Simple Field Extensions and the Construction of the Complex Numbers

A subfield $L\subseteq \mathrm{\Omega }$ contains $E\cup E^{\mathrm{\prime }}$ if and only if it contains $E$ and every element of $E^{\mathrm{\prime }}$.

By minimality, $E(E^{\mathrm{\prime }})$ is the intersection of precisely the subfields described in step 1.1.

Interchanging $E$ and $E^{\mathrm{\prime }}$ shows that $E^{\mathrm{\prime }}(E)$ is the same intersection.

Hence both generated fields equal the displayed intersection; denote it by $E{E}^{\mathrm{\prime }}$.

Apply [F1] to the inclusion $F\hookrightarrow K$ and $a$; this gives the stated evaluation homomorphism and its uniqueness.

By [F3], $a$ is transcendental exactly when $\ker ({\mathrm{ev}}_a)=0$.

Suppose $a$ is algebraic. Then the kernel is a nonzero proper ideal, so [F2] gives $\ker ({\mathrm{ev}}_a)=(m)$ for a nonzero nonconstant $m$.

Multiplying $m$ by the inverse of its leading coefficient does not change its principal ideal, so choose the generator $m$ monic.

For any $f\in F[x]$, $f(a)=0$ if and only if $f$ belongs to the kernel, which is equivalent to $f\in (m)$ and hence to $m\mid f$.

If $m=uv$ with both $u$ and $v$ nonconstant, then $0=m(a)=u(a)v(a)$; since $K$ is a field, one factor evaluates to zero and lies in $(m)$, impossible because its degree is smaller than $\mathrm{deg}m$. Thus $m$ is irreducible.

If $m^{\mathrm{\prime }}$ is another monic polynomial with the same property, then $m\mid m^{\mathrm{\prime }}$ and $m^{\mathrm{\prime }}\mid m$ by step 3.2; equal degree and monicity give $m=m^{\mathrm{\prime }}$.

Let $R$ be the set on the right. By [F1], $g\mathrm{\ne }0$ implies $g(a)\mathrm{\ne }0$, so every displayed quotient is defined.

The choices $(f,g)=(c,1)$ and $(x,1)$ show that $F\cup \{a\}\subseteq R$.

Common denominators show that $R$ is closed under addition, subtraction, and multiplication.

If $f(a)g(a{)}^{-1}\mathrm{\ne }0$, then $f\mathrm{\ne }0$ by [A1], and its inverse is $g(a)f(a{)}^{-1}\in R$.

Conversely, every subfield containing $F$ and $a$ contains $f(a)$, $g(a)$, and $g(a{)}^{-1}$ for every $f,g\in F[x]$ with $g\mathrm{\ne }0$; hence it contains $R$.

Thus $R$ is a subfield of $K$ containing $F$ and $a$, so $F(a)\subseteq R$ by [F2].

In particular $R\subseteq F(a)$, and step 3.1 gives equality.

Define $\mathrm{\Phi }(f(a)g(a{)}^{-1})=f(b)g(b{)}^{-1}$. The denominators are nonzero by [A1].

If $f(a)g(a{)}^{-1}=r(a)s(a{)}^{-1}$, cross-multiplication gives $(fs-rg)(a)=0$; [A1] gives $fs=rg$, and evaluation at $b$ proves that the two proposed images agree. Thus $\mathrm{\Phi }$ is well-defined.

The formula preserves sums and products, fixes $F$, and sends $a$ to $b$.

The same construction with $a$ and $b$ interchanged is inverse to $\mathrm{\Phi }$, so $\mathrm{\Phi }$ is an $F$-isomorphism.

Any $F$-homomorphism sending $a$ to $b$ must send $f(a)g(a{)}^{-1}$ to $f(b)g(b{)}^{-1}$; [F1] therefore forces it to equal $\mathrm{\Phi }$.

Irreducibility makes $p$ nonconstant, and [F1] makes $K$ a field.

The constant-class map $F\to K$ is injective: if a constant $c$ lies in $(p)$, then $c=qp$; uniqueness in [F2], comparing $c=0p+c$ with $c=qp+0$, forces $c=0$.

