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Surprisingly, it's not clearly defined online. The first thing that comes up is Abel-Ruffini theorem, which only refers to "radicals" and not RUFFINI radicals.

Ian Stewart's book has it appear out of thin air as if it's prior knowledge and common to all readers. Unfortunately, I have no memory of learning this.

Given its fancy name, I am sure it's different from radicals in general(otherwise, it's very un-math-like to do something meaningless like giving something a fancy name just for the sake of it).

What is the concrete definition of it?

Melba1993
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  • How is it first described and used in Ian Stewart's book? – fleablood Mar 24 '16 at 23:27
  • "The general polynomial equation $F(t)=0$ is soluble by Ruffini radicals if there exists a finite tower of subfields $\mathbb{C}(s_1,...,s_n) = K_0 \subseteq K_1...K_r = \mathbb{C}(t_1,...,t_n)$ such that for $j=1,,,.r$, $K_j=K_{j-1}(\alpha_j)$, $\alpha_j^{nj} \in K_j$ for $n_j \geq 2, n_j \in \mathbb{N}$" – Melba1993 Mar 24 '16 at 23:34
  • Don't see it as a straightforward definition of what a Ruffini radicals at all. It assume prior knowledge of what it is quite clearly... which is irritating me. The web seems to tell me there's no such thing. The only hit i get is on google books for Stewart's book. – Melba1993 Mar 24 '16 at 23:35
  • evidently https://books.google.com/books?id=JnsZBwAAQBAJ&pg=PA117&lpg=PA117&dq=stewart+soluble+by+Ruffini+radicals+if+there+exists+a+finite+tower+of+subfields&source=bl&ots=S2KIBuIwuK&sig=RehjPANL_KasZxZf0TV9fHOfAR8&hl=en&sa=X&ved=0ahUKEwiF2JqEv9rLAhVCzWMKHaUgD18Q6AEIHTAA#v=onepage&q=stewart%20soluble%20by%20Ruffini%20radicals%20if%20there%20exists%20a%20finite%20tower%20of%20subfields&f=false – Will Jagy Mar 24 '16 at 23:39
  • It really is unprofessional in math not to define something before mentioning it in another theorem/definition really, so i guess this isn't a common term to people as well? Can I just take it to be the radicals of rationals? And yes it's first appearance is in definition 8.8 as given by Will Jagy – Melba1993 Mar 24 '16 at 23:41
  • Okay, thanks to Will Jagy link, my interpretation is that Ruffini radicals are defined by context. A radical is a Ruffini radical when it is a radical that defines one of the "tower of finite subfields". In other words, a Ruffini radical is one that is in the extended field with the roots of the polynomial. A plain ol' radical need not be. – fleablood Mar 24 '16 at 23:51
  • I think he was defining the Ruffini radical in the definition of the term "soluable by Ruffini radicals". I'd have prefered it to be more explicit but I can see how that'd be seen as a legitimate definition. – fleablood Mar 24 '16 at 23:52
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    I'm deleting my comment of radicals of rationals. That was a completely wrong misconception. I'm now pretty sure Stewarts was defining "Ruffini Radicals" and "soluble by Ruffini radicals" in the same definition. It's soluble if there is that finite tower of field extensions and the Ruffini radicals are the radicals that extend a field in that tower. – fleablood Mar 25 '16 at 00:05

2 Answers2

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Sigh. He is defining a phrase "soluble by Ruffini radicals." That is what is in italics. It is not clear that he will have any use for the shorter phrase "Ruffini Radicals."

He begins with

The next definition is not standard, but its name is justified because it reflects the assumptions made by Ruffini in his attempted proof that the quintic is insoluble.

and then provides the definition

DEFINITION 8.8. The general polynomial equation $F(t)=0$ is soluble by Ruffini radicals if there exists a finite tower of subfields $$\mathbb{C}(s_{1},\ldots,s_{n})=K_{0}\subseteq K_{1}\subseteq\cdots\subseteq K_{r}=\mathbb{C}(t_{1},\ldots,t_{n})\tag{8.6}$$ such that for $j=1,\ldots,r$, $$K_{j}=K_{j-1}(\alpha_{j}) \qquad\text{and}\qquad \alpha_{j}^{n_{j}}\in K_{j} \qquad\text{for}\qquad n_{j}\geq2,\ n_{j}\in\mathbb{N}$$

Source: Stewart, N. I., Galois Theory. Fourth edition. CRC Press (2015).

