9.5. A FEW GENERALIZATIONS 237

For example, consider 3x5−25x3+45x+1 or equivalently x5− 253 x3+15x+ 1

3 ∈Q(x) .It clearly has no rational roots and a graph will show it has 5 real roots. Let F=Q(ω) whereω denotes all kth roots of unity for k≤ 5. Then some computations show that none of theseroots of the polynomial are in F and they are all distinct. Thus the polynomial cannot besolved by radicals involving kth roots for k ≤ 5 of numbers in Q. In fact, it can’t be solvedby radicals involving kth roots for k ≤ 5 of numbers in Q(ω) .

Recall that Q(√

2)

can be written as a+b√

2 where a,b are rational. However, alge-braic numbers are roots of polynomials having rational coefficients. Can each of these bewritten in this way in terms of radicals. It was just shown that, surprisingly, this is not thecase. It is a little like the fact from real analysis that it is extremely difficult to give an ex-plicit description of a generic Borel set, except that the present situation seems even worsebecause in the case of Borel sets, you can sort of do it provided you use enough hard settheory. Thus you must use the definition of algebraic numbers described above. It is alsopointless to search for the equivalent of the quadratic formula for polynomials of degree 5or more.

9.5 A Few GeneralizationsSometimes people consider things which are more general. Also, it is worthwhile identify-ing situations when all polynomials are separable to generalize Theorem 9.4.51.

9.5.1 The Normal Closure of a Field ExtensionAn algebraic extension F(a1,a2, · · · ,am) is contained in a field which is a normal extensionof F. To begin with, recall the following definition.

Definition 9.5.1 When you have F(a1, · · · ,am) with each ai algebraic so F(a1, · · · ,am) isa field, you could consider f (x) ≡∏

mi=1 fi (x). where fi (x) is the minimum polynomial of

ai. Then if K is a splitting field for f (x) , this K is called the normal closure. It is at leastas large as F(a1, · · · ,am) and it has the advantage of being a normal extension.

Let G(K,F) ={

η1,η2, · · · ,ηq}. The conjugate fields are defined as the fields

η j (F(a1, · · · ,am)) .

Thus each of these fields is isomorphic to any other and they are all contained in K. Let K′denote the smallest field contained in K which contains all of these conjugate fields. Notethat if k ∈ F(a1, · · · ,am) so that η i (k) is in one of these conjugate fields, then η jη i (k) isalso in a conjugate field because η jη i is one of the automorphisms of G(K,F). Let

S ={

k ∈K′ : η j (k) ∈K′ each j}.

Then from what was just shown, each conjugate field is in S. Suppose k ∈ S. What aboutk−1?

η j (k)η j(k−1)= η j

(kk−1)= η j (1) = 1

and so(η j (k)

)−1= η j

(k−1). Now

(η j (k)

)−1 ∈ K′ because K′ is a field. Therefore,η j(k−1)∈K′. Thus S is closed with respect to taking inverses. It is also closed with respect

to products. Thus it is clear that S is a field which contains each conjugate field. However,