# Do the field of complex numbers arise necessarily and uniquely as the only field of pairs of ordered real numbers retaining some desired proprieties?

Let me first assess that I'm not an expert on the subject, so I galdly welcome edits or suggestion, and don't be too mad at me if my assumptions are mistaken.
The field of complex number is a set of ordered pair of real numbers equipped with some additional proprieties ( which makes it a field indeed).
Now, let's say that we didn't come up with the idea of complex numbers through the study of polynomials.
Instead we want to create (for our own fun) a set of ordered pair of real numbers (x,y) with some additional structure/proprieties that makes it behave nicely as our field of real numbers and in addition, it has the property that the subset of all ordered pair (x,0) behave exactly as our beloved field of real numbers under any operation we take.
So, we want a field of ordered pair of real numbers such as:

1. It has all the proprieties of a field.
2. Its subset of all the ordered pairs (x,0) is indeed the field of real numbers.

I don't know if these assumptions are enough to make the field of complex numbers arise naturally(necessaarly and uniquely) or I'm neglecting some other conditions.

Am I missing out some desired proprieties? If yes, which?
Is legitimate to ask yourself this question as a consequnce of considering complex numbers an extension of real numbers?

• You have to be careful thinking about complex numbers as just "pairs of real numbers." In fact, $\mathbb{C}$ and $\mathbb{R}^{2}$ are not isomorphic (as rings, for example). – pwerth Dec 15 '18 at 17:17
• @pwerth Um... that was the entire point of the question, wasn't it. – fleablood Dec 15 '18 at 17:18
• I don't know enough about definitions in set theory to give a definite answer, but the fact that we can replace $i$ with, say $j=-i$ and get exactly the same theorems seems to me to mitigate against uniquenes, unless we're going to define $i$ and $j$ as "effectively the same" or simething. – timtfj Dec 15 '18 at 17:26
• The nasty thing is that the only distinction between $i$ and $j$ in my previous comment is that $i\neq j$. All other properties are the same for both. – timtfj Dec 15 '18 at 17:31
• The algebra of dual numbers comes pretty close to satisfying your requirements. They don't form a field, but: "Division of dual numbers is defined when the real part of the denominator is non-zero." Another near-miss is the algebra of split-complex numbers. – Calum Gilhooley Apr 2 at 22:48

A reasonable assumption should be that this field is an $$\mathbb{R}$$-algebra, so every element can be written as $$a(1,0)+b(0,1)$$ and we can identify $$(1,0)$$ with $$1$$. Set $$u=(0,1)$$. Then, in order to have a multiplication that defines the structure of (associative) $$\mathbb{R}$$-algebra we just need to decide what's $$u^2=p+qu$$.

In particular, we need that, for every $$a,b$$ (not both $$0$$), $$a+bu$$ is invertible, so there should exist $$x$$ and $$y$$ so that $$1=(a+bu)(x+yu)=ax+byu^2+(bx+ay)u=(ax+pby)+(bx+ay+qby)u$$ The linear system $$\begin{cases} ax+pby=1 \\[4px] bx+(a+qb)y=0 \end{cases}$$ must have a unique solution, so $$a(a+qb)-pb^2\ne0$$ It follows that the discriminant of $$z^2+qz-p$$ must be negative: $$q^2+4p<0$$.

Now let's try and find $$i=r+su$$ such that $$i^2=-1$$: $$-1=(r+su)^2=r^2+2rsu+s^2u^2=(r^2+ps^2)+(2rs+qs^2)u$$ Note that $$s\ne0$$, because $$r^2=-1$$ has no solution.

If $$q=0$$, we get $$r=0$$ and $$s=\pm\sqrt{-1/p}$$. If $$q\ne0$$, then $$s=-2r/q$$ and $$r^2+p\frac{4r^2}{q^2}=-1$$ yields $$r=\pm\sqrt{\frac{-q^2}{q^2+4p}}$$ In any case such $$i$$ exists.

Now it is clear that $$\{1,i\}$$ is a basis for $$\mathbb{R}^2$$ and that the field we get is isomorphic to the complex numbers.

