ABEL’S THEOREM AND ABELIAN FUNCTIONS. 
325 
If we consider two equations in two variables say 
f n =x Q +»! y+xtfj* + . . . +x n y* =0 
<£>«=X 0 -|-X 1 ?/d-X 3 ?/~'+ . . . 4~X m y m = 0 
and if X be the eliminant of f n and <f) m with regard to y, then we have 
X=A/+Bf 
Now X being of degree mn in x, A must be of degree mn—n and B of mn — m ; 
while it is sufficient that the highest power of y in A be the (m—l) th and in B the 
(n — l) th . Write then 
A=A 0 +A 1 y+A 2 y®+ . . . + A m _{y m ~ l 
B = B 0 + Bp/ + Bp/ 3 + . . . +B„_ 1 y"~h 
Substitute in X; since X is explicitly free from y all the coefficients of powers of y 
in the result must be zero. This then gives 
A 0 .Tj -f- A 
+B 0 X 1 +B 1 X 0 
= 0 
AqX 3 ~b A A^Xq 
+ B 0 X 3 +B 1 X 1 + B 3 X 0 
= 0 
A 0 a; 3 4- A]CC 3 -|~ A 3 aq A 3 Xq 
+ B 0 x 3 + B i X 3 +B 3 X 1 +b 3 x 0 
= 0 
wi+w—1 equations to determine the ratios of the m+n quantities A, B. Let 
E— 
1 
y 
0 
y - • • 
yin—l 
0 
0 
. . . 
X Q 
o ... 
0 
X x 
Xo 
t 
X 1 
%o • • 
0 
Xo 
X L 
X () . • • 
F = 
0 
0 . 
0 
1 
y 
y* ■ ■ • 
x 1 
x 0 
o... 
X, 
X 0 
o ... 
X 2 
Xi 
x n . . . 
x 3 
x L 
x 0 • • • 
Then A r bears to the minor of y r in E the same ratio as B, bears to the minor of 
y* in F : thus 
A_B 
E F 
X n 
X * 
But the diagonal term in E is 
