Is $i$ well defined?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.