# Talk:Fundamental theorem of algebra

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## Is analysis required for proof?

"All proofs of the fundamental theorem of algebra involve some analysis, at the very least the concept of continuity of real or complex functions. This is unavoidable because the statement of the theorem depends on analysis; the sets of real and complex numbers are analytic objects. Some proofs also use differentiable or even analytic functions."

I don't think this makes much useful sense -- it's either trivial (if "some analysis" means "just the analysis to define C"), or untrue (meaning that the "because" doesn't hold above).
The algebraic complex numbers are algebraically closed, whereas the algrebraic real numbers are not. That difference is the essence of FTA, and the algebraic numbers are not inherently "analytic objects". A proof for the algebraic complex numbers would be easy to extend to the whole of C, as C is just a transcendental extension of the algebraic complex numbers, and transcendental extensions are very simple and boring in that they introduce no non-trivial polynomial equalities, by definition. Besides, FTA is a meaningful statement even if we define no metric at all on C (so there is no concept of a limit). (What sorts of proofs exist that the algebraic complex numbers are algebraically closed is a question of mathematical logic, though) 80.68.82.115 (talk) 23:49, 6 January 2008 (UTC)
Now please define what a real field R is and under what conditions one considers a real algebraic field to be algebraically closed. The one definition I know involves the algebraic statement of the intermediate value theorem in the form that every polynomial with coefficients in R and odd degree has at least one root in R. Which is a sufficient setup to proceed via the Euler-Gauss proof (see Basu-Pollack-Roy). But that the "real" real numbers have this intermediate value property requires some basic real analysis.--LutzL (talk) 19:59, 7 January 2008 (UTC)
This quickly becomes a question not so much about which proofs are possible but about which proof steps one would consider "analysis", and how much of the reasoning back towards the axioms one wants to consider "part of" the proof. I tend to consider the "message" of the FTA to be one about the relation between R and C (essentially, that all we need to adjoin is i). Therefore, if one can derive the algebraic closedness of F[x]/(x²+1) assuming some purely algebraic properties of F, and then afterwards plug in F=R, then I would personally consider that to be an algebraic proof of the FTA even if one needed analysis to prove that R has the assumed properties in the first place. (Within reason, of course: it is easy to find a formal loophole in this description. However, the notion of what the "idea" in a proof is is an intuitive rather than a formal one in the first place). –Henning Makholm 19:09, 8 January 2008 (UTC)
I agree that the "because" from the sentence quoted above is untrue. Take, for instance, the assertion "For each $x$ ∈ R, $-x^2$ has a square root in C". It is about the set R of real numbers and about the set C of complex numbers. However, it can be proved without using any analysis.JCSantos (talk) 14:46, 8 January 2008 (UTC)

### Old discussion

"All proofs of the fundamental theorem necessarily involve some analysis,...

Agreed

"...since the very definition of the complex numbers requires the use of limits."

Is this statement true? Complex numbers are defined as ordered pairs of real numbers, with appropriate definitions for complex arithmetic operation. The crucial analytical property of the real numbers is Completeness, which can be defined in terms of the existence of Least Upper Bounds without reference to limits. Daran 04:06, 14 Sep 2003 (UTC)

