Talk:Fundamental theorem of algebra/Archive 1

Archive 1

factorizing

i read from somewhere something thats in direct contridiction with this article. in here it says any polynomial of degree n can be factorized as p(a)=(a-an)(a-an-1)...(a-a0) (sorry ima noob to wiki) either they are complex or real, but from that place it says it works only for n distinct real zeros. i think this article is correct but not sure, so can someone tell me the exact detail of this factorizing? also, i dont understand how the solution of general quintic is algebraic, like it cannot be solved exactly without using infinty series

I assume that in your undisclosed article there is referred to a specific method for factorising. Ofcourse it is possible that a certain method only works for real roots. This does not exclude the possibility to use other methods in case of complex roots. Bob.v.R 17:14, 15 September 2005 (UTC)

3 = 4?

for instance, he shows that the equation $x^{4}=4x-3$, although incomplete, has four solutions: $1$, $1$, $-1+i$√$2$, and $-1-i$√$2$.

This looks more like three true solutions to me... --Abdull 13:31, 9 June 2006 (UTC)

As is traditional, they're counted with multiplicity. That should be explicit though Algebraist 21:37, 18 February 2007 (UTC)

Generalizations

Can this theorem be generalized to (complex) matrix polynomials, or even to Banach algebras?

One generalization which has just been proven -- any field where all polynomials of prime degree have roots is algebraically closed. This improves on the proof given in the article, which requires roots for polynomials whose degree is 2 or odd. 24.149.215.73 16:15, 24 January 2006 (UTC) J. Shipman

It would be really worthwhile to include this generalization in the article, but your FOM posts are not enough to satisfy the WP:NOR criteria. Did you submit the result for publication in a journal? -- EJ 16:14, 8 January 2007 (UTC)

Generalisation to ANY other complex Banach algebra with unit fail at that first hurdle since a solution to aX-1=0 is an inverse to a and the only complex Banach algebra which is a field is C A Geek Tragedy 11:47, 18 February 2007 (UTC)

Since the FToA is equivalent to the statement that every complex matrix as an eigenvalue, one can view the nonemptiness of spectrum (for general Banach algebras) as a generalisation, especially since the standard way of proving it is the same as one of the proofs of the FT. Algebraist 21:36, 18 February 2007 (UTC)

OOO Nice :) A Geek Tragedy 19:15, 19 February 2007 (UTC)

Vagueness in winding number proof

To me, a non-methematician, the winding number proof seems incomplete. Our article on winding number states that the winding number is a property of a given curve with respect to a given point. However, in the present article, we never indicate what point we're referring to. It also doesn't say why the existence of a zero of $p(z)$ makes the statement any less absurd. Could someone familiar with this area please explain? --Doradus 17:49, 19 December 2004 (UTC)

The point referred to is the origin. (The article does say "counter-clockwise around 0".) In the proof, the loop is transformed in a continuous manner while the winding number changes from n to 0. In order for this to be a contradiction, there has to be no opportunity for the winding number to change as you transform the loop. But if p(z) = 0 somewhere, then the loop can at some stage pass through the origin, allowing the winding number to change. --Zundark 22:17, 19 December 2004 (UTC)

Ok, I made it "winding number with respect to zero". However, I think the proof is still unconvincing. What is the winding number of the constant circle |z|=0? Given what you just said, I presume it is zero. Are we sure of that? If so, it should be included in the proof; something like "... from the original circle to the constant circle |z|=0, whose winding number is 0, not n". I guess what I'm saying is that if we're going to claim something is absurd, it should be explicitly absurd. --Doradus 19:02, 20 December 2004 (UTC)

It's not talking about the winding number of |z|=0, but rather the winding number of p(z) when |z|=0. This is 0, as it's a non-zero constant function. --Zundark 20:48, 20 December 2004 (UTC)

Aha! Thanks for the reminder. --Doradus 04:04, 21 December 2004 (UTC)

Hang on... One of the premises of this proof is that if p has no zeros, then one can choose a value for z such that the zⁿ term dominates, and hence p will have a winding number of n with respect to zero. However, they then proceed to consider z values closer and closer to zero. Who says the winding number doesn't change once zⁿ no longer dominates? --Doradus 04:33, 21 December 2004 (UTC)

There is no possibility for it to change if the loop never crosses 0. This is sort of intuitively "obvious", especially for what we actually need in this proof (you can't take a loop encircling the origin and shrink it down to a point without hitting the origin). But it's not so easy to prove rigorously. If you really want to see all the details, you should consult an appropriate book. (If you don't have such a book you could download chapter 1 of Hatcher's book on algebraic topology. The Fundamental Theorem of Algebra is Theorem 1.8, but you will need to read the earlier parts of the chapter to understand the proof. The proof uses the fundamental group of the circle rather than winding numbers, but it amounts to the same thing - the elements of the fundamental group are essentially winding numbers.) --Zundark 14:42, 21 December 2004 (UTC)

Ah. And if it does cross 0, then the function has at least one zero. I think I get the picture now. Thanks for your patience. I think I may try to clarify some of these points in the article itself. --Doradus 17:13, 21 December 2004 (UTC)

I actually agree with Doradus in that the proof is not always very clear. I thought the sentence "Since p(z) has no zeros, the path can never cross over 0 as it deforms, and hence its winding number with respect to 0 will never change" at first sight may not seem to be a trivial fact, so I constructed an actual homotopy between p(z) and p(0) to illustrate that the curve couldn't cross zero during the homotopy.LkNsngth (talk) 03:34, 7 April 2008 (UTC)

Algebraic/analytic stuff

Surely talk of a "purely algebraic" proof, makes little or no sense? Real number is a term that requires a rigorous grounding or definition of some kind. It seems silly to say we can avoid the analysis in the definition of real numbers (and hence avoid analysis in the definition of complex numbers, and possibly avoid analysis in the proof of the fundamental theorem of algebra) by merely stipulating axioms for real numbers. One might as well stipulate axioms for complex numbers, and include the fundamental theorem in this list of axioms. Woohoo, we avoided a limiting process!

That depends on your point of view. At least one of the proofs (Euler-Gauß) is valid for any real, algebraically closed field. The real algebraic closure of the rationals is less than the real numbers, it is still countable. The fact that the real numbers are real-algebraically closed is of course analytical, it is essentially the intermediate value theorem.--LutzL (talk) 06:59, 25 August 2008 (UTC)

Bounds

I've added a small section about a priori bounds on the zeroes of a polynomial. It seems a kind of worth information; I'm only not sure if this article is the right place to it... --PMajer (talk) 18:24, 25 October 2008 (UTC)

I think there is already an article, yes there it is, Properties of polynomial roots, containing a section on bounds.--I took the liberty to restrict the p-norms to p>=1.--LutzL (talk) 19:09, 25 October 2008 (UTC)

Yes, of course, it's p>=1, thank you. So at the moment I added a link to "Properties of polynomial roots"; I also added a short proof (it seems to me worth, because it is quite short and elementary). Then one can also decide to move it. --PMajer (talk) 11:19, 27 October 2008 (UTC)