The Fundamental Theorem of Algebra

Put $S:={\sum }_{i<n}\mathrm{\mid }{a}_i\mathrm{\mid }$ and $R:=1+\frac{S}{\mathrm{\mid }{a}_n\mathrm{\mid }}\mathrm{.}$ Then $R>1$, and therefore ${\sum }_{i<n}\mathrm{\mid }{a}_i\mathrm{\mid }{R}^i\le S{R}^{n-1}<\mathrm{\mid }{a}_n\mathrm{\mid }{R}^n\mathrm{.}$

The estimate in step 1.1 gives $\mid {\sum }_{i<n}{a}_i{R}^i\mid \le {\sum }_{i<n}\mathrm{\mid }{a}_i\mathrm{\mid }{R}^i<\mathrm{\mid }{a}_n\mathrm{\mid }{R}^n\mathrm{.}$ Hence $f(R)=a_n{R}^n+{\sum }_{i<n}{a}_i{R}^i$ has the same sign as $a_n$.

Because $n$ is odd, $(-R{)}^n=-R^n$. Also $\mid {\sum }_{i<n}{a}_i(-R{)}^i\mid \le {\sum }_{i<n}\mathrm{\mid }{a}_i\mathrm{\mid }{R}^i<\mathrm{\mid }{a}_n\mathrm{\mid }{R}^n,$ so $f(-R)=a_n(-R{)}^n+{\sum }_{i<n}{a}_i(-R{)}^i$ has the opposite sign from $a_n$. Thus $f(-R)$ and $f(R)$ have opposite signs.

By [L1], the polynomial function $f$ is continuous on $[-R,R]$. Since step 2.2 shows that $0$ lies between $f(-R)$ and $f(R)$, [L2] gives a point $c\in [-R,R]$ with $f(c)=0$.

Suppose $\mathrm{deg}f$ is odd. Then [L1] gives $a\in \mathbb{R}$ with $f(a)=0$.

By [L2], the linear polynomial $x-a$ divides $f$. Since $f$ is irreducible and $x-a$ is nonconstant, this forces $f$ to be associated to $x-a$, so $\mathrm{deg}f=1$.

Therefore an irreducible real polynomial can have odd degree only in the linear case. If it is not linear, its degree is not odd and hence is even.

Write $g=u+iv$ with $u,v\in \mathbb{R}[x]$, and put $\bar{g}:=u-iv$. Then $g\bar{g}=u^2+v^2\in \mathbb{R}[x]\mathrm{.}$ Because $g$ is nonconstant, so is $g\bar{g}$.

By the standing hypothesis and [L2], the real polynomial $g\bar{g}$ splits over $\mathbb{C}$, so it has a complex root $z$. In particular, $0=(g\bar{g})(z)=g(z)\bar{g}(z)\mathrm{.}$

Since $\mathbb{C}$ is a field by [F1], step 2.1 implies either $g(z)=0$ or $\bar{g}(z)=0$.

If $g(z)=0$ then $g$ already has a root in $\mathbb{C}$. If $\bar{g}(z)=0$, then conjugating that equality and using the conjugation law from [F1] gives $g(\bar{z})=0$. Thus $g$ has a root in $\mathbb{C}$ in either case.

The polynomial $g\in \mathbb{C}[x]$ was arbitrary. Therefore every nonconstant polynomial over $\mathbb{C}$ has a root in $\mathbb{C}$, and [L1] makes $\mathbb{C}$ algebraically closed.

Choose $\alpha \in E\setminus F$. Then $F⊊F(\alpha )\subseteq E$. Since $[E:F]=2$, fact [L2] forces $[F(\alpha ):F]=2$ and hence $E=F(\alpha )$. Therefore [L1] gives $m_{\alpha }(x)=x^2+bx+c$ with $b,c\in F$.

