TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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$Frac (D)$ is a field and $d \mapsto d / 1$ 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 $Frac (D) = (D ∖ {0})^{- 1} D$ 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 $1 / 1 \neq 0 / 1$ , so the localisation is nonzero.

F1F2

1.2

Let $a / b$ be nonzero. By the vanishing criterion in [F2], $a \neq 0$ , because otherwise $a / b = 0$ . Thus $a \in D ∖ {0}$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 6 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

The CRing Project, Chapter 13: Fields and Extensions (standard reference, not scraped)

Frac⁡(D)Frac(D) is a field and d↦d/1d↦d/1 embeds the integral domain DD