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Homomorphisms out of a complete ordered field are order-preserving

Statement

Let $F$ be a complete ordered field and $G$ an ordered field, and let $φ : F \to G$ be a field homomorphism (Field homomorphism and embedding). Then $φ$ is injective and order-preserving: $x > 0$ in $F$ implies $φ (x) > 0$ in $G$ , and consequently $a < b$ implies $φ (a) < φ (b)$ .

Facts & Assumptions

Given: A complete ordered field $F$ , an ordered field $G$ , and a field homomorphism $φ : F \to G$ .

[L1]

$φ (1_{F}) = 1_{G}$ , $φ (x + y) = φ (x) + φ (y)$ , $φ (x y) = φ (x) φ (y)$ , $φ (- x) = - φ (x)$ ; and every field homomorphism is injective, its kernel being an ideal of the field $F$ with $φ (1_{F}) = 1_{G} \neq 0_{G}$ (Field homomorphism and embedding).

[L2]

In a complete ordered field every $a \geq 0$ is a square $a = y^{2}$ ; the positive elements are exactly the nonzero squares (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L3]

In any ordered field a nonzero square is positive: $y \neq 0 \Rightarrow y^{2} > 0$ (Squares of nonzero elements are positive).

[L4]

Order via the positive cone: $x > 0$ means $x \in P$ , $a < b$ means $b - a \in P$ ; trichotomy holds (Ordered field).

Proof

technique · direct

1.1

$φ$ is injective: by [L1] its kernel is an ideal of the field $F$ , and since $φ (1_{F}) = 1_{G} \neq 0_{G}$ the kernel is ${0_{F}}$ .

1.2

Fix $x \in F$ with $x > 0$ ; then $x \geq 0$ , so by [L2] there is $y \in F$ with $x = y^{2}$ , and $y \neq 0$ since $y = 0$ would give $x = 0$ , against $x > 0$ by [L4].

L2L4

2.1

Applying $φ$ , $φ (x) = φ (y^{2}) = φ (y)^{2}$ , and $φ (y) \neq 0_{G}$ because $y \neq 0$ and $φ$ is injective.

step 1.1step 1.2L1

3.1

By [L3] in $G$ , the nonzero square $φ (y)^{2}$ is positive, so $φ (x) = φ (y)^{2} > 0$ ; as $x > 0$ was arbitrary, $x > 0 \Rightarrow φ (x) > 0$ for all $x \in F$ .

step 2.1L3

4.1

If $a < b$ then $b - a > 0$ , so $φ (b - a) > 0$ ; since $φ (b - a) = φ (b) - φ (a)$ by [L1], we get $φ (b) - φ (a) > 0$ , i.e. $φ (a) < φ (b)$ .

step 3.1L1L4

5.1

Hence $φ$ is an injective, order-preserving field homomorphism.

step 1.1step 4.1∎

Depends on

Used by

A field homomorphism of ordered fields need not preserve order Counterexample
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)
M. Spivak, Calculus, 4th ed., Ch. 8 (standard reference, not scraped)