## Introduction to ordered fields

Let \(K\) be a field. An **ordering** of \(K\) is a subset \(P\) of \(K\) having the following properties:

- ORD 1
- Given \(x \in K\), we have either \(x \in P\), or \(x=0\), or \(-x \in P\), and these three possibilities are mutually exclusive. In other words, \(K\) is the disjoint union of \(P\), \(\{0\}\), and \(-P\).
- ORD 2
- If \(x, y \in P\), then \(x+y\) and \(xy \in P\).

We shall also say that \(K\) is **ordered by \(P\)**, and we call \(P\) the set of **positive elements**.

Let’s notice following properties of an ordered field:

- Since \(1 \neq 0\) and \(1=1^2=(-1)^2\) we see that \(1 \in P\).
- More generally, a non zero square is always positive because for \(x \neq 0\), \(x^2=(-x)^2\) and either \(x \in P\) or \(-x \in P\).

## A finite field cannot be ordered

Let us assume that \(K\) is ordered by \(P\). By **ORD 2**, it follows that \(p= 1+ \cdots + 1 \in P\), whence \(K\) has characteristic \(0\) and therefore cannot be finite.

## \(\mathbb{Q}\) is an infinite ordered field

We verify at once that \(\mathbb{Q}\) is ordered if we take as \(P\) the set of positive rational numbers.

This is the only possible ordering of \(\mathbb{Q}\). Suppose that \(K\) is ordered by \(P\). We have seen above that \(1 \in P\). Therefore \(n=1 + \cdots +1 \in P\), i.e. \(\mathbb{N^*} \subset P\).

If \(m\) is a positive integer, \(\frac{1}{m}=m \cdot \frac{1}{m^2} \in P\) by **ORD 2** and consequently \(\mathbb{Q}^*_+ \subset P\). A negative rational cannot belong to \(P\) because \(0\) would also belong to \(P\) in contradiction with **ORD 1**. Finally \(P=\mathbb{Q}^*_+\).

## \(\mathbb{C}\) is an infinite field that cannot be ordered

While \(\mathbb{Q}\) is an ordered field included in \(\mathbb{C}\), \(\mathbb{C}\) cannot be ordered. To show that this is the case, assume that \(\mathbb{C}\) is ordered by a subset \(P\). If \(\imath \in P\), then by **ORD 2**, \(\imath^2 = -1 \in P\) and \(\imath^4 = 1 \in P\). Still according to **ORD 2**, \(\imath^2 + \imath^4=0 \in P\) contradicting **ORD 1**. And if \(-\imath \in P\), we have the same contradiction as before because \((-\imath)^2 = \imath^2\).

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