Characterisation of Ordered Fields

Theorem
Let $(k, +, \cdot)$ be a field with unity $1$ and zero $0$.

Then the following are equivalent:


 * 1) There exists a total order $\leq$ on $k$ such that $(k,\leq)$ is a totally ordered field
 * 2) $-1$ cannot be written as a sum of squares of elements of $k$
 * 3) $0$ cannot be written as a non-empty sum of squares of non-zero elements of $k$

Proof
2. $\Leftrightarrow$ 3.:

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


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

Then


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

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

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


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

Then for any $j \in I$,


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

Dividing through by $x_j^2$ we find that


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

1. $\Rightarrow$ 2.:

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

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


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

2. $\Rightarrow$ 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. $\Leftrightarrow$ 1., $0, 1 \notin S$.

Trivially, $S$ is closed under addition. Also, for any subsets $\{x_i : i \in I\}$, $\{y_j : j \in J\} \subseteq k$,


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

so $S$ is closed under multiplication.

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

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 $k \backslash \{0\}$

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


 * $\displaystyle \bigcup_{i \in \N} M_i \in \Gamma $

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

Since clearly $0 \notin M$, if we define


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

then $M$, $\{0\}$ and $-M$ are pairwise disjoint, for if $x, -x \in M$ then $x - x = 0 \in M$, a contradiction.

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

The following two claims justify this statement.

Claim 1: $k = M \cup \{0\} \cup (-M)$

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


 * $ M' = \{ x + ay : x, y \in M \cup \{0\}, (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 $x + ay$, $z + aw \in M'$. Then


 * $(x + ay)(z + aw) = (xz + a^2yw) + a(yz + xw)$

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 $x$, $y$ are nonzero, $0 \notin M'$.

Since $1 \in S \subseteq M'$, $1 \in M'$.

If $t = x + ay \in M'$, then

Therefore every $t \in M'$ has a multiplicative inverse in $M'$, so $M'$ is a multiplicative subgroup of $k \backslash \{0\}$.

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

This and the fact that $M \subseteq M'$ imply that $M = M'$.

Therefore $a = 1 + 1a \in M$.

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

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

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

Equivalently, $x \leq y \Leftrightarrow (x < y) \lor (x = y)$

Claim 2: $(k, \leq)$ is a totally ordered field.

Proof: The proof is an elementary check of the relevant axioms.

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

Reflexivity : trivial from the definition.

Antisymmetry : Suppose that $x \leq y$ and $y \leq x$.

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 $y = x$.

Transitivity : Suppose $x \leq y$ and $y \leq z$.

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

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

Therefore $x \leq z$.

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

This is because if $x \nleq y$ then $y - x \notin M$ and $y - x \neq 0$. Then $y - x \in (-M)$, so $x - y \in M$ and $y \leq x$.

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

Let $x,y,z \in k$ with $x < y$.

Then $(y + z) - (x + z) = y - x \in M$.

So $x + z \leq y + z$.

Now suppose further that $z > 0$

Then $yz - xz = z(y - x) \in M$, because $z \in M$ and $y - x \in M$ by hypothesis, and $M$ is closed under multiplication.

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

Note
This proof is important not only for the result above, but for the discussion of the partition $M \cup \{0\} \cup (-M)$ 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


 * $(1)$: A field $k$ is a totally ordered field and $N \subseteq k$ is it's set of positive elements.


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

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