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Math Help - Fundamental Theorem of AlgebravProof

  1. #1
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    Fundamental Theorem of AlgebravProof

    Why do many books go through "downsizing" arguments to prove the Fundamental Theorem of Algebra? Isn't this the simplest proof?

    Theorem. Every non-constant polynomial with complex coefficients has a zero in  \mathbb{C} .


    Proof. Let  P(z) be any polynomial. If  P(z) \neq 0 for all  z \in \mathbb{C} ,  f(z) = 1/P(z) is an entire function. Furthermore, if  P is non-constant,  P \to \infty as  z \to \infty and  f is bounded. By Liouville's Theorem,  f is constant and so  P , contrary to our assumption.


    Yet most books make this so complicated. Why?
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    Quote Originally Posted by manjohn12 View Post
    Why do many books go through "downsizing" arguments to prove the Fundamental Theorem of Algebra? Isn't this the simplest proof?

    Theorem. Every non-constant polynomial with complex coefficients has a zero in  \mathbb{C} .


    Proof. Let  P(z) be any polynomial. If  P(z) \neq 0 for all  z \in \mathbb{C} ,  f(z) = 1/P(z) is an entire function. Furthermore, if  P is non-constant,  P \to \infty as  z \to \infty and  f is bounded. By Liouville's Theorem,  f is constant and so  P , contrary to our assumption.


    Yet most books make this so complicated. Why?
    there are many proofs of the fundamental theorem of algebra [most of them require knowledge of topics outside of abstract algebra].

    the proof learned in algebraic topology is, in my opinion, the most straightforward. It is actually a very quick proof.
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    Quote Originally Posted by manjohn12 View Post
    Why do many books go through "downsizing" arguments to prove the Fundamental Theorem of Algebra? Isn't this the simplest proof?

    Theorem. Every non-constant polynomial with complex coefficients has a zero in  \mathbb{C} .


    Proof. Let  P(z) be any polynomial. If  P(z) \neq 0 for all  z \in \mathbb{C} ,  f(z) = 1/P(z) is an entire function. Furthermore, if  P is non-constant,  P \to \infty as  z \to \infty and  f is bounded. By Liouville's Theorem,  f is constant and so  P , contrary to our assumption.


    Yet most books make this so complicated. Why?
    In order to understand this proof you need to be familar with complex analysis while this result is an algebraic result. Thus, some books feel that algebraic results deserve algebraic proofs. I am familar with another proof that uses Galois theory, however it is little bit more involved, but still nice.
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    Quote Originally Posted by ThePerfectHacker View Post
    In order to understand this proof you need to be familar with complex analysis while this result is an algebraic result. Thus, some books feel that algebraic results deserve algebraic proofs. I am familar with another proof that uses Galois theory, however it is little bit more involved, but still nice.

    But most complex analysis books still use a really algebraic approach to proving the Fundamental Theorem of Algebra.
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    Quote Originally Posted by manjohn12 View Post
    But most complex analysis books still use a really algebraic approach to proving the Fundamental Theorem of Algebra.
    How? What books?
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    Quote Originally Posted by ThePerfectHacker View Post
    How? What books?
    Bak and Newman use above proof. I don't recall the other books which use the complicated proofs.

    But I know triangle inequalities and stuff like that.
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    Quote Originally Posted by manjohn12 View Post
    Bak and Newman use above proof. I don't recall the other books which use the complicated proofs.

    But I know triangle inequalities and stuff like that.
    That proof is analytic! You are using a fact about analytic functions on the complex plane which is proven using the knowledge of contour integration in the complex plane. This is not in any way an algebraic proof, it is 100% analytic.
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    What you post is a "complicated" proof: its complications are just hidden in "Liouville's Theorem".

    The simplest proof that I know only requires knowing that a complex number can be written in exponential form. Of course it takes quite a lot computation. Is that what you mean by "complicated"?
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    Quote Originally Posted by ThePerfectHacker View Post
    That proof is analytic! You are using a fact about analytic functions on the complex plane which is proven using the knowledge of contour integration in the complex plane. This is not in any way an algebraic proof, it is 100% analytic.

    Theorem. Let  P_{n}(z) = a_{z}z^{n}+ a_{n-1}z^{n-1} + \ldots + a_{1}z + a_0 . Then  P_{n}(z) has at least one zero.


