238 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

K′ was defined as the smallest field which contains the conjugate fields. Therefore, S =K′and so this shows that each η j maps K′ to itself while fixing F. Thus G(K,F)⊆ G(K′,F)because each of the η i is in G(K′,F). This is what was just shown. However, sinceK′ ⊆K,it follows that also G(K′,F) ⊆ G(K,F) . Therefore, G(K′,F) = G(K,F) and by the oneto one correspondence between the intermediate fields and the Galois groups, it followsthat K′ =K. If K′ is a proper subset of K then you would need to have G(K′,F) a propersubgroup of G(K,F) but these are equal. This proves the following lemma.

Lemma 9.5.2 Let K denote the normal extension of F(a1, · · · ,am) with each ai algebraicso that F(a1, · · · ,am) is a field. Thus K is the splitting field of the product of the minimumpolynomials of the ai. Then K is also the smallest field containing the conjugate fieldsη j (F(a1, · · · ,am)) for

{η1,η2, · · · ,ηq

}= G(K,F).

Lemma 9.5.3 In Definition 9.4.48, you can assume that Fk is a normal extension of F.

Proof: First note that Fk = F [a1,a2, · · · ,ak]. Let G be the normal extension of Fk. ByLemma 9.5.2, G is the smallest field which contains the conjugate fields

η j (F(a1,a2, · · · ,ak)) = F(η ja1,η ja2, · · · ,η jak

)for {η1,η2, · · · ,ηm}= G(Fk,F). Also,

(η jai

)ki = η j

(aki

i

)∈ η jFi−1,η jF= F. Then

G= F(η1 (a1) ,η1 (a2) , · · · ,η1 (ak) ,η2 (a1) ,η2 (a2) , · · · ,η2 (ak) · · ·)

and this is a splitting field so is a normal extension. Thus G could be the new Fk withrespect to a longer sequence of ai but would now be a splitting field. ■

9.5.2 Conditions for SeparabilitySo when is it that a polynomial having coefficients in a field F is separable? It turns out thatthis is always the case for fields which are enough like the rational numbers. It involvesconsidering the derivative of a polynomial. In doing this, there will be no analysis used, justthe rule for differentiation which we all learned in calculus. Thus the derivative is definedas follows. (

anxn +an−1xn−1 + · · ·+a1x+a0)′

≡ nanxn−1 +an−1 (n−1)xn−2 + · · ·+a1

This kind of formal manipulation is what most students do anyway, never thinking aboutwhere it comes from. Here nan means to add an to itself n times. With this definition, it isclear that the usual rules such as the product rule hold. This discussion follows [26].

Definition 9.5.4 A field has characteristic 0 if na ̸= 0 for all n ∈N and a ̸= 0. Otherwise afield F has characteristic p if p ·1 = 0 for p ·1 defined as 1 added to itself p times and p isthe smallest positive integer for which this takes place.

Note that with this definition, some of the terms of the derivative of a polynomial couldvanish in the case that the field has characteristic p. I will go ahead and write them anyway.For example, if the field has characteristic p, then (xp−a)′ = 0. because formally it equalsp ·1xp−1 = 0xp−1, the 1 being the 1 in the field.