Is every field of characteristic zero whose nonzero elements are squares or the additive inverse of squares but not both a Euclidean ordered field?

A Euclidean ordered field is an ordered field $$F$$ for which every positive element of $$F$$ is a square.

Now, call a field $$F$$ "quasi-Euclidean" if for any nonzero $$x \in F$$, exactly one of $$x$$ and $$-x$$ is a square. Then, this definition leads to the following question:

Is every "quasi-Euclidean" field of characteristic zero in fact a Euclidean ordered field?

If a "quasi-Euclidean" field $$F$$ is not a Euclidean ordered field, then there must be two nonzero elements $$x, y \in F$$ for which $$x^2+y^2$$ is not a square, and hence there must be a third nonzero element $$z \in F$$ for which $$x^2+y^2+z^2=0$$.

Note that there do exist "quasi-Euclidean" fields of nonzero characteristic, and of course, they cannot be ordered fields. For example, $$\mathbb{F}_p$$ (the Galois field of order $$p$$) is "quasi-Euclidean" for any prime $$p \equiv 3 \pmod 4$$.

The existence of lots of counterexamples follows from the following lemma.

Lemma: Let $$k$$ be a field in which $$-1$$ is not a square and let $$x\in k$$ be such $$-x$$ is not a square. Then $$-1$$ is not a square in $$k(\sqrt{x})$$.

Proof: If $$x$$ is a square then $$k(\sqrt{x})=k$$ so we are done, so let us assume $$x$$ is not a square. If $$(a+b\sqrt{x})^2=-1$$ for $$a,b\in k$$ then we must have $$2ab=0$$ and $$a^2+xb^2=-1$$. We cannot have $$b=0$$ since $$-1$$ is not a square in $$k$$, so we must have $$a=0$$ so $$xb^2=-1$$. But then $$-x=1/b^2$$ is a square in $$k$$, a contradiction.

Using this lemma, you can start with any field in which $$-1$$ is not a square and repeatedly adjoin square roots to preserve this property. In particular, you can iterate transfinitely to eventually get a field in which for all $$x$$, either $$x$$ or $$-x$$ is a square. Since $$-1$$ is not a square, both options cannot be true at once if $$x$$ is nonzero.

For instance, you could start with $$k=\mathbb{Q}$$ and then adjoin a square root of $$-2$$ as the first step (or a square root of $$-n$$ for any $$n>0$$ that is not a square in $$\mathbb{Q}$$). In the end you will get a quasi-Euclidean field in which $$-2$$ is a square.

• Nice! It's been a while since I went through transfinite inductions of this type, but it seems to work. In order not to leave my comfort zone I may prefer to do the Zorn's lemma thingy inside an algebraic closure of $\Bbb{Q}(\sqrt{-2})$, but I do believe that it is possible to define a suitable compatibillity condition in a more abstract setting, removing the need for extra comfort :-) Feb 11, 2022 at 5:23
• In the case of a countable field, you don't need to do it transfinitely--you can just do a dovetailing recursion indexed by $\mathbb{N}$ where you make sure you eventually hit every element and adjoin a square root of it or its negative. Feb 11, 2022 at 5:53
• @EricWofsey Yes, but you probably want to first embed in a larger field like the algebraic closure, so you can index all the elements of every extension that appears. (In fact I would hit each element infinitely many times, so you're sure to hit it after it's added to the field if it ever is.) Feb 11, 2022 at 12:57
• @WillSawin: Or to put it another way, algebraic closure of countable fields is easy. Feb 11, 2022 at 13:05