# Talk:Abel–Ruffini theorem

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## §Too much emphasis

Good article but there is a lot of emphasis used. I'd suggest that the author remove all but the most important italics. —Preceding unsigned comment added by Jormundgard (talkcontribs) 21:27, 29 December 2005 (UTC)

I think this article contains a major historical inaccuracy. It says that Ruffini (and, independently, Abel) proved that the solution of the general polynomial equation of degree ≥ 5 in radicals is impossible. I think what these two showed is that the solution in radicals of the general polynomial equation of degree = 5 is impossible. Evariste Galois is generally recognized as the guy who extended the result to degrees > 5. Isn't that right? DavidCBryant 18:18, 10 January 2007 (UTC)

If you have a proof for degree 5, the result for degree >5 follows immediately. Specifically, if there were a solution in radicals for ax6 + bx5 + cx4 + dx3 + ex2 + fx + g = 0, then you could just put g=0 and you immediately have a solution in radicals for degree-5 polynomials, which contradicts the Abel-Ruffini theorem. -- Dominus 18:44, 10 January 2007 (UTC)
Yeah, I was definitely asleep at the switch when I wrote my first post. Thanks for clearing that up. But I still think something's not quite right, in between this article, the biography of Galois, and the biographies of Ruffini and Abel. I was reading all that stuff yesterday, and I got the distinct impression that the (admittedly confusing) history of the "quintic" problem is not described clearly on Wikipedia. I'll try reading it all over again so I can explain what bugged me a little more precisely. (Oh ... I think this is part of it. Abel-Ruffini establishes the impossibility of a general solution, but does not completely characterize the special cases, such as x8 - 2x4 + 16 = 0, where a solution in radicals is possible. Didn't it take Galois field theory to complete that characterization? I'm not real big on algebra.) DavidCBryant 12:25, 11 January 2007 (UTC)
I too take issue with the historical accuracy. Far as I've gathered, Galois only found a method to answear whether a given quintic equation could be solved by radicals, not if it was solvable by other means should the radicals fail. The general result, using what had then become known as "galois theory", came ca 60 years after the proof by Niels Henrik Abel. But I'm not sure who got it first of Abel or Ruffini, though there was a noticeable delay between Abel finishing his proof and having it printed, posthumously. EverGreg 20:35, 3 August 2007 (UTC)

It appears to me that there are two misunderstandings with the misinterpretation section. Firstly, a polynomial being solvable by radicals means that there is an expression of one of the roots involving the field operations +,-,*,/ together with extraction of roots, starting from the base field. I interpret this differently from root extraction on the coefficients, in the sense that we allow taking a cube root of an expression involving a square root, as in Cardano's formula for the cubic. The second, more important, point is that the Abel-Ruffini Theorem does not say that 'not all higher degree polynomials can be solved by radicals'. It only asserts that there is no single formula which works, say, for all quintics. In particular, it was still possible that each quintic had its own formula. It was the work of Galois which finally showed that there are quintics which are not solvable, since a polynomial is solvable by radicals if and only if it's Galois group is a solvable group. In fact, if p and q are primes, then $f=x$$5$$-pqx+p$ is irreducible and has Galois group $S$$5$, so is not solvable by radicals. (It is irreducible by Eisenstein's criterion. It has two turning points and precisely three real roots. Therefore its Galois group contains a transposition - complex conjugation - and is transitive, hence $S$$5$.) This is an easier example than that given. One final comment: the proof given is of a statement logically stronger than the theorem - it gives an explicit polynomial which is not solvable by radicals. --A Hubery (talk) 12:23, 13 January 2009 (UTC)

Talking with someone over lunch led to the following. If each quintic had its own formula giving a radical expression of a root, then in particular, the generic quintic would as well. This formula would then, via specialisation, have to work for almost all quintics. However, that still leaves open the possibility that all rational quintics have their own formulae. I guess, though, that neither Abel nor Ruffini considered the generic quintic at all. --A Hubery (talk) 14:07, 13 January 2009 (UTC)
That argument is wrong, as can be seen by starting with an algebraically closed field like the complex numbers (the argument does not mention the field, so it should still work). Then indeed every quintic trivially has its own formula giving a root, namely just a constant; yet there is still no formula for the generic quintic. This is because the generic quintic is not a particular case of a quintic (which must have its coefficients in the base field), but rather has its coefficients in a (transcendental) extension field of the base field. Concretely, the generic quintic over the complex numbers has its coefficients in a field of rational functions $\C(c_4,c_3,c_2,c_1,c_0)$; here even the extended field cannot express the roots as constants. This detail indeed makes the proof given in this article a but subtle (the current formulation is certainly not sufficient). Marc van Leeuwen (talk) 20:31, 12 January 2010 (UTC)