Because $\mathrm{char}F\mathrm{\ne }2$, the element $2$ is invertible in $F$. Put $\delta :=2\alpha +b\mathrm{.}$ Using $m_{\alpha }(\alpha )=0$, ${\delta }^2=4{\alpha }^2+4b\alpha +b^2=b^2-4c\in F\mathrm{.}$

Also $\alpha =\frac{\delta -b}{2},$ so $F(\alpha )=F(\delta )$. Therefore $E=F(\alpha )=F(\delta )=F(\sqrt{d})$ with $d:={\delta }^2\in F$. If $d$ were already a square in $F$, then $\delta \in F$ and the displayed formula would give $\alpha \in F$, contradicting step 1.1. Hence $d$ may be chosen nonsquare.

Assume $F$ is algebraically closed, and let $f\in F[x]$ be nonconstant. By [L1] it has a root $a\in F$, so [L3] gives $f=(x-a)q$ for some $q\in F[x]$. Repeating the same argument on $q$ while it remains nonconstant writes $f$ as a product of linear factors, so $f$ splits over $F$ in the sense of [L2].

Assume every nonconstant polynomial in $F[x]$ splits over $F$, and let $E\mathrm{/}F$ be finite. For any $\alpha \in E$, fact [L6] makes $\alpha$ algebraic over $F$. If $\alpha \mathrm{\notin }F$, let $m_{\alpha }$ be its minimal polynomial over $F$. Then [L7] makes $m_{\alpha }$ a nonconstant irreducible polynomial with $m_{\alpha }(\alpha )=0$. By the hypothesis, $m_{\alpha }$ splits over $F$ and therefore has a root $a\in F$. Fact [L3] gives $x-a$ as a linear factor of $m_{\alpha }$, contradicting irreducibility. Hence every $\alpha \in E$ already lies in $F$, so $E=F$.

Assume $F$ has no nontrivial finite extension, and let $f\in F[x]$ be nonconstant. By [L5] there is an extension $K\mathrm{/}F$ and an element $\alpha \in K$ with $f(\alpha )=0$. Then $\alpha$ is algebraic over $F$, so [L4] makes $F(\alpha )\mathrm{/}F$ finite. By the hypothesis this finite extension must be trivial, hence $\alpha \in F$. Therefore every nonconstant polynomial over $F$ has a root in $F$, and [L1] says that $F$ is algebraically closed.

Steps 1.1, 1.2, and 1.3 prove $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$, so the three conditions are equivalent.

By [L1], it is enough to prove that every nonconstant polynomial in $\mathbb{R}[x]$ splits over $\mathbb{C}$. Fix such a polynomial $f\in \mathbb{R}[x]$.

Put $h:=f(x)(x^2+1)$. By [L2], choose a splitting field $E\mathrm{/}\mathbb{R}$ of $h$. The roots of $h$ are algebraic over $\mathbb{R}$, and $E$ is generated by finitely many of them, so [L13] makes $E\mathrm{/}\mathbb{R}$ finite. Because $E$ is a splitting field of $h$, [L3] makes it normal, and [L4] makes it separable. Thus $E\mathrm{/}\mathbb{R}$ is finite Galois.

Let $G:=\mathrm{Gal}(E\mathrm{/}\mathbb{R})$. By [L7], choose a Sylow $2$-subgroup $H\le G$, and put $M:=E^H$. Then [L5] gives $[M:\mathbb{R}]=[G:H],$ which is odd by the definition of a Sylow $2$-subgroup.

Since $x^2+1$ splits in $E$, choose $j\in E$ with $j^2=-1$. By [L10], there is a unique real-field embedding $\iota :\mathbb{C}\to E$ sending $i$ to $j$. Identify $\mathbb{C}$ with its image, so that $\mathbb{R}\subseteq \mathbb{C}\subseteq E\mathrm{.}$

