Difference between revisions of "Fundamental Theorem of Algebra"
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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. | 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.) | (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]]. | 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]]. | ||
Revision as of 16:39, 24 June 2006
Introduction
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 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 
were a complex polynomial with no complex roots. Then 
would be an entire function. To see that it is bounded, we must show that 
for some 
. It can easily be shown that outside a sufficiently large disc, is dominated by its leading term, which gets large in absolute value, so it suffices to work inside a disc of radius 
for some to find the minimum value. But the (closed) disc of radius is a compact set, so 
must achieve a minimum inside the disc. This minimum value is . Hence 
, so would be a bounded entire function and hence constant. This proves the theorem.
See also
(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.