## §Proof Correct

There was a question in the article itself as to whether the proof was correct. It noted that the proof asserted that [E:F] is less than or equal to 5!, whereas what is needed is that |G(E/F)| is less than or equal to 5!. But G(E/F) is the Galois group of the extension E/F, so these are equivalent statements. Thus, the proof is correct as written, and I edited out your concern. --LamilLerran 18:46, 14 February 2007 (UTC)

## §Which field

I was reading this page and I think it would be helpful, to myself and other readers, if the author included: 1.) In the proof section, that x is an element of the reals (it is not explicitly mentioned), 2.) Which field F is specifically.- Nmech (talk) 18:44, 22 February 2008 (UTC)

## §Inaccuracy in article

There was an inaccuracy in the article which I have just corrected. It said that Ruffini's proof had a minor gap. As far as I understand, the only person who believed that the gap was minor was Cauchy. In an excellent article on the subject, Michael Rosen (Niels Hendrik Abel and Equations of the Fifth Degree Author(s): Michael I. Rosen Source: The American Mathematical Monthly, Vol. 102, No. 6 (Jun. - Jul., 1995), pp. 495-505 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2974763) calls the gap "significant." I have tried to represent his view with an update in the article. —Preceding unsigned comment added by 216.103.214.64 (talk) 16:11, 19 September 2008 (UTC)

## §Non-Plagiarism from E2

The similarities between the proof in this article and the one on Everything2 are most likely unmissable, but that should surprise no one, as I wrote both of them, and they're based on the exposition given in Fraleigh's Abstract Algebra book. Stormwyrm (talk) 10:30, 9 October 2008 (UTC)

CAn you tell me why the number have to be trancendentals? I guess irrational algebraics independents will be enought. — Preceding unsigned comment added by 79.157.195.138 (talk) 23:45, 9 January 2015 (UTC)

## §Ruffini's gap

In the history section it's written "Ruffini assumed that a solution would necessarily be a function of the roots (in modern terms, he failed to prove that the splitting field is contained in the tower of radicals which corresponds to a solution expressed in radicals)." I admit to be a little confuse, but this sentence surely is not very clear: what's the meaning of root? n-th root or zero? And solution means zero, right? Or perhaps it refers to the solution in the sense of the required radical tower? Also, is the sentence in the brackets correct? I think it shouldn't be "is contained", but "it's equal" or "contains" (I guess those two are equivalent). I probably don't remember much about Galois theory, but this sentence needs to ber clarified.--Sandrobt (talk) 03:37, 3 September 2010 (UTC)

Since you asked on my talk page, I'll give my reading on the sentence, though I'm not its author. If one has a hypothetical formula that expresses the roots of the equation in terms of the coefficients and (nested) radicals (or several formulas for several roots), then one can form a tower of fields, starting with the coefficient field and successively adding radicals; one works inside-out so as to start with the simplest radicals whose argument only involve coefficients, then radicals whose arguments contain those radicals and so forth, until finally one has a field that contains all the values given by the formulas. Now over this field the polynomial can be factored (or split) into linear factors (namely $x-r_i$ for all roots ri) but it is (a priori) not necessarily the smallest field for which this is possible: the splitting field of the polynomial. Presumably Ruffini's proof draws a contradiction from the fact that the splitting field can be obtained merely by adjoining radicals (so that it occurs in some tower of fields as described above, in fact as the topmost field), but he failed to show that it cannot happen that adjoining one radical to a field over which the polynomial does not yet split produces a field in which it does, but which strictly contains the splitting field. In fact I find only the part of the cited sentence before the parentheses unclear: I think it should say "radical" where it says "root", as a root of a polynomial is the same thing as a solution of the corresponding equation, and anyway it is not clear what "a function of" means here. Marc van Leeuwen (talk) 08:37, 7 September 2010 (UTC)
If I got what you mean, Ruffini failed to prove that the splitting field contains the field genereted adding the radicals (or, i.e. the splitting field is equal to the field genereted adding the radicals, since the splitting field is contained in the other one), but from the sentence "he failed to prove that the splitting field is contained in the tower of radicals which corresponds to a solution expressed in radicals" I understand exactely the oposite: Ruffini failed to prove that the splitting field contained in the field genereted adding the radicals. Where am I wrong?--Sandrobt (talk) 15:11, 7 September 2010 (UTC)
The tower is a set of fields; the splitting field is contained in the tower if it is equal to (not: is contained in) one of those fields. Admittedly "is member of" would be clearer than "is contained in", but I guess language is not always logical. Marc van Leeuwen (talk) 15:47, 7 September 2010 (UTC)
Ok, now I get what that sentence mean! For some reason I was always thinking it was referring to the last (biggest) field of the tower (since if the splitting field is member of the tower, it must be equal to the last one). Thanks a lot!--Sandrobt (talk) 12:16, 8 September 2010 (UTC)