In $K$, $p(a)=p(x)+(p)=0$ by the quotient arithmetic; equivalently this is evaluation at $a$ from [F3].

By [F2], write $f=qp+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}p$; hence $f+(p)=r+(p)$, so every class has a reduced representative.

Thus the constant-class map supplies the field extension $K\mathrm{/}F$.

If two reduced representatives $r,s$ give the same class, then $r-s=qp$. Applying uniqueness in [F2] to $r-s$ shows $q=0$ and $r=s$.

Writing the unique reduced polynomial coefficientwise yields the displayed unique expression; when $n=1$ it consists only of $c_0$, and the zero class is represented by the zero polynomial.

A nonconstant polynomial is nonzero and not a unit, so [F1] supplies an irreducible factor $q$ of $f$.

Divide $q$ by its nonzero leading coefficient to obtain a monic irreducible factor $p$; $(p)=(q)$ and still $p\mid f$.

By [F2], $K=F[x]\mathrm{/}(p)$ is a field extension and $a=x+(p)$ satisfies $p(a)=0$.

Since $f=ph$ for some $h\in F[x]$, evaluation gives $f(a)=p(a)h(a)=0$.

By [F1], evaluation at $b$ is a homomorphism ${\mathrm{ev}}_b:F[x]\to L$ fixing $F$.

Since $p(b)=0$, the ideal $(p)$ lies in $\ker ({\mathrm{ev}}_b)$; [F2] therefore gives a unique homomorphism $\phi :K\to L$ with $\phi (f+(p))=f(b)$.

The formula fixes constant classes and sends $a=x+(p)$ to $b$; its image is exactly the set $F[b]$ of polynomial values.

Because [F3] makes $K$ a field and $\phi (1)=1$, its kernel is not all of $K$ and hence is zero; thus $\phi$ is a field homomorphism.

Any homomorphism fixing $F$ and sending $a$ to $b$ sends every $f(a)$ to $f(b)$; since every element is such an $f(a)$ by [F3], it equals $\phi$.

Evaluation has image $F[a]$ and kernel $(m_a)$ by [F1]; [F2] therefore induces $F[x]\mathrm{/}(m_a)\cong F[a]$.

Since $m_a$ is irreducible, [F4] makes the quotient and hence $F[a]$ a field.

The field $F[a]$ contains $F$ and $a$, while every subfield containing them contains all polynomial values; minimality in [F5] gives $F[a]=F(a)$.

Division by $m_a$ in [F3] gives each quotient class a representative of degree below $n$, hence gives every element of $F(a)$ a displayed power expression.

If two such expressions agree, their difference is a polynomial of degree below $n$ in the kernel $(m_a)$; uniqueness of the remainder in [F3] makes the difference zero coefficientwise.

The existence and uniqueness in steps 4.1--5.1 are exactly the assertion that the displayed powers form a basis; by definition, its number $n$ is $[F(a):F]$.

By [F1], the minimal polynomials of $\alpha$ and $\beta$ divide $p$; monic irreducibility of all three makes both minimal polynomials equal to $p$.

Using [F2] and the quotient identifications in [F3], obtain $F$-homomorphisms $\phi :L\to M$ and $\psi :M\to L$ with $\phi (\alpha )=\beta$ and $\psi (\beta )=\alpha$.

The composite $\psi \phi$ fixes $F$ and $\alpha$, so uniqueness in [F2] makes it the identity on $L$; similarly $\phi \psi$ is the identity on $M$.

Hence $\phi$ is an $F$-isomorphism, and the same uniqueness clause shows that no other such isomorphism can send $\alpha$ to $\beta$.

For $r\in \mathbb{R}$, either $r=0$, so $r^2=0$, or [F2] gives $r^2>0$. Thus $r^2+1>0$ by the ordered-field laws in [F1].

Hence $x^2+1$ has no real root.

Since it has degree two, [F3] now gives its irreducibility over $\mathbb{R}$.