Changed to "...since the construction of complex numbers from rational numbers requires the use of limits." Still not satisfactory, but more accurate. wshun 04:33, 14 Sep 2003 (UTC)
Changed again, and probably still not right. -- Daran 05:02, 14 Sep 2003 (UTC)
Maybe it would have been clearer to say that the construction of the complex numbers requires the use of limits, since every construction of R (infinite decimals, Cauchy sequences, Dedekind cuts, nested sequences of closed intervals, etc.) explicitly or implicitly makes use of limits. Proof that there exists a field satisfying the axiomatic characterization of R by the least upper bound principle requires the use of limits.
In any case, the current version of the article is genuinely wrong. The statement that the "complex rationals" (which is not standard terminology, by the way) satisfy the same algebraic properties as C is not merely wrong. Juxtaposed with the statement that one is algebraically closed and the other is not, it is contradictory. Michael Larsen 17:53, 14 Sep 2003 (UTC)
I meant the algebraic axioms, though of course you're quite right. I think I'm out of my depth here, and will bow out gracefully. -- Daran 19:57, 14 Sep 2003 (UTC)
OK, I've fixed it up according to my lights. Take a look and see if you're satisfied with the new version.Michael Larsen
Er..., you talked me back into it.  :-)
No I'm not. It is possible to prove the FTA from 1. The axioms of the real numbers and 2. The construction of the complex numbers from the reals. It is not necessary to refer to any particular construction of the reals. Moreover the axioms of the reals, specifically the LUB property can be stated without reference to limits. Finally your statement that "Proof that there exists a field satisfying the axiomatic characterization of R by the least upper bound principle requires the use of limits." itself requires justification.
I'm not convinced that the statement "All proofs of the fundamental theorem necessarily involve the use of limits" is true. I agree with the statement "All proofs of the fundamental theorem necessarily involve some analysis" because I consider arguments involving the LUB property to be "analysis" even if they don't make use of limits. (My old textbook on calculus defined "analysis" as "limits, etc.") The definition given in mathematical analysis to which I've linked the article, is inadequate. -- Daran 03:37, 15 Sep 2003 (UTC)
Hmm, I'm happy with the article as it now stands, after your most recent revision. If you are, too, great! If not, of course you can rewrite it. If your new version annoys me, I may change it---or more likely, I'll just give up. If I give up leaving a substantively incorrect article behind, I will consider it a failure of the wikipedia process. I think wikipedia needs both professionals and amateur enthusiasts, but it may be that the two can't easily coexist... Michael Larsen
I'm not happy with it, but I don't know how to improve it, so I guess I'll just have to remain unhappy. I think we can coexist... -- Daran 05:31, 15 Sep 2003 (UTC)
Since "use of limits", "algebra", "analysis" are not precise terms, the statement under contention is not mathematical or even metamathematical, but philosophical. Do all the arguments have an "analytic flavor" or a "limit-using flavor"? I would say yes, and I think most mathematicians would agree. But this is an area of legitimate disagreement, unlike the statement I removed in my last edit. What I would like to do in my contributions to wikipedia, besides helping build up a database of accurate definitions and correct statements of theorems, is to convey some sense of mathematical culture, the kind of thing one imbibes by hanging around a math department common room listening to the grad students talking. But I won't have much time to do this in the near future. Michael Larsen
I have rewritten the sentence. I think "continuity of polynomials" is easier to understand than "limit" in construction of complex numbers, and it fits better for our three proofs in the article. -wshun 23:59, 15 Sep 2003 (UTC)
OK, I can live with the current version---I think the wording is a little awkward, but I have no mathematical or philosophical quarrel with it. Michael Larsen 03:40, 16 Sep 2003 (UTC)
I'm happy with that too. -- Daran 06:51, 16 Sep 2003 (UTC)
Remove the "less algebraic" argument. Not necessary. -wshun 21:34, 16 Sep 2003 (UTC)

(now considered something of a misnomer by many mathematicians) - some mention why should probably be made? Dysprosia 06:47, 30 May 2004 (UTC)

Agreed! As is stands now, the reader may only guess why the remark is written there. My guess would be that 'many mathematicians' would nowadays replace 'algebra' by 'analysis'. But my guess may in fact be wrong. So, yes, a why would be a good addition here. Bob.v.R 05:24, 26 Jun 2004 (UTC)

I'm definitely not sure why: it was never called The Fundamental Axiom Of Algebra. It's fundamental in the sense that it validates much of the general analysis in algebra, the same way the fundamental theorem of calculus does there. Mark Hurd 15:43, 15 Oct 2004 (UTC)

On the page misnomer it says now: 'The Fundamental theorem of algebra can be proved from the axioms and is therefore not fundamental.' However, if it a theorem, it is obvious that it can be proven. So why can't it be a fundamental theorem? As it stands now, I would propose to delete the frase about the misnomer from this article. Bob.v.R 18:56, 11 Dec 2004 (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, Dec 19, 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 Dec 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, Dec 20, 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 Dec 2004 (UTC)

Aha! Thanks for the reminder. --Doradus 04:04, Dec 21, 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, Dec 21, 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 Dec 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, Dec 21, 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)

## 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)

## 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)

## 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)

## Algebraic Proof

The algebraic proof needs to state what exactly the induction hypothesis is, since there are different ways of expressing the theorem, which are non-equivalent when used as induction hypotheses. Also, the statement

