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Field homomorphisms between ordered fields fix $Q$

Statement

Let $F$ and $G$ be ordered fields with canonical rational embeddings $ι_{F} : Q \to F$ and $ι_{G} : Q \to G$ (The unique embedding of ℚ into an ordered field). Then every field homomorphism $φ : F \to G$ (Field homomorphism and embedding) fixes $Q$ , meaning $φ \circ ι_{F} = ι_{G}, i.e. φ (ι_{F} (q)) = ι_{G} (q) for all q \in Q .$

Facts & Assumptions

Given: Ordered fields $F, G$ , a field homomorphism $φ : F \to G$ , and the canonical embeddings $ι_{F}, ι_{G}$ .

[L1]

The canonical embedding $ι : Q \to F$ acts on $Z$ by $n \mapsto n \cdot 1_{F}$ and on $Q$ by $p / q \mapsto (p \cdot 1_{F}) (q \cdot 1_{F})^{- 1}$ (likewise for $ι_{G}$ ) (The unique embedding of ℚ into an ordered field).

[L2]

The canonical natural $n \cdot 1$ is the $n$ -fold sum $1 + \dots + 1$ ; the integers embed with $q \cdot 1 \neq 0$ for $q \neq 0$ (Canonical naturals are positive and strictly increasing).

[L3]

$φ$ is a field homomorphism: $φ (1_{F}) = 1_{G}$ , $φ (x + y) = φ (x) + φ (y)$ , $φ (x y) = φ (x) φ (y)$ , $φ (0_{F}) = 0_{G}$ , $φ (- x) = - φ (x)$ , and $φ (x^{- 1}) = φ (x)^{- 1}$ for $x \neq 0$ (Field homomorphism and embedding).

Proof

technique · direct

1.1

By [L3], $φ (1_{F}) = 1_{G}$ and $φ$ preserves sums, products, negation, and inversion of nonzero elements.

1.2

By [L1], $ι_{F}$ and $ι_{G}$ send each $n \in Z$ to $n \cdot 1_{F}$ and $n \cdot 1_{G}$ , and each $p / q \in Q$ to $(p \cdot 1) (q \cdot 1)^{- 1}$ in the respective field.

1.3

Because $n \cdot 1_{F}$ is the $n$ -fold sum of $1_{F}$ ([L2]) and $φ$ is additive with $φ (1_{F}) = 1_{G}$ , we get $φ (n \cdot 1_{F}) = n \cdot 1_{G}$ for every canonical natural $n$ .

L2L3

2.1

For each integer $n$ this extends by sign: $φ (ι_{F} (0)) = φ (0_{F}) = 0_{G} = ι_{G} (0)$ and $φ (ι_{F} (- n)) = φ (- (n \cdot 1_{F})) = - (n \cdot 1_{G}) = ι_{G} (- n)$ , so $φ (ι_{F} (m)) = ι_{G} (m)$ for all $m \in Z$ .

step 1.2step 1.3L3

3.1

For a rational $p / q$ with integers $p, q \neq 0$ , $ι_{F} (p / q) = (p \cdot 1_{F}) (q \cdot 1_{F})^{- 1}$ , so $φ (ι_{F} (p / q)) = φ (p \cdot 1_{F}) φ (q \cdot 1_{F})^{- 1} = (p \cdot 1_{G}) (q \cdot 1_{G})^{- 1} = ι_{G} (p / q)$ .

step 1.2step 2.1L3

4.1

Since $φ \circ ι_{F}$ and $ι_{G}$ agree on every rational, $φ \circ ι_{F} = ι_{G}$ : the homomorphism $φ$ fixes $Q$ .

step 3.1∎

Depends on

Used by

Uniqueness of the complete ordered field: ℝ up to a unique isomorphism Theorem

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed. (standard reference, not scraped)
University of Wisconsin Math 521 notes: Real analysis (standard reference, not scraped)

Field homomorphisms between ordered fields fix QQ