Theory and applications of finite groups, by G.A. Miller, H. F. Blichfeldt [and] L. E. Dickson.

~ 143] REDUCIBLE AND IRREDUCIBLE FUNCTIONS

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domains. Indeed, its linear factors are X+e, x e-1; while every quadratic factor is of one of the forms x2 —i, x'+ax41 (a2= ~-2).
EXAMPLE 3. X3-2 is irreducible in R(1). For, if it were reducible,
it would have a linear factor x-a/b, where a and b are relatively prime
integers, of which b may be taken to be positive. Then a3-2b3=O. If
b has a prime factor 0(3> 1), then 3 divides a3 and hence divides a, whereas
a and b have no common factor 3. Thus b = 1, a3 =2. Hence the positive
integer a divides 2, so that a= 1 or 2, and a3= 1 or 8, whereas a3= 2. Hence
x3-2 is irreducible in R(1).
If f(x) is reducible in R, f(x) =0 is called a reducible equation in R; in the contrary case, an irreducible equation in R.
143. Irreducible Binomials. If p is a prime number and
if A is a number of a domain R, but A is not the pth power of a
number of R, then x -A is irreducible in R.
The roots of xP =A may be denoted by
r, cor, Sir,... WP-,l1
where w is an imaginary pth root of unity. Let there be a factor
with coefficients in R of xP-A. It has a constant term of
the form + -owrt, where 0<t<p. By the theory of numbers,
there exist integers b, c such that bt-cp= 1. Hence R contains
(WosrVt)b == orby+ = corAC = r'Ae,
where r' is one of the above roots. Thus r' is in R, so that A
is the pth power of a number r' in R, contrary to hypothesis.
144. Theorem. Let the coefficients of the polynomials f(x)
and g(x) be numbers of a domain R and let f(x) be irreducible
in R. If one root a of f(x)=0 satisfies g(x)=0, every root of
f(x) =0 satisfies g(x) = 0 and f(x) is a divisor of g(x) in R.
The greatest common divisor h(x) of f(x) and g(x) is not
a constant, since it has the factor x-a. The usual process for
finding h(x) involves only rational operations; hence its coefficients are numbers of the domain R. Since f(x) is irreducible
in R, its divisor h(x), with coefficients in R, is of the same
degree as f(x), and hence equals cf(x), where c is a number in
R. But h(x) divides g(x). Hence f(x) divides g(x) in R.