    Proof. Assume to the contrary. Then  f(z) = \frac{1}{P_{n}(z)} is a bounded entire function, hence constant by Liouville's Theorem. Contradiction. We have:

     P_{n}(z) = a_{n}z^{n}+ a_{n-1}z^{n-1} + \cdots + a_{1}z + a_0

     \therefore |P_{n}(z)| = |z^n| \cdot \left|a_{n}+ \frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + \cdots + \frac{a_{1}}{z^{n-1}} + \frac{a_0}{z^n} \right|

     \geq |z^n| \left( |a_n| - \left|\frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + \cdots + \frac{a_{1}}{z^{n-1}} + \frac{a_0}{z^n} \right| \right)

     \geq |z^n| \left( |a_n|- \left(\underbrace{\frac{|a_{n-1}|}{|z|} + \frac{|a_{n-2}|}{|z^2|} + \cdots + \frac{|a_{1}|}{|z^{n-1}|} + \frac{|a_0|}{|z^n|}}_{Q} \right) \right)


    Consider two regions in the complex plane. The region inside and on the circle  |z| = R and the region outside the circle  |z| = R . We choose  R so large such that  Q \leq \frac{|a_n|}{2} for all  z .

     \geq |z^n| \left(|a_n|- \frac{|a_n|}{2} \right) = |z^n| \frac{|a_2|}{2}

     \therefore |f(z)| = \frac{1}{P_{n}(z)|} \leq \frac{1}{|z^n| \frac{|a_n|}{2}}  \leq \frac{1}{R^{n} \frac{|a_n|}{2}} = M

    So  f(z) is a bounded entire function and constant.


    Why don't Bak and Newman do this?
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    Quote Originally Posted by manjohn12 View Post
    Theorem. Let  P_{n}(z) = a_{z}z^{n}+ a_{n-1}z^{n-1} + \ldots + a_{1}z + a_0 . Then  P_{n}(z) has at least one zero.


    Proof. Assume to the contrary. Then  f(z) = \frac{1}{P_{n}(z)} is a bounded entire function, hence constant by Liouville's Theorem. Contradiction. We have:

     P_{n}(z) = a_{n}z^{n}+ a_{n-1}z^{n-1} + \cdots + a_{1}z + a_0

     \therefore |P_{n}(z)| = |z^n| \cdot \left|a_{n}+ \frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + \cdots + \frac{a_{1}}{z^{n-1}} + \frac{a_0}{z^n} \right|

     \geq |z^n| \left( |a_n| - \left|\frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + \cdots + \frac{a_{1}}{z^{n-1}} + \frac{a_0}{z^n} \right| \right)

     \geq |z^n| \left( |a_n|- \left(\underbrace{\frac{|a_{n-1}|}{|z|} + \frac{|a_{n-2}|}{|z^2|} + \cdots + \frac{|a_{1}|}{|z^{n-1}|} + \frac{|a_0|}{|z^n|}}_{Q} \right) \right)


    Consider two regions in the complex plane. The region inside and on the circle  |z| = R and the region outside the circle  |z| = R . We choose  R so large such that  Q \leq \frac{|a_n|}{2} for all  z .

     \geq |z^n| \left(|a_n|- \frac{|a_n|}{2} \right) = |z^n| \frac{|a_2|}{2}

     \therefore |f(z)| = \frac{1}{P_{n}(z)|} \leq \frac{1}{|z^n| \frac{|a_n|}{2}}  \leq \frac{1}{R^{n} \frac{|a_n|}{2}} = M

    So  f(z) is a bounded entire function and constant.


    Why don't Bak and Newman do this?
    This is my third time telling you, this is still not an algebraic proof. You are using the theory of entire functions - analytic results. Bak and Newman do this, not exactly in the way you wrote it, but their proof is standard.
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    Quote Originally Posted by ThePerfectHacker View Post
    This is my third time telling you, this is still not an algebraic proof. You are using the theory of entire functions - analytic results. Bak and Newman do this, not exactly in the way you wrote it, but their proof is standard.
    Ok. So I guess they dont to the trouble to show that it is bounded.
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  12. #12
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    The interesting thing about the fundamental theorem of algebra is that it is not so fundamental.
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    Quote Originally Posted by ThePerfectHacker View Post
    The interesting thing about the fundamental theorem of algebra is that it is not so fundamental.
    It is fundamental in the sense like the fundamental theorem of arithmetic. E.g. the building blocks of numbers are primes.
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