Section 2: Properties of algebraic functions which satisfy a given equation: Part 1

Translated by me as:

                                                                                  II

                                   Properties of algebraic functions which satify a given equation.

Let
                                                 c₀ + c₁ y + c₂ y ² ... c_r-1 y ^r-1 + c_r y ^r = 0.   . . . . . . .  1.

be an equation of degree r, where  c₀ , c₁ ... are rational functions of x', x'' ...,  x', x'' ... being independent of any quantities. Suppose we can satisfy this equation by putting in place of  y an algebraic function of x', x'' ...
Let

                                          y =   q₀ +  p ^{1 / n} + q₂  . p ^{2 / n} + ... q _n-1. p ^{( n-1) / n}  . . . . . . . . . .  2.

be this function. Substituting this expression for y in the proposed equation, we obtain , by virtue of the forgoing, an expression of the form.

                                         r₀ +  p ^{1 / n} + r₂  . p ^{2 / n}  ...+ r _n-1 . p ^{( n-1) / n} = 0.  . . . . . . . . 3.

 where  r₀ , r₁ , r₂ ,.., r_n-1 are rational functions of  p , q₀ , q₂ ... q_n-1. [original text corrected]


Commentary:

Once again there is a typo where q₁ should be q₂ since q₁ has previously been set to 1. Here y is a general algebraic function so based on the preceding sections we could say that y an algebraic function of order μ and degree m.

                                  y =   q₀ +  p ^{1 / n} + q₂  . p ^{2 / n} + ... q _n-1. p ^{( n-1) / n} ; 

where n is a prime number, q_0, q_{2 ...} q _n-1 are algebraic functions of order μ and degree m - 1 and also, p is an algebraic function of order μ - 1 and unrestricted degree and  p ^{1 / n} cannot be expressed as a rational function of  q_0, q_{1 ...} q _n-1.   Abel does not address the issue that y is a multi-valued function. Presumably not all the possible values of y are expected to be solutions of the given equation. In the case of the well know solution to the quadratic equation both values of the square root do yield solutions of the given equation. In the case of the solution of the cubic equation the picture is not so clear. The algebraic function which is a solution of a given cubic equation will have multiple terms with square roots under cube roots. It seems more than 3 different values would result yet we know the cubic equation cannot have more than 3 solutions.