View source for Fundamental Theorem of Algebra

== Introduction ==
The '''fundamental theorem of algebra''' states that every [[nonconstant]] [[polynomial]] with [[complex number|complex]] coefficients has a complex [[root]]. In fact, every known proof of this theorem involves a bit of [[analysis]], since a purely algebraic construction of the complex numbers is very hard to work with.

== Proof ==
One of the shortest proofs of the fundamental theorem of algebra uses [[Liouville's Theorem]] of [[complex analysis]], which says that a [[bounded]] [[entire function]] is [[constant]]. Suppose that <math>p(x)</math> were a complex polynomial with no complex roots. Then <math>1/p(x)</math> would be an entire function. To see that it is bounded, we must show that <math>|p(x)|\ge c>0</math> for some <math>c</math>. It can easily be shown that outside a sufficiently large [[disc]], <math>p(x)</math> is dominated by its [[leading term]], which gets large in [[absolute value]], so it suffices to work inside a disc of radius <math>R</math> for some <math>R</math> to find the minimum value. But the (closed) disc of radius <math>R</math> is a [[compact set]], so <math>|p(x)|</math> must achieve a minimum inside the disc. This minimum value is <math>c</math>. Hence <math>\left|\frac{1}{p(x)}\right|\le\frac{1}{c}</math>, so <math>1/p(x)</math> would be a bounded entire function and hence constant. This proves the theorem.

== See also ==
* [[Algebra]]
(This ought to be cleaned up.)

There is another proof that uses [[group theory]] and [[Galois theory]] that is very nice, and someone (possibly me) should write it here at some point. The advantage of this proof is that it uses only a very small amount of analysis (just the [[Intermediate Value Theorem]]), and the rest is all actually [[algebra]].