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

~ 1471

GAL0SIAN RESOLVENTS28

285

are merely permuted amongst themselves by any substitution on Xi,..., x,. Thus the coefficients of the polynomial
X(V) are rational integral symmetric functions of xi,.., x"
with coefficients in the domain R, and hence equal numbers
of R. Taking V = IV, we obtain X(V) =s,F''(V1). Since
F1 (Vs) 70, we get (4).
EXAMPLE. Recurring to the special equation (2), we shall obtain
the explicit expressions (4) for the case O=X2, V1=X2-x1. Then
(3') F(V) = V6 +4V4 +4V2 +16,
(J Xi X1 X3 X3 xi
VVI V+V1 v-yb V-V0 V+V1, V+V0
-2V' -4V4- _12V' - 8V2-116VJ-48,
as shown by inserting the values of x1,..., V, given at the end of ~ 145.
Hence
-2V,5-4V,4- 12V13 8V,2 - 16V1-48
GV15 +16V1' +8V1
In view of the theorem, we have
Xi=P( Va), X2=P(Vb), X3=P(Vc), X3=P(Vd), Xi=P(Ve).
These results may be verified by evaluating the expressions.
The numerator and denominator of the above fraction for X2 may be
expressed as linear functions of V1 by means of the relation VI,2 - 2 V1 +2 = 0
of ~ 146. We get
-48V1 +32 (-3 V1+2)(V,+2) -3V,2-4V,~4 -1OV,-l-10
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ - = 1~~~~~~~ 1- 1.
16V1-64 - (V -4)(VI+2) VJ12 J7V, 8 -10
While therefore X2 is numerically equal to VI- 1 (each being i), it is not
admissible to take this reduced function r(Vi) VI- 1 as the function
p(Vi) of the theorem, since it would no longer be true that, by applying
the substitution a= (xX2), we would have xi = r(Va)). Indeed, if we apply
a to V1- 1X2-X1-1, we obtain x,-X2_-1#,-x,. The explanation is that
we should reduce the second member of the true relation xi=p(Va) by
means of V2a +2Va+2=0; we thus obtain
P ya)-16Va-64- -1,
16 Va + 64
which is the correct value of xi. Since V0, satisfies the same quadratic
equation as V1, our first reduction yields also the true relation X1= VI- 1.