Tag Archives: ordered-fields

A non Archimedean ordered field

Let’s recall that an ordered field \(K\) is said to be Archimedean if for any \(x \in K\) there exists a natural number \(n\) such that \(n > x\).

The ordered fields \(\mathbb Q\) or \(\mathbb R\) are Archimedean. We introduce here the example of an ordered field which is not Archimedean. Let’s consider the field of rational functions
\[\mathbb R(x) = \left\{\frac{S(x)}{T(x)} \ | \ S, T \in \mathbb R[x] \right\}\] For \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) we can suppose that the polynomials have a constant polynomial greatest common divisor.

Now we define \(P\) as the set of elements \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) in which the leading coefficients of \(S\) and \(T\) have the same sign.

One can verify that the subset \(P \subset \mathbb R(x)\) satisfies following two conditions:

ORD 1
Given \(f(x) \in \mathbb R(x)\), we have either \(f(x) \in P\), or \(f(x)=0\), or \(-f(x) \in P\), and these three possibilities are mutually exclusive. In other words, \(\mathbb R(x)\) is the disjoint union of \(P\), \(\{0\}\) and \(-P\).
ORD 2
For \(f(x),g(x) \in P\), \(f(x)+g(x)\) and \(f(x)g(x)\) belong to \(P\).

This means that \(P\) is a positive cone of \(\mathbb R(x)\). Hence, \(\mathbb R(x)\) is ordered by the relation
\[f(x) > 0 \Leftrightarrow f(x) \in P.\]

Now let’s consider the rational fraction \(h(x)=\frac{1}{x} \in \mathbb R(x)\). \(h(x)\) is a positive element, i.e. belongs to \(P\). And for any \(n \in \mathbb N\), we have
\[h(x)-n=\frac{1}{x}-n=\frac{x-n}{x} \in P\] as the leading coefficients of \(x\) and \(x-n\) are both equal to \(1\). Therefore, we have \(h(x) > n\) for all \(n \in \mathbb N\), proving that \(\mathbb R(x)\) is not Archimedean.

A field that can be ordered in two distinct ways

For a short reminder about ordered fields you can have a look to following post. We prove there that \(\mathbb{Q}\) can be ordered in only one way.

That is also the case of \(\mathbb{R}\) as \(\mathbb{R}\) is a real-closed field. And one can prove that the only possible positive cone of a real-closed field is the subset of squares.

However \(\mathbb{Q}(\sqrt{2})\) is a subfield of \(\mathbb{R}\) that can be ordered in two distinct ways. Continue reading A field that can be ordered in two distinct ways

An infinite field that cannot be ordered

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. Continue reading An infinite field that cannot be ordered