• "A reasonable assumption should be that this field is an R-algebra" Why is that a reasonable assumption? Tomek Kania gives an answer where that is not at true. – fleablood Dec 15 '18 at 17:50
• @fleablood But $\{(x,0):x\in\mathbb{R}\}$ is not a subfield, of course, in that example. And, even if it were, it wouldn't be isomorphic to $\mathbb{R}$. – egreg Dec 15 '18 at 18:05
• You seem to be correct. – fleablood Dec 15 '18 at 18:06
• Since this is easier to understand for me, I like it. Anyway I'd like to see an answer where assumption are outlined in list ( If we reasonably assume this and this), then the conclusion is presented ( Given that, the only possibile field arising is that of complex numbers). Or if this is not possible motivating it. Thank you btw. – Gabriele Scarlatti Dec 15 '18 at 18:15
• @GabrieleScarlatti I'm guessing that the main missing assumption is a definition of something equivalent to $i$, maybe with some tweak to your existing assumptions. (I've voted up your comment in the hope that people see what kind of answer you're after.) – timtfj Dec 15 '18 at 19:03

As stated, the question is too vague to be precisely answered but let me give you some intuition that without further clarifications, one may construct multiplications on $$\mathbb R^2$$ that give fields not isomorphic to the complex numbers. Actually, you may get a field isomorphic to the real numbers.

Note that $$\mathbb{R}$$ is a vector space over $$\mathbb{Q}$$ of infinite dimension, so as an Abelian group (or even rational vector space) $$\mathbb{R}$$ is isomorphic to $$\mathbb{R}^2$$ because $$\mathbb{R}$$ and $$\mathbb{R}^2$$ have Hamel bases of the same cardinality when regared as rational vector spaces. Thus, you may take a group isomorphism $$\varphi\colon \mathbb R^2 \to \mathbb{R}$$ and define multiplication in $$\mathbb{R}^2$$ by $$(x_1, y_1) \ast (x_2, y_2) = \varphi((x_1, y_1))\cdot \varphi((x_1, y_1)).$$

Consequently, $$(\mathbb{R}^2, +, *)$$ is a field isomorphic to the field of real numbers, which is obviously not isomorphic to the field of complex numbers.

• Although I didn't understand the details of your answer, I understood your point: My assumptions are not enough to make the complex field arise. But I would like to know what is needed, and if it's a legitimate question to ask :) – Gabriele Scarlatti Dec 15 '18 at 17:30
• "one may construct multiplications on R2 that give fields not isomorphic to the complex numbers". Doesn't that answer the question with a definitive "No"? – fleablood Dec 15 '18 at 17:47
• @fleablood, it seems so :-) – Tomek Kania Dec 15 '18 at 17:49
• While this gives a different field, it doesn't quite fit condition 2 in the question. Namely, after twisting the structure with $\varphi$ the pairs $(x,0)$ are no longer isomorphic to the usual reals. They aren't even a subfield. – Jyrki Lahtonen Dec 15 '18 at 17:50
• The last claim follows from Artin-Schreier: If $L/K$ is a finite extension of fields, and $L$ is algebraically closed, then $[L:K]=2$, and you get $L$ by adjoining a square root of $-1$ to $K$. – Jyrki Lahtonen Dec 15 '18 at 17:53

Suppose we want to define a field $$\mathbb{X}$$ where the equation "$$x^2+1=0$$" has solutions. Quotation marks remind us that, in order for the above writing to be meaningful, we have to firstly define in $$\mathbb{X}$$ an addition "$$+$$", a multiplication "$$\times$$" (after all, "$$x^2$$" stands for "$$x \times x$$"), and their neutral elements ("$$0$$" and "$$1$$"), respectively.

We already know that $$(\mathbb{R},+,\times)$$ fails the task. So -with a certain hindsight, actually-, we address $$\mathbb{X}=\mathbb{R}^2$$ (as underlying set), and follow the program hereafter:

1. Define the addition between ordered pairs of real numbers, $$+$$, in the usual way.
2. Prove that $$(\mathbb{R}^2,+)$$ is an abelian group with neutral element $$0:=(0,0)$$.
3. Define the multiplication by scalars, $$*$$, in the usual way.
4. Prove that $$(\mathbb{R}^2,+)$$ is a vector space over $$\mathbb{R}$$; the set $$\lbrace 1:=(1,0),i:=(0,1) \rbrace$$ is a basis, then any $$z∈(\mathbb{R}^2,+)$$ can be written as $$z=x*1+y*i$$.
5. Define the (inner) multiplication $$\times$$:

-) by requesting:

a) it is distributive with $$+$$ and bilinear,

b) $$1$$ is its neutral element, and

-) by setting unrestrictedly: $$i×i=α*1+β*i$$ for some $$α,β∈\mathbb{R}$$ (after all, $$i×i$$ is by definition an element of the vector space $$(\mathbb{R}^2,+)$$).