Let $\alpha \in M$. Since $\mathbb{R}\subseteq \mathbb{R}(\alpha )\subseteq M$, the tower law [L6] shows that $[\mathbb{R}(\alpha ):\mathbb{R}]$ divides $[M:\mathbb{R}]$, so $[\mathbb{R}(\alpha ):\mathbb{R}]$ is odd. Because $\alpha \in E$ and step 1.2 made $E\mathrm{/}\mathbb{R}$ finite, the element $\alpha$ is algebraic over $\mathbb{R}$. Fact [L14] therefore makes the minimal-polynomial degree of $\alpha$ over $\mathbb{R}$ equal to the same odd number $[\mathbb{R}(\alpha ):\mathbb{R}]$; by [L12] it must have degree $1$. Hence $\alpha \in \mathbb{R}$. Since every element of $M$ lies in $\mathbb{R}$, we get $M=\mathbb{R}$.

Because $M=E^H=\mathbb{R}=E^G$, the bijection in [L5] forces $H=G$. Therefore $G$ is a finite $2$-group.

Put $J:=\mathrm{Gal}(E\mathrm{/}\mathbb{C})$. Then $J\le G$, so step 3.1 makes $J$ a finite $2$-group. Suppose $J$ is nontrivial.

By [L8], the nontrivial $2$-group $J$ has a subgroup $K$ of index $2$. Put $L:=E^K$. Then $E^J=\mathbb{C}$ by [L5], and [L5] together with [L6] gives $[L:\mathbb{C}]=[J:K]=2.$

The field $\mathbb{C}$ has characteristic zero, hence not $2$, so [L11] makes the quadratic extension $L\mathrm{/}\mathbb{C}$ equal to $\mathbb{C}(\sqrt{d})$ for some $d\in \mathbb{C}$. By [L9], choose $e\in \mathbb{C}$ with $e^2=d$. Then $L=\mathbb{C}(\sqrt{d})=\mathbb{C}(e)=\mathbb{C},$ contradicting step 5.1. Therefore $J$ is trivial.

Since $\mathrm{Gal}(E\mathrm{/}\mathbb{C})=1$, the fixed field of the trivial subgroup is $E$; by [L5], that fixed field is also $\mathbb{C}$. Thus $E=\mathbb{C}$. Because $E$ is a splitting field of $h$ and $f$ divides $h$, the polynomial $f$ splits over $\mathbb{C}$. By step 1.1 and [L1], this proves that $\mathbb{C}$ is algebraically closed.

By [L1], the field $\mathbb{C}$ is algebraically closed.

Applying [L2] to the field $\mathbb{C}$ and the given polynomial $f$ shows that $f$ splits over $\mathbb{C}$.

Since $f$ was arbitrary, every nonconstant polynomial in $\mathbb{C}[x]$ splits over $\mathbb{C}$.

Fact [L1] gives the algebraically closed part of the definition in [L3].

Fact [L2] gives the algebraic-extension part of the definition in [L3].

Therefore [L3] makes $\mathbb{C}\mathrm{/}\mathbb{R}$ an algebraic closure of $\mathbb{R}$.

By [L1], the set $A$ is a subfield of $\mathbb{C}$ containing $\mathbb{Q}$.

Let $f(x)=a_n{x}^n+\cdots +a_0\in A[x]$ be nonconstant. Since $\mathbb{C}$ is algebraically closed by [L2], the polynomial $f$ has a root $z\in \mathbb{C}$.

The coefficients $a_0,\dots ,a_n$ are algebraic over $\mathbb{Q}$, so $F:=\mathbb{Q}(a_0,\dots ,a_n)$ is a finite extension of $\mathbb{Q}$ by [L3]. The element $z$ is a root of a nonzero polynomial over $F$, so it is algebraic over $F$; therefore [L4] makes $F(z)\mathrm{/}F$ finite. By [L5], the extension $F(z)\mathrm{/}\mathbb{Q}$ is finite, and then [L6] makes $z$ algebraic over $\mathbb{Q}$. Hence $z\in A$.

Step 2.1 shows that every nonconstant polynomial in $A[x]$ has a root in $A$, so $A$ is algebraically closed. Since every element of $A$ is algebraic over $\mathbb{Q}$ by definition, [L7] makes $A$ an algebraic closure of $\mathbb{Q}$.

