November 14th, 2017, 09:34 AM  #1 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Fundamental Theorem of Algebra Proof 01
1) Let w = z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz = z(z$\displaystyle ^{2}$+az+b) 2) Is z$\displaystyle ^{3}$ a subset of z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz? Yes, if z$\displaystyle ^{2}$ is a subset of z$\displaystyle ^{2}$+az+b, which it is by induction.* 3) Since z$\displaystyle ^{3}$ maps to all of complex plane, so does w = z$\displaystyle ^{3}$+az$\displaystyle ^{2}$+bz. 4) The extension should be obvious: P$\displaystyle _{n1}$ $\displaystyle \rightarrow$ P$\displaystyle _{n}$. FTA 5) * z$\displaystyle ^{2}$ is a subset of z$\displaystyle ^{2}$+az+b if z_{0}^{2} = z$\displaystyle ^{2}$+az+b has a solution for all z$\displaystyle _{0}$. 
November 14th, 2017, 09:50 AM  #2  
Senior Member Joined: Aug 2012 Posts: 2,308 Thanks: 706  Quote:
Quote:
Quote:
No. Not obvious. Not even apparently related to FTA. Unclear what you mean.  
November 15th, 2017, 06:46 AM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
The point of the proof is to show z$\displaystyle ^{n}$+a$\displaystyle _{n1}$z$\displaystyle ^{n1}$+..+a$\displaystyle _{1}z$ maps to entire complex plane: FTA It is shown by induction that {z$\displaystyle ^{n}$} is a subset of {z$\displaystyle ^{n}$+a$\displaystyle _{n1}$z$\displaystyle ^{n1}$+..+a$\displaystyle _{1}z$} and z$\displaystyle ^{n}$ maps to entire complex plane. z$\displaystyle ^{n}$+a$\displaystyle _{n1}$z$\displaystyle ^{n1}$+..+a$\displaystyle _{1}z$ maps to entire complex plane iff z$\displaystyle ^{n}$+a$\displaystyle _{n1}$z$\displaystyle ^{n1}$+..+a$\displaystyle _{1}z$+a$\displaystyle _{0}$ does. It is convenient to use both. Last edited by zylo; November 15th, 2017 at 07:02 AM. 
November 15th, 2017, 09:02 AM  #4 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  OP Clarified
1) Let $\displaystyle w=z^{3}+az^{2}+bz=z(z^{2}+az+b) $ 2) {$\displaystyle z^{2}$} $\displaystyle \equiv$ {$\displaystyle z^{2}+az+b$} by induction##, ie, with each $\displaystyle z^{2}$ I can associate $\displaystyle z^{2}+az+b$, ie $\displaystyle z^{2} \leftrightarrow z^{2}+az+b \rightarrow z^{3} \leftrightarrow z^{3}+az^{2}+bz \rightarrow$ {$\displaystyle z^{3}$} $\displaystyle \equiv$ {$\displaystyle z^{3}+az^{2}+bz$} and $\displaystyle z^{3}$ maps to all of z. Therefore 3) $\displaystyle w=z^{3}+az^{2}+bz$ maps to all of z. 4) The extension should be obvious: P$\displaystyle _{n1}$ $\displaystyle \rightarrow$ P$\displaystyle _{n}$. FTA ## Let $\displaystyle z_{A}=z_{B}^{2}+az_{B}+b$, for each $\displaystyle z_{A}$ you can find z$\displaystyle _{B}$ INDUCTION premise, and for each $\displaystyle z_{B}$ you can find $\displaystyle z_{A}$, ie, {$\displaystyle z^{2}$} $\displaystyle \equiv$ {$\displaystyle z^{2}+az+b$} 
November 15th, 2017, 09:36 AM  #5 
Senior Member Joined: Aug 2012 Posts: 2,308 Thanks: 706  
November 15th, 2017, 03:01 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics 
\popcorn

November 16th, 2017, 07:28 AM  #7 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Ex for n=4
