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

NORMAL DOMAINS

159

None of the roots II are alike, for suppose ki(ph)= ' (ph),
that is, h(p ) — (p) =
then III is an equation having p,, as a root. But the irreducible
equation f(x) - 0 has also Ph as a root. Hence III must be
satisfied by all the roots of f(x) 0; for instance, by p. Consequently, )i() - () =0
This equation by I may be written p - pk = 0, which cannot be
true, since p is a primitive number.
Ex. 1. In Ex. 5, ~ 133, we have given an irreducible equation with the
roots p, pi, P2, pa, conjugate to p in the normal domain D(p). We have
P1 = p2, P2 = p3 p3 = p4. Hence the roots may be represented by the
series 4
P, p2,, p 4. I
If in I we write p3 for p, we get
p3, p32, p33, p34,
where P32 p2, P33 = P, p34 = p. Hence the transposition (pp3) only
changed the order of the roots.
Ex. 2. What is the order of the roots, if in Ex. 1, we apply the
transposition (pp2)?
148. Theorem. Every transposition (ppk) in the normal
domain Q(p) is equal to some one of the transpositions (ppl),
(pp2), ", (ppn,-1)
We have Ph = h(P), I
where <h(p) is a root of the normal equation f() =. Upon
c,(p) perform the transposition (ppa), and we get Oh(pi). This
is a number conjugate to +,,(p), and is, therefore, one of the
other roots of f(x) = 0, say Pi (~ 138), so that
pAk -= (p,). II
Since the transposition (php,) changes Ph to p,, and the transposition (ppi) changes 4h(p) to Oh(p,), we have from equations
I and II that (papk) = (ppi)'