Characterisation of Ordered Fields

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Let $\struct {k, +, \cdot}$ 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 $\struct {k, \le}$ is an 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 $k$


$(2)$ iff $(3)$

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

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


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

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

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

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

Then for any $j \in I$:

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

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

$\ds -1 = -\paren {\frac {x_j} {x_j} }^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } \paren {\frac {x_i} {x_j} }^2$

$(1)$ implies $(2)$

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

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

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

$(2)$ 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 $\set {x_i: i \in I}, \set {y_j: j \in J} \subseteq k$:

$\ds \paren {\sum_{i \mathop \in I} x_i^2} \cdot \paren {\sum_{j \mathop \in J} y_j^2} = \sum_{i, j} \paren {x_i y_j}^2 \in S$

so $S$ is closed under multiplication.

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

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 $k \setminus \set 0$

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

$\ds \bigcup_{i \mathop \in \N} M_i \in \Gamma$

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

Clearly $0 \notin M$.


$\paren {-M} := \set {x \in k : -x \in M}$

Suppose $x, -x \in M$.


$x - x = 0 \in M$

and this contradicts $0 \notin M$.

Hence $M$, $\set 0$ and $-M$ are pairwise disjoint.

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

The following two claims justify this statement.

Lemma 1

$k = M \cup \set 0 \cup \paren {-M}$

Proof of Lemma 1

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

$M' = \set {x + a y: x, y \in M \cup \set 0, \paren {x = 0 \lor y = 0} }$

We have:

$S \subseteq M \subseteq M'$

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

Let $x + a y$, $z + a w \in M'$.


$\paren {x + a y} \paren {z + a w} = \paren {x z + a^2 y w} + a \paren {y z + x w}$


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


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


\(\ds t \paren {\frac x {t^2} + a \frac y {t^2} }\) \(=\) \(\ds \frac x t + a \frac y t\)
\(\ds \) \(=\) \(\ds \frac {x + a y} {x + a y}\)
\(\ds \) \(=\) \(\ds 1\)

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

So $M'$ is a multiplicative subgroup of $k \setminus \set 0$.

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

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

$M = M'$


$a = 1 + 1 a \in M$

Let $-a \in M$.


$a \in -M$

so every $a \in k$ lies in exactly one of $M$, $\set 0$ and $-M$.

Thus $k = M \cup \set 0 \cup \paren {-M}$.


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


$x \le y \iff \paren {x < y} \lor \paren {x = y}$

Lemma 2

$\struct {k, \le}$ is an 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 $z - x = \paren {y - x} + \paren {z - y} \in M$.

Therefore $x \le z$.


The comparability of each pair of elements of $\struct {k, \le}$ is immediate from the above mentioned fact that $M$, $\set 0$ and $\paren {-M}$ partition $k$:

Suppose $x \nleq y$.


$y - x \notin M$


$y - x \ne 0$


$y - x \in \paren {-M}$


$x - y \in M$


$y \le x$

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 $x < y$.


$\paren {y + z} - \paren {x + z} = y - x \in M$


$x + z \le y + z$.

Now suppose further that $z > 0$.

We have that:

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


$M$ is closed under multiplication.


$y z - x z = z \paren {y - x} \in M$

This establishes that $k$ is an ordered field.




This proof is important not only for the result above, but for the discussion of the partition $M \cup \set 0 \cup \paren {-M}$ 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, $k = N \cup \set 0 \cup \paren {-N}$ and $k$ is ordered by $x < y \iff y - x \in N$.

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