An introduction to the modern theory of equations, by Florian Cajori.

GALOIS THEORY OF ALGEBRAIIC NUMlBERS 139

Consider the case when the product G(x) - H(x) has only
integral coefficients. Then it is evident that it -1 must be
divisible by m. n. Since k is relatively prime to m, and I to n,
it follows that
lc = pn1, 1 = 2qm,
where p and q are integers. We may now write
G(Sx)_ njl g1(x), H(x) q -h1(x),
where the functions g1(x) and h1(x) have only integral coefficients.
Consequently, if f(x) is resolvable into two rational factors
G(x) and H(x), which have fractional coefficients, so that
we have
f()e = f (x) *. 1(x),
then we have also f(x) _ y)q - /,(z) * 71(x),
where the coefficients are integral throughout. Hence, if f(x)
is resolvable into rational factors, it is resolvable into such
factors with integral coefficients.
128. Reducibility of f(x). Whether the function f(x), in
which the coefficients are integers and the degree n does not
exceed 4 or 5, is reducible or not in the domain 02(), can readily
be ascertained by the aid of Gauss's lelmma and ordinary
algebra.
We assume that, in f(x), the coefficient co, of xZ' is unity. If
a0 is not unity, we can change the function so that it will be
unity by taking x = -/, and multiplying by ac,"(to0
For every integral value a of x, which causes f(x) to vanish,
we have a factor x- a of f(x), ~ 3. Here a must be a factor
of c,. This consideration enables us always to determine the
reducibility or irreducibility of functions f(x) of the second or
third degree.
If f(x) is of the fourth degree, then, if there is no linear
rational factor, there can be no cubic rational factor. To test