394 CHAPTER 16. VECTOR SPACES

Now notice that for a an algebraic number, F [a] is a vector space with field of scalarsF. Similarly, for {a1, · · · ,am} algebraic numbers, F [a1, · · · ,am] is a vector space with fieldof scalars F. The following fundamental proposition is important.

Proposition 16.4.31 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 minimal polynomial 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. Thus F [a] is spannedby {

1,a,a2, · · · ,an−1}Since p(x) has smallest degree of all polynomial 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 linearcombinations of products of

{ak1

1 ,ak22 , · · · ,akn

n

}where the ki are nonnegative integers. From

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

m

∏i=1

deg(ai)

entries. By Theorem 16.3.5 this proves the first claim.Finally consider the last claim. Let g(a1, · · · ,am) be a polynomial in

F [a1, · · · ,am]

Since

dimF [a1, · · · ,am]≡ p≤m

∏j=1

deg(a j)< ∞,

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