# Contents

## Statement

Classically, the fundamental theorem of algebra states that

• The field of complex numbers $\mathbb{C}$ is algebraically closed. In other words, every nonconstant polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$.

Many proofs of this theorem are known; some use complex analysis (the reciprocal of a polynomial cannot be bounded), some use algebraic topology (the degree of a map is invariant with respect to homotopy), and some use advanced calculus (polynomial functions on the complex numbers are open mappings). All of these proofs involve, at some level, the fact that the real numbers are Dedekind complete, which has as a classical consequence the fact that the real numbers are archimedean.

## Algebraic proof via real closed fields

Despite its name, the fundamental theorem of algebra makes reference to a concept from analysis (the field of complex numbers). However, the analytic part may be reduced to a minimum: that the field of real numbers is real closed. This has been known essentially forever, and is easily proved using (for example) the intermediate value theorem.

The rest of the proof is algebraic and, unlike the other proof methods, applies to all real closed fields, which need not be archimedean. It is due to Emil Artin, and forms a basic chapter in the Artin–Schreier theory of real closed fields.

We recall that a real closed field is an ordered field such that every positive element has a square root, and every polynomial of odd degree has a root. Clearly the polynomial $x^2 + 1$ has no root in a real closed field.

###### Theorem

If $F$ is real closed, then $K = F[\sqrt{-1}]$ is algebraically closed.

###### Proof

We must show that any irreducible polynomial $p$ of degree greater than $0$ with coefficients in $K$ has a root in $K$.

The splitting field? of $p$ is a finite Galois extension $L$ of $F$, with Galois group $G$. If $G(2)$ is the Sylow 2-group? of $G$, then the fixed field? of $G(2)$ is an odd degree extension of $F$, given by adjoining a root of an odd degree irreducible polynomial $q$ over $F$. But since $F$ is real closed, $q$ has a root in $F$; by irreducibility, the degree must be $1$, so that in fact $G = G(2)$. We have ${|G|} \gt 1$ since the splitting field contains $K$.

So $G$ is a $2$-primary group. But for any prime number $p$, a nontrivial finite $p$-group has nontrivial center, and is therefore solvable by an inductive argument. Therefore the extension $L/F$ arises from a tower of non-trivial quadratic extension?s

$F \subseteq L_1 \subseteq \ldots \subseteq L_n = L$

By the quadratic formula, the first field $L_1$ arises by adjoining roots to $F$ of a polynomial $x^2 + a x + b$,

$\frac{-a \pm \sqrt{a^2 - 4b}}{2},$

where $a^2 - 4b$ is negative. Since $F$ is real closed, the positive element $4b - a^2$ has a square root in $F$, so that the roots displayed above belong to $K = F[\sqrt{-1}]$. So $L_1 = K$. But $K$ has no nontrivial quadratic extensions by the lemma that follows, so in fact $L_1 = L_n = K$ and the theorem is proved.

###### Lemma

Every element of $K = F[\sqrt{-1}]$ has a square root in $K$.

###### Proof

The proof is most easily apprehended by analogy with polar coordinate representations of complex numbers and half-angle formulas, where a square root of $r e^{i\theta}$ is given by $r^{1/2}e^{i\theta /2}$. Let $i$ be a fixed square root of $-1$, and let $a + b i$ be an arbitrary element of $K$, with $a, b \in F$. We must solve $(x + y i)^2 = a + b i$, i.e., find $x, y \in F$ that solve

$x^2 - y^2 = a, \qquad 2x y = b$

Since $a^2 + b^2$ has a square root in $F$, we may assume by homogeneity in $x, y$ that $(a, b)$ is on the unit circle: $a^2 + b^2 = 1$. By interchanging $x$ and $y$ if need be, we may assume $0 \leq a \leq 1$; replacing $y$ by $-y$ if need be, we may assume $b \geq 0$. Taking $x, y \geq 0$ such that

$x^2 = \frac{1+a}{2}, \qquad y^2 = \frac{1-a}{2},$

we obtain a solution (since $x^2 - y^2 = a$ and $4 x^2 y^2 = b^2$).