Note that "soluble by Ruffini radicals" as a whole is italic.

Frenzy Li
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Will Jagy
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  • I am presuming that in the definition "of soluble by Ruffini radicals" he states the condition that there must exist some $\alpha_i^{n_j}$, I am presuming that THOSE $\alpha_i$ are the Ruffini radicals. It's a contextual definition and it is defined then and there. – fleablood Mar 25 '16 at 00:00
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    "It is not clear that he will have any use for the shorter phrase "Ruffini Radicals." but he does use it quite a few times. But I think it is clear that in defining "soluble by Ruffini radicals" for F(t) there exist radicals $\alpha_i$ that he is defining "Ruffini radicals" to be these radicals specifically associated with F(t). I think its a fair definition. – fleablood Mar 25 '16 at 00:11
  • @fleablood sure, I was not going to bother to check. This is the sort of thing I write when I am pretty sure that no other mathematician will use my definition; it is temporary, used to make sense of something in my article, but sufficiently strange that it is not going to become popular. – Will Jagy Mar 25 '16 at 01:07
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    Yeah, that makes sense. It's used in Stewart's book. Stewart's book only. And it's defined solely for a process. – fleablood Mar 25 '16 at 05:24
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    Just saw this reply. Isn't there a typo above (and in some editions of the book). Shouldn't it be $K_{j+1}=K_j(α_j)$ ? – Jap88 Sep 12 '18 at 16:18
  • By the way, for the record this is a nice blog posting related to this and that explains Ruffini radicals. Stewart does use them as a defined concept in the exercises. – Jap88 Sep 14 '18 at 01:49
  • @Jap88 two or three things. You did not include a link for any blog posting. Let's see, I did not have the book then and I don't now; I was answering as one who has also published mathematics. Last, this question is still open, you are free to post your own answer. Doing that would move the question to the front of the "active" list, people well might notice your answer. – Will Jagy Sep 14 '18 at 01:59
  • Sorry. Here is that blog: https://jmanton.wordpress.com/2016/08/11/a-snippet-of-galois-theory/ I think that blog + knowing about the typo would do the trick. – Jap88 Sep 19 '18 at 01:27
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A Ruffini radical in this context means that it isn't allowed to be of a form so that it "brings in" any radicals (n-th roots) from anywhere else than the field generated by the roots (and the polynomial coefficients, which is a subfield). The non-Ruffini radical fields that Stewart implicitly talks about don't have that limitation but can be pulled from anywhere. They don't have the limitation $K_j \subseteq \mathbb{C}(t_1,...,t_n)$.

Stewart says on page 118 in the 4th edition (about Ruffini radicals): "The aim of this definition is to exclude possibilities like $\sqrt{-121}$ in Cardano's solution of the quartic equation $t^4-15 t-4=0$, which does not lie in the field generated by the roots, but is used to express them by radicals."

A section or two forward in the book, after the mentioning of Ruffini radicals, Stewart has included a theorem and proof of Abel (difficult) that shows that as a matter of fact bringing in roots from "anywhere else" doesn't bring any more power to the game. So Ruffini radicals turns out to be all you need. This blog talks about this: https://jmanton.wordpress.com/2016/08/11/a-snippet-of-galois-theory/

Doing Abel's proof using Galois theory (which didn't exist when Abel did his proof) is much easier and is covered by, for example, Wikipedia: https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

Stewart shows how to do it "the hard way" without Galois theory, the way Abel did it. This is what makes it necessary to introduce "Ruffini radicals".

(Note that some editions of Stewart's book seem to have a typo. In the definition of Ruffini solubility it should say $K_{j+1}=K_j(α_j)$)

Jap88
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