$K$ is an ordered field $\leftrightarrow$ Subset $P \subseteq K$ exists with certain characteristics

Question

I need help for the following task. Unfortunately I don't have any idea how to start.

a) Prove: A field $K$ is an ordered $\leftrightarrow$ Subset $P \subseteq K$ exists with following features:

1. $\forall x,y \in P$ : $x+y \in P$

2. $\forall x,y \in P : x\cdot y \in P$

3. $\forall x \in K$ applies exactly one of these relationships:

$x = 0$, $x\in -P$, $x \in P$

Edit: A field is called ordered if it satisfies all these properties:

1. $\forall x,y \in \mathbb {R}$ applies exactly one of these relationships: $x=y, x<y$, or $x>y$ (trichotomy)

2. $\forall x,y,z \in \mathbb{R}$ applies $x<y,y<z \rightarrow x<z$ (transitive relation)

3. $\forall x,y,z \in \mathbb{R}$ applies: $x<y \rightarrow x+z<y+z$

4. $\forall x,y,z \in \mathbb{R}$ applies: $x<y,0<z \rightarrow xz<yz$ (I don't know the english names for these properties)

b) Let K be a field of rational expressions with real coefficients, that means expressions of the form $\frac{p}{q}$ where $p$ and $q\neq0$ are polynoms with real coefficients. Show that $K$ is ordered. (Hint: Use a) with $P$={$\frac{a_0 +a_1x+...+a_{n-1}x^{n-1}+a_nx^n}{b_0+b_1x+...+b_{n-1}x^{m-1}+b_mx^m}$ $\in K$ : $a_n \cdot b_m > 0, m, n \in \mathbb{N}$

My approach for b: https://i.stack.imgur.com/tafOz.png

I am thankful for any advice.

Answer 1

The question has essentially been answered in the comments, but I shall write it out explicitly for the sake of completeness.

It seems very likely that your professor wanted you to demonstrate that there are two different ways to understand ordered fields: firstly by means of a total order compatible with the field structure, and secondly by means of a 'positive cone' $P$ that tells you what the 'positive' elements should be. It looks like you (or your professor) missed out a crucial detail. It doesn't really make much sense to say that '$K$ is an ordered field if and only if' it satisfies some given properties, since there is no indication of what exactly $K$ is to start with.

It's a bit like saying that a 'thing' is an ordered field if it has said properties, without defining what this 'thing' is and what it what it means for it to have or not have these properties. There always has to be a clear, well-defined starting point for such questions, and in this case your starting point is assuming that $K$ is a field (which rules out $\mathbb{Z}$, as I mentioned in the comments).

Here is the 'field order' definition (field order is my own terminology and not in standard use so far as I am aware):

A field order on a field $K$ is a total order $\leq$ on $K$ satisfying

(i) If $x,y,z\in K$ and $x\leq y$, then $x+z\leq y+z$

(ii) If $x,y,z\in K$ and $0\leq z$, then $xz\leq yz$.

An ordered field is a field $K$ equipped with a field order.

Here is the 'positive cone' definition:

A positive cone for a field $K$ is a subset $P\subseteq K$ satisfying

(i) $x,y\in P\implies x+y\in P$

(ii) $x,y\in P\implies xy\in P$

(iii) If $x\in P$ and $-x\in P$, then $x=0$.

An ordered field is a field $K$ equipped with a positive cone $P$.

It is then fairly routine to check that there is a one-to-one correspondence between field orders on $K$ and positive cones of $K$:

The field order $\leq$ corresponds to the positive cone $\{x\in K:0\leq x\}$. The positive cone $P$ corresponds to the field order defined by declaring that $x\leq y$ if and only if $y-x\in P$.

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EDIT: If $K$ is an ordered field (by the first definition), then we show that $P:=\{x\in K:0\leq x\}$ is closed under addition as follows:

Let $x,y\in P$. Then $0\leq y$ and using (i), we have that $0\leq x\implies 0+y=y\leq x+y$. So $0\leq x+y$ by transitivity and then $x+y\in P$.

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See $\S$ 23.1 of Stewart, I., Galois Theory (Fourth Edition), CRC Press, 2015 and the Wikipedia article Ordered Field.