How do you build the complex numbers from foundations?

Question

Id like to know how the complex numbers are derived from axioms. Most textbooks talk of them as starting from the complex plane. Is there a way to derive the complex properties from say the properties of fields or is there some peano axiom like system for complex numbers.

Answer 1

You can construct the real numbers $\mathbb{R}$ then define complex numbers to be the set $\mathbb{R}\times\mathbb{R}$ equipped with component-wise addition $$(a,b)+(c,d)=(a+c,b+d)$$

and multiplication given by

$$(a,b)\cdot(c,d)=(ac-bd,ad+bc)$$

Answer 2

Another way that has not yet been given, although it isn't so from first principles, is to "construct" the complex numbers as a quotient ring $\mathbb{R}[x]/\langle x^2 + 1\rangle$, where $\langle x^2+1\rangle$ is an ideal in $\mathbb{R}[x]$. Then $1$ is the ideal $ 1 + \langle x^2+1\rangle$ and $i$ is the ideal $x + \langle x^2+1\rangle$. Multiply the first ideal by $a$ and the second by $b$ and add them to form a number, repeat with a separate $c,d$ and you get the rules in the answer given by Alex. A little more work is needed, since the quotient ring must be a field like $\mathbb{C}$. To do this, you simply need to show that the ideal $\langle x^2+1\rangle$ is maximal. This particular treatment is given in Galois Theory, 2nd Edition by David A. Cox.

The set of reals in this construction are all values $r + \langle x^2+1\rangle$ where $r \in \mathbb{R}$. Consequently the field $\mathbb{R}$ is a subfield of $\mathbb{C}$.

The proper statement for this is $R[x]/\langle x^2+1\rangle \simeq \mathbb{C}$, or that the two are equivalent up to isomorphism.

I am assuming here you are happy with the existence of $\mathbb{R}$ and at least reasonably comfortable with the concept of Rings and Ideals. If not, I like Rings, Fields and Groups by Allenby and Classic Set Theory by Derek Goldrei because it covers this from a set-theoretic perspective and is written in a style suitable for self study (full disclosure: I'm an OU student and have exchanged a few forum posts inside the OU with the author).

Answer 3

The natural way to obtain complex numbers is Field extension, and $\mathbb{C}$ is the only possible finite extension of reals. Simply talking, each extension is obtained by inviting new numbers that extend the area of usage of our algebraic operations (seeking algebraic closure of the set). We have derived $\mathbb{Z}$ from $\mathbb{N}$ by solving equation $x + a = 0$, then have derived $\mathbb{Q}$ from $\mathbb{Z}$ by solving $x*y=a$, then extensions of $\mathbb{Q}$ (algebraic numbers) appear when we try to solve $x^n - a = 0$ and $\mathbb{R}$ with its irrational numbers is a closure of $\mathbb{Q}$ (as $\mathbb{Q}$ is dense in $\mathbb{R}$ and we can approach any irrational number with a sequence of rationals). Finally, $\mathbb{C}$ is derived from the solutions of equation $x^n + a = 0$ and $\mathbb{C}$ is algebraically closed, according to the fundamental theorem of algebra (ok, the field of complex algebraic numbers, but we have already invented transcendentals in $\mathbb{R}$).