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A field homomorphism of ordered fields need not preserve order

Statement refuted

Refuted claim: every field homomorphism $φ : F \to G$ between ordered fields is order-preserving, that is, $a < b$ in $F$ implies $φ (a) < φ (b)$ in $G$ .

The witness is the conjugation map on $Q (\sqrt{2}) = {a + b \sqrt{2} : a, b \in Q} \subseteq R$ , which is a field homomorphism from an ordered field to itself yet sends the positive element $\sqrt{2}$ to the negative element $- \sqrt{2}$ .

Facts & Assumptions

Given: The reals $R$ , a complete ordered field, with the positive square root $\sqrt{2}$ of $2$ .

[L1]

In $R$ the element $2 = 1 + 1$ has a positive square root $\sqrt{2}$ with $(\sqrt{2})^{2} = 2$ and $\sqrt{2} > 0$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L2]

No rational number squares to $2$ , so $\sqrt{2}$ is irrational (FALSE: some rational number squares to 2).

[L3]

A field homomorphism $φ$ satisfies $φ (x + y) = φ (x) + φ (y)$ , $φ (x y) = φ (x) φ (y)$ , and $φ (1) = 1$ (Field homomorphism and embedding).

[L4]

In an ordered field, $x > 0$ means $x$ lies in the positive cone, and $a < b$ means $b - a > 0$ ; exactly one of $x > 0$ , $x = 0$ , $x < 0$ holds (Ordered field).

[L5]

A field homomorphism from a complete ordered field into an ordered field is order-preserving (Homomorphisms out of a complete ordered field are order-preserving).

Counterexample

technique · direct

1.1

Working inside $R$ , let $K = {a + b \sqrt{2} : a, b \in Q}$ ; then $(a + b \sqrt{2}) + (c + d \sqrt{2}) = (a + c) + (b + d) \sqrt{2}$ and $(a + b \sqrt{2}) (c + d \sqrt{2}) = (a c + 2 b d) + (a d + b c) \sqrt{2}$ both lie in $K$ , so $K$ is closed under addition and multiplication.

L1algebra

1.2

Each nonzero $a + b \sqrt{2} \in K$ is invertible in $K$ , with $(a + b \sqrt{2})^{- 1} = \frac{a - b \sqrt{2}}{a^{2} - 2 b^{2}}$ , where $a^{2} - 2 b^{2} \neq 0$ since otherwise $(a, b) \neq (0, 0)$ forces $b \neq 0$ and the rational $(a / b)^{2} = 2$ , contradicting [L2].

L2algebra

1.3

The representation of an element of $K$ as $a + b \sqrt{2}$ with $a, b \in Q$ is unique, for $a + b \sqrt{2} = a^{'} + b^{'} \sqrt{2}$ with $b \neq b^{'}$ would give $\sqrt{2} = (a^{'} - a) / (b - b^{'}) \in Q$ .

L2algebra

1.4

In $K$ the element $\sqrt{2} = 0 + 1 \cdot \sqrt{2}$ satisfies $\sqrt{2} > 0$ .

L1L4

2.1

By steps 1.1 and 1.2, $K$ is a subfield of $R$ , hence an ordered field under the positive cone inherited from $R$ .

step 1.1step 1.2L4

2.2

By the uniqueness in step 1.3, the map $σ : K \to K$ given by $σ (a + b \sqrt{2}) = a - b \sqrt{2}$ is well defined.

step 1.3

2.3

The real number $2^{1 / 4} = \sqrt{\sqrt{2}}$ is not in $K$ , for $2^{1 / 4} = a + b \sqrt{2}$ would square to $\sqrt{2} = (a^{2} + 2 b^{2}) + 2 a b \sqrt{2}$ , whence step 1.3 forces $a^{2} + 2 b^{2} = 0$ and $2 a b = 1$ , impossible for real $a, b$ .

step 1.3algebra

3.1

$σ$ is additive: $σ ((a + b \sqrt{2}) + (c + d \sqrt{2})) = (a + c) - (b + d) \sqrt{2} = (a - b \sqrt{2}) + (c - d \sqrt{2}) = σ (a + b \sqrt{2}) + σ (c + d \sqrt{2})$ .

step 2.2algebra

3.2

$σ$ is multiplicative: $σ ((a + b \sqrt{2}) (c + d \sqrt{2})) = (a c + 2 b d) - (a d + b c) \sqrt{2} = (a - b \sqrt{2}) (c - d \sqrt{2}) = σ (a + b \sqrt{2}) σ (c + d \sqrt{2})$ .

step 2.2algebra

3.3

$σ$ fixes the identity: $σ (1) = σ (1 + 0 \cdot \sqrt{2}) = 1$ .

step 2.2

3.4

$σ (\sqrt{2}) = σ (0 + 1 \cdot \sqrt{2}) = - \sqrt{2}$ , and $- \sqrt{2} < 0$ in $K$ by step 1.4.

step 2.2step 1.4L4

4.1

By steps 3.1, 3.2, and 3.3, $σ$ satisfies the three homomorphism identities, so $σ : K \to K$ is a field homomorphism between ordered fields.

step 3.1step 3.2step 3.3L3

5.1

Steps 2.1 and 4.1 exhibit a field homomorphism $σ$ between ordered fields with $\sqrt{2} > 0$ in the domain yet $σ (\sqrt{2}) = - \sqrt{2} < 0$ by step 3.4, so $σ$ is not order-preserving, refuting the claim that every field homomorphism between ordered fields is order-preserving.

step 2.1step 4.1step 3.4L4

6.1

There is no conflict with [L5]: if $K$ were complete, [L5] would make $σ$ order-preserving, contrary to step 5.1. Hence $K$ is not complete.

step 5.1L5∎

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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)