Fundamental Theorem of Algebra. « Felix’ Math Place

Fundamental Theorem of Algebra.

Abstract.
We want to give a proof of the Fundamental Theorem of Algebra using methods from Complex Analysis, in particular Liouville’s Theorem.

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

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

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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 $\displaystyle 0 = f(\alpha_{n_1}) = q(\alpha_{n+1}) \cdot (\alpha_{n+1} - \alpha_{n+1}) + r = 0 + r,$ 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 $\displaystyle f = q \cdot (x - \alpha_{n+1}) = \beta \prod_{i=1}^{n+1} (x - \alpha_i),$ i.e. the induction hypothesis holds for $f$ , too.

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Tags: fundamental theorem of algebra

This entry was posted on Monday, May 4th, 2009 at 03:33 Europe/Berlin and is filed under Algebra, Beautiful Proofs, Complex Analysis. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

one response to “Fundamental Theorem of Algebra.”

spielwiese. » Blog Archive » another project. says:

04.05.2009 at 05:52 Europe/Berlin

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