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A field homomorphism of ordered fields need not preserve order

Statement refuted

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

The witness is the conjugation map on $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\} \subseteq \mathbb{R}$ , 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 $\mathbb{R}$ , a complete ordered field, with the positive square root $\sqrt{2}$ of $2$ .

[L1]

In $\mathbb{R}$ 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 $\sqrt{a} \ge 0$ 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 $\varphi$ satisfies $\varphi(x + y) = \varphi(x) + \varphi(y)$ , $\varphi(xy) = \varphi(x)\varphi(y)$ , and $\varphi(1) = 1$ (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 $\mathbb{R}$ , let $K = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\}$ ; then $(a + b\sqrt{2}) + (c + d\sqrt{2}) = (a + c) + (b + d)\sqrt{2}$ and $(a + b\sqrt{2})(c + d\sqrt{2}) = (ac + 2bd) + (ad + bc)\sqrt{2}$ 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 $(a + b\sqrt{2})^{-1} = \frac{a - b\sqrt{2}}{a^2 - 2b^2}$ , where $a^2 - 2b^2 \ne 0$ since otherwise $(a, b) \ne (0, 0)$ forces $b \ne 0$ 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 $a, b \in \mathbb{Q}$ is unique, for $a + b\sqrt{2} = a' + b'\sqrt{2}$ with $b \ne b'$ would give $\sqrt{2} = (a' - a)/(b - b') \in \mathbb{Q}$ .

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 $\mathbb{R}$ , hence an ordered field under the positive cone inherited from $\mathbb{R}$ .

step 1.1step 1.2L4

2.2

By the uniqueness in step 1.3, the map $\sigma : K \to K$ given by $\sigma(a + b\sqrt{2}) = a - b\sqrt{2}$ 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 $\sqrt{2} = (a^2 + 2b^2) + 2ab\sqrt{2}$ , whence step 1.3 forces $a^2 + 2b^2 = 0$ and $2ab = 1$ , impossible for real $a, b$ .

step 1.3algebra

3.1

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

step 2.2algebra

3.2

$\sigma$ is multiplicative: $\sigma((a + b\sqrt{2})(c + d\sqrt{2})) = (ac + 2bd) - (ad + bc)\sqrt{2} = (a - b\sqrt{2})(c - d\sqrt{2}) = \sigma(a + b\sqrt{2})\,\sigma(c + d\sqrt{2})$ .

step 2.2algebra

3.3

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

step 2.2

3.4

$\sigma(\sqrt{2}) = \sigma(0 + 1 \cdot \sqrt{2}) = -\sqrt{2}$ , 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, $\sigma$ satisfies the three homomorphism identities, so $\sigma : K \to K$ 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 $\sigma$ between ordered fields with $\sqrt{2} > 0$ in the domain yet $\sigma(\sqrt{2}) = -\sqrt{2} < 0$ by step 3.4, so $\sigma$ 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 $\sigma$ 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)