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

116 Chapter 6. Algebraic insolubility of the quintic
In the published paper, ABEL involved himself in a difficult reductio ad absurdum
to rule out this case. However, because the proof given in the letter to KÜLP is more
detailed, it is presented before the differences between the two proofs are sketched.
Besides the symmetric function (6.13), there is another obvious symmetric function
under permutations of x2, . . . , x5,
f (x1) = v1v2.
The function f (x1) is of the form (6.11). ABEL introduced
and found that it must divide
(z − v1) (z − v2) = z 2 − φ (x1) z + f (x1) = R, (6.14)
5
5
∏ (z − vk) = ∑ pkz k=1
k=0
k = R ′ ,
in which p0, . . . , p5 were symmetric functions of x1, . . . , x5 by the theorem 1 on LA-
GRANGE resolvents. Since R ′ was unaltered by transpositions ( 1u
u1 ) it followed that all
the polynomials derived from (6.14) through this transposition,
z 2 − φ (xu) z + f (xu) = ρu for 1 ≤ u ≤ 5,
would divide R ′ . However, as R ′ was a polynomial of the fifth degree, some polyno-
mials among ρ1, . . . , ρ5 had to share a common factor. Assuming that ρ1 and ρ2 had a
factor in common ABEL concluded
z = f (x1) − f (x2)
φ (x1) − φ (x2) .
This value of z must be one of the values of v and thus the left hand side had five differ-
ent values. However, the right hand side had 10 different values, and a contradiction
had been reached, ruling out the case µ = 2.
The published argument in Beweis der Unmöglichkeit 38 followed the one given in
the letter to KÜLP until ABEL had demonstrated that
φ (x1) = v1 + v2 =
4
∑ rkx k=0
k 1
and had recognized that φ had five different values under permutations of
x1, . . . , x5. Whereas the proof in the letter then explicitly constructed the polynomi-
als R and R ′ , the original argument was much more roundabout. Substituting any one
x k of x2, . . . , x5 for x1, ABEL obtained the value φ (x k) as the sum of two of the five
values of v.
38 (N. H. Abel, 1826a).