# 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 has complex roots, counting multiplicities. In other words, every polynomial over splits over , 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 is a complex polynomial of degree with no complex roots; without loss of generality, suppose that is monic. Then is an entire function; we wish to show that it is bounded. It is clearly bounded when ; we now consider the case when .

Let be the sum of absolute values of the coefficients of , so that . Then for , It follows that is a bounded entire function for . On the other hand, by the Heine-Borel Theorem, the set of for which is a compact set so its image under is also compact; in particular, it is bounded. Therefore the function is bounded on the entire complex plane when .

Now we apply Liouville's theorem and see that is constant, so is a constant polynomial. The theorem then follows.

### Algebraic Proof

Let be a polynomial with complex coefficients. Since is a polynomial with real coefficients such that the roots of are also roots of , it suffices to show that every polynomial with *real* coefficients has a complex root. To this end, let the degree of be , where is odd. We induct on the quantity .

For , we note that for sufficiently large negative real numbers , ; for sufficiently large positive real numbers , . It follows from the Intermediate Value Theorem that has a real root.

Now suppose that . Let be a splitting field of over , and let be the roots of in .

Let be an arbitrary real number, and let for . Let The coefficients of are symmetric in . Therefore they can be expressed as linear combinations of real numbers times the elementary symmetric polynomials in ; thus they are real numbers. Since the degree of is , it follows by inductive hypothesis that has a complex root; that is, for some .

Now, since there are infinitely many real numbers but only finitely many integer pairs with , it follows that for two distinct numbers , . It follows that and are both complex numbers, so and satisfy a quadratic equation with complex coefficients. Hence they are complex numbers. Therefore has a complex root, as desired.

## References

- Samuel, Pierre (trans. A. Silberger),
*Algebraic Theory of Numbers*, Dover 1970, ISBN 978-0-486-46666-8 . - Proofs of the Fundamental Theorem of Algebra on Cut the Knot