Total ordering on a field defined by a subset

Question

This is an exercise in my analysis class, which I only could finish partially.

Let $(F,+,\cdot)$ be a field and let $\mathbb{P} \subset F$ be a set with the following properties: 1) $0 \in \mathbb{P}$, 2) $x + y\in \mathbb{P}$ and $xy\in \mathbb{P}$ for all $x,y \in \mathbb{P}$ and 3) if $x, -x \in \mathbb{P}$ then $x = 0$.

I need to prove that there exists a unique order relation on this field such that the field becomes an ordered field and such that $F^+ = \mathbb{P}$ (i.e. the 'positive' elements are $\mathbb{P}$).

Question: I was able to prove everything, except that the order relation is unique. I defined $x \leq y$ if and only if $y - x \in \mathbb{P}$. This makes $\mathbb{P}$ the set of positive elements, defines a total order and hence makes $F$ an ordered field.

To prove that this relation is unique, do I have to assume that there is another such relation and show that they give exactly the same results? Or does uniqueness follow from the construction?

Answer 1

It is certainly not the unique order relation without the essential part “and such that $F^+ = \mathbb P$”. Once you have that, it is immediate. By definition, for any order relation $\leq$ on $F$ with positive elements in $\mathbb P$ and for any $x, y\in F$, we have $y-x\in\mathbb P \iff y-x\geq 0$, so the set of positive elements uniquely defines the order on $F$.

Answer 2

Suppose $\le_1$ and $\le_2$ are two orders such that $F^+=\Bbb P$.

Then for all $x,y \in F$:

$$x \le_1 y \iff y-x \ge_1 0 \iff y-x \in \Bbb P \iff y-x \ge_2 0 \iff x \le_2 y$$

where the first and last step use properties of ordered fields and the other steps use that $\Bbb P$ is the set of positive elements of $\le_1$ resp. $\le_2$.

So the order is unique (if $\Bbb P$ given).