Talk:Abel–Ruffini theorem/Archive 1

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 (talk • contribs) 21:27, 29 December 2005 (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)

Is this article accurate?
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 &ge; 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 &gt; 5. Isn't that right? DavidCBryant 18:18, 10 January 2007 (UTC)


 * If you have a proof for degree 5, the result for degree &gt;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)
 * 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)
 * 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 $$\Complex(c_4,c_3,c_2,c_1,c_0)$$; 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)
 * 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 $$\Complex(c_4,c_3,c_2,c_1,c_0)$$; 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)