206 CHAPTER 7. VECTOR SPACES AND FIELDS

Proof: Let the minimal 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)

where r (x) has degree less than the degree of p (x) if it is not zero. Hence q (a) = r (a).Thus F [a] is spanned by {

1, a, a2, · · · , an−1}

Since p (x) has smallest degree of all polynomials which have a as a root, the above set isalso linearly independent. This proves the second claim.

Now consider the first claim. By definition, F [a1, · · · , am] is obtained from all linear

combinations of products of{ak11 , a

k22 , · · · , akn

n

}where the ki are nonnegative integers. From

the first part, it suffices to consider only kj ≤ deg (aj). Therefore, there exists a spanningset for F [a1, · · · , am] which has

m∏i=1

deg (ai)

entries. By Theorem 7.2.4 this proves the first claim.Finally consider the last claim. Let g (a1, · · · , am) be a polynomial in {a1, · · · , am} in

F [a1, · · · , am]. Since

dimF [a1, · · · , am] ≡ p ≤m∏j=1

deg (aj) <∞,

it follows1, g (a1, · · · , am) , g (a1, · · · , am)

2, · · · , g (a1, · · · , am)

p

are dependent. It follows g (a1, · · · , am) is the root of some polynomial having coefficientsin F. Thus everything in F [a1, · · · , am] is algebraic. Why is F [a1, · · · , am] a field? Letg (a1, · · · , am) be as just mentioned. Then it has a minimal polynomial,

p (x) = xq + aq−1xq−1 + · · ·+ a1x+ a0

where the ai ∈ F. Then a0 ̸= 0 or else the polynomial would not be minimal. Therefore,

g (a1, · · · , am)(g (a1, · · · , am)

q−1+ aq−1g (a1, · · · , am)

q−2+ · · ·+ a1

)= −a0

and so the multiplicative inverse for g (a1, · · · , am) is

g (a1, · · · , am)q−1

+ aq−1g (a1, · · · , am)q−2

+ · · ·+ a1−a0

∈ F [a1, · · · , am] .

The other axioms of a field are obvious. ■Now from this proposition, it is easy to obtain the following interesting result about the

algebraic numbers.

Theorem 7.3.32 The algebraic numbers A, those roots of polynomials in F [x] which arein G, are a field.