Now consider the rational functions. When f(x', x'' ...) and φ(x', x'' ...) are two entire functions, it is evident that the quotient
f(x', x'' ...)
φ(x', x'' ...)
is a special case of the result of the first three operations, which are rational functions. It can therefore be considered a rational function as the result of the repetition of these operations. If we denote by v', v'', v''' etc several functions of the form
f(x', x'' ...)
φ(x', x'' ...)
we can easily see that the function
f(v', v'' ...)
φ(v', v'' ...)
can be reduced to the same form. It follows that any rational function of several quantities x', x'' ... can always be reduced to the form
f(x', x'' ...)
φ(x', x'' ...)
where the numerator and denominator are entire functions.
Commentary:
Abel derives a general expression for a rational function. For him this result is easy so his explanation is not detailed. His idea seems to be that whenever two expressions of this form are combined with any of the three allowed operations it results in an expression of this same form. So no matter how many times or in what combinations you use the three allowed operations you will always be left with an expression of this same form.
I do not know whether you are still posting Abel's original paper. I also read the blog named "Fermat Last Theorem" and Peter's book "Abel's Proof", I saw your comment on this proposition which Ruffini failed to show:
ReplyDeleteall the irrational functions like R^(1/m) are rational functions in the roots
In Peter's book and Abel's paper, they showed this for the top level of radical nested which is not hard to understand
Now the problem is to keep this going down to the next floor and they proceed as:
suppose R=S+v^(1/n)+A v^(2/n)+B v^(3/n)+.....
then R_1, R_2,R_3 etc are different values that R can assume when change the choice of v^(1/n)
then (As the previous blog said) R_2= w* R_1 where w is a root of unity.
I think this is the mistake made by the bloger and this is also a blank point in Peter's book.
I think we can proceed as follows:
when choose another choice of v^1/n, we can get another value for R and hence different R^1/m
but the key point is even if we replace
y=p+R^1/m+p2 R^2/n+......which is the original and top level of the expression , with the completely different R and hence different R^1/m
y is still the root in the quintic equation!
what i said means when we explicitly write out an radical expression for y, there are nested radicals. Then I claim from the innermost radical, every time we face the problem to choose an possible value for S^1/k
, we can choose an arbitary one!
This can be understand when thinking of Cardon Formula where you can choose any cubic root or square root when you have to! (some of them identified so you would not get 6 roots instead of 3)
Now the problem is gone, since for any R_1. R_2 or whatever you choose, y=p+R^1/m+.... are roots for the quntic, so can be represented by rational function of roots.
Then v^1/n can be represented and hence any lower cases.