## §Statement that "the automorphisms $\sigma'$ also leave $E$ fixed" seems incorrect

I am not going to attempt to edit the article because I do not understand it, but I would like to raise a question about a step in the argument that appears to be incorrect or is at least unclear (to me).

In the section "Proof" it is stated that "the automorphisms $\sigma'$ also leave $E$ fixed." As I understand it, the earlier statement that "every permutation $\sigma$ ... induces an automorphism $\sigma'$ on $E$ that leaves $Q$ fixed" means that $\sigma'$ leaves $Q$ point-wise fixed. This seems right. But as for $\sigma'$ leaving $E$ fixed, it is either trivially true that $\sigma'$ leaves $E$ non-point-wise fixed (that is, $\sigma'$ maps $E$ into itself), as $\sigma'$ is an automorphism, or it is false that $\sigma'$ leaves $E$ point-wise fixed, because, for example, it might swap $y_1$ with $y_2$. (It is true that some induced $\sigma''$ acting on $E[x]$ leaves certain elements of $E[x]$, such as $f(x)$, fixed.)

So that's my point of confusion. I hope this feedback is helpful.

71.183.88.158 (talk) 14:43, 2 March 2012 (UTC)

## §The proof is broken

The proof seems to show that Q adjoin five transcendentals has degree 25 over Q, which is absurd. The base field in the argument needs to be changed to the transcendental extension of Q generated by the elementary symmetric functions (i.e. the field the coefficients of f lies in). — Preceding unsigned comment added by 68.40.207.129 (talk) 03:24, 30 July 2012 (UTC)

## §Abel's proof

This article uses Galois' approach to prove the theorem which was first proved by Abel. Why not include Abel's original proof, or at least a description of it? I have never seen Abel's proof and I am curious about it and would like to see it if it is not terribly complicated. If it is similar to that of Galois, then perhaps a description of the differences would suffice. For that matter, I would like to know about Ruffini's incomplete proof as well. — Anita5192 (talk) 05:38, 13 October 2013 (UTC)

See Pesic's book (which I've just added to bibliography). It contains a translation of Abel's paper, among other things. 5.12.29.230 (talk) 09:36, 12 March 2015 (UTC)
Thank you! I will look it up. — Anita5192 (talk) 03:01, 13 March 2015 (UTC)

## §Inaccuracy

I think there is something inaccurate (or even inexact) in the section "Interpretation". The sentence : "The theorem says that not all solutions of higher-degree equations can be obtained by starting with the equation's coefficients and rational constants, and repeatedly forming sums, differences, products, quotients, and radicals (n-th roots, for some integer n) of previously obtained numbers." It is clear from the context that this sentence is relative to equations with numerical coefficients, not to those with litteral coefficients. But Abel's theorem does not say this. It only says that there is no general formula to solve the general equation of degree n (with litteral coefficients). Suppose that for each polynomial P with numerical coefficients, there would exist a particular solution by radicals (of different form for each polynomial). It is not at all obvious, if possible, to prove that this would imply the existence of a general formula (this would if it is supposed that the form of the solution is the same for an infinite number of polynomials). Whence my question. Michael Bensimhoun (talk) 08:38, 9 July 2014 (UTC)

I agree. This mistake is repeated later in the article: "The Abel–Ruffini theorem says that there are some fifth-degree equations whose solution cannot be so expressed." From the statement that the general quintic has Galois group S_5 and hence is unsolvable, it does not follow that some quintics with numerical coefficients are unsolvable or even irreducible, as this would require Hilbert's irreducibility theorem. David Brink (talk) 08:15, 24 September 2014 (UTC)