## Classical FTA via advanced calculus

As noted above, many proofs of the fundamental theorem are known. The following proof has the advantage that it requires very little machinery; it hardly uses anything not known by Gauss?1.

Let $f\colon \mathbb{C} \to \mathbb{C}$ be a nonconstant polynomial mapping, and suppose $f$ has no zero.

1. First, ${|f(z)|}$ attains an absolute (positive) minimum. For, choose any $z' \in \mathbb{C}$. Since $\lim_{z \to \infty} f(z) = \infty$, there exists some compact ball $B$ containing $z'$ so that ${|f(z)|} \gt {|f(z')|}$ whenever $z \notin B$. By compactness, ${|f(z)|}$ attains an absolute minimum for $z$ ranging over $B$; by choice of $B$, it is the same minimum as for $z$ ranging over all of $\mathbb{C}$.

2. Suppose ${|f|}$ attains an absolute minimum at $z = z_0$. The polynomial $f$ may be uniquely written in the form

$f(z) = f(z_0) + (z - z_0)^n g(z)$

where $g$ is polynomial and $g(z_0) \neq 0$. Put

$F(z) = f(z_0) + g(z_0)(z - z_0)^n$

and choose $\delta \gt 0$ small so that

${|z - z_0|} = \delta \Rightarrow {|g(z) - g(z_0)|} \lt {|g(z_0)|}$
3. $F$ maps the circle $C = \{z : {|z - z_0|} = \delta\}$ onto a circle of radius $r = {|g(z_0)|}\delta^n$ centered at $f(z_0)$. (This uses the fact that any complex number has an $n^{th}$ root, which one can prove using polar coordinate representations. We omit the details.) Choose $z' \in C$ so that $F(z')$ is on the line segment between the origin and $f(z_0)$. Then

${|F(z')|} = {|f(z_0)|} - r$

We also have

${|f(z') - F(z')|} = {|g(z') - g(z_0)|} {|z' - z_0|^n} \lt {|g(z_0)|} \delta^n = r$

according to how we chose $\delta$. We conclude by observing

${|f(z')|} \leq {|F(z')|} + {|f(z') - F(z')|} \lt {|f(z_0)|} - r + r = {|f(z_0)|},$

## In weak foundations

Many proofs rely explicitly on the double negation rule by first supposing that a polynomial $p$ has no root and deriving a contradiction. However, the algebraic proof is almost entirely constructive. (Some general results on splitting fields are problematic in constructive algebra, as is the intermediate value theorem in constructive analysis, but their usage in this proof is fine.)

In fact, the only problem is Lemma 1. This may fail in a topos (such as sheaves over $\mathbb{C}$), since we not be able to find a square root of a complex number $x$ (or element of $K[\sqrt{-1}]$ more generally) if we do not whether or not $x$ is apart from zero (because there is no continuous square-root function).

Most varieties of constructive mathematics (including that in Errett Bishop's book) nevertheless accept the FTA, because the needed square roots follow from weak countable choice ($WCC$, which is a consequence of either excluded middle or countable choice). A fully choice-free constructive proof also exists for the Cauchy complex numbers (which agree with the Dedekind complex numbers by $WCC$).

Fred Richman (1998) has proposed that, in the absence of $WCC$, the FTA should be interepreted as a statement about sets of roots rather than about individual roots. He constructs a complete metric space $\hat{M}_n(\mathbb{C})$ which, classically, is the space of $n$-element multisets of complex numbers (and constructively is the completion of that space) and proves that every complex polynomial $p$ of degree $n$ may be associated with a point in this space in such a way that the $n$ elements of that point (when viewed as a multiset, if possible, and morally in any case) are the $n$ roots of $p$.

## References

1. Maybe the bit involving compactness wasn’t available in Gauss’s time. It would be interesting to write this bit out so that the entire proof would be understood by an eighteenth-century mathematician.

Revised on August 11, 2011 07:16:22 by Todd Trimble (69.118.58.208)