Construction 1$$\div$$5 leads to the following definition for the multiplication $$\times$$ we are building up:

$$((x_1,y_1),(x_2,y_2))↦(x_1,y_1 )×(x_2,y_2 ):=(x_1x_2+αy_1y_2,x_1 y_2+y_1x_2+βy_1y_2) \tag 0$$

1. Prove that actually $$\times$$ is distributive with $$+$$ and bilinear, and $$1$$ is neutral element.
2. Prove that, moreover, $$\times$$ is associative (and abelian).
3. Prove that the zero-product property -that we want- holds if and only if: $$β^2+4α<0 \tag 1$$

4. Prove that, if the constraint $$(1)$$ is fulfilled, then $$|z|:= \sqrt{x^2-αy^2+βxy}$$ defines a norm in $$\mathbb{R}^2$$.

5. Define conjugate of $$z=(x,y)$$ the element $$\bar z := (x+βy,-y)$$.

6. Prove that , $$∀z≠0$$, the (only) multiplicative inverse is given by $$z^{-1} := \bar z /|z|^2$$.

7. Prove that, moreover, the following hold:

a) $$\bar{\bar z} := \overline{(\bar z)}=z$$

b) $$\overline{z_1+z_2} = \bar z_1 + \bar z_2$$

c) $$\overline{z_1×z_2} = \bar z_1 × \bar z_2$$

d) $$|\bar z|=|z|$$

e) $$|z_1×z_2|=|z_1 ||z_2 |$$

f) $$γ∈\mathbb{R} \Rightarrow \overline{γz}=γ \bar z$$

g) $$\overline{z^{-1}} = \bar z^{-1}, ∀z≠0$$

(Incidentally, this implies that $$z$$ is root of a polynomial with real coefficients if and only if $$\bar z$$ is.)

13. Prove that, if the constraint $$(1)$$ is fulfilled, then the equation $$z^2+w=0$$ has solutions in $$(\mathbb{R}^2,+,×)$$ whatever the constant term $$w∈(\mathbb{R}^2,+,×)$$ is: this was precisely the starting point that has triggered the whole.

Finally, there are uncountably many fields $$\mathbb{C}'(\alpha,\beta):=(\mathbb{R}^2,+,\times_{\alpha,\beta})$$ -where $$\times_{\alpha,\beta}$$ is defined by $$(0)$$ with $$\alpha, \beta$$ constrained by $$(1)$$- doing the job as well as $$\mathbb{C}:=\mathbb{C}'(-1,0)$$. The key point is that they are actually "one", being all of them isomorphic to each other and then to the "standard" $$\mathbb{C}$$.

I'm not qualified to give a full answer, but one way to proceed is to start off with the complex plane but instead of defining $$i$$ as $$\sqrt{-1}$$, define it as an operator which performs rotations: eg multiplying $$3$$ by $$i$$ moves it round by $$90°$$ to $$3i$$ on the imaginary axis.

I think that might be enough to get your field, without a polynomial in sight.

We'll have to arbitrarily decide whether the rotation is clockwise or anticlockwise. But that's no different from arbitrarily "choosing" which square root of $$-1$$ is called $$i$$.

I'll leave others to decide on the uniqueness or otherwise. (My gut feeling is that even the usual version of the complex numbers isn't fully defined since we can replace $$i$$ with $$-i$$ and everything still works; $$i$$ itself doesn't seem completely defined to me.)

• But there's no reason to assume that is the only possible field. – fleablood Dec 15 '18 at 17:46
• @fleablood I've edited to clarify that I'm not claiming uniqueness, just giving an alternative way to get the complex numbers. (The questioner wants to know what might need adding to his assumptions to achieve this without going via $\sqrt{-1}$, and something equivalent to $i$ is obviously necessary. I did use the word might.) – timtfj Dec 15 '18 at 18:21