Apply [F2] to [F1]. With the construction in [F5], this proves that $\mathbb{C}$ is a field, that the constant-class map embeds $\mathbb{R}$, and that every class has a unique representative $a+bx$, hence a unique form $a+bi$.

Quotient addition and multiplication, followed by $i^2=-1$, give the two displayed coordinate formulas. All field axioms are inherited from the field in step 1.1.

If $a+bi\mathrm{\ne }0$, uniqueness in step 1.1 gives $a\mathrm{\ne }0$ or $b\mathrm{\ne }0$. The corresponding square is positive by [F4], the other square is nonnegative, and therefore $a^2+b^2>0$ in the ordered field [F3].

Direct multiplication using step 2.1 gives $(a+bi)(a-bi)=a^2+b^2$. Since the real denominator is nonzero by step 2.2, division proves the inverse formula.

Uniqueness in [F1] makes $\mathrm{\Phi }$ well-defined and injective.

For every $(a,b)\in {\mathbb{R}}^2$, the complex number $a+bi$ maps to $(a,b)$, so $\mathrm{\Phi }$ is surjective.

Applying $\mathrm{\Phi }$ to the addition formula in [F1] gives the first coordinate identity.

Applying $\mathrm{\Phi }$ to the multiplication formula in [F1] gives the second coordinate identity.

By [F3], $i$ is a root of $x^2+1$, and [F1] and [F4] make that polynomial its minimal polynomial.

The unique coordinate form in [F3] gives $\mathbb{C}=\mathbb{R}(i)$.

Apply [F2] to step 1.1: the power basis is $1,i$ and the degree is $2$.

The equation $j^2=-1$ says exactly that $j$ is a root of $x^2+1$.

Apply [F3] to [F1], [F2], and step 1.1. It gives the unique homomorphism fixing $\mathbb{R}$, sending $i$ to $j$, and having image $\mathbb{R}[j]$.

Coordinate expansion using [F1] and [F2] proves the two homomorphism laws, that conjugation fixes every real number, and that applying it twice is the identity. Thus conjugation is an involutive real-field automorphism.

Direct multiplication gives $z\bar{z}=a^2+b^2=\mathrm{\mid }z{\mathrm{\mid }}^2$.

Lagrange's identity gives $(a^2+b^2)(u^2+v^2)-(au+bv{)}^2=(av-bu{)}^2\ge 0.$ The final inequality follows from [F5], with the zero case included.

By [F2] and [F3], $\mathrm{\mid }z\mathrm{\mid }\ge 0$. If $z=0$, then $\mathrm{\mid }z\mathrm{\mid }=0$; conversely, $\mathrm{\mid }z\mathrm{\mid }=0$ makes $a^2+b^2=0$, and [F5] forces $a=b=0$.

Apply step 1.2 to $zw$ and use step 1.1: $\mathrm{\mid }zw{\mathrm{\mid }}^2=(zw)\overline{zw}=(z\bar{z})(w\bar{w})=\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\mid }w{\mathrm{\mid }}^2$. Both sides of $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ are nonnegative, so uniqueness in [F3] proves multiplicativity.

Hence $au+bv\le \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$: if $au+bv\le 0$ this follows from $\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }\ge 0$; otherwise both quantities are nonnegative, step 1.3 and [F4] give the inequality after squaring.

Expanding with [F1] and [F2], then using step 3.2, gives $\mathrm{\mid }z+w{\mathrm{\mid }}^2=a^2+b^2+u^2+v^2+2(au+bv)\le (\mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }{)}^2\mathrm{.}$

Both sides of $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ are nonnegative, so [F4] turns the squared inequality in step 4.1 into the triangle inequality.

By [F1], $i^2=-1$; since $\sigma$ is a field homomorphism fixing $-1$, applying it gives $\sigma (i{)}^2=-1$.

In the complex field, $t^2+1=(t-i)(t+i)$; hence a root has $t=i$ or $t=-i$. Thus $\sigma (i)\in \{i,-i\}$.

If $\sigma (i)=i$, uniqueness in [F3] makes $\sigma$ the identity. If $\sigma (i)=-i$, [F2] and uniqueness in [F3] make $\sigma$ conjugation.