Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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Every injective ring map from a domain into a field factors uniquely through its field of fractions

Statement

Let D be an integral domain, K a field, and f:D→K an injective unital ring homomorphism. There is a unique unital ring homomorphism f~:Frac⁡(D)⟶K such that f~(d/1)=f(d) for all d∈D. It is injective and satisfies f~(a/b)=f(a)f(b)−1.

Facts & Assumptions

Given: An injective unital ring homomorphism f:D→K from an integral domain to a field.

[F1]

Every nonzero element of a field is a unit (Field).

[F2]

A map that sends every localisation denominator to a unit factors uniquely through the localisation, by the displayed fraction formula (Universal property of localisation: maps that invert S factor uniquely through S−1R).

[F3]

The canonical map embeds D in Frac⁡(D) (Frac⁡(D) is a field and d↦d/1 embeds the integral domain D).

Proof

technique · direct
1.1

If b≠0 in D, injectivity gives f(b)≠0, so [F1] makes f(b) a unit. Since the denominators defining Frac⁡(D) are exactly the nonzero elements, [F2] gives the unique extension and its formula.

F1F2
2.1

If f~(a/b)=0, multiply f(a)f(b)−1=0 by the unit f(b) to obtain f(a)=0. Injectivity of f gives a=0, hence a/b=0. Thus f~ is injective.

step 1.1F3algebra∎

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