By [L1], complex conjugation fixes every coefficient of $f$.

Apply conjugation to the equality $f(z)=0$. Because conjugation is a field automorphism and fixes the coefficients of $f$, this gives $0=\overline{f(z)}=f(\bar{z})\mathrm{.}$

Therefore $\bar{z}$ is also a root of $f$. Since $z\mathrm{\notin }\mathbb{R}$, one has $\bar{z}\mathrm{\ne }z$, so the two roots form a conjugate pair.

By [L1], choose a root $z\in \mathbb{C}$ of $f$. If $z\in \mathbb{R}$, then the minimal polynomial of $z$ over $\mathbb{R}$ is $x-z$, and [L3] makes $x-z$ divide $f$. Since $f$ is irreducible, this forces $\mathrm{deg}f=1$.

Suppose instead that $z\mathrm{\notin }\mathbb{R}$. Then [L2] gives $f(\bar{z})=0$. Put $q(x):=(x-z)(x-\bar{z})=x^2-(z+\bar{z})x+z\bar{z}\in \mathbb{R}[x]\mathrm{.}$ Because $q(z)=0$, fact [L3] makes the minimal polynomial of $z$ over $\mathbb{R}$ divide $q$ and also divide $f$. Since $f$ is irreducible, that minimal polynomial is associated to $f$, so $\mathrm{deg}f\le \mathrm{deg}q=2.$ The degree cannot be $1$ in the nonreal case, hence $\mathrm{deg}f=2$.

The real-root and nonreal-root cases are exhaustive, so every irreducible polynomial in $\mathbb{R}[x]$ has degree $1$ or $2$.

If $f$ is a nonzero constant, then it already has the required form: it is itself the nonzero scalar multiplying the empty product. Now assume that $f$ is a nonzero nonunit polynomial. By [L1], it factors as $f=u{p}_1\cdots p_r$ with $u\in {\mathbb{R}}^{\times }$ and each $p_j$ irreducible in $\mathbb{R}[x]$.

By [L2], each irreducible factor $p_j$ has degree $1$ or $2$. Therefore each $p_j$ is either linear or an irreducible quadratic, so the factorization of step 1.1 is exactly the required one.

The constant and nonconstant cases together prove the statement for every nonzero polynomial in $\mathbb{R}[x]$.

By [L1] and [L2], there are $c\in {\mathbb{C}}^{\times }$ and complex numbers ${\beta }_1,\dots ,{\beta }_n$ such that $f(x)=c{\prod }_{k=1}^n(x-{\beta }_k)\mathrm{.}$ Let ${\alpha }_1,\dots ,{\alpha }_r$ be the distinct values among the ${\beta }_k$, and let $m_j$ be the number of indices $k$ with ${\beta }_k={\alpha }_j$. Then $f(x)=c{\prod }_{j=1}^r(x-{\alpha }_j{)}^{m_j},$ and by construction $m_1+\cdots +m_r=n$.

Suppose also that $f(x)=c^{\mathrm{\prime }}{\prod }_{j=1}^s(x-{\gamma }_j{)}^{n_j}$ with $c^{\mathrm{\prime }}\in {\mathbb{C}}^{\times }$, distinct ${\gamma }_j$, and positive integers $n_j$. In the UFD $\mathbb{C}[x]$, each linear factor $x-\alpha$ is irreducible, hence prime. Therefore the exponent with which $x-\alpha$ appears in a factorization of $f$ is uniquely determined. After reordering, this gives $r=s,{\gamma }_j={\alpha }_j,n_j=m_j\text{ for every }j\mathrm{.}$ So the multiplicities are well defined.

Step 1.1 gives a factorization whose exponents sum to $n$, and step 2.1 gives uniqueness of those exponents. This is exactly the statement that a degree-$n$ complex polynomial has exactly $n$ roots counted with multiplicity.