Examples of non-orderable fields.

Question

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

Answer 1

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.