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I know, it may sound as nothing but a provocative question, and probably it is. However I've been thinking about it for a while, despite being aware that the question itself may not have much sense.

Consider the field $\mathbb{R}$. Each element can be defined univocally. First $0$ and $1$, then the integers, so the rationals and then all the others (for instance as equivalence classes of Cauchy sequences on $\mathbb{Q}$).

Now we can define the complex field $\mathbb{C}$ as $$\mathbb{C} = \mathbb{R}[X]/(X^2+1)$$ where $\mathbb{R}[X]$ is the ring of polynomials with real coefficient. However here it becomes impossible to univocally define a root of the polynomial $X^2+1$ since it has two roots (which we will eventually call $\pm i$) and they are totally indistinguishable. I know that in practice it's not a problem, we just decide to call one of the two roots $i$ and the other $-i$. But what's going on exactly? Is it some kind of "axiom" the fact that we are allowed to choose one out of a set of two identical elements?

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    $\begingroup$ $\mathbb R[i]$ is isomorphic to $\mathbb R[-i]$ $\endgroup$ – J. W. Tanner Aug 3 at 20:49
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    $\begingroup$ I don't really understand your question. The image of $X$ in $\mathbb{R}[X]/(X^2+1)$ "is" $i$; the image of $-X$ "is" $-i$. $\endgroup$ – Richard D. James Aug 3 at 20:51
  • $\begingroup$ I think OP might be asking why something like $i,-i$ might be less well-defined than something like $1,-1$. $\endgroup$ – Integrand Aug 3 at 20:52
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    $\begingroup$ I think the question does make sense. At least, it does in terms of Galois theory. If you just look inside the field $\{a+b\sqrt2\mid a,b\in\mathbb{Q}\}$ you cannot distinguish between $\sqrt2$ and $-\sqrt2$ either. $\endgroup$ – David A. Craven Aug 3 at 21:15
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    $\begingroup$ @ECL No, not really. There is a canonical quotient map $\mathbb{R}[X] \to \mathbb{R}[X]/(X^2+1)$ that maps $f(X)$ to the class of $f(X)$ mod $(X^2+1)$. There is no choice involved: that's more or less the meaning of "natural" or "canonical". There is another ring homomorphism $\mathbb{R}[X] \to \mathbb{R}[X]/(X^2+1)$, namely the one mapping $f(X)$ to $f(-X)$ mod $(X^2+1)$, but that's not the canonical quotient map. $\endgroup$ – Richard D. James Aug 3 at 21:48
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In the plane with an orientation, we can distinguish $i$ from $-i$. So with that additional structure, $i$ is well defined.

In the field $\mathbb Q[\sqrt2]$, can we distinguish the two square roots of $2$ from each other? Not unless we add additional structure to do it.

In the group $\mathbb Z$, can we distinguish the two generators $1$ and $-1$ from each other? Not unless we add additional structure to it.

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  • $\begingroup$ I disagree a bit with your answer as $\mathbb R$ has a well defined positive cone ($1$ is a multiplicative unit so is distinguishable from $-1$ etc) so one can distinguish between $\sqrt 2$ and $-\sqrt 2$ for example in an intrinsic coordinate free way, namely by definition $\sqrt 2$ is the unique real solution of $x^2=2$ that lies in the positive cone of $\mathbb R$; on the other hand there is no coordinate free way to distinguish between $i$ and $-i$ $\endgroup$ – Conrad Aug 3 at 22:35
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    $\begingroup$ Of course "positive cone of $\mathbb R$" is an example of what I mean by "additional structure". We can say "the rotation about $0$ that transforms $1$ to $i$ in the plane is counterclockwise". This is just as acceptable (to me) as specifying one of the two embeddings of $\mathbb Q[\sqrt2]$ into the reals, or choosing an ordering for $\mathbb Z$. $\endgroup$ – GEdgar Aug 3 at 23:09
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It's well-defined in the sense that you can define $\mathbb{C}$ perfectly well without any reference to the "square root of $-1$", just by defining a complex number to be a pair of real numbers $(a,b)$ with the operations $(a,b) + (c,d) = (a+b, c+d)$ and $(a,b)(c,d) = (ac - bd, ad + bc)$. If we then decide to write the pair $(a,b)$ as $a + bi$ for syntactic sugar, then the number written as $i$ is perfectly well-defined as the pair $(0,1)$.

Of course, as the other answers have noted, the fact that $a + bi \mapsto a-bi$ is a field automorphism of $\mathbb{C}$ means there's no "principled", algebraic way of telling the two apart.

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No, it is not well-defined. The reason is that complex conjugation is a field automorphism of $\mathbb{C}$. This means that the act of complex conjugation respects multiplication and addition. So any statement using field operations and the real numbers that holds for $\mathrm{i}$ also holds for $-\mathrm{i}$.

If you want to make it well defined, you need something that breaks complex conjugation, and thus separates $\mathrm{i}$ from $-\mathrm{i}$. Putting an orientation on the complex plane will do that for you, but that is putting the cart before the horse somewhat, because it presupposes that you have chosen $\mathrm{i}$.

Edit: there appears to be some issue around the definition of 'well-defined'. I am taking as my definition that there is a description of it that uniquely determines it using properties of the field. Any definition of $\mathrm{i}$ that you can come up with will equally apply to $-\mathrm{i}$, and in that sense it is not well-defined.

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    $\begingroup$ Just because something is not an isomorphic invariant doesn't mean it's not well-defined. $i = X$ is a perfectly well-defined element of $\mathbb{R}[X] / (X^2 + 1)$. $\endgroup$ – Jair Taylor Aug 3 at 21:05
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    $\begingroup$ But there is no canonical isomorphism $\mathbb{R}[X]/(X^2+1)\to \mathbb{C}$, so that doesn't help. $\endgroup$ – David A. Craven Aug 3 at 21:07
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    $\begingroup$ It need not be canonical to be well-defined. It only needs to be unambiguous. $\endgroup$ – Jair Taylor Aug 3 at 21:08
  • $\begingroup$ How do you define an isomorphism between the two structures without first defining $\mathrm{i}$? I have edited my question to define what I mean by well-defined. I guess this comes down to which definition of that term you would use. $\endgroup$ – David A. Craven Aug 3 at 21:11
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    $\begingroup$ Yes, it's just a discussion of the terminology, other than that we agree. I'd just define $i$ (arbitrarily) as $X$ in $\mathbb{C} = \mathbb{R}[x]/(X^2+1)$. Why not? Alternately, define $\mathbb{C} = \mathbb{R}^2$ with $i = (0,1)$ and appropriate definitions for addition and multiplication. You have to have some definition of $\mathbb{C}$ and some unambiguous definition of $i$. Once you have that defined, then you can discuss its various automorphisms. $\endgroup$ – Jair Taylor Aug 3 at 21:18
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If by "well-defined" you mean "distinguishable from -i without making a choice," then the answer is no. But this is true of many things on some level, is it not? "Right" is not well-defined, and for that reason the cross-product is not well-defined in this sense. Someone, at some point, had to create a convention. When the complex plane was defined, it likely made sense to make the positive imaginary numbers "up."

Sign conventions are notoriously annoying, in particular in electromagnetism.

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