Fundamental Theorem of Algebra

The fundamental theorem of algebra states that every nonconstant polynomial with complex coefficients has a complex root. In fact, every known proof of this theorem involves some analysis, since the result depends on certain properties of the complex numbers that are most naturally described in topological terms.

It follows from the division algorithm that every complex polynomial of degree $n$ has $n$ complex roots, counting multiplicities. In other words, every polynomial over $\mathbb{C}$ splits over $\mathbb{C}$, or decomposes into linear factors.

Proofs

Proof by Liouville's Theorem

We use Liouville's Boundedness Theorem of complex analysis, which says that every bounded entire function is constant.

Suppose that $P(z)$ is a complex polynomial of degree $n$ with no complex roots; without loss of generality, suppose that $P$ is monic. Then $1/P(z)$ is an entire function; we wish to show that it is bounded. It is clearly bounded when $n=0$; we now consider the case when $n>0$.

Let $R$ be the sum of absolute values of the coefficients of $P$, so that $R \ge 1$. Then for $\lvert z \rvert \ge S$, \[\lvert P(z) \rvert \ge \lvert z^n \rvert - (R-1) \lvert z^{n-1} \rvert = \lvert z^{n-1} \rvert \cdot \bigl[ \lvert z \rvert - (R-1) \bigr] \ge R^{n-1} .\] It follows that $1/P(z)$ is a bounded entire function for $\lvert z \rvert > R$. On the other hand, by the Heine-Borel Theorem, the set of $z$ for which $\lvert z \rvert \le R$ is a compact set so its image under $1/P$ is also compact; in particular, it is bounded. Therefore the function $1/P(z)$ is bounded on the entire complex plane when $n>0$.

Now we apply Liouville's theorem and see that $1/P(z)$ is constant, so $P(z)$ is a constant polynomial. The theorem then follows. $\blacksquare$

Algebraic Proof

Let $P(x)$ be a polynomial with complex coefficients. Since $F(x) = P(x) \overline{P(x)}$ is a polynomial with real coefficients such that the roots of $P$ are also roots of $F$, it suffices to show that every polynomial with real coefficients has a complex root. To this end, let the degree of $F$ be $d = 2^n q$, where $q$ is odd. We induct on the quantity $n$.

For $n=0$, we note that for sufficiently large negative real numbers $x$, $F(x) < 0$; for sufficiently large positive real numbers $x$, $F(x) > 0$. It follows from the Intermediate Value Theorem that $F(x)$ has a real root.

Now suppose that $n > 0$. Let $C$ be a splitting field of $F$ over $\mathbb{C}$, and let $x_1, \dotsc, x_d$ be the roots of $F$ in $C$.

Let $c$ be an arbitrary real number, and let $y_{c,i,j} = x_i + x_j + cx_ix_j$ for $1 \le i \le j \le d$. Let \[G_c(x) = \prod_{1 \le i \le j \le d} (x-y_{c,i,j}) .\] The coefficients of $G$ are symmetric in $x_1, \dotsc, x_d$. Therefore they can be expressed as linear combinations of real numbers times the elementary symmetric polynomials in $x_1, \dotsc, x_n$; thus they are real numbers. Since the degree of $G_c$ is $\binom{d+1}{2} = 2^{n-1}q(d+1)$, it follows by inductive hypothesis that $G_c$ has a complex root; that is, $y_{c,i(c),j(c)} \in \mathbb{C}$ for some $1 \le i(c) \le j(c) \le d$.

Now, since there are infinitely many real numbers but only finitely many integer pairs $(i,j)$ with $1 \le i \le j \le d$, it follows that for two distinct numbers $c,c'$, $(i(c),j(c)) = (i(c'),j(c')) = (i,j)$. It follows that $x_i + x_j$ and $x_ix_j$ are both complex numbers, so $x_i$ and $x_j$ satisfy a quadratic equation with complex coefficients. Hence they are complex numbers. Therefore $F$ has a complex root, as desired. $\blacksquare$

References

See also

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