16.4. VECTOR SPACES AND FIELDS∗ 391
Proposition 16.4.22 Let F ⊆ K ⊆ L be fields. Then [L : F ] = [L : K] [K : F ].
Proof: Let {li}ni=1 be a basis for L over K and let
{k j}m
j=1 be a basis of K over F . Thenif l ∈ L, there exist unique scalars xi in K such that l = ∑
ni=1 xili. Now xi ∈ K so there exist
f ji such that xi = ∑mj=1 f jik j. Then it follows that
l =n
∑i=1
m
∑j=1
f jik jli
It follows that{
k jli}
is a spanning set. Is it linearly independent? Suppose
n
∑i=1
m
∑j=1
f jik jli = 0.
Then, since the li are independent, ∑mj=1 f jik j = 0 and since
{k j}
j is independent, eachf ji = 0 for each j for a given arbitrary i. Therefore,
{k jli}
is a basis. ■Note that if p(x) were not irreducible, then you could find a field extension G such that
[G : F]≤ n. You could do this by working with an irreducible factor of p(x).Usually, people simply write b rather than [b] if b ∈ F. Then with this convention,
[bφ (x)] = [b] [φ (x)] = b [φ (x)] .
This shows how to enlarge a field to get a new one in which the polynomial has a root.By using a succession of such enlargements, called field extensions, there will exist a fieldin which the given polynomial can be factored into a product of polynomials having degreeone. The field you obtain in this process of enlarging in which the given polynomial factorsin terms of linear factors is called a splitting field.
Theorem 16.4.23 Let p(x) = xn + an−1xn−1 + · · ·+ a1x+ a0 be a polynomial with coeffi-cients in a field of scalars F. There exists a larger field G such that there exist {z1, · · · ,zn}listed according to multiplicity such that
p(x) =n
∏i=1
(x− zi)
This larger field is called a splitting field. Furthermore,
[G : F]≤ n!
Proof: From Proposition 16.4.18, there exists a field F1 such that p(x) has a root, z1(= [x] if p is irreducible.) Then by the Euclidean algorithm
p(x) = (x− z1)q1 (x)+ r
where r ∈ F1. Since p(z1) = 0, this requires r = 0. Now do the same for q1 (x) that wasdone for p(x) , enlarging the field to F2 if necessary, such that in this new field
q1 (x) = (x− z2)q2 (x) .
and sop(x) = (x− z1)(x− z2)q2 (x)