# Is $i$ well defined? [duplicate]

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?

• $\mathbb R[i]$ is isomorphic to $\mathbb R[-i]$ – J. W. Tanner Aug 3 at 20:49
• 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" $-i$. – Richard D. James Aug 3 at 20:51
• I think OP might be asking why something like $i,-i$ might be less well-defined than something like $1,-1$. – Integrand Aug 3 at 20:52
• 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. – David A. Craven Aug 3 at 21:15
• @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. – Richard D. James Aug 3 at 21:48

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.

• 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 $-i$ – Conrad Aug 3 at 22:35
• 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 $\mathbb Z$. – GEdgar Aug 3 at 23:09

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.

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.

• 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 $\mathbb{R}[X] / (X^2 + 1)$. – Jair Taylor Aug 3 at 21:05
• But there is no canonical isomorphism $\mathbb{R}[X]/(X^2+1)\to \mathbb{C}$, so that doesn't help. – David A. Craven Aug 3 at 21:07
• It need not be canonical to be well-defined. It only needs to be unambiguous. – Jair Taylor Aug 3 at 21:08
• 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. – David A. Craven Aug 3 at 21:11
• 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. – Jair Taylor Aug 3 at 21:18

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.