# A field without a canonical square root of $-1$

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.

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Am I missing something here? What is the canonical choice for $\sqrt{-1}$ in $\mathbb{C}$? – nullUser Aug 12 '12 at 0:30
The definition of $\mathbb{C}$ I had in mind was the set of formal expressions $a+ib$, with $a,b\in\mathbb{R}$. The element called $i$ is then the canonical choice. Did you have a different definition of $\mathbb{C}$ in mind? – Julian Rosen Aug 12 '12 at 0:56
$\mathbb{R}$ has many (isomorphic) algebraic closures. I was thinking of $\mathbb{C}$ as a particular choice of algebraic closure, given by a concrete description. – Julian Rosen Aug 12 '12 at 1:20
@Pink: then (as I suggested in my comment to Pete Clark's answer) you aren't talking about fields but about descriptions of fields. – Qiaochu Yuan Aug 12 '12 at 1:37
All the answers so far work with a somewhat informal definition of “canonical” — hence their somewhat unsatisfactory debatability. One possible formalisation of the question is “Is there a definition in ZFC of a field, ZFC-provably isomorphic to C, but with no ZFC-definable square root of –1?” (Feel free to replace ZFC with your preferred foundational system, of course.) – Peter LeFanu Lumsdaine Aug 12 '12 at 17:01

$\overline{\mathbb{Q}_5}$. The square roots of $-1$ actually already lie in $\mathbb{Z}_5$ and can be computed using Hensel's lemma but there's no reason to prefer the one congruent to $2 \bmod 5$ over the one congruent to $3 \bmod 5$ or vice versa.

I am not sure this would pass the test as it is possible that most mathematicians would give you the one congruent to $2 \bmod 5$ just because it is the first square root of $-1$ in $\mathbb{Z}/5\mathbb{Z}$ you find as you start from $0$ and add $1$, so take a larger prime congruent to $1 \bmod 4$ instead of $5$, maybe, to force them to try something smarter (say a prime large enough that it is not feasible to compute $\left( \frac{p-1}{2} \right)! \bmod p$).

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Note that the poster asks for a field "abstractly isomorphic to $\mathbb{C}$". – hardmath Aug 11 '12 at 22:51
@hardmath: $\overline{\mathbb{Q}_p}$ is abstractly isomorphic to $\mathbb{C}$. (More generally, any two algebraically closed fields of characteristic $0$ with the same uncountable cardinality are isomorphic.) – Qiaochu Yuan Aug 11 '12 at 23:03
Thanks, I was jumping to a wrong conclusion from the mention of $\mathbb{Z}_5$ that you'd have nonzero characteristic. It's a very interesting construction since the isomorphism as fields does not give a homeomorphism. – hardmath Aug 11 '12 at 23:13
@hardmath: it's much worse than that; my recollection is that it cannot even be made measurable (so its existence is probably independent of ZF). – Qiaochu Yuan Aug 11 '12 at 23:15
I hadn't noticed this answer until today! Very good! – paul garrett Aug 18 '12 at 21:00

You can model the complex numbers by linear combinations of the $2\times 2$ unit matrix $\mathbb{I}$ and a real $2\times 2$ skew-symmetric matrix with square $-\mathbb I$, of which there are two, $\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$. I see no obvious reason to prefer one over the other.

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This one seems very persuasive. I would expect a majority for $-1$ at the top right corner, but not necessarily an overwhelming majority. – André Nicolas Aug 12 '12 at 0:16

I don't feel that the question is really a coherent one. Let me try to illustrate that as follows.

Suppose I hand you the field $\overline{\mathbb{C}} = \{a - bi \ | \ a+bi \in \mathbb{C} \}$. It seems that in your sense of canonical, the canonical square root of $-1$ in this field is $-i$.

But wait: $\overline{\mathbb{C}}$ is not a(n even set-theoretically) different field from $\mathbb{C}$! It is just being presented differently: namely starting with the incarnation you usually have of $\mathbb{C}$ (which you have not specified in your question, but is rather implicit in its formulation that you have one) and compose it with the complex conjugation automorphism. The point here is that it doesn't make sense to distinguish between the fields $\mathbb{C}$ and $\overline{\mathbb{C}}$, only to distinguish between different identifications of these fields.

In summary, whenever your field has an automorphism which moves an element $x$ such that $x^2 = -1$, there is no canonical choice between $x$ and $-x$, at least not in any mathematically robust sense.

