TITLE: Is $i$ well defined?
QUESTION [9 upvotes]: 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?

REPLY [13 votes]: 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.

REPLY [4 votes]: 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.