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Fundamental Theorem of Algebra

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The Fundamental Theorem of Algebra is a mathematical theorem stating that every nonconstant polynomial whose coefficients are complex numbers has at least one complex number as a root. In other words, given any polynomial



(where  is any positive integer), we can find a complex number  so that



One important case of the Fundamental Theorem of Algebra is that every nonconstant polynomial with real coefficients must have at least one complex root. Since it is not true that every such polynomial has to have at least one real root (as the example  demonstrates), many mathematicians feel that the complex numbers form the most natural setting for working with polynomials.

In fact, a stronger version of the Fundamental Theorem of Algebra is also true: a polynomial of degree  can be factored completely into a product of  linear polynomials:



A way of saying this is that every polynomial of degree  has exactly  complex roots, "counting multiplicity".

Carl Friedrich Gauss is generally credited with the first satisfactory proof of this theorem, his proof being the principal result in his Ph.D. thesis finished in 1799.

Proving the Fundamental Theorem of Algebra

Using complex analysis

A startlingly simple proof is based on Liouville's theorem: If  is a polynomial function of a complex variable then both  and  will be holomorphic in any domain where . But, by the triangle inequality, we know that outside a neighborhood of the origin , so if there is no  such that , we know that  is a bounded entire (i.e., holomorphic in all of ) function. By Liouville's theorem, it must be constant, so  must also be constant.

Using algebra (and a bit of real analysis)

There are also proofs that do not depend on complex analysis, but they require more algebraic or topological machinery.

We need to show that any algebraic extension of  has degree one. Since , any such field extension also extends . Now, any  algebraic over  must also be algebraic over , but its minimal polynomial cannot be of odd degree, because any such polynomial must have a real root by the intermediate value theorem, so the splitting field of  over  must have degree a power of 2. Its Galois group must have a normal subgroup of index 2, but a generating element must already be in  by the quadratic formula. This shows that the extension has, at most, degree 2, but appealing once again to the quadratic formula, we see that  is closed under quadratic extensions, so  must itself be algebraically closed.

Using the fundamental group of the punctured plane 

We can assume without loss of generality that the leading coefficient of  is 1. Now assume that  has no roots. Then, the maps  given by  are homotopic in  for all , and hence they are all null homotopic there (take ). However, for large enough , the maps  given by  and  (where  is the degree of the polynomial ) are homotopic, and thus  is not null homotopic.

Using the second homotopy group of the Riemann sphere 

Without loss of generality the leading coefficient of  is 1. We consider  as a map from the Riemann sphere to itself (taking infinity to infinity). By considering the homotopy , where , this map is homotopic to the map . Hence it suffices to show that the map  is not null homotopic. However, in the homotopy group  we have , and so it suffices to show that  is not null homotpic, which is equivalent to the fact that the sphere is not contractible.