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I will need a help with a sentence of Gilles Deleuze" in Difference and Repetition: "An equation is solvable ... by radicals, when the partial resolvents are binomial equations and the indices of the groups are prime numbers". Thank you very much in advance!

dan tausk
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    That doesn't seem to make much sense and/or probably it is taken out of context. Can you give a complete reference? If it is a book mention also the page, or give a link to it, and the same for a paper. – DonAntonio Nov 02 '16 at 13:55
  • "Solvability by radicals" is at the historical roots of Galois theory (main keyword); see (http://math.stackexchange.com/q/720630). I didn't know Gilles Deleuze was interested in that... – Jean Marie Nov 02 '16 at 14:06
  • Thanks Christopher L. I am interested in how Abel produced the radical and revolutionary reversal of the problem-solution relation, by studying the conditions of solvability rather than the working on direct solutions, claimed by Deleuze. I am following Abel's reasoning and use of resolvents from Pesic: Abel's Proof: An Essay on the Sources and Meaning of Mathematical Unsolvability that concentrate in Abel's work (rather than Galois's). I've got a copy and will read Rica da Silva's article. I have a (now rather rusted) understanding of abstract algebra from University. – dan tausk Dec 04 '16 at 12:13
  • Thanks Jean Marie: The source of Deleuze’s statement is Difference and Repetition, page 180, Columbia University Press, 1994 (Transl. by Paul Patton) or Différence et Répétition, page 233, Presses Universitaires de France, 1968. He was mostly interested in the problematic rather in the axiomatic thinking, in topology, rather than in algebra, In continuous variations rather in point relations, in relationability rather in relations, etc. – dan tausk Dec 04 '16 at 12:29

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(I do not have enough reputation to comment, and I feel that I can give an 'approximation' to an answer anyway.)

First of all, without making any direct judgements, it's fair to say that the question can seem unclear. It would for example help to know how much you do know, and more precisely what part you need help understanding. However, I will try to give directions for making what is said in the quotation you gave as clear as possible, instead of giving a concise and formal answer.

Deleuze seems to take a lot of knowledge of abstract algebra for granted. As noted from JeanMarie's comment, Galois theory is definitely the field of which Deleuze assumes the reader has knowledge. A very comprehensible book on all abstract algebra you need for Galois theory, including a short chapter on Galois theory it self is Abstract Algebra by Beachy and Blair.

In short and rather informally, you can say that resolvents are functions of the roots to the original equation. You then instead consider equations involving these resolvents. If these equations are binomial equations, i.e. of the form $x^p=a$, then Gauss had shown these could be solved by radicals, and thus it would follow that the original equation could be solved by radicals.

An article that might make things a lot clearer than I can do here is Galois Groups in the work of Mira Fernandes by Amaro Rica da Silva. Some prerequisites from abstract algebra may help, but hopefully you should be able to grasp the general idea enough to feel satisfied that you know what Deleuze is talking about. Especially from the quote on p.4 that reads:

"Galois discovered a method of finding the group of a given equation, the successive partial resolvent equations and their associated groups that result from extending the field of coefficients with the roots of these resolvents. These groups turn out to be subgroups of the original group. Galois shows that when the group of an equation with respect to a given field is the identity, then the roots of the equation are members of that field.

Application of Galois' theory to the solution of polynomial equations by rational operations and radicals then follows. When the partial resolvent that serves to reduce to a subgroup $G_2$ the group $G_1$ of an equation is of the form of a binomial equation $x^p=a$ with $p$ prime, then $G_2$ is a normal subgroup of index $p$ of $G_1$. Conversely a normal subgroup $G_2$ of prime index $p$ of $G_1$ yields a binomial resolvent $x^p=a$."

If not considered an answer to your question, it might be an aid for others to give a more concise one.

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