So, using the induction hypothesis, qt has, at least, one real root; in other words, zi + zj + tzizj is real for two distinct elements i and j from {1,…,n}.

has too many commas and does not make much sense, since the induction hypothesis probably does not guarantee that a real root exists (unless something false is being proved, as there are many real polynomials without any real roots). —The preceding unsigned comment was added by 152.157.78.172 (talkcontribs) 15:35, 25 September 2006 (UTC)

I think you are right. I hope the new version is better. -- EJ 14:15, 8 January 2007 (UTC)

broken link: C. F. Gauss, “New Proof of the Theorem That Every Algebraic Rational Integral Function In One Variable can be Resolved into Real Factors of the First or the Second Degree”, 1799 ~~

• Is this proof somehow circular? It supposes that the roots are all in $\mathbb{C}\,$. OTOH, I think that the proof could be made somehow more general than that, namely, that for any field F (is characteristic 0 required?) satisfying: (a) for every non-zero element x in F either $\sqrt{x} \in F\,$ or $\sqrt{-x} \in F\,$ (but not both) and (b) every odd-degree polynomial in F has a root in F, then F[i] is the algebraic closure of F. Albmont (talk) 16:36, 18 March 2009 (UTC)
Where do you see any circularity? Neither of the two proofs assumes the roots to be in C, you must have misread something. As for the generalization, the proofs (and indeed, the result) only works for real-closed fields (choose definition #3 from the list), i.e., on top of the conditions you mentioned, you also need to assume that F is formally real. (In both proofs, this condition shows up in the omitted proof that every number from F[i] has a square root.) — Emil J. 16:55, 18 March 2009 (UTC)
I got the circularity from here:
As mentioned above, it suffices to check the statement “every non-constant polynomial p(z) with real coefficients has a complex root”. This statement can be proved by induction on the greatest non-negative integer k such that 2k divides the degree n of p(z). Let a be the coefficient of zn in p(z) and let F be a splitting field of p(z) (seen as a polynomial with complex coefficients); in other words, the field F contains C and there are elements z1, z2, …, zn in F such that (...)
that is, the argument seems to imply that the splitting field of p(z) is C - or am I misreading something? As for the generalization, I guess that if F has a square root for (every element or its additive inverse) implies that F can be ordered, which implies that it's a formally real field. Albmont (talk) 20:16, 19 March 2009 (UTC)
First, you are misreading it. The text you emphasized just says that p has complex coefficients, it says nothing about its roots. In fact, the next sentence talks about F containing C and other stuff, which clearly suggests that F is bigger than C (at least a priori; the whole point of the proof is to show that, after all, F = C).
Second, it is not at all true that a field in which every number or its inverse has a square root (and −1 does not) can be ordered. Any finite field GF(q) with q ≡ −1 (mod 4) is a counterexample. By a more sophisticated construction, you can also find a counterexample where additionally every polynomial of odd degree has a root, and of characteristic 0. So no, you cannot omit the formally real condition. You can simplify it though: under the other condition it is equivalent to requiring that −1 is not a sum of two squares, you don't need an arbitrary finite number of them. — Emil J. 11:34, 20 March 2009 (UTC)
In fact, the generalization is mentioned in article real closed field: F is a formally real field such that every polynomial of odd degree with coefficients in F has at least one root in F, and for every element a of F there is b in F such that a = b2 or a = −b2. Maybe the algebraic proof of this fact could be placed in that article. Albmont (talk) 20:20, 19 March 2009 (UTC)
The article is not very clean on that, but I don't know how to change it without making it confusing. In the general case, the complex field should be taken as the quadratic extension C=R[i]=R[U]/(U2+1) of the real algebraic field R. Then the splitting field F is an algebra over C and the proof shows that it is of dimension one.--LutzL (talk) 10:09, 20 March 2009 (UTC)
• Thank you all for making things clear for me. However, one bit of counterexample seems challenging: By a more sophisticated construction, you can also find a counterexample where additionally every polynomial of odd degree has a root, and of characteristic 0. I think this could be done by failing to have for some "positive" (an element is positive if it's non-zero and it has a square root) x and y a "negative" x + y. In other words, three "positive" elements x, y and z whose sum is zero. One such construction probably would have to take the Axiom of Choice and start with, for example, $\mathbb{Q}[\sqrt{\pi}, \sqrt{e}, i \sqrt{\pi + e}]\,$. Or am I missing some obvious construction? Albmont (talk) 13:22, 23 March 2009 (UTC)
