# Examples of non-orderable fields.

I wish to find examples of non-orderable fields. We know that fields with finite characteristics cannot be ordered, especially finite fields. Also $$\mathbb{C}$$ - the field of complex numbers cannot be ordered. Every time whenever $$0$$ can be written as a sum of squares, then the field is not orderable. If $$0=\sum_{i=1}^n x_i^2$$, then, every square is positive in an ordered field, and this would imply $$0>0$$, thus $$>$$ would not be an order. What are some more examples of non-orderable fields? Explicitly: We wish to find $$(F,+,\cdot,0,1)$$ on which we can't put a relation $$<$$ such that $$\forall x,y,z\in F: x $$\forall x,y\in F: (x>0 \wedge y > 0) \Rightarrow x\cdot y > 0$$

• According to the theory of formally real fields, a field admits an ordering if and only if $-1$ is not a sum of squares (this requires Zorn's lemma). This is equivalent to saying that $0$ is not a sum of nonzero squares. – egreg Mar 14 at 23:26

A field is orderable if and only if $$-1$$ is not a sum of squares (equivalently, if and only if $$0$$ is not a sum of nonzero squares).

These fields are called formally real.

Let $$F$$ be a formally real field. We will show that it is orderable. Call a subset $$P\subseteq F$$ a prepositive cone if it is closed under addition and multiplication, contains all squares, and does not contain $$-1$$.

Since $$F$$ is formally real, the set of all sums of squares is a prepositive cone. It is easy to see that this implies that the family of all prepositice cones in $$F$$ satisfies the hypothesis of Zorn's lemma (because it is nonempty, and the upper bound of a chain is simply the union), so there is a maximal prepositive cone $$P$$.

We claim that $$P$$ is a positive cone, i.e. it has the property that for every $$a\in F$$, either $$a\in P$$ or $$-a\in P$$. Indeed, suppose $$a,-a\notin P$$. Then by maximality of $$P$$, there is no prepositive cone containing $$P$$ with $$a$$ in it, so there are some $$p_1,p_2\in P$$ such that $$p_1+ap_2=-1$$ (note that since $$P$$ already contains all squares, the set of these expressions is closed under multiplication and addition) and likewise, we have $$p_1',p_2'$$ such that $$p_1'-ap_2'=-1$$. Then $$\begin{split} (-1)\cdot p_2 + (-1)\cdot p_2'&=p_1p_2'+ap_2p_2'+p_2p_1'-ap_2p_2'\\ & =p_1p_2'+p_2p_1'\in P. \end{split}$$

On the other hand, $$p_2+p_2'\in P$$ and the square $$(p_2+p_2')^{-2}\in P$$, so their product $$(p_2+p_2')^{-1}\in P$$, so this implies that $$(-p_2-p_2')\cdot (p_2+p_2')^{-1}=-1\in P$$, a contradiction.

Now, using the fact that $$P$$ is a positive cone, you can show that the relation $$a\leq b$$ iff $$b-a\in P$$ is an ordering.

• Thanks for these facts, I will find those useful, but I was rather looking for some explicit examples of non-orderable fields other than $\mathbb{C}$ and fields of finite characteristics. – Michal Dvořák Mar 14 at 23:35
• @MichalDvořák: For one, any extension of ${\mathbf Q}[i]$. But I suppose you are interested in examples where $-1$ is not a square. – tomasz Mar 14 at 23:49
• just for clarification, does prepositive cone contain $0$? – Michal Dvořák Mar 14 at 23:55
• $0$ is a square, so yes. To get a strict ordering just say that $a<b$ if $a\neq b$ and $b-a\in P$. Or just replace the notion of a prepositive cone with the notion of a strictly prepositive cone, as one which does not contain $0$, and contains all nonzero squares. – tomasz Mar 14 at 23:58
• @MichalDvořák: another funny characterisation of formally real fields is that a field $F$ is formally real iff there is some $K$ such that $F\subseteq K\subseteq \overline F$ (where $\overline F$ is the algebraic closure of $F$) and $[\overline{F}:K]$ is finite. – tomasz Mar 15 at 0:05