Is it useful that $k(X)$ is an ordered field if $k$ either is?

Question

I have recently found that if $k$ is an ordered field, $k(X)$ is also an ordered field. Proof below, in the proof that order is defined.

My question:

I have been never told that a field of algebraic fractions is ordered. Nor in High School or in University. So I have never seen a use of this fact. Is it really useless?

The proof:

I'll use the positive and negative subsets way. So an ordered field $k$ has a subset $P$ such that $0$ is not in $P$, $P+P\subset P$, $P\cdot P\subset P$ and $(k\setminus P)\cdot(k\setminus P)\subset P\cup\{0\}$. And the order is defined $$x\ge y\iff x-y\in P\cup\{0\}$$ Reflexive, antisymmetric and transitive properties are straightforwardly proven. It is clear that $x\in P\iff x>0$. We have also that if $a,x,y\in k$,
$$\text{If }x-y, y-x\notin P\cup\{0\}\text{ then }(x-y)^2+(x-y)(y-x)=0\notin P\text{, a contradiction, so the order is total}$$ $$\text{If }x\ge y\text{ then }(a+x)-(a+y)=x-y\in P\cup\{0\}\text{ so }a+x\ge a+y$$ $$\text{If }x,y>0\text{ then }xy>0$$ Now, assuming that such a $P$ is defined in $k$, define $$P_0'=\left\{ \sum_{j=0}^n a_jX^j: n\in\Bbb Z_{\ge 0}\wedge \big(\forall j((j\in\Bbb Z\wedge 0\le j\le n) \to a_j\in k\big)\wedge a_n\in P\right\}$$ That is, $P_0'$ is the set of polynomials with positive leading coefficient. Now define $$N_0'=k[X]\setminus(P_0'\cup\{0\})$$ Now define $P'$ as the set of fractions $p(X)/q(X)$ where $p$ and $q$ are both in $P_0'$ or both in $N_0'$. It is easy to see that the conditions for $P'$ are satisfied, and this completes the proof.

Answer 1

You have the definition of a "positive cone" wrong. The positive cone $P$ must be closed under addition and multiplication, must exclude $-1$ (and not $0$!) and all squares in the field must belong to $P$.

In fact, your conditions:

• $P$ is closed under multiplication and addition.
• $P$ excludes zero.

are satisfied by $\mathbb K\setminus\{0\}$ for any field $\mathbb K$.

I don't know if your original statement (the field of fractions of an ordered field is ordered) is true or false.

Answer 2

Your statement is known, I don't know any interesting consequence of this.

An alternate way to see this is the following:

Assume by contradiction that $k(X)$ is not ordered. Then, there exists rational fractions $Q_1,..,Q_n$ such that $$-1=(Q_1)^2+...+(Q_n)^2$$

Next, since $k$ is ordered, it has infinite characteristic. Therefore, there exists some $a \in k$ such that all the denominators in $Q_1,..,Q_n$ are non-zero at $a$.

This gives $$-1=(Q_1(a))^2+...+(Q_n(a))^2$$ which shows that $k$ is NOT ordered.

Answer 3

For instance for $k=\mathbb{Q}$, this gives the "simplest" example of non-archimedean ordered field. In fact, for every cut in $k$, there is an order on $k(X)$ or one of its quotient fields such that $X$ is made to lie in the cut.