Linear Algebra, Theory And Applications, 2012a

F.11. SOLVABILITY BY RADICALS 483
At this point, it is a good idea to recall the big fundamental theorem mentioned above
which gives the correspondence between normal subgroups and normal field extensions since
it is about to be used again.
F ≡ F 0 ⊆ F 1 ⊆ F 2 ··· ⊆F k−1 ⊆ F k ≡ K
G (F, F) ={ι} ⊆ G (F 1 , F) ⊆ G (F 2 , F) ··· ⊆G (F k−1 , F) ⊆ G (F k , F)
(6.26)
Theorem F.11.4 Let K be a splitting field of any polynomial p (x) ∈ F [x] where F is either
of characteristic 0 or of characteristic p with F p = F. Let{F i } k i=0
be the increasing sequence
of intermediate fields between F and K. Then each of these is a normal extension of F and
the Galois group G (F j−1 , F) is a normal subgroup of G (F j , F). In addition to this,
G (F j , F) ≃ G (K, F) /G (K, F j )
where the symbol ≃ indicates the two spaces are isomorphic.
Theorem F.11.5 Let f (x) be a polynomial in F [x] where F is a field of characteristic 0
which contains all n th roots of unity for each n ∈ N. LetK be a splitting field of f (x) . Then
if f (x) is solvable by radicals over F, then the Galois group G (K, F) is a solvable group.
Proof: Using the definition given above for f (x) to be solvable by radicals, there is a
sequence of fields
F 0 = F ⊆ F 1 ⊆···⊆F k , K ⊆ F k ,
where F i = F i−1 [a i ], a ki
i ∈ F i−1 , and each field extension is a normal extension of the preceding
one. You can assume that F k is the splitting field of a polynomial having coefficients
in F j−1 . This follows from the Lemma F.11.3 above. Then starting the hypotheses of the
theorem at F j−1 rather than at F, it follows from Theorem F.11.4 that
G (F j , F j−1 ) ≃ G (F k , F j−1 ) /G (F k , F j )
By Lemma F.11.1, the Galois group G (F j , F j−1 ) is Abelian and so this requires that
G (F k , F) is a solvable group.
Of course K is a normal field extension of F because it is a splitting field. By Theorem
F.10.5, G (F k , K) is a normal subgroup of G (F k , F). Also G (K, F) is isomorphic to
G (F k , F) /G (F k , K) andsoG (K, F) is a homomorphic image of G (F k , F) which is solvable.
Here is why this last assertion is so. Define θ : G (F k , F) /G (F k , K) → G (K, F) by
θ [σ] ≡ σ| K . Then this is clearly a homomorphism if it is well defined. If [σ] =[α] this
means σα −1 ∈ G (F k , K) andsoσα −1 fixes everything in K so that θ is indeed well defined.
Therefore, by Theorem F.10.5, G (K, F) mustalsobesolvable.
Now this result implies that you can’t solve the general polynomial equation of degree 5
or more by radicals. Let {a 1 ,a 2 , ··· ,a n }⊆G where G is some field which contains a field
F 0 .Let
F ≡ F 0 (a 1 ,a 2 , ··· ,a n )
the field of all rational functions in the numbers a 1 ,a 2 , ··· ,a n . I am using this notation
because I don’t want to assume the a i are algebraic over F. Now consider the equation
p (t) =t n − a 1 t n−1 + a 2 t n−2 + ···±a n .
and suppose that p (t) has distinct roots, none of them in F. Let K be a splitting field for
p (t) overF so that
n∏
p (t) = (t − r i )
k=1