RePoSS #11: The Mathematics of Niels Henrik Abel: Continuation ...

7.1. Solubility of Abelian equations 145
Thus, the original problem of solving the equation φ (x) = 0 of degree µ had
been reduced to solving certain equations, (7.5) and (7.6), of lower degrees. Gener-
ally, the equation of degree m would not be solvable by radicals, but as ABEL went on
to demonstrate, the m equations of degree n could always be solved algebraically.
7.1.2 Algebraic solubility of Abelian equations
If all the roots of the equation φ (x) = 0 fell into the same orbit of θ (one chain), i.e. are
of the form
x1, θ (x1) , θ 2 (x1) , . . . , θ n−1 (x1) ,
the situation was equivalent to assuming m = 1 above. In this case, ABEL let α denote
a primitive µ th root of unity and formed the rational expression
ψ (x) =
�
µ−1
∑ α
k=0
k θ k �µ
(x) . (7.7)
Through direct calculations, he proved that
�
ψ θ k �
(x) = ψ (x) for all k = 0, 1, . . . , µ − 1,
which showed that ψ was a symmetric function of the roots of φ (x) = 0. Thus, ψ (x)
could be expressed rationally in known quantities. Next, ABEL introduced µ radicals
of (7.7),
µ√ vu =
µ−1
∑ α
k=0
k uθ k (x) for 0 ≤ u ≤ µ − 1, (7.8)
by attributing to αu the different µ th roots of unity 1, α, α 2 , . . . , α µ−1 . From these radi-
cals, it was a routine procedure for ABEL to obtain the expression
where A was a constant.
θ k (x) = 1
�
µ−1
−A +
µ
∑
u=1
α uk µ√ vu
�
, k = 0, 1, . . . , µ − 1, (7.9)
The expression (7.9), however, contained µ − 1 extractions of roots with exponent
µ which seemed to indicate that a total of µ µ−1 different values could be obtained al-
though the degree of φ (x) = 0 was only µ. ABEL resolved this apparent contradiction,
similar to one noticed by L. EULER (1707–1783) (see section 5.1), by an elegant argu-
ment prototypic of his approach to the theory of equations. In the deduction, ABEL
proved that all the root extractions depended on one of them by considering
µ√ vk ( µ√ v1) µ−k =
�
µ−1
∑ α
u=0
ku θ u �
(x) ×
�
µ−1
∑ α
u=0
u θ u �µ−k
(x) .