Fundamental Theorem of Algebra

The Fundamental theorem of algebra states that any nonconstant polynomial with complex coefficients has at least one complex root. The theorem implies that any polynomial with complex coefficients of degree $n$ has $n$ complex roots, counted with multiplicity. A field $F$ with the property that every nonconstant polynomial with coefficients in $F$ has a root in $F$ is called algebraically closed, so the fundamental theorem of algebra states that

The field $\mathbb C$ of complex numbers is algebraically closed.

The polynomial $x^2+1$ has no real roots, but it has two complex roots $i$ and $-i.$

The polynomial $x^2+i$ has two complex roots, namely $\pm \dfrac{1-i}{\sqrt{2}}.$

One might expect that polynomials with complex coefficients have issues with nonexistence of roots similar to those of real polynomials; that is, it is not unreasonable to guess that some polynomial like $$x^3+ix^2-(1+\pi i)x-e$$ will not have a complex root, and finding such a root will require looking in some larger field containing the complex numbers. The fundamental theorem of algebra says that this is not the case: all the roots of a polynomial with complex coefficients can be found living inside the complex numbers already.

This section gives a more precise statement of the different equivalent forms of the fundamental theorem of algebra. This requires a definition of the multiplicity of a root of a polynomial.

The multiplicity of a root $r$ of a polynomial $f(x)$ is the largest positive integer $k$ such that $(x-r)^k$ divides $f(x).$ Equivalently, it is the smallest positive integer $k$ such that $f^{(k)}(r) \ne 0,$ where $f^{(k)}$ denotes the $k^\text{th}$ derivative of $f.$

To prove that the different forms of the fundamental theorem of algebra are equivalent, we'll begin by proving the following theorem.

Let $F$ be a field. The following are equivalent:

(1) Every nonconstant polynomial with coefficients in $F$ has a root in $F.$
(2) Every nonconstant polynomial of degree $n$ with coefficients in $F$ has $n$ roots in $F,$ counted with multiplicity.
(3) Every nonconstant polynomial with coefficients in $F$ splits completely as a product of linear factors with coefficients in $F.$

Clearly (3)$\Rightarrow$(2)$\Rightarrow$(1), so the only nontrivial part is (1)$\Rightarrow$(3). To see this, induct on the degree $n$ of $f(x).$ The base case $n=1$ is clear. Now suppose the result holds for polynomials of degree $n-1.$ Then let $f(x)$ be a polynomial of degree $n.$ By (1), $f(x)$ has a root $a.$ A standard division algorithm argument shows that $x-a$ is a factor of $f(x)$:

Divide $f(x)$ by $x-a$ to get $f(x) = (x-a)q(x)+r,$ where $r$ is a constant polynomial. Plugging in $a$ to both sides gives $0 = (a-a)q(a)+r,$ so $r=0.$ So $f(x) = (x-a)q(x).$ But $q(x)$ is a polynomial of degree $n-1,$ so it splits into a product of linear factors by the inductive hypothesis. Therefore $f(x)$ does as well. So the result is proved by induction. $_\square$

The fundamental theorem of algebra says that the field $\mathbb C$ of complex numbers has property (1), so by the theorem above it must have properties (1), (2), and (3).

If $f(x) = x^4-x^3-x+1,$ then complex roots can be factored out one by one until the polynomial is factored completely: $f(1) = 0,$ so $x^4-x^3-x+1 = (x-1)(x^3-1).$ Then $1$ is a root of $x^3-1,$ so $$x^4-x^3-x+1 = (x-1)(x-1)(x^2+x+1),$$ and now $x^2+x+1$ has two complex roots, namely the primitive third roots of unity $\omega$ and $\omega^2,$ where $\omega = e^{2\pi i/3}.$ So $$x^4-x^3-x+1 = (x-1)^2(x-\omega)(x-\omega^2).$$ There are three distinct roots, but four roots with multiplicity, since the root $1$ has multiplicity $2.$

The ability to factor any polynomial over the complex numbers reduces many difficult nonlinear problems over other fields (e.g. the real numbers) to linear ones over the complex numbers. For example, every square matrix over the complex numbers has a complex eigenvalue, because the characteristic polynomial always has a root. This is not true over the real numbers, e.g. the matrix $$\begin{pmatrix} 0&1\\-1&0\end{pmatrix},$$ which rotates the real coordinate plane by $90^{\circ},$ has no real eigenvalues.

Another general application is to the field of algebraic geometry, or the study of solutions to polynomial equations. The assumption that the coefficients of the polynomial equations lie in an algebraically closed field is essential for simplifying and strengthening the theory, as it guarantees that the field is "big enough" to contain roots of polynomials. For example, the set of complex solutions to a polynomial equation with real coefficients often has more natural and useful properties than the set of real solutions.