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I think "field" in the title should be replaced by "description of a field." – Qiaochu Yuan Aug 12 '12 at 1:01
If I were handed the field $\widehat{\mathbb{C}}$, I would compute that the two square roots of -1 in this field were called $i$ and $-i$, and I think I would still choose $i$ as the canonical one. – Julian Rosen Aug 12 '12 at 1:17
@Pink Elephants: I think you are missing the point. For instance, suppose I change notation and refer to $\mathbb{C}$ as the set of all numbers $a+bI$ such that $a,b \in \mathbb{R}$ and $I^2 = -1$. There is no way to tell from this description whether $I$ means $i$ or $-i$. Maybe your question has a psychological coherence to it, but mathematically there is no distinction to be made. – Pete L. Clark Aug 12 '12 at 7:07

A field abstractly isomorphic to $\mathbb{C}$ but does not have a canonical choice of square root of $-1$ is $\mathbb{C}$ itself. There is nothing canonical about which square root of $-1$ to call $i$ and which square root to call $-i.$ By calling one of the square roots $i$ you have just set up notation, nothing more. The fact that there is a field automorphism of $\mathbb{C}$ which interchanges the two square roots confirms this, as it preserves all algebraic structure. Even more, it preserves the natural metric structure of $\mathbb{C}.$

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$@$Geoff: yes, that is exactly what I was trying to say in my answer. – Pete L. Clark Aug 12 '12 at 0:27
@Pete: Yes, I think your answer appeared as I was writing mine. We are saying essentially the same thing, I agree. – Geoff Robinson Aug 12 '12 at 7:57

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.

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One such field is $\Bbb R[X]/(X^2+1)$. There's no good reason to choose $X$ over $-X$ (or vice versa) to map to $i$ in the automorphism to $\mathbb{C}$.

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I think almost everybody would pick $X$ as the "canonical" square root of -1 in this field – Julian Rosen Aug 11 '12 at 22:13
I don't think I would :) – Old John Aug 11 '12 at 22:17
I've always thought $-X$ was the more obvious choice. – Potato Aug 11 '12 at 22:19
@Potato: I know you are playing Devil's Advocate here, but I am going to play Devil's Devil's Advocate (God's Advocate?). When you write down $\mathbb{R}[x]/(x^2 + 1)$ you are really writing down the functor $\text{Hom}(\mathbb{R}[x]/(x^2 + 1), -)$ it represents, which sends an $\mathbb{R}$-algebra to the set of square roots of $-1$ it contains. This interpretation of the functor relies on the fact that we distinguish the element $x$ in the presentation; $\mathbb{R}[x]/(x^2 + 1)$ is precisely the universal $\mathbb{R}$-algebra containing a square root of $-1$ and $x$ is precisely the... – Qiaochu Yuan Aug 11 '12 at 22:33
Maybe this is silly, but how about $\mathbb{R}[\hat{x},\overline{x}]/(\hat{x}^2+1,\hat{x}+\overline{x})$ (I would call the variables $x$ and $y$, but maybe then $x$ is distinguished) – Julian Rosen Aug 11 '12 at 23:27

I think "Potato"'s answer is good, and "Pink Elephants"' response is also good, noting that "canonical" is not literally canonical. :) To manufacture another example that prevents an arguably natural choice of this sort, we might try to construct a field that contains a copy of $\mathbb C$, but which is "hidden" a little. For example, real matrices of the form $\pmatrix{a & b \cr -b & a}$?

While it is defensible to choose $a=1$, $b=1$ there, then what about $\pmatrix{a&-b\cr b & a}$? :)

Or, just varying Potato's example, $\mathbb R[x]/\langle x^2+2\rangle$?

Or from among the simple factors of $\mathbb R[x]/\langle x^4+4\rangle$?

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These matrices can be constructed abstractly as follows: let $V$ be a real $2$-dimensional inner product space. Then we obtain these matrices as the endomorphism algebra of $V$ as a representation of $\text{SO}(V)$. Choosing a square root of $-1$ more or less corresponds to choosing an orientation on $V$. – Qiaochu Yuan Aug 11 '12 at 22:38

The splitting field for $x^4+1$ over the reals.

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Would you care to elaborate a bit? I'm tired and not overly bright - don't make me think... – user1729 Sep 11 '12 at 10:58
@user1729 "splitting field" means the smallest field containing the real numbers and the roots of the polynomial $x^4+1$. – i. m. soloveichik Sep 11 '12 at 22:25
I know what it means, I just want you to explain where the ambiguity arises, and perhaps why the case of $x^4+1$ is different from $x^2+1$. – user1729 Sep 12 '12 at 11:11
It's the same as for $x^2+1$. You can't tell $i$ from $-i$. My example just shows that you could use the splitting field of another polynomial different from $x^2+1$ (in fact any polynomial which doesn't have just real roots). The roots of the polynomial x^4+1 are $\frac{1}{2}\pm\sqrt{2}\pm i\sqrt{2}$. – i. m. soloveichik Sep 12 '12 at 13:04
Then say it in your answer! – user1729 Sep 12 '12 at 13:06