Do you mean that you start with $\mathbb Q(\sqrt\pi,\sqrt e,i\sqrt{\pi+e})$, and find its maximal extension contained in C which does not include i using Zorn's lemma? Well, I don't see how to prove that $i\notin\mathbb Q(\sqrt\pi,\sqrt e,i\sqrt{\pi+e})$ (or equivalently, $\sqrt{\pi+e}\notin\mathbb Q(\sqrt\pi,\sqrt e)$) in the first place, as e and π are not known to be algebraically independent over Q. However, if you replace e and π with any algebraically independent pair of real numbers, then I think that this approach could work.
There are probably many different ways how to construct such a field, but what I had in mind was the following. First, fix any prime p ≡ −1 (mod 4), and let Kp be the union of all finite fields GF(pe) with odd e. (More precisely: construct the algebraic closure H of GF(p). Then H contains a unique isomorphic copy of each GF(pe), and the union of these copies for odd e is a subfield of H.) Then Kp has characteristic p, every polynomial of odd degree over Kp has a root, every number or its opposite has a square root, −1 has no square root, but −1 is the sum of two squares (indeed, every element of Kp is). Now, in order to construct a field with all these properties but with characteristic 0, we take an ultraproduct of all these Kp over a nonprincipal ultrafilter (or simply, we apply the compactness theorem), which works as there are infinitely many primes p ≡ −1 (mod 4). With a bit of care, the argument goes through in plain ZF without the axiom of choice (as the compactness theorem for countable theories does not need AC). — Emil J. 15:02, 23 March 2009 (UTC)
This ultraproduct is a very interesting construction... But there's no way to avoid the AC, since AC is equivalent to "every field has an algebraic closure" and "every filter is a subset of an ultrafilter". But my intuition breaks down even in the simplest cases; I can't figure out what is any ultraproduct of the $\mathbb{Z}_p\,$, just that it has characteristic 0 (maybe those ultraproducts aren't isomorphic; probably we can chose ultrafilters where -1 has a square root and ultrafilters where -1 doesn't). Albmont (talk) 12:21, 24 March 2009 (UTC)
First, AC is not equivalent to either of the two statements you mention, since both of them follow from the Boolean prime ideal theorem, which is provably weaker than AC (as shown by Halpern and Levy). In any case, this is irrelevant. You do not need the algebraic closure of every field, but only of a very specific field (a finite one), and in this particular case you can construct the algebraic closure explicitly. (More generally, the algebraic closure of any countable field can be constructed without AC.) The ultraproduct construction does indeed need some choice, but as I wrote, you only need the compactness theorem (the ultraproduct is only a way of doing a compactness argument in a more "algebraic" way), and the compactness theorem for countable theories is provable without AC. In fact, there is also another reason why the result is automatically provable without AC, if provable at all. The existence of the counterexample is equivalent to consistency of a particular recursively axiomatized first-order theory, hence it is an arithmetical $\Pi^0_1$-statement, and ZFC is conservative over ZF for such statements. For example, you can do the construction of the counterexample inside L, where AC always holds, and it will still work as a counterexample in the outer universe, as satisfaction of first-order formulas in a structure is absolute for transitive models of ZF.
As for properties of the ultraproduct, the key to that is Łoś's theorem. In particular, it implies that any first-order formula in the language of fields which holds in Kp for all but finitely many p also holds in $\prod_pK_p/U\,$. For example, that −1 has no square root is expressible by the formula $\neg\exists x\,x^2+1=0$, hence it holds in the ultraproduct. The other properties mentioned above (i.e., that every odd degree polynomial has a root, and every element is a sum of two squares) are also given by a set of first-order sentences, and are valid in all Kp, hence they are valid in the ultraproduct. That the characteristic is 0 is expressed by the set of sentences $\underbrace{1+\dots+1}_{n\text{ times}}\ne0$ for n > 0, and each one of these is valid in all but finitely many of the Kp (namely, it can only fail for p | n), hence the ultraproduct has characteristic 0. — Emil J. 14:12, 24 March 2009 (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)

