# Characterisation of Totally Ordered Fields

## Contents

## Theorem

Let $\left({k, +, \cdot}\right)$ be a field with unity $1$ and zero $0$.

Then the following are equivalent:

- $(1): \quad$ There exists a total ordering $\le$ on $k$ such that $\left({k, \le}\right)$ is a totally ordered field
- $(2): \quad$ $-1$ cannot be written as a sum of squares of elements of $k$
- $(3): \quad$ $0$ cannot be written as a non-empty sum of squares of non-zero elements of $k$

## Proof

### $(2)$ iff $(3)$

Suppose there exist $\left\{ {x_i: i \in I}\right\}$ such that

- $\displaystyle -1 = \sum_{i \mathop \in I} x_i^2$

Then

- $\displaystyle 0 = 1^2 + \sum_{i \mathop \in I} x_i^2$

a non-empty sum of non-zero squared of $k$.

Conversely, suppose that there is a set $\left\{ {x_i: i \in I}\right\} \ne \varnothing$ with $x_i \ne 0$ for all $i \in I$ such that:

- $\displaystyle 0 = \sum_{i \mathop \in I} x_i^2$

Then for any $j \in I$:

- $\displaystyle -x_j^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } x_i^2$

Dividing through by $x_j^2$ we find that:

- $\displaystyle -1 = -\left(\frac{x_j}{x_j}\right)^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } \left({\frac {x_i} {x_j} }\right)^2$

### $(1)$ implies $(2)$

By Properties of Totally Ordered Field, in a totally ordered field we have, $-1 < 0$, and squares are non-negative.

Therefore for any subset $\{x_i : i \mathop \in I\} \subseteq k$,

- $\displaystyle -1 < 0 \le \sum_{i \mathop \in I} x_i ^2$

### $(2)$ implies $(1)$

Suppose that $-1$ is not a sum of squares in $k$.

Let $S$ be the set of non-empty sums of squares of non-zero elements of $k$.

Then by supposition and $(2) \iff (1)$:

- $0, 1 \notin S$

Trivially, $S$ is closed under addition.

Also, for any subsets $\left\{ {x_i: i \in I}\right\}, \left\{ {y_j: j \in J}\right\} \subseteq k$:

- $\displaystyle \left({\sum_{i \mathop \in I} x_i^2}\right) \cdot \left({\sum_{j \mathop \in J} y_j^2}\right) = \sum_{i, j} \left({x_i y_j}\right)^2 \in S$

so $S$ is closed under multiplication.

It follows that $S$ is a multiplicative subgroup of the set difference $k \setminus \left\{{0}\right\}$.

Now let $\Gamma$ be the collection of all subsets $M$ of $k$ such that:

- $S \subseteq M$
- $M$ is closed under addition
- $M$ is a multiplicative subgroup of $k \setminus \left\{ {0}\right\}$

Then every chain $\mathscr C = \left\{{M_i : i \in \N}\right\}$ has an upper bound:

- $\displaystyle \bigcup_{i \mathop \in \N} M_i \in \Gamma$

So by Zorn's Lemma there is a maximal element $M$ with these properties.

Clearly $0 \notin M$.

Let:

- $\left({-M}\right) := \left\{{x \in k : -x \in M}\right\}$

Suppose $x, -x \in M$.

Then:

- $x - x = 0 \in M$

and this contradicts $0 \notin M$.

Hence $M$, $\left\{ {0}\right\}$ and $-M$ are pairwise disjoint.

At this point the reader should think of $M$ as a partition of $k$ into positive elements, $\left\{ {0}\right\}$ and negative elements.

The following two claims justify this statement.

#### Lemma 1

$k = M \cup \left\{{0}\right\} \cup \left({-M}\right)$

#### Proof of Lemma 1

Let $a \in k$, $a \ne 0$, $-a \notin M$, and:

- $M' = \left\{ {x + a y: x, y \in M \cup \left\{ {0}\right\}, \left({x = 0 \lor y = 0}\right)}\right\}$

We have:

- $S \subseteq M \subseteq M'$

and it is trivial to check that $M'$ is closed under addition.

