domain D

Statement

For every integral domain D, the localisation Frac⁡(D) is a field. Its canonical map D⟶Frac⁡(D),d⟼d/1, is an injective unital ring homomorphism.

Facts & Assumptions

Given: An integral domain D.

[F1]

The field of fractions is the localisation at D∖{0} (The field of fractions Frac⁡(D)=(D∖{0})−1D of an integral domain).

[F2]

A localisation map is injective exactly when every denominator has trivial annihilator (Equality, vanishing, and the kernel of the localisation map).

[F3]

Localisation is a commutative ring with the stated fraction arithmetic (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

[F4]

A field is a nonzero commutative ring in which every nonzero element is a unit (Field).

Proof

technique · direct
1.1

Every nonzero element of the domain D has trivial annihilator. Hence [F2], applied to S=D∖{0}, makes the canonical homomorphism injective. In particular 1/1≠0/1, so the localisation is nonzero.

F1F2
1.2

Let a/b be nonzero. By the vanishing criterion in [F2], a≠0, because otherwise a/b=0. Thus a∈D∖{0}, and [F3] gives (a/b)(b/a)=1.

F1F2F3
2.1

Every nonzero element is therefore a unit, and [F3] supplies the commutative-ring structure. By [F4], Frac⁡(D) is a field.

F3F4step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

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Sources