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

GALOIS RESOLVENT BY ADJUNCTION 1

175

For suppose s,.ti and s,.t yielded the same value for M; that
is, suppose
iis = M 1operated upon by s,.t,
and M, = M operated upon by s,.t,
then, operating with (s,.ti)-1 upon Mi would give MS = M operated
upon by (s,tk)(s,ti) —l
That is, (s,.tk)(sti)-l is a substitution contained in the group Q
and is equal to, say s,,. If s, _ (s,.t)(s,.t)-1, then, operating with
st,, we get stk =sr S t
$rtk -- 8mSr'i -- $,Otti?
where sm' is a substitution in Q. Since the effects of s,.tk and
sM't, upon M2 are the effects of tk and ti alone, it follows that
t = ti, which is contrary to supposition. Hence s,.ti and s,.tk must
yield different values when applied to M.1.
The function S((y) (y — M)(y- MD) *. (y- Mj_) is now
seen to be invariant under any substitution of P.
The coefficients of y in <(/), obtained by performing the indicated multiplications, are symmetric functions of M11, M1,,
MJ_-1 and, therefore, by the definition of iM, functions in 0~ of
the roots of f(x) = 0 functions which admit of the substitutions of the Galois group P. Hence these coefficients are
numbers in 2 (~ 154).
To prove the irreducibility of >(/), assume that O(y) is any
function of y in Q, which vanishes for y = M. Then 0(11) = 0.
Since O(M) must acdmit all the substitutions of the Galois
group (~ 153), we must have 0(l1.i)= 0, where i has any value
0, 1, 2,., (j- 1). Hence 0(y) cannot be of lower degree than
the jth. As all the roots 3,M, M, M, of ~(y) = 0 satisfy
O(y) = 0, 0(yJ) is divisible by 0(y)).
Now, if ~(y) were reducible, one of its factors would vanish
for y =M. Since 0(y) may be any algebraic function in 52
which vanishes for y = M, let 0(y) represent this factor. Then
it would follow that this factor would be divisible by the whole
product 0(y), which is impossible. Hence <S(y) is irreducible.