9.2. TRANSCENDENTAL NUMBERS 199
and each pk (anα1, · · · ,anαn) is an integer. Thus f (anα1, · · · ,anαn) is an integer. Fromthis, it is obvious that f (α1, · · · ,αn) is rational. Indeed, from 9.3,
f (α1, · · · ,αn) = ∑k1+···+kn=d
ak1···kn pk11 (α1, · · · ,αn) · · · pkn
n (α1, · · · ,αn)
Now multiply both sides by aMn , an integer where M is chosen large enough that
aMn f (α1, · · · ,αn) =
∑k1+···+kn=d
ah(k1,...,kn)n ak1···kn pk1
1 (anα1, · · · ,anαn) · · · pknn (anα1, · · · ,anαn) an integer.
where h(k1, ...,kn) is some nonnegative integer. Thus f (α1, · · · ,αn) is rational. If the fhad rational coefficients, then m f would have integer coefficients for a suitable m and som f (α1, · · · ,αn) would be rational which yields f (α1, · · · ,αn) is rational. ■
Nothing would change in the last claim of this theorem if Q were a general field. Youwould get f (α1, · · · ,αn) is in the general field.
Corollary 9.1.7 Let α1, · · · ,αn be roots of the polynomial equation
p(x)≡ xn +an−1xn−1 + · · ·+a1x+a0 = 0
where each ai is in a field F. Then any symmetric polynomial in α1, · · · ,αn which hascoefficients in F is in F.
Proof: Let f (x1, ...,xn) be a symmetric polynomial. Then by the symmetric polynomialtheorem,
f (α1, · · · ,αn) = ∑k
bksk11 sk2
2 · · ·sknn
where the sk (α1, · · · ,αn) is ± the coefficient of xk in ∏ni=1 (x−α i) . Thus ak =±sk and so
the above sum is in F. ■
9.2 Transcendental NumbersMost numbers are like this, transcendental. Here the algebraic numbers are those whichare roots of a polynomial equation having rational numbers as coefficients, equivalentlyinteger coefficients. By the fundamental theorem of algebra, all these numbers are in Cand they constitute a countable collection of numbers in C. Therefore, most numbers in Care transcendental. Nevertheless, it is very hard to prove that a particular number is tran-scendental. Probably the most famous theorem about this is the Lindermannn Weierstrasstheorem, 1884.
Theorem 9.2.1 Let the α i be distinct nonzero algebraic numbers and let the ai be nonzeroalgebraic numbers. Then ∑
ni=1 aieα i ̸= 0.
I am following the interesting Wikepedia article on this subject. You can also look at thebook by Baker [5], Transcendental Number Theory, Cambridge University Press. There arealso many other treatments which you can find on the web including an interesting articleby Steinberg and Redheffer, already mentioned, which appeared in about 1950.