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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25
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The unique embedding of ℚ into an ordered field

Statement

Let F be an ordered field (Ordered field). There is a unique field homomorphism ι:Q→F (Field homomorphism and embedding). On the integers it is given by n↦n⋅1F (with −n↦−(n⋅1F) and 0↦0), and on a rational written as p/q with q≥1 by ι(p/q)=ι(p) (q⋅1F)−1. Moreover ι is injective and order-preserving, so it is an embedding of Q as an ordered subfield of F, and it is the only field homomorphism Q→F.

Facts & Assumptions

Given: An ordered field F; the field Q of The rationals form a totally ordered field, every element of which is 0 or ±p/q with integers p,q≥1. For an integer p write ι(p) for p⋅1F if p≥0 and −(∣p∣⋅1F) if p<0.

[L1]

Q is an ordered field; a nonzero p/q with q≥1 is positive exactly when p≥1 (The rationals form a totally ordered field).

[L2]

The canonical naturals satisfy n⋅1F>0 for n≥1, n↦n⋅1F is injective, (m+n)⋅1F=m⋅1F+n⋅1F, and (mn)⋅1F=(m⋅1F)(n⋅1F) (Canonical naturals are positive and strictly increasing).

[L4]

Sign rules: a product of positives is positive, and for c>0 one has a<b iff ac<bc (Sign rules for products and monotonicity of multiplication).

[L5]

A field homomorphism preserves +, ⋅, and 1, and hence 0, negation, and inverses (Field homomorphism and embedding).

Proof

technique · direct
1.1

Define ι on the integers by ι(n)=n⋅1F for n≥0 and ι(−n)=−(n⋅1F); by [L2] this is additive and multiplicative on Z and sends 1↦1F.

L2
1.2

For a rational x=p/q with q≥1 define ι(x)=ι(p) (q⋅1F)−1, which makes sense because q⋅1F>0≠0 has an inverse.

L2
2.1

Well-defined: if p/q=p′/q′ with q,q′≥1, then pq′=p′q in Z, so [L2] gives ι(p)(q′⋅1F)=ι(p′)(q⋅1F), and multiplying by the positive (q⋅1F)−1(q′⋅1F)−1 yields ι(p)(q⋅1F)−1=ι(p′)(q′⋅1F)−1; thus ι(x) is independent of the representative.

step 1.1step 1.2L2L3
2.2

Multiplicativity: for x=p/q, y=r/s one has xy=(pr)/(qs), and ι(xy)=ι(pr)((qs)⋅1F)−1=ι(p)ι(r)(q⋅1F)−1(s⋅1F)−1=ι(x)ι(y), using (mn)⋅1F=(m⋅1F)(n⋅1F) and (uv)−1=u−1v−1.

step 1.2L2
2.3

Additivity: with x+y=(ps+rq)/(qs), ι(x+y)=(ι(p)(s⋅1F)+ι(r)(q⋅1F))(q⋅1F)−1(s⋅1F)−1=ι(p)(q⋅1F)−1+ι(r)(s⋅1F)−1=ι(x)+ι(y), using the additive and multiplicative identities of [L2].

step 1.2L2
2.4

Positivity: if x=p/q>0 in Q with q≥1, then p≥1 by [L1], so ι(p)=p⋅1F>0 and q⋅1F>0 by [L2], whence (q⋅1F)−1>0 by [L3] and ι(x)=ι(p)(q⋅1F)−1>0 by [L4].

step 1.2L1L2L3L4
2.5

Uniqueness on Z: let ψ:Q→F be any field homomorphism; then ψ(1)=1F, additivity forces ψ(n)=n⋅1F=ι(n) for n≥1, and ψ(0)=0, ψ(−n)=−(n⋅1F), so ψ=ι on Z.

step 1.1L5
3.1

Unit: ι(1)=ι(1/1)=ι(1)(1⋅1F)−1=1F; hence ι is a field homomorphism Q→F.

step 2.2step 2.3L2L5
3.2

Order: for x<y in Q we have y−x>0, so ι(y)−ι(x)=ι(y−x)>0 by 2.3 and 2.4, that is ι(x)<ι(y); thus ι is order-preserving.

step 2.3step 2.4
4.1

Injectivity: if x≠y then x<y or y<x, and 3.2 forces ι(x)≠ι(y); so ι is injective, an embedding of ordered fields.

step 3.2
5.1

Uniqueness on Q: for p/q∈Q, ψ(p/q)=ψ(p)ψ(q)−1=ι(p)(q⋅1F)−1=ι(p/q) since ψ preserves products and inverses; hence ψ=ι, so ι is the unique field homomorphism Q→F.

step 2.5step 1.2L5∎

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