70 CHAPTER 3. VECTOR SPACES

Theorem 3.4.19 Let a ∈A. Then there exists a unique monic irreducible polynomial p(x)having coefficients in F such that p(a) = 0. This polynomial is the minimum polynomial.

Proof: Let p(x) be a monic polynomial having smallest degree such that p(a) = 0.Then p(x) is irreducible because if not, there would exist a polynomial having smallerdegree which has a as a root. Now suppose q(x) is monic with smallest degree such thatq(a) = 0. Then q(x) = p(x) l (x)+ r (x) where if r (x) ̸= 0, then it has smaller degree thanp(x). But in this case, the equation implies r (a) = 0 which contradicts the choice of p(x).Hence r (x) = 0 and so, since q(x) has smallest degree, l (x) = 1 showing that p(x) = q(x).■

Definition 3.4.20 For a an algebraic number, let deg(a) denote the degree of the minimumpolynomial of a.

Also, here is another definition.

Definition 3.4.21 Let a1, · · · ,am be in A. A polynomial in {a1, · · · ,am} will be an expres-sion of the form

∑k1···kn

ak1···knak11 · · ·a

knn

where the ak1···kn are in F, each k j is a nonnegative integer, and all but finitely many of theak1···kn equal zero. The collection of such polynomials will be denoted by

F(a1, · · · ,am) .

The splitting field of g(x) ∈ F [x] is F(a1, · · · ,am) where the {a1, · · · ,am} are the roots ofg(x) in A.

Now notice that for a an algebraic number, F(a) is a finite dimensional vector spacewith field of scalars F. Similarly, for {a1, · · · ,am} algebraic numbers, F(a1, · · · ,am) is afinite dimensional vector space with field of scalars F. The following fundamental propo-sition demonstrates this observation. This is a remarkable result.

Proposition 3.4.22 Let {a1, · · · ,am} be algebraic numbers. Then

dimF(a1, · · · ,am)≤m

∏j=1

deg(a j)

and for an algebraic number a,

dimF(a) = deg(a)

Every element of F(a1, · · · ,am) is in A and F(a1, · · · ,am) is a field.

Proof: Let the minimum polynomial of a be

p(x) = xn +an−1xn−1 + · · ·+a1x+a0.

If q(a) ∈ F(a) , thenq(x) = p(x) l (x)+ r (x)