Let $x + ay$, $z + aw \in M'$.

Then:

- $\left({x + a y}\right) \left({z + a w}\right) = \left({x z + a^2 y w}\right) + a \left({y z + x w}\right)$

Since:

- $a^2 \in S \subseteq M'$

and:

- $M$ is closed under multiplication and addition

it follows that:

- $M'$ is also closed under multiplication.

Since $a$ and at least one of $x, y \ne 0$, it follows that:

- $0 \notin M'$

Since $1 \in S \subseteq M'$:

- $1 \in M'$

Let $t = x + a y \in M'$.

Then:

\(\displaystyle t \left({\frac x {t^2} + a \frac y {t^2} }\right)\) | \(=\) | \(\displaystyle \frac x t + a \frac y t\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {x + a y} {x + a y}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |

Therefore every $t \in M'$ has a multiplicative inverse in $M'$

So $M'$ is a multiplicative subgroup of $k \setminus \left\{{0}\right\}$.

Therefore $M'$ satisfies all the conditions subject to which we chose $M$.

This and the fact that $M \subseteq M'$ imply that:

- $M = M'$

Therefore:

- $a = 1 + 1 a \in M$

Let $-a \in M$.

Then:

- $a \in -M$

so every $a \in k$ lies in exactly one of $M$, $\left\{ {0}\right\}$ and $-M$.

Thus $k = M \cup \left\{ {0}\right\} \cup \left({-M}\right)$.

$\Box$

Now define a binary relation on $k$ by $x < y \iff y - x \in M$.

Equivalently:

- $x \le y \iff \left({x < y}\right) \lor \left({x = y}\right)$

#### Lemma 2

$\left({k, \le}\right)$ is a totally ordered field.

#### Proof of Lemma 2

The proof is an elementary check of the relevant axioms.

First we check that $\le$ is a total order on $k$.

Trivial from the definition.

Suppose that $x \le y$ and $y \le x$.

We cannot have $x - y \in M$ and $y - x \in M$, because since $M$ is closed under addition, this would imply $0 \in M$.

Therefore, $x - y = y - x = 0$ and $y = x$.

Suppose $x \le y$ and $y \le z$.

Then $y - x \in M$ and $z - y \in M$.

$M$ is closed under addition, so $z - x = \left({y - x}\right) + \left({z - y}\right) \in M$.

Therefore $x \le z$.

The comparability of each pair of elements of $\left({k, \le}\right)$ is immediate from the above mentioned fact that $M$, $\left\{ {0}\right\}$ and $\left({-M}\right)$ partition $k$:

Suppose $x \nleq y$.

Then:

- $y - x \notin M$

and:

- $y - x \ne 0$

Then:

- $y - x \in \left({-M}\right)$

so:

- $x - y \in M$

and:

- $y \le x$

Thus $x$ and $y$ are comparable.

It remains to show that the total order $\le$ is compatible with the field structure.

Let $x, y, z \in k$ such that $x < y$.

Then:

- $\left({y + z}\right) - \left({x + z}\right) = y - x \in M$

So:

- $x + z \le y + z$.

Now suppose further that $z > 0$.

We have that:

- $z \in M$ and $y - x \in M$ by hypothesis

and:

- $M$ is closed under multiplication.

Therefore:

- $y z - x z = z \left({y - x}\right) \in M$

This establishes that $k$ is a totally ordered field.

$\Box$

$\blacksquare$

## Note

This proof is important not only for the result above, but for the discussion of the partition $M \cup \left\{ {0}\right\} \cup \left({-M}\right)$ into positive elements, zero and negative elements.

In the proof above we have implicitly shown the following proposition:

$(1)$ if and only if $(2)$, where:

- $(1): \quad$ A field $k$ is a totally ordered field and $N \subseteq k$ is its set of positive elements.

- $(2): \quad$ $0 \notin N$, $N$ is closed under addition and multiplication, $k = N \cup \left\{ {0}\right\} \cup \left({-N}\right)$ and $k$ is ordered by $x < y \iff y - x \in N$.

and as above, $-N = \left\{ {x \in k : -x \in N}\right\}$.