November 14th, 2017, 09:34 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  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: 1,919 Thanks: 533  Quote:
Quote:
Quote:
No. Not obvious. Not even apparently related to FTA. Unclear what you mean.  
November 15th, 2017, 06:46 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  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: 1,919 Thanks: 533  
November 15th, 2017, 03:01 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 383 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics 
\popcorn

November 16th, 2017, 07:28 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
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: 383 Thanks: 205 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!


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