Proof of Existence and Uniqueness of the Complex Numbers

Throughout, $\mathbb{R}$ denotes the real numbers, an ordered field, and conditions (a), (b), (c) are those in the statement. We use the field axioms freely, together with the identities $u\cdot0=0$ and $u\cdot(-v)=-(u\cdot v)$, valid in any field and proved from the distributive law as in the proof of Existence and Uniqueness of the Square Root of a Sum of Two Squares. For a real number $x$ we write $x^{2}=x\cdot x$.

Part A (consequences of condition (a)). Let $K$ be a field satisfying (a), with additive identity $0_{K}$ and multiplicative identity $1_{K}$. Since $0+0=0$ holds in $\mathbb{R}$, it holds in $K$ by (a); adding the additive inverse of $0$ in $K$ gives $0=0_{K}$. Since $1\cdot1=1$ holds in $\mathbb{R}$, it holds in $K$; and $1\neq0=0_{K}$, so $1$ has a multiplicative inverse in $K$, and multiplying by it gives $1=1_{K}$. For $a\in\mathbb{R}$ the equation $a+(-a)=0=0_{K}$ holds in $K$, so the real number $-a$ is the additive inverse of $a$ in $K$; and for $a\neq0$ the equation $a\cdot(1/a)=1=1_{K}$ holds in $K$, so the real number $1/a$ is the multiplicative inverse of $a$ in $K$. In particular the element $-1$ appearing in (b) is the real number $-1$.

Part B (uniqueness of the representation). Let $(K,j)$ be a complex pair and let $a,b,c,d\in\mathbb{R}$ satisfy $a+bj=c+dj$. Adding to both sides the additive inverses of $c$ and of $bj$ and using commutativity and associativity of addition together with the distributive law in the form $dj+(-(bj))=(d+(-b))j$, we get

where $a+(-c)$ and $d+(-b)$ are real numbers by Part A and (a). Suppose $d+(-b)\neq0$. Multiplying both sides by the multiplicative inverse of $d+(-b)$ in $K$, which by Part A is the real number $1/(d+(-b))$, expresses $j$ as a product of real numbers formed in $K$, hence by (a) as a real number. But then, by Existence and Uniqueness of the Square Root of a Sum of Two Squares applied to the real numbers $j$ and $0$, we would have $0\le j\cdot j$, while (b) and Part A give $j\cdot j=-1$; applying the same lemma to $1$ and $0$ gives $0\le1$, hence $-1\le0$ after adding $-1$ to both sides, and antisymmetry of the total order of $\mathbb{R}$ would force $-1=0$, i.e. $1=0$, contradicting Field. Therefore $d+(-b)=0$, that is $d=b$, and then $a+bj=c+bj$ gives $a=c$. Combining with (c): for every $z\in K$ there is exactly one pair $(a,b)$ of real numbers with $z=a+bj$.

Part C (arithmetic in a complex pair). Let $(K,j)$ be a complex pair and $a,b,c,d\in\mathbb{R}$. Commutativity and associativity of addition and the distributive law give $(a+bj)+(c+dj)=(a+c)+(b+d)j$; and the distributive law, commutativity and associativity of multiplication, condition (b) and the identity $u\cdot(-v)=-(u\cdot v)$ give

where by (a) all displayed operations on real numbers are those of $\mathbb{R}$.

Part D (proof of claim 1: existence).

Step 1: a model. Let $\mathbb{C}_{0}=\mathbb{R}\times\mathbb{R}$ be the Cartesian product, and define, for real $a,b,c,d$,

We verify the axioms of Field. Addition is associative and commutative and has identity $(0,0)$ and additive inverses $(-a,-b)$, since all of this holds componentwise in $\mathbb{R}$. Multiplication is commutative, since interchanging the two arguments turns $(ac-bd,ad+bc)$ into $(ca-db,cb+da)$, the same pair. Expanding both sides with the distributive law of $\mathbb{R}$ shows that $((a,b)\odot(c,d))\odot(e,f)$ and $(a,b)\odot((c,d)\odot(e,f))$ are both equal to

so multiplication is associative. Also $(a,b)\odot(1,0)=(a\cdot1-b\cdot0,\;a\cdot0+b\cdot1)=(a,b)$, and $(1,0)\neq(0,0)$ because $1\neq0$ in $\mathbb{R}$, so $(1,0)$ is a multiplicative identity distinct from the additive identity. For distributivity,

which by the distributive law of $\mathbb{R}$ equals $\bigl((ac-bd)+(ae-bf),\;(ad+bc)+(af+be)\bigr)=\bigl((a,b)\odot(c,d)\bigr)\oplus\bigl((a,b)\odot(e,f)\bigr)$. Finally, let $(a,b)\neq(0,0)$. By Existence and Uniqueness of the Square Root of a Sum of Two Squares applied to $a$ and $0$, and to $b$ and $0$, we have $0\le a^{2}$ and $0\le b^{2}$; if $a^{2}+b^{2}=0$ then adding $b^{2}$ to $0\le a^{2}$ gives $b^{2}\le a^{2}+b^{2}=0$, so $b^{2}=0$ and hence $b=0$ because a field has no zero divisors, and symmetrically $a=0$, contradicting $(a,b)\neq(0,0)$. So $a^{2}+b^{2}\neq0$, and with $t=\bigl(a/(a^{2}+b^{2}),\,-b/(a^{2}+b^{2})\bigr)$ we get

Hence $\mathbb{C}_{0}$ with $\oplus$ and $\odot$ is a field.

