AND LINES IN THE INTERSECTIONS OF A SYSTEM OF SURFACES. 
171 
(B.) (i.) If £ y, l are the co-ordinates of any 'point on the locus (f> ( x , y, z) — 0 
[where <f> is a rational integral function of x, y, z which contains no repeated 
factors ), and if the substitutions x = £, y = rj, z = £ make xp (x, y, z) = 0 [where 
xp is a rational integral function of x, y, z), then xp contains the first power of <f> as 
a factor. 
(ii.) If x = y = 77, z = £ make xp = 0 , ~~ = 0, then \ Jj contains the second power of 
<p as a factor. 
0-v/r 0^ • 
(iii.) If x = y = rj, z = £ make xp = 0,-fi = 0, . . . = 0 , then xp contains <f> m 
as a factor. 
To prove (i.) suppose first that (p is indecomposable. It is obvious that xfj cannot 
be of lower dimensions than <f> in any one of the variables ; if it were, then all the 
values of x, y, z which make (p = 0 would not make xp = 0. 
It may happen that cf> does not contain all the variables x, y, z. But it must 
contain