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 $$\forall x,y,z\in F: x<y \Rightarrow x+z<y+z$$ $$\forall x,y\in F: (x>0 \wedge y > 0) \Rightarrow x\cdot y > 0$$

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    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Michal Dvořák Mar 14 at 23:35
  • $\begingroup$ @MichalDvořák: For one, any extension of ${\mathbf Q}[i]$. But I suppose you are interested in examples where $-1$ is not a square. $\endgroup$ – tomasz Mar 14 at 23:49
  • $\begingroup$ just for clarification, does prepositive cone contain $0$? $\endgroup$ – Michal Dvořák Mar 14 at 23:55
  • $\begingroup$ $0$ 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. $\endgroup$ – tomasz Mar 14 at 23:58
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    $\begingroup$ @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. $\endgroup$ – tomasz Mar 15 at 0:05

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