Illustration of OP for n=4. The objective is to show that $\displaystyle z^{4}+az^{3}+bz^{2}+cz$ maps to all of complex plane (FTA) if $\displaystyle z^{3}+az^{2}+bz$ does. {$\displaystyle z^{3}$} $\displaystyle \subset$ {$\displaystyle z^{3}+az^{2}+bz+c$} ** $\displaystyle \therefore z^{3} \rightarrow z^{3}+az^{2}+bz+c$ $\displaystyle \therefore z^{4} \rightarrow z^{4}+az^{3}+bz^{2}+cz$ $\displaystyle \therefore$ {$\displaystyle z^{4}$} $\displaystyle \subset$ {$\displaystyle z^{4}+az^{3}+bz^{2}+cz$}. z$\displaystyle ^{4}$ maps to all of complex plane, $\displaystyle \therefore z^{4}+az^{3}+bz^{2}+cz$ maps to all of complex plane, *** which is same as $\displaystyle z^{4}+az^{3}+bz^{2}+cz+d$ maps to all of complex plane, which is used to prove n=5. **$\displaystyle z_{0}^{3}=z^{3}+az^{2}+bz+c$ has a solution for every z$\displaystyle _{0}$ because n=3 given, ($\displaystyle z^{3}+az^{2}+bz+c$ maps to entire complex plane). *** If this step bothers you, see proof in post #4 which has an extra step. 
November 16th, 2017, 08:57 AM  #8 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
The following is a generalization illustrating some steps in previous post. Theorem If f(z) maps to entire complex plane, so does zf(z). Proof $\displaystyle z_{0} = f(z)$ has a solution for all of $\displaystyle z_{0}$ because f(z) maps to entire complex plane. $\displaystyle \therefore$ {z} $\displaystyle \subset$ {f(z)} $\displaystyle \therefore$ z $\displaystyle \rightarrow$ f(z), maps to $\displaystyle \therefore z^{2} \rightarrow zf(z)$ $\displaystyle \therefore$ {$\displaystyle z^{2}$} $\displaystyle \subset$ {zf(z)} z$\displaystyle ^{2}$ maps to entire complex plane. $\displaystyle \therefore$ zf(z) maps to entire complex plane. Alternate proof: Ref Post #4 $\displaystyle z_{0}$ = f(z) has a solution for all of $\displaystyle z_{0}$ because f(z) maps to entire complex plane, and every z determines a $\displaystyle z_{0}$. $\displaystyle \therefore$ {z} $\displaystyle \equiv$ {f(z)} $\displaystyle \therefore z \leftrightarrow f(z)$ $\displaystyle \therefore z^{2} \leftrightarrow zf(z)$ $\displaystyle \therefore$ {$\displaystyle z^{2}$} $\displaystyle \equiv$ {$\displaystyle zf(z)$} $\displaystyle \therefore$ zf(z) maps to entire complex plane. 
November 16th, 2017, 10:02 AM  #9 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
The pattern for FTA utilizing above theorem is: $\displaystyle P_{n1}(z) \rightarrow zP_{n1}(z) \rightarrow P_{n}(z)$ It is essential to note that $\displaystyle z^2 +az$ maps to entire complex plane iff $\displaystyle z^2 +az +b$ does, and the n degree generalization. 
November 16th, 2017, 10:54 AM  #10 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics 
Great job! You just proved that since $f(z) = \bar{z}$ maps to the entire complex plane, then so does $g(z) = z\bar{z} = z^2$. Keep up the great work proving solid mathematical theorems!


Tags 
algebra, fundamental, proof, theorem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Fundamental Theorem of Algebra Proof  zylo  Complex Analysis  50  December 5th, 2017 08:10 AM 
proof of fundamental theorem of arithmetic  AspiringPhysicist  Abstract Algebra  3  October 2nd, 2014 12:19 AM 
Is this proof of the fundamental theorem of arithmetic ok?  king.oslo  Algebra  2  September 19th, 2013 10:48 PM 
Herstein: Abstract Algebra Proof (Sylow's Theorem)  ThatPinkSock  Abstract Algebra  1  November 11th, 2011 04:25 AM 
Fundamental Theorem of Algebra  johnny  Algebra  11  December 21st, 2007 09:51 PM 