What are the axioms that differentiate a complex field from a real field?

Question

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 $\forall z, \exists \bar z / z\cdot\bar z > 0$.

Answer 1

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.

Answer 2

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 $i$.

Answer 3

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.

Answer 4

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.