216 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

Definition 9.4.8 Recall that the symbol [K : F] whereK is a field extension of F means thedimension of the vector space K with field of scalars F.

F η→≃

F̄

p(x) η p(x) = p̄(x) p̄(x)

F(r1, · · · ,rn)ζ i→≃

F̄(r̄1, · · · , r̄n)

i = 1, · · · ,m,

{m≤ [K : F]m = [K : F] , r̄i ̸= r̄ j

In the next theorem, the polynomials p(x) , p̄(x) are not necessarily irreducible.

Theorem 9.4.9 Let η be an isomorphism from F to F̄ and let K,K̄ be splitting fields ofp(x) and p̄(x) . If the roots of p(x) are {r1, · · · ,rm} , recall that F(r1, · · · ,rm)≡K is a fieldsince each root of the polynomials is algebraic. Thus, this must be the splitting field ofp(x), the smallest field which contains each of the roots of p(x). The case is similar for K̄.Then

1. There exist at most [K : F] isomorphisms ζ i :K→ K̄ which extend η .

2. If {r̄1, · · · , r̄n} are distinct, then there exist exactly [K : F] isomorphisms of the abovesort.

3. In either case, the two splitting fields K,K̄ are isomorphic with any of these ζ i serv-ing as an isomorphism.

Proof: Suppose [K : F] = 1. Say a basis for K is {r} . Then {1,r} is dependent and sothere exist a,b ∈ F, not both zero such that a+br = 0. Then it follows that r ∈ F and so inthis case F=K. Then the isomorphism which extends η is just η itself and there is exactly1 isomorphism.

Next suppose [K : F] > 1. Then p(x) has a factor q(x) irreducible over F which hasdegree larger than 1. If not, you could factor p(x) as linear factors and so all the rootswould be in F so the dimension [K : F]would equal 1. Without loss of generality, let theroots of q(x) in K be {r1, · · · ,rm}. Thus

q(x) =m

∏i=1

(x− ri) , p(x) =n

∏i=1

(x− ri)

Now q̄(x)≡ η (q(x)) defined analogously to p̄(x) , also has degree at least 2. Furthermore,it divides p̄(x) all of whose roots are in K̄. This is obvious because η is an isomorphism.You have

l (x)q(x) = p(x) so l̄ (x) q̄(x) = p̄(x) .

Denote the roots of q̄(x) in K̄ as {r̄1, · · · , r̄m} where they are counted according to multi-plicity.

Recall why [F(r1) : F] = m. It is because q(x) is irreducible and monic so by Lemma9.4.7, it is the minimum polynomial for each of the ri. Since q(x) is irreducible, it followsthat 1,r1,r2

1, ...,rm−11 must be independent so the dimension is at least m. However, it is not

more than m because q(x) is of degree m. Thus, using the division algorithm, everythingin F(r1) is expressible as a polynomial in r1 of degree less than m.