What are the axioms that differentiate a complex field from a real field? Excluding one from the other.

Is it fair to say that:

a. A complex field is a field with the extra (?) axiom has an element $i$ such that $i^2 + 1 = 0$

b. A real field is a field where such element does not exists (?)

I am trying to find a definition that is mutually exclusive between a real and a complex field. However I have two problems:

  1. It is weird to define a field (case b) by something that it doesn't have. Also it can't be that complex filed adds an axiom, if so the definition doesn't seem to be exclusive.

  2. When defining the field of reals, the non existence of element $i$, is not mentioned (although I guess one can demonstrate by some means that $\forall a \neq 0, a^2 > 0$, which in turn is like saying $\nexists i / i^2 = -1$).

Finally, how does conjugation enters in the picture? can the existence of a conjugation itself belong to an axiom? For example $\forall z, \exists \bar z / z\cdot\bar z > 0$.

  • 2
    $\begingroup$ The key points about the real numbers is they are totally ordered (unlike complex numbers) and Dedekind-complete (unlike rational numbers) $\endgroup$ – Henry May 10 '18 at 22:14
  • $\begingroup$ I didn't want to be specific of "real numbers" or "complex numbers" because in principle the abstract question can be applied to prime-fields (e.g. finite field). $\endgroup$ – alfC May 10 '18 at 22:20
  • $\begingroup$ Defintion (b) is really bad. There are lots of fields without the square root of $-1$, including finite fields, which have nothing to do with reals. Definition (a) is not good as well because of exactly the same reason: there are lots of fields extending reals with a solution to $x^2+1=0$. $\endgroup$ – freakish May 10 '18 at 22:22
  • $\begingroup$ One possible axiomatic approach is to define reals axiomatically and then define complex numbers as an algebraic closure of reals. $\endgroup$ – freakish May 10 '18 at 22:25
  • $\begingroup$ @Henry, I am not saying that it is incorrect, but order is odd also because it is not part of the definition of either field. By the way, does a prime field have order? $\endgroup$ – alfC May 11 '18 at 1:30

In short: The reverse way is more natural, indeed. A complex field is an extension of real field containing element $i$. It is also a historical reason. Complex numbers were introduced later than reals. So a., not exactly, by yes, although b. would be very strange. There doesn't exist also a turkey, a pineaple, an oak, but they definitely do not define reals.

  • $\begingroup$ Ok, but still, what is the axiom of a real field that forbids the existence of $i$ (in the real field)? In other words, if the complex field is defined as the axioms of the real field plus the existence of $i$, then the definitions are not mutually exclusive. $\endgroup$ – alfC May 10 '18 at 22:21
  • $\begingroup$ @alfC But the field means, in short, two abelian groups and distributivity of multiplication over addition. If we add $i$ to reals, preserving axioms of a field, we obtain an extension of reals, which is the set of complex numbers. $\endgroup$ – Przemysław Scherwentke May 10 '18 at 22:27

We define $\mathbb{C}$ as a field extension of $\mathbb{R}$, i.e. $\mathbb{C}:=\mathbb{R}[x]/(x^2+1)$, where $x^2+1$ is $\mathbb{R}$-irreducible i.e. there is no root to this polynomial in $\mathbb{R}$. So it may help to think in terms of the existence of roots of polynomials here. The conjugation is just an arithmetic result of the definition of $i$.


I am a bit unsure what you mean by a complex field and a real field, but I will assume you mean the field of real numbers $(\mathbb{R}, +, \times)$ and the field of complex numbers $(\mathbb{C}, +, \times)$ (which I will henceforth refer to as $\mathbb{R}$ and $\mathbb{C}$, respectively).

The key distinction is that $\mathbb{R}\subset\mathbb{R}(i)=\mathbb{C}$, that is the field of complex numbers is equal to the field of real numbers adjoin $i$, where $i^2+1=0$, as you mentioned. This is both necessary and sufficient to define $\mathbb{C}$.

In response to your hesitance to differentiate $\mathbb{C}$ from $\mathbb{R}$ based on $i$ alone, I would say that you have to think about defining the sets (and the the fields based on sets) inductively, from particular terms to (more) general terms. You start by understanding $\mathbb{N}$ and defining $\mathbb{Z}$ based on $\mathbb{N}$, and likewise with $\mathbb{Q}$ and $\mathbb{R}$, building upon what you already know in a slow progression. This implies you understand what $\mathbb{R}$ is prior to any concept of what a complex number is, and should therefore approach this problem as defining/constructing $\mathbb{C}$ from $\mathbb{R}$, not vice versa.

  • $\begingroup$ You don't have to start from naturals and build each structure at each step to reach reals. There's the axiomatic approach and I assume that OP is looking for something similar to complex numbers. And thus I don't find your answer helpful. $\endgroup$ – freakish May 10 '18 at 22:29
  • $\begingroup$ I am trying to build an vector space from a field. An depending on the field determine if it is a real space or a complex space. I am also under the impression that a prime finite field is considered a real field. $\endgroup$ – alfC May 10 '18 at 23:20
  • $\begingroup$ @freakish I understand that one can define the reals without visiting each previous step, I was attempting to point out that one understands what the reals are based on what the quotients are, and what the quotients are... etc. and that the author should be using the reals as a base from which to define the complex numbers, since (b) gave me the impression the author was instead defining the reals based on the complex numbers. $\endgroup$ – simonv May 11 '18 at 0:54
  • $\begingroup$ @simonv, yes, I am looking for an axiomatic independent definition of a complex field and a real field that do not depend on each other and are exclusive. It could be that it is not possible since the existence of a complex field may imply the existence of a real subfield. (also think of prime fields as well, not necessary the "reals"). $\endgroup$ – alfC May 11 '18 at 1:29
  • $\begingroup$ @alfC My bad then, I misunderstood the question. $\endgroup$ – simonv May 11 '18 at 8:51

There is extensive literature on formally real fields, which basically are fields that can be ordered (in the sense of becoming an ordered field). Artin and Schreier proved that this condition is equivalent to $-1$ not being a sum of squares.

A formally real field is called real closed if it doesn't admit algebraic extensions which are formally real fields. A consequence of Zorn's lemma is that any formally real field has an algebraic extension which is real closed. Moreover, if $F$ is real closed, then $x^2+1$ is irreducible in $F[x]$ and, if we adjoin a root $i$ of $x^2+1$, then $F[i]$ is algebraically closed.

The simplest example of a real closed field is the field of real numbers, but also the field of real numbers which are algebraic over $\mathbb{Q}$ is real closed.

  • $\begingroup$ Do you really mean "can be ordered" or do you mean "can be ordered under certain conditions" (like order preserved under addition). I would guess that any pair of comparables can be ordered just by comparing the first coordinate, and if that results in equivalency then compare the second. $\endgroup$ – DanielV May 10 '18 at 22:22
  • $\begingroup$ @DanielV “Ordered field” has a very precise meaning; I'll add a reference. $\endgroup$ – egreg May 10 '18 at 22:42
  • $\begingroup$ prime field are also ordered in the same sense? $\endgroup$ – alfC May 11 '18 at 1:26
  • $\begingroup$ @alfC An ordered field must have characteristic $0$. $\endgroup$ – egreg May 11 '18 at 6:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.