Talk:Abel–Ruffini theorem

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)
 * 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)

There would be nothing wrong with this article IF it were for a different audience
There would be nothing wrong with this article IF it were for a different audience, like a class of college math majors.

But that is not the audience. So even if every word of the proof of the Abel-Ruffini theorem is correct, this needs much more detail filled in before it becomes satisfactory for Wikipedia.

To give just one example, consider this sentence:

"'' Expanding $$P(x)$$ out yields the elementary symmetric functions of the $$y_i$$:
 * $$s_{1}=y_{1}+y_{2}+y_{3}+y_{4}+y_{5}$$,
 * $$s_{2}=y_{1}y_{2}+y_{1}y_{3}+y_{1}y_{4}+y_{1}y_{5}+y_{2}y_{3}+y_{2}y_{4}+y_{2}y_{5}+y_{3}y_{4}+y_{3}y_{5}+y_{4}y_{5}$$,
 * $$s_{3}=y_{1}y_{2}y_{3}+y_{1}y_{2}y_{4}+y_{1}y_{2}y_{5}+y_{1}y_{3}y_{4}+y_{1}y_{3}y_{5}+y_{1}y_{4}y_{5}+y_{2}y_{3}y_{4}+y_{2}y_{3}y_{5}+y_{2}y_{4}y_{5}+y_{3}y_{4}y_{5}$$,
 * $$s_{4}=y_{1}y_{2}y_{3}y_{4}+y_{1}y_{2}y_{3}y_{5}+y_{1}y_{2}y_{4}y_{5}+y_{1}y_{3}y_{4}y_{5}+y_{2}y_{3}y_{4}y_{5}$$,
 * $$s_{5}=y_{1}y_{2}y_{3}y_{4}y_{5}$$.''"

A typical reader would have no idea what the word "yields" means here. The very next sentence is this:

"The coefficient of $$x^n$$ in $$P(x)$$ is thus $$(-1)^{5-n} s_{5-n}$$."

Even here with the crucial information that the coefficients of a polynomial are plus or minus the elementary symmetric functions of its roots, this important general fact is still not stated clearly at all.

Other parts of the proof are no better for the readers of Wikipedia.50.205.142.35 (talk) —Preceding undated comment added 22:49, 9 February 2020 (UTC)
 * The proof is bad for everybody. I have tagged it as confusing. D.Lazard (talk) 17:40, 24 May 2021 (UTC)
 * I have completely rewritten the proof. D.Lazard (talk) 21:25, 26 May 2021 (UTC)
 * Thank you! — Anita5192 (talk) 21:33, 26 May 2021 (UTC)
 * Thank you! — Anita5192 (talk) 21:33, 26 May 2021 (UTC)

Poor rendering with Chrome is confusing
Viewing this article with Chrome 87.0.4280.88, the minus signs in the superscripted and subscripted expression "5-n" do not render in the expression for the coefficients of x^n in P(x) difficult to interpret. (This follow the expansion of P(x).) I happened to open the same page in Firefox, and suddenly it made sense. I have no idea whether there's an alternate way to write it that would render properly.Captain Puget (talk) 16:57, 20 December 2020 (UTC)


 * This is happening in several articles. I just put in a help request at Village pump (proposals).—Anita5192 (talk) 17:17, 20 December 2020 (UTC)

Wich?
I think "wich a cyclic quotient group." is wrong, should it be "with a cyclic quotient group."?  Ϣere Spiel  Chequers  13:13, 23 June 2021 (UTC)
 * Typo fixed, with a minor clarification. D.Lazard (talk) 15:05, 23 June 2021 (UTC)

Somehow
"On the other hand, for n ≤ 4, the symmetric group and all its subgroups are solvable. Somehow, this explains the existence of the quadratic, cubic, and quartic formulas."

How do we know that explains the existence of them, if we don't know how? How do we know that without proving it? Isn't that the whole point of mathematics? Anonymous7002 (talk) 04:58, 20 May 2022 (UTC)
 * I have completed the sentence for clarification. D.Lazard (talk) 08:31, 20 May 2022 (UTC)