360 APPENDIX A. CLASSIFICATION OF REAL NUMBERS

q(x) can only be a scalar. q(x) must be 1 because both p(x), and p̂(x) have leadingcoefficient equal to 1.

If m≥ n, then by the division algorithm for polynomials, xm = p(x)q(x)+ r (x) wherer (x) is either 0 or has degree less than n the degree of p(x). Then am = r (a) and so thisshows that if q(x) is a polynomial then q(a) can always be written in the form r (a) whereeither r (a)= 0 or r (x) has smaller degree than p(x) the minimum polynomial. Incidentally,this works for linear transformations in place of a for exactly the same reasons. We denoteby k≡

(k1 · · · kr

)an ordered list of nonnegative integers.

Let Q [a1, ...,ar] be all finite sums of the form ∑k akak11 · · ·akr

r where the ki are nonneg-ative integers, the ak are rational numbers, and the ai are nonzero algebraic numbers. Thenfrom what was just observed, this is always of the form ∑{k such that ki≤ni} akak1

1 · · ·akrr where

the ni are degrees of the algebraic numbers ai. Then Q [a1, ...,ar] is a vector space over thefield of scalars Q or more generally, you would have F [a1, ...,ar] a vector space over thefield of scalars F. It follows that Q [a1, ...,ar] has a spanning set{

ak11 · · ·a

krr ,0≤ ki ≤ ni−1

}Therefore, the dimension ofQ [a1, ...,ar] , as such a vector space, is no more than ∏

ri=1 ni ≡

m, the product of the degrees of the algebraic numbers. Letting g(a1, ...,ar) be a polynomialin which the ai are algebraic, it follows that with m as just defined,

1,g(a1, ...,ar) ,g(a1, ...,ar)2 , ...,g(a1, ...,ar)

m

cannot be linearly independent. There are too many of them. It follows that there existscalars bk in Q or the field of scalars, such that b0 + ∑

mk=1 bkg(a1, ...,ar)

k = 0 and so,g(a1, ...,ar) is an algebraic number. This shows that if you have algebraic numbers, youcan multiply them, add them, raise them to positive integer powers and so forth, and theresult will still be an algebraic number. Sloppily expressed, doing algebra to algebraicnumbers yields algebraic numbers.

In fact, you can exploit the existence of a polynomial of minimum degree for whicha number in Q [a1, ...,ar] is a root, to show that Q [a1, ...,ar] is actually a field. However,this is not needed here. It is enough to note that Q [a1, ...,ar] is a commutative ring, dis-cussed below. Note how this compares with the previous section about extending Q to R.This extends Q to a larger field Q [a1, ...,ar] ⊆ C but not anywhere near all the way to thefield C = R+ iR. In fact, as shown in an early problem the set of algebraic numbers, iscountable whereas R and C are not. Of course, if you had a which is not a root of anypolynomial having rational coefficients, then Q [a] would not be finite dimensional and theabove process will fail. When this occurs, we say that a is transcendental.

A.2 The Symmetric Polynomial TheoremFirst here is a definition of polynomials in many variables which have coefficients in a com-mutative ring. A commutative ring would be a field except it lacks the axiom which givesmultiplicative inverses for nonzero elements of the ring. A good example of a commutativering is the integers. In particular, every field is a commutative ring. Thus, a commutativering satisfies the following axioms. They are just the field axioms with one omission justmentioned. You might not have x−1 if x ̸= 0. We will assume that the ring has 1, themultiplicative identity.