## Reference

" Shouldn't the theory itself be referenced? " — Preceding unsigned comment added by OceanEngineerRI (talkcontribs) 22:28, 20 August 2013 (UTC)

" In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of completeness), which is not an algebraic concept. Additionally, it is not fundamental for modern algebra; its name was given at a time in which algebra was mainly about solving polynomial equations with real or complex coefficients " where was this statement taken from? it has no reference attached. —Preceding unsigned comment added by 189.221.32.18 (talk) 06:11, 17 October 2010 (UTC)

## Proofs section - "Cauchy's theorem" links to the disambiguation page rather than the specific theorem

I've tried to fix the two links in the complex-analytic proof using Cauchy's theorem - I think the link was supposed to point to Cauchy's integral theorem rather than this disambiguation page. Can someone double-check this is the right place? Thanks.

--Schauspieler (talk) 09:53, 18 March 2012 (UTC)

## Abel's impossibility theorem

I think Abel's impossibility theorem should be worked into this article somehow, but I couldn't find a spot that I liked. There are a couple of places in the article that say things like "it follows from the fundamental theorem of algebra that every non-constant polynomial with real coefficients can be written as a product of polynomials with real coefficients whose degree is either 1 or 2" which can mislead people into thinking that you can always actually write down those coefficients/roots via some formula, but of course you can't. Abel's theorem says that there can be some roots that can only be approximated and not directly written down. Wrs1864 (talk) 12:19, 26 April 2013 (UTC)

Abels theorem can sensibly only be applied to the case where the coefficients have a finite representation. If the polynomial coefficients are general reals, then we are dealing with the impossibility of the solution of a non-realizable problem. Since we have no access to a faithful finite representation of the coefficients, all we can talk about are approximations of the polynomial and its roots, and their eventual limit. This is what Weierstraß was doing in 1889 when he invented the Durand-Kerner method in his homotopy proof of FTA.--LutzL (talk) 11:21, 10 March 2014 (UTC)

## A constructive topological proof

I do not feel any of the current theorems satisfy how I think of the result. This seems a special case of properties of continuous and smooth transformations of spheres into themselves. Imagine an elastic sheet wrapped around a basketball. Now, if we stretch and re-arrange the sheet, that's a transformation of the surface of the basketball into itself. Now, the elastic naturally removes any wrinkles and folds, but we can get two layers by slicing the sheet, grabbing an edge, wrapping it all the way around the basketball, and then sewing it back to together. We can do this process as times as we want, each time, getting one more layer of the elastic sheet around the ball. Now, if we try to puncture the basketball with a nail, we'll have to go through as many layers as we wrapped around it. So, if f(z) is the mapping represented by the elastic sheet, f(z) = a will have as many solutions as there were wrappings.

The key to the proof, then, is realizing that every polynomial over the complex numbers represents such a mapping of the sphere into itself, with the degree of the polynomial being the number of times the elastic sheet was wrapped around the basketball. So, we need to show that the polynomial is continuous and doesn't fold the complex numbers (the core concept of an analytic function), and that the degree is the winding number. So the theorem primarily a statement about the special properties of complex analysis. — Preceding unsigned comment added by Uscitizenjason (talkcontribs) 03:10, 16 June 2013 (UTC)

This is only another, but nice, visualization of the winding number proof.--LutzL (talk) 11:04, 10 March 2014 (UTC)

## Topological proofs section

There seems to be an error in the topo proofs section. It's stated:

  ... if a is a kth root of −p(z0)/ck and ... 

Since z0 and ck are both constants this doesn't seem to make a lot of sense. I think it's supposed to say

  ... if a is a kth root of −p(z0 + ta)/ck and ... 

If I can convince myself that the proof makes sense that way and actually proves something I'll fix it but I'm still puzzling over it at this point (call me slow, whatever, I'm reading this page to try to grok the proof of the Fund. Theo. which I never really learned). — Preceding unsigned comment added by Salaw (talkcontribs) 17:13, 9 March 2014 (UTC)

Actually what I wrote above doesn't make a lot of sense, clearly the reading should be something else. Anyhow after another cup of coffee maybe it'll all come clear. And this time I'll actuall sign my comment. Salaw (talk) 17:17, 9 March 2014 (UTC)

Never mind; I was being dense. I see it now. Sorry for the confusion. Salaw (talk) 17:27, 9 March 2014 (UTC)