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

232

THEOIRY OF EQUATIONS

root y in common. Then II must be satisfied by all the roots of I.
Multiplying together the members of the m equations thus obtained, we get
f(x)m= F (;) * F.(x),
where Fi(x)-= fi(x, 7).fi (x, wy) *. t;(x, w..-ly),
F2() = fo (x, a) *~ 2(x, 7) '.. f-('a,..-17).
Fi(x) and F2(x) are respectively of the degrees mu1 and mn2; their
coefficients, being symmetric functions of the roots of I, lie in t2. Since
J(x) is irreducible and m and i are both prime, we must have
Fi (a',) - /(:), F.:(x)-f(xc)q,
mlii= 3il, q2 -m u21, 2l1 -t- it _ 0, 7- =.
Ex. 14. Show that in Ex. 13f(x) -f (x, y)- fi(, wy) **/fi(x, on-l1),
where fix, y) is irreducible in the domain l2(w, y), and is linear with
respect to x.
Ex. 15. Show that if fi(x, ) = 0 yields in Ex. 14
ao = Co + c1y + C272 + *.. Cn-ly1-l,
then (1 = Co + c1W' + C2Wo22 + *-. C-7_lw'-ly1-1,
etc., where ao, ax, etc., are roots of f(x)= 0, and Co, cl,., n_1 are numbers in Q. Show that the difference of two roots of '(x)= 0 cannot be a
number in Q.
Ex. 16. Prove that an irreducible solvable quintic with real coefficients
cannot have three real roots and two complex roots.
Show that the Galois group (1) must be of the fifth degree; (2) cannot be G12(5), G6(5)I, G(5)II (Ex. 5, ~ 104); (3) cannot be G5(5), ~ 171;
(4) to test G20(5), take y2 in Ex. 1, which admits it. If any two roots,
say ao and a,, are assumed to be conjugate imaginaries, then
y = aoA + alB + C,
where A, B, C are real values. Since A - a4 -- a3, B = a2 - as - a4,
we cannot have A = B, because that would make a2 = 0t4. Thus, we see
that y cannot be real. Consequently y2 cannot be real, unless y is a pure
imaginary. Hence y= (eto-a1) (a4 —a 2). That y.2 may lie in 2, we must
have y = i/f. Vg and go - al = iVf, e4 - = vg, where f and g are
positive numbers in 0. But by Ex. 1 5, f and g cannot be perfect squares.
By Exs. 13, 14, 15 we see that the roots of the given quintic are numbers
in the domain 9(,0, y, where w is a complex fifth root of unity and y is a
root of the irreducible equation /5 - a = 0. Hence Vf and Vg do not
lie in 2(, y) and the equations ao - a = i v/, a4 - t2 = vg are impossible.
Consequently G20(5) is not the group, ~ 155, B.