Put $\iota(a)=(a,0)$ for $a\in\mathbb{R}$ and $j_{0}=(0,1)$. Then $\iota$ is injective, $\iota(a)\oplus\iota(b)=\iota(a+b)$ and $\iota(a)\odot\iota(b)=(ab-0,\,0+0)=\iota(ab)$; moreover

which is the additive inverse of $\iota(1)=(1,0)$; and $\iota(b)\odot j_{0}=(b,0)\odot(0,1)=(0,b)$, so that

We refer to the last identity as the decomposition identity of the model.

Step 2: transport of structure. The field $\mathbb{C}_{0}$ contains a copy of $\mathbb{R}$ rather than $\mathbb{R}$ itself, so we replace that copy by $\mathbb{R}$. Let $N=\mathbb{C}_{0}\setminus\{\iota(a):a\in\mathbb{R}\}$. There exist a set $D$ with $D\cap\mathbb{R}=\varnothing$ and a bijection $\theta:N\to D$. Indeed, with the usual encoding of an ordered pair as the set $\{\{x\},\{x,y\}\}$, take $D=\{(z,\mathbb{R}):z\in N\}$ and $\theta(z)=(z,\mathbb{R})$; if some element $(z,\mathbb{R})$ of $D$ were a real number, then $\mathbb{R}\in\{z,\mathbb{R}\}\in(z,\mathbb{R})\in\mathbb{R}$ would be a membership cycle, which the axiom of foundation forbids. Only the existence of such a $D$ and $\theta$ is used, so with another encoding of ordered pairs any such choice serves.

Set $K=\mathbb{R}\cup D$ and define $\Phi:\mathbb{C}_{0}\to K$ by $\Phi(\iota(a))=a$ for $a\in\mathbb{R}$ and $\Phi(z)=\theta(z)$ for $z\in N$. Since $\iota$ is injective, each element of $\mathbb{C}_{0}$ is either $\iota(a)$ for exactly one real $a$ or lies in $N$, so $\Phi$ is well defined; and since $\theta$ is a bijection onto $D$ and $\mathbb{R}\cap D=\varnothing$, $\Phi$ is a bijection. Define operations on $K$ by

By construction $\Phi(x\oplus y)=\Phi(x)+\Phi(y)$ and $\Phi(x\odot y)=\Phi(x)\cdot\Phi(y)$ for all $x,y\in\mathbb{C}_{0}$. Since $\Phi$ is a bijection, each field axiom for $K$ follows by transporting the corresponding axiom of $\mathbb{C}_{0}$: associativity, commutativity and distributivity are preserved because $\Phi$ carries the operations to the operations; $\Phi((0,0))$ and $\Phi((1,0))$ are the additive and multiplicative identities of $K$ and are distinct because $\Phi$ is injective; and $\Phi$ carries additive and multiplicative inverses in $\mathbb{C}_{0}$ to such inverses in $K$. Hence $K$ is a field.

For real $a,b$ we have $\Phi^{-1}(a)=\iota(a)$, so the sum of $a$ and $b$ formed in $K$ is $\Phi(\iota(a)\oplus\iota(b))=\Phi(\iota(a+b))=a+b$, the sum formed in $\mathbb{R}$, and likewise the product formed in $K$ is $\Phi(\iota(ab))=ab$. Since also $\mathbb{R}\subseteq K$, condition (a) holds. Put $j=\Phi(j_{0})$. Then $\Phi((1,0))=1$ is the multiplicative identity of $K$, its additive inverse in $K$ is the real number $-1$ by Part A, and

so condition (b) holds. Finally, let $z\in K$ and let $(a,b)=\Phi^{-1}(z)$. Applying $\Phi$ to the decomposition identity of Step 1 gives $z=a+b\cdot j$, so condition (c) holds. Thus $(K,j)$ is a complex pair, which proves claim 1.

Part E (proof of claim 2: uniqueness). Let $(K,j)$ and $(K',j')$ be complex pairs. By Part B, every $z\in K$ has exactly one representation $z=a+bj$ with $a,b\in\mathbb{R}$, and likewise in $K'$ with $j'$ in place of $j$. Define $\varphi:K\to K'$ by $\varphi(a+bj)=a+bj'$; this is well defined by that uniqueness. If $z=a+bj$ and $w=c+dj$, then by Part C, applied in $K$ and then in $K'$,

For $a\in\mathbb{R}$ we have $a=a+0\cdot j$ because $0\cdot j=0$, so $\varphi(a)=a+0\cdot j'=a$; and $j=0+1\cdot j$ gives $\varphi(j)=j'$. Defining $\psi:K'\to K$ by $\psi(a+bj')=a+bj$ in the same way, the uniqueness of representations gives $\psi(\varphi(z))=z$ and $\varphi(\psi(z'))=z'$ for all $z\in K$ and $z'\in K'$, so $\varphi$ is a bijection with inverse $\psi$, and $\psi$ preserves both operations, fixes every real number and satisfies $\psi(j')=j$ by the same computations with the roles exchanged.

It remains to prove uniqueness of $\varphi$. Let $\chi:K\to K'$ preserve addition and multiplication, fix every real number and satisfy $\chi(j)=j'$. For $z=a+bj$ with $a,b\in\mathbb{R}$,

Hence $\chi=\varphi$, completing the proof of claim 2.