Anyone familiar with Abel's insolubility of the quintic proof

  • Thread starter icantadd
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  • #1
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Greetings PF,

I would like to sketch the proof that Abel gave for the insolubility of the quintic. This is a question of general interest to me, and one of those things, that I probably should see once. I am going off a translation of Abel's proof, and I hope to get very general guidance to the nature of the proof, i.e. what Abel was going for. Ultimately, I want to distill the proof into the elementary steps in an intuitive (?ahem?) manner, so that 1) a good sketch of the proof could be given, and 2) from the sketch, some form of the proof could be given that does NOT use full blown Galois theory, or any facts like Sym_5 is insoluble (although this does implicitly come up in the proof since ultimately the contraction that 120 = 10, is ultimately what causes the reductio to unravel.), and meanwhile, fully deconstruct the proof for understanding's sake.


So to start we assume that the quintic is soluble, so that
[tex]y^5 + ay^4 + by^3 + cy^2 + dy + e[/tex] is solvable precisely in terms of radicals alone.


Next we must use a consideration about the general form that any polynomial p(x) soluble by radicals must have. If it is soluble, it must be the sum of (possibly nested) algebraic manipulations (roots, exp, mult,div,plus,sub) of the coefficients of p(x)
[tex] y = c_0 + c_1R^{\frac{1}{m} + \ldots + c_{m-1} [/tex],
where in the above [tex]c_i, R[/tex] are all functions containing (nested algebraic) rational functions of only the coefficients. However, I don't understand the step where Abel claims that one can assume m to be prime. What realization have I missed. I know that ultimately, I am only desiring a deconstruction into a sketch, but I would like to do a full deconstruction in the process.

So far, the steps are
1) Goal is to prove the quintic is in general insolvable by radicals.
2) Assume it is to derive a contradiction
3) Prove that the general form of the solution must be as expressed above.

And for
3a) Assert that y must be definable as a rational function of algebraic expressions of coefficients
3b) Consider the coefficients a,b,c,d,e, and f(a,b,c,d,e) as a rational function of the coeff (as zero order)
3c) consider[tex]f({p_0}^{\frac{1}{m}})[/tex] as a "first order" algebraic function, where p_0 is zero order.
3d) similarly [tex]f({p_m}^{\frac{1}{m}})[/tex] is an m'th order function.
3e) See that factoring m, and writing [tex]R^{\frac{1}{m}}[/tex] as a sequence of successive radicals, we are allowed to assume m is prime (my problem)
 

Answers and Replies

  • #2
Hurkyl
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I don't understand what you're having trouble with; did you just forget that
[tex]\sqrt[p]{\sqrt[q]{R}} = \sqrt[pq]{R}[/tex]​
?
 
  • #3
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Ooh, thank you. So if the function is, say,
[tex] \sqrt[4]{x} [/tex], it can be written as [tex] \sqrt{\sqrt{x}} [/tex]. So assuming m is prime is fine.
 

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