Skip to main content.

Fundamental Theorem of Algebra.

.

As a warm-up, I want to give probably the most beautiful proof of the Fundamental Theorem of Algebra which I know, using the theory of one complex variable. In case you don't know the theorem:

Theorem (Fundamental Theorem of Algebra).

Let f \in \C[x] be a polynomial, f \neq 0. If n = \deg f, there exist constants \alpha_1, \dots, \alpha_n, \beta \in \C, \beta \neq 0 such that

f = \beta \prod_{i=1}^n (x - \alpha_i).

The main ingredient of the proof is the following statement, which is in fact eqiuvalent to the Fundamental Theorem:

Lemma.

Let f \in \C[x] \setminus \C be a non-constant polynomial. Then there exists an \alpha \in \C with f(\alpha) = 0.

Proof.

Assume that on the contrary, f is zero-free. In that case, g : z \mapsto \frac{1}{f(z)} defines a entire function, i.e. a function defined on \C which is holomorphic everywhere. We will show that g is bounded, whence it follows by Liouville's Theorem that g is constant. This implies that f is constant, a contradiction.

First, write f = \sum_{i=0}^n a_i x^i with a_n \neq 0; then

g(z) = z^{-n} \cdot \frac{1}{a_n + \sum_{i=0}^{n-1} a_i z^{i - n}}.

Clearly, for z \to \infty, a_n + \sum_{i=0}^{n-1} a_i z^{i - n} \to a_n \neq 0 uniformly, whence g(z) \to 0 uniformly for z \to \infty. Therefore, g is bounded on B_1 := \{ z \in \C \mid \abs{z} > R \} for some R > 0.

Now consider g on B_2 := \{ z \in \C \mid \abs{z} \le R \}. We have that g is continuous on B_2, and since B_2 is compact, we know that |g| attains its maximum on B_2, whence g is bounded on B_2.

Therefore, g is bounded on B_1 \cup B_2 = \C, and we can conclude.

Now we are able to prove the Fundamental Theorem:

Proof (Fundamental Theorem).

We proceed by induction on \deg f, the degree of f. If \deg f = 0, then f \in \C, whence we can set n := 0 and \beta := f \in \C \setminus \{ 0 \}.

Now assume that the statement holds for polynomials of degree n. Let f be a polynomial of degree n + 1. By the lemma, there exists some \alpha_{n+1} \in \C with f(\alpha_{n+1}) = 0. Now, using the Division Algorithm, write f = q \cdot (x - \alpha_{n+1}) + r with polynomials q, r \in \C[x] satisfying \deg r < \deg (x - \alpha_{n+1}) = 1, i.e. r \in \C. Now

0 = f(\alpha_{n_1}) = q(\alpha_{n+1}) \cdot (\alpha_{n+1} - \alpha_{n+1}) + r = 0 + r,

whence we have r = 0 and, therefore, f = q \cdot (x - \alpha_{n+1}). As \deg f = \deg q + \deg (x - \alpha_{n+1}) = \deg q + 1, we have \deg q = n.

Therefore, by the induction hypothesis, there exist \alpha_1, \dots, \alpha_n, \beta \in \C, \beta \neq 0 with q = \beta \prod_{i=1}^n (x - \alpha_i), whence

f = q \cdot (x - \alpha_{n+1}) = \beta \prod_{i=1}^{n+1} (x - \alpha_i),

i.e. the induction hypothesis holds for f, too.

Categories: Algebra Beautiful Proofs Complex Analysis
Tags:

Comments.

spielwiese. » Blog Archive » another project. wrote on May 4, 2009:

[...] to my research. the formulae will be rendered with latex; mathml is simply unuseable so far. an example post shows a feature with i added to my wp-latex enhancer plugin: (primitive) environments for [...]