TITLE: A field without a canonical square root of $-1$
QUESTION [12 upvotes]: The following is a question I've been pondering for a while.  I was reminded of it by a recent dicussion on the question How to tell $i$ from $-i$?

Can you find a field that is abstractly isomorphic to $\mathbb{C}$, but that does not have a canonical choice of square root of $-1$?

I mean canonical in the following sense: if you were to hand your field to one thousand mathematicians with the instructions "Pick out the most obvious square root of -1 in this field. Your goal is to make the same choice as most other mathematicians," there should be be a fairly even division of answers.

REPLY [6 votes]: The answer to one possible formalization of the question is yes.  It is consistent that there is a definable field isomorphic to $\mathbb{C}$ with no definable imaginary unit.  To construct such a thing, we need a pair of sets $\{X,Y\}$ that is definable, but such that neither $X$ nor $Y$ is definable.  It is consistent for such a strange object to exist: see François's answer to the question Does every nonempty definable finite set have a definable member?)
Now let $\mathbb{C}$ be your favorite construction of the complex field.  One of its imaginary units, $i$, is definable, or else we are done. Define the structure $F = (\mathbb{C} \times \{X,Y\}) / \mathord{\sim}$ where $(a+bi, X) \sim (a-bi, Y)$ for all $a,b \in \mathbb{R}$.  Then $F$ is a definable structure isomorphic to $\mathbb{C}$ and its imaginary units are $\{(i,X),(-i,Y)\}$ and $\{(-i,X),(i,Y)\}$.  Neither of these imaginary units can be definable, or else $X$ and $Y$ would be definable from $i$ and therefore definable.