68 CHAPTER 3. VECTOR SPACES

There are some easy things to observe about this.

Proposition 3.4.13 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 =∑

ni=1 ∑

mj=1 f jik jli. It follows that

{k jli}

is a spanning set. If ∑ni=1 ∑

mj=1 f jik jli = 0. Then, since the li are independent, it follows that

∑mj=1 f jik j = 0 and since

{k j}

is independent, each f ji = 0 for each j for a given arbitraryi. Therefore,

{k jli}

is a basis. ■You will see almost exactly the same argument in exhibiting a basis for L (V,W ) the

linear transformations mapping V to W .Note that if p(x) were of degree n and not irreducible, then there still exists an extension

G containing a root of p(x) such that [G : F] ≤ n. You could do this by working with anirreducible factor of p(x).

Theorem 3.4.14 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 and {z1, · · · ,zn} contained in G,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 3.4.9, there exists a field F1 such that p(x) has a root, z1 (= [x])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 was done for p(x) , enlarging thefield to F2 if necessary, such that in this new field q1 (x) = (x− z2)q2 (x) and so p(x) =(x− z1)(x− z2)q2 (x) . After no more than n such extensions, you will have obtained thenecessary field G.

Finally consider the claim about dimension. By Proposition 3.4.9, there is a larger fieldG1 such that p(x) has a root a1 in G1 and [G1 : F] ≤ n. Then p(x) = (x−a1)q(x) . Con-tinue this way until the polynomial equals the product of linear factors. Then by Proposition3.4.13 applied multiple times, [G : F]≤ n!. ■

Example 3.4.15 The polynomial x2 + 1 is irreducible in R(x) , polynomials having realcoefficients. To see this is the case, suppose ψ (x) divides x2 +1. Then x2 +1 = ψ (x)q(x) .If the degree of ψ (x) is less than 2, then it must be either a constant or of the form ax+b.In the latter case, −b/a must be a zero of the right side, hence of the left but x2 +1 has noreal zeros. Therefore, the degree of ψ (x) must be two and q(x) must be a constant. Thusthe only polynomial which divides x2 +1 are constants and multiples of x2 +1. Therefore,this shows x2+1 is irreducible. Find the inverse of

[x2 + x+1

]in the space of equivalence

classes, R(x)/(x2 +1

).

You can solve this with partial fractions.

1(x2 +1)(x2 + x+1)

=− xx2 +1

+x+1

x2 + x+1