Another application worth mentioning briefly is to integration with partial fractions. Over the real numbers, there are awkward cases involving irreducible quadratic factors of the denominator. The algebra is simplified by using partial fractions over the complex numbers (with the caveat that some complex analysis is required to interpret the resulting integrals).

Let $p(x)$ be a polynomial with real coefficients. It is true that $p(x)$ can be factored into linear factors over the complex numbers, but the factorization is slightly more complicated if the factors are required to have real coefficients.

For instance, the polynomial $x^2+1$ can be factored as $(x-i)(x+i)$ over the complex numbers, but over the real numbers it is irreducible: it cannot be written as a product of two nonconstant polynomials with real coefficients.

Every polynomial $p(x)$ with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients.

Induct on $n.$ The base cases are $n=1$ and $n=2,$ which are trivial. Now suppose the theorem is true for polynomials of degree $n-2$ and $n-1.$ Let $f(x)$ be a polynomial of degree $n,$ and let $a$ be a complex root of $f(x)$ (which exists by the fundamental theorem of algebra). There are two cases.

If $a$ is real, then $f(x) = (x-a)q(x)$ for a polynomial $q(x)$ with real coefficients of degree $n-1.$ By the inductive hypothesis, $q(x)$ can be factored into a product of linear and irreducible quadratic factors, so $f(x)$ can as well.

If $a$ is not real, then let ${\overline a}$ be the complex conjugate of $a.$ Note that ${\overline a} \ne a.$ Write $f(x) = c_nx^n+\cdots+c_1x+c_0,$ then

$$\begin{aligned} {\overline{f(x)}} &= \overline{c_nx^n+\cdots+c_1x+c_0} \\ &= {\overline{c_nx^n}} + \cdots + \overline{c_1x} + \overline{c_0} \\ &= {\overline{c_n}} \, {\overline{x}}^n+\cdots + {\overline{c_1}} \, {\overline{x}} + \overline{c_0} \\ &= c_n {\overline x}^n +\cdots + c_1{\overline x} + c_0\\ &= f({\overline{x}}) \end{aligned}$$ by properties of the complex conjugate (and because the $c_i$ are real numbers). So if $f(a) = 0,$ then $f({\overline{a}}) = {\overline{f(a)}} = {\overline{0}} = 0.$ The conclusion is that non-real roots of polynomials with real coefficients come in complex conjugate pairs.

Write $f(x) = (x-a)q(x),$ where $q(x)$ has complex coefficients, and plug in $\overline{a}$ to both sides. Then $q({\overline a}) = 0.$ (This is where the argument uses that ${\overline{a}} \ne a.)$ So $q(x) = (x-{\overline{a}})h(x),$ so $f(x) = (x-a)(x-{\overline{a}})h(x).$ Write the product of the first two factors as $g(x),$ then $g(x)$ is a quadratic irreducible polynomial with real coefficients. Since $g(x)$ divides $f(x)$ over the complex numbers, and both polynomials are real, $g(x)$ must divide $f(x)$ over the real numbers. (Proof: use the division algorithm over the real numbers, $f = gj+k,$ with $k=0$ or $\text{deg}(k)<\text{deg}(g),$ and then over the complex numbers $g$ divides $f$ and $gj,$ so must divide $k;$ so $k=0.)$

So $h(x)$ is a polynomial of degree $n-2$ with real coefficients, which factors as expected by the inductive hypothesis, so $f(x)$ does as well. This completes the proof. $_\square$

There are no "elementary" proofs of the theorem. The easiest proofs use basic facts from complex analysis. Here is a proof using Liouville's theorem that a bounded holomorphic function on the entire plane must be constant, along with a basic fact from topology about compact sets.

Let $p(z) = a_nz^n + \cdots + a_0$ be a polynomial with complex coefficients, and suppose that $p(z) \ne 0$ everywhere. (So of course $a_0 \ne 0.)$ Then $\frac1{p(z)}$ is holomorphic everywhere.

Now $\lim\limits_{z\to\infty} p(z) = \infty,$ for instance, because $$|p(z)| \ge |a_n||z|^n - (|a_{n-1}||z|^{n-1} + \cdots + |a_0|)$$ by the triangle inequality. So for large enough $|z|$, say $|z|>R,$ $|p(z)| > |a_0|.$

But in the disc $|z|\le R,$ the function $|p(z)|$ attains its minimum value (because the disc is compact). Call this value $m.$ Note that $m > 0.$

Then $|p(z)|>\text{min}(m,|a_0|)$ for all $z,$ so $$\left| \frac1{p(z)} \right| < \frac1{\text{min}(m,|a_0|)}$$ for all $z,$ so it is a bounded holomorphic function on the entire plane, so it must be constant by Liouville's theorem. But then $p(z)$ is constant.

So the argument has shown that any nonconstant polynomial with complex coefficients has a complex root, as desired.