October 3rd, 2017, 10:10 AM  #21  
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  Quote:
 
October 3rd, 2017, 10:18 AM  #22 
Senior Member Joined: Aug 2012 Posts: 1,972 Thanks: 550  As I understand it, Gauss gets credit for the first complete (no pun intended) proof, but by modern standards of rigor he didn't have a proof either. In fact the last hole in Gauss's proof was not resolved till as recently as 1920. https://en.wikipedia.org/wiki/Fundam...rem_of_algebra Or are you saying that I shouldn't brag about posting Zylo's proof sketch earlier in this thread? Well you're right I guess. Last edited by Maschke; October 3rd, 2017 at 10:21 AM. 
October 3rd, 2017, 11:00 AM  #23  
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  Quote:
Perhaps I was being a bit snarky with Zylo for proposing a proof that may not have met the standards of rigor of 200 years ago.  
October 4th, 2017, 06:01 AM  #24 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Post #19, the "algebraic proof," is wrong. Consider the following algebraic proof: General third degree complex polynomial: $\displaystyle z^{3}+pz^{2}+qz=c$ $\displaystyle z^{3}$ maps to all points in the complex plane $\displaystyle pz^{2}$ maps to all points in the complex plane 1) $\displaystyle qz$ maps to all points in the complex plane. Therefore $\displaystyle z^{3}+pz^{2}+qz$ maps to all points in the complex plane. Therefore $\displaystyle z^{3}+pz^{2}+qz=c$ has a solution for any c. Same argument for ndegree polynomial. EDIT: 1) does not follow has to be proven. Back to proof I was working on when above suddenly occurred to me. I note none of the replies to previous proof saw what was wrong with it, so they just contained irrelevant vaque generalities. Last edited by zylo; October 4th, 2017 at 06:20 AM. 
October 4th, 2017, 06:19 AM  #25  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 211 Thanks: 64 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
1) here is fine so long as $q \neq 0$ (and if $q = 0$, then you just have fewer terms to worry about). Indeed, for any $a \in \mathbb{C}$, $qz = a$ has a solution $z = \frac{a}{q}$. The issue with this argument is the one I've pointed out above. Last edited by cjem; October 4th, 2017 at 06:31 AM.  
October 5th, 2017, 07:21 AM  #26 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Quote from post # 19 "Previous post #18 gave topological proof of FTA. An algebraic proof follows: Any 3rd degree complex polynomial $\displaystyle z^{3}+pz^{2}+qz+c =0$ can be written in the form $\displaystyle f(z)=z(z^{2}+pz+q)=C$ Assuming induction for the second degree polynomial, $\displaystyle f(z)=z(za)(zb)=\rho\rho_{a}\rho_{b}e^{i(\theta +\theta _{a}+\theta _{b})}$ $\displaystyle z=\rho e^{i\theta} \quad za=\rho _{a}e^{i\theta _{a}}, \quad zb=\rho _{b}e^{i\theta _{b}}$ $\displaystyle a=ae^{i\alpha},\quad b=be^{i\alpha}$  Argument exactly same for nth degree polynomial." The above was incomplete but contained core concept of proof: Objective: Show that $\displaystyle z(za)(zb)$ maps to entire complex plane, ie, $\displaystyle z=\rho e^{i\theta}$ exists for any value of $\displaystyle z(za)(zb)=c.$ The argument is easier to follow for $\displaystyle z(za)=\rho \rho_{a}e^{i(\theta+\theta_{a})}$. 1) Hold $\displaystyle \theta$ constant. Then $\displaystyle z(za) = \rho \rho_{a}$ varies from 0 to $\displaystyle \infty$ as $\displaystyle \rho$ goes from to 0 to $\displaystyle \infty$, variation depending on $\displaystyle \theta$. 2) Hold $\displaystyle \rho$ constant. Then $\displaystyle arg [z(za)]$ varies from $\displaystyle 0+\theta_{a}(0)$ to $\displaystyle 2\pi +\theta_{a}(0)$, variation depending on $\displaystyle \rho$. 3) Think of the process as creating a polar map consisting of rays $\displaystyle \theta$ and circles $\displaystyle \rho$. On each ray label points $\displaystyle z(za)=\rho \rho_{a}$. On each circle label points $\displaystyle arg [z(za)]$. 4) Then, given $\displaystyle c=Re^{i\Theta}$, locate intersection of R and $\displaystyle \Theta$ on the map to determine $\displaystyle \rho$ and $\displaystyle \theta$. Visualize by sketching $\displaystyle za$: draw $\displaystyle a$ and a circle of radius $\displaystyle \rho$ at the end of $\displaystyle a$. Line from origin to point on the circle is $\displaystyle za$. The proof is geometric/algebraic and requires nothing beyond elementary complex algebra. Last edited by zylo; October 5th, 2017 at 07:48 AM. Reason: Change quote to direct copy 
October 5th, 2017, 08:13 AM  #27  
Senior Member Joined: Sep 2016 From: USA Posts: 416 Thanks: 230 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
2. Above you are attempting to prove this fact. However, your argument is far from rigorous. One problem is that $\theta_a$ varies as $\rho$ varies even for fixed $\theta$. You have yet to give a reasonable argument for why a polynomial can't "skip" values. Furthermore, you seem to think it maps circles to circles and rays to rays. Neither of these are true and counterexamples have already been given. The most you can say is that a simple closed curve will map to a closed curve. Even this image need not be simple and a counterexample of this has also been given. Even if it were, this is far from sufficient to show what you are trying to show. 3. Your induction argument still runs into the problems pointed out earlier in this thread if you try to proceed for arbitrary $n$. If you can't address this, then the best you can hope for is a proof of FTA for degrees no greater than 3. I suggest looking at Rouche's theorem if you are really interested in proving FTA via complex analysis.  
October 5th, 2017, 10:48 AM  #28 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  FTA Quickie
SDK, I apply FTA by induction to the lower order polynomial f(z), regardless of degree. I map on to circles and lines, not to circles and lines. Sorry you didn't understand my post #26. You may prefer the following: FTA Quickie: 1) Given 3rd degree complex polynomial $\displaystyle z(z^{2}+pz+q)=zf(z)=c$ By induction applied to the second degree polynomial, $\displaystyle f(z)$ maps z to entire complex plane. 2) $\displaystyle zf(z)$ maps continuously to entire complex plane (from 0 to infinity). 3) $\displaystyle zf(z)=c$ has a solution (root) z=d for any c. 4) Same argument for any degree. I have googled FTA. The complexity and abstraction are not my cup of tea. I prefer a simple, transparent, comprehensible, easy to remember, proof, where possible. If it bothers you, be consoled by fact that it is virtually unnoticed and writ in water, and the others will be around a long time. Anyhow, thanks for your interest. 
October 5th, 2017, 01:08 PM  #29 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Even simpler: Given any n degree complex polynomial written as: $\displaystyle F_{n}(z)=a_{0}$ $\displaystyle F_{n}(z)$ is continuous for all $\displaystyle z$ and varies continuously from complex $\displaystyle 0$ to complex $\displaystyle \infty$ (eventually $\displaystyle z^{n}$ predominates). It follows $\displaystyle F_{n}(z)=a_{0}$ has a root for any $\displaystyle a_{0}$. 
October 5th, 2017, 01:24 PM  #30 
Senior Member Joined: Aug 2012 Posts: 1,972 Thanks: 550  

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