Linear Algebra, Theory And Applications, 2012a

484 FIELDS AND FIELD EXTENSIONS
Then it follows that
a i = s i (r 1 , ··· ,r n )
where the s i are the elementary symmetric functions defined in Definition F.1.2. For σ ∈
G (K, F) you can define ¯σ ∈ S n by the rule
¯σ (k) ≡ j where σ (r k )=r j .
Recall that the automorphisms of G (K, F) take roots of p (t) torootsofp (t). This mapping
σ → ¯σ is onto, a homomorphism, and one to one because the symmetric functions s i
are unchanged when the roots are permuted. Thus a rational function in s 1 ,s 2 , ··· ,s n is
unaffected when the roots r k are permuted. It follows that G (K, F) cannot be solvable if
n ≥ 5 because S n is not solvable.
For example, consider 3x 5 − 25x 3 +45x + 1 or equivalently x 5 − 25
3 x3 +15x + 1 3
. It clearly
has no rational roots and a graph will show it has 5 real roots. Let F be the smallest field
contained in C which contains the coefficients of the polynomial and all roots of unity. Then
probably none of these roots are in F and they are all distinct. In fact, it appears that the
real numbers which are in F are rational. Therefore, from the above, none of the roots are
solvable by radicals involving numbers from F. Thus none are solvable by radicals using
numbers from the smallest field containing the coefficients either.