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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 $ι : Q \to F$ (Field homomorphism and embedding). On the integers it is given by $n \mapsto n \cdot 1_{F}$ (with $- n \mapsto - (n \cdot 1_{F})$ and $0 \mapsto 0$ ), and on a rational written as $p / q$ with $q \geq 1$ by $ι (p / q) = ι (p) (q \cdot 1_{F})^{- 1}$ . 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 $Q \to F$ .

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 $\pm p / q$ with integers $p, q \geq 1$ . For an integer $p$ write $ι (p)$ for $p \cdot 1_{F}$ if $p \geq 0$ and $- (∣ p ∣ \cdot 1_{F})$ if $p < 0$ .

[L1]

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

[L2]

The canonical naturals satisfy $n \cdot 1_{F} > 0$ for $n \geq 1$ , $n \mapsto n \cdot 1_{F}$ is injective, $(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ , and $(m n) \cdot 1_{F} = (m \cdot 1_{F}) (n \cdot 1_{F})$ (Canonical naturals are positive and strictly increasing).

[L3]

If $a > 0$ then $a^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

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

[L5]

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

Proof

technique · direct

1.1L2

Define $ι$ on the integers by $ι (n) = n \cdot 1_{F}$ for $n \geq 0$ and $ι (- n) = - (n \cdot 1_{F})$ ; by [L2] this is additive and multiplicative on $Z$ and sends $1 \mapsto 1_{F}$ .

1.2L2

For a rational $x = p / q$ with $q \geq 1$ define $ι (x) = ι (p) (q \cdot 1_{F})^{- 1}$ , which makes sense because $q \cdot 1_{F} > 0 \neq 0$ has an inverse.

2.1step 1.1step 1.2L2L3

Well-defined: if $p / q = p^{'} / q^{'}$ with $q, q^{'} \geq 1$ , then $p q^{'} = p^{'} q$ in $Z$ , so [L2] gives $ι (p) (q^{'} \cdot 1_{F}) = ι (p^{'}) (q \cdot 1_{F})$ , and multiplying by the positive $(q \cdot 1_{F})^{- 1} (q^{'} \cdot 1_{F})^{- 1}$ yields $ι (p) (q \cdot 1_{F})^{- 1} = ι (p^{'}) (q^{'} \cdot 1_{F})^{- 1}$ ; thus $ι (x)$ is independent of the representative.

2.2step 1.2L2

Multiplicativity: for $x = p / q$ , $y = r / s$ one has $x y = (p r) / (q s)$ , and $ι (x y) = ι (p r) ((q s) \cdot 1_{F})^{- 1} = ι (p) ι (r) (q \cdot 1_{F})^{- 1} (s \cdot 1_{F})^{- 1} = ι (x) ι (y)$ , using $(m n) \cdot 1_{F} = (m \cdot 1_{F}) (n \cdot 1_{F})$ and $(u v)^{- 1} = u^{- 1} v^{- 1}$ .

2.3step 1.2L2

Additivity: with $x + y = (p s + r q) / (q s)$ , $ι (x + y) = (ι (p) (s \cdot 1_{F}) + ι (r) (q \cdot 1_{F})) (q \cdot 1_{F})^{- 1} (s \cdot 1_{F})^{- 1} = ι (p) (q \cdot 1_{F})^{- 1} + ι (r) (s \cdot 1_{F})^{- 1} = ι (x) + ι (y)$ , using the additive and multiplicative identities of [L2].

2.4step 1.2L1L2L3L4

Positivity: if $x = p / q > 0$ in $Q$ with $q \geq 1$ , then $p \geq 1$ by [L1], so $ι (p) = p \cdot 1_{F} > 0$ and $q \cdot 1_{F} > 0$ by [L2], whence $(q \cdot 1_{F})^{- 1} > 0$ by [L3] and $ι (x) = ι (p) (q \cdot 1_{F})^{- 1} > 0$ by [L4].

2.5step 1.1L5

Uniqueness on $Z$ : let $ψ : Q \to F$ be any field homomorphism; then $ψ (1) = 1_{F}$ , additivity forces $ψ (n) = n \cdot 1_{F} = ι (n)$ for $n \geq 1$ , and $ψ (0) = 0$ , $ψ (- n) = - (n \cdot 1_{F})$ , so $ψ = ι$ on $Z$ .

3.1step 2.2step 2.3L2L5

Unit: $ι (1) = ι (1 / 1) = ι (1) (1 \cdot 1_{F})^{- 1} = 1_{F}$ ; hence $ι$ is a field homomorphism $Q \to F$ .

3.2step 2.3step 2.4

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.

4.1step 3.2

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

5.1step 2.5step 1.2L5∎

Uniqueness on $Q$ : for $p / q \in Q$ , $ψ (p / q) = ψ (p) ψ (q)^{- 1} = ι (p) (q \cdot 1_{F})^{- 1} = ι (p / q)$ since $ψ$ preserves products and inverses; hence $ψ = ι$ , so $ι$ is the unique field homomorphism $Q \to F$ .

The unique embedding of ℚ into an ordered field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources