prove that $\mathbb C$ is a field
Andrew Henderson 
I would like to know how to prove that $\mathbb C$ is a field, $( \mathbb C, +, \times)$
I have found out that the additive identity is $0_\mathbb C=0=0+0i$,
multiplicative identity is $1_\mathbb C = 1+0i=1$,
additive inverses $-(x+iy)=(-x)+i(-y)$
and multiplicative inverses $(x+iy)^{-1}:=\frac{x}{x^{2}+y^{2}}+ i(\frac{-y}{x^2+y^2})$
but after many hours of research online i cannot find a proof of this or how to get to these additive/multiplicative identities and inverses. Please could someone show me how to do this, I have only ever looks at finite sets within a field and not the whole set of complex numbers and none of my complex analysis books use an axiomatic approach to this subject.
Many thanks for your help
$\endgroup$ 53 Answers
$\begingroup$To show that it is a field, you need to show that these elements you listed really do what they're supposed to do. For example, you note that $1$ is your multiplicative identity. To show that this is true, you have to verify that if you multiply that number by any complex number $z$, the product is $z$. Similarly, each of these complex numbers you list has to satisfy the appropriate property.
Do that for each field axiom, and you have your proof.
For example, here's what you might write down for the additive inverse axiom:
"The element $0+0i\in\mathbb{C}$ is an additive identity: Taking any other $a+bi\in\mathbb{C}$, we see that
$(0+0i)+(a+bi) = (0+a)+(0+b)i = a+bi$
Thus, $\mathbb{C}$ has an identity for addition."
It's mostly about verifying that the definitions of complex addition and multiplication give rise to operations that do what addition and multiplication are supposed to do, and the work hardly feels like you're doing anything, which can be weird at first.
Does that help?
$\endgroup$ 5 $\begingroup$This proof is simply and eloquently furnished in Walter Rudin's Principles of Mathematical Analysis, pp 5-13, with the meat of the proof on the last two pages.
I won't reproduce the entire proof here. It suffices to show that $\mathbb C$ satisfies the field axioms, which is easy once it is established that $\mathbb R$ is a field.
The field axioms, as given by Rudin, are as follows. $F$ is a field if:
- (additive closure) If $x \in F$ and $y \in F$, then their sum $x+y$ is in $F$.
- (additive commutativity) $x + y = y + x$ for all $x, y \in F$.
- (additive associativity) $(x + y) + z = x + (y + z)$ for all $x, y, z \in F$.
- (additive identity) $F$ contains an element $0$ such that $0 + x = x$ for every $x \in F$.
(additive inverse) To every $x \in F$ there corresponds an element $-x \in F$ such that $x + (-x) = 0$.
(multiplicative closure) If $x \in F$ and $y \in F$, then their product $xy$ is in $F$.
- (multiplicative commutativity) $xy = yx$ for all $x, y \in F$.
- (multiplicative associativity) $(xy)z = x(yz)$ for all $x,y,z \in F$.
- (multiplicative identity) $F$ contains an element $1 \ne 0$ such that $1x = x$ for all $x \in F$.
(multiplicative inverse) If $x \in F$ and $x \ne 0$ then there exists an element $1/x \in F$ such that $x \cdot (1/x) = 1$.
(Distributive law) $x(y+z) = xy + xz$ for all $x, y, z \in F$.
Then the complex numbers are defined as follows: A complex number is an ordered pair $x = (a,b)$ of real numbers $a, b \in \mathbb R$ such that the following hold:
- If $x = (a,b)$ and $y = (c,d)$, then $x = y$ if and only if $a = c$ and $b = d$.
- (complex addition) $x + y = (a+c, b+d)$.
- (complex multiplication) $x \cdot y = (ac - bd, ad + bc)$.
Then go through and check each of the axioms. For example, Rudin proves the distributive law by writing $$\begin{align*} x(y+z) &= (a,b)(c+e,d+f) \\ &= (ac + ae - bd - bf, ad + af + bc + be) \\ &= (ac - bd, ad + bc) + (ae - bf, af + be) \\ &= xy + xz. \end{align*}$$
$\endgroup$ 1 $\begingroup$To find the additive inverse, just look at the definiton. For the multiplicative inverse note that for $z=x+iy, \bar z = x-iy$ you have $$x^2+y^2 = |z|^2 = z\bar z$$ and thus $$z^{-1} = \frac1z = \frac{\bar z}{z\bar z} = \frac1{|z|^2} \bar z = \frac1{x^2+y^2}(x-iy)$$ as claimed.
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