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 $p \equiv 3 \pmod 4$.


1 Answer 1


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.

  • $\begingroup$ 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 :-) $\endgroup$ Feb 11, 2022 at 5:23
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    $\begingroup$ 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. $\endgroup$ Feb 11, 2022 at 5:53
  • $\begingroup$ @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.) $\endgroup$
    – Will Sawin
    Feb 11, 2022 at 12:57
  • $\begingroup$ @WillSawin: Or to put it another way, algebraic closure of countable fields is easy. $\endgroup$
    – user21820
    Feb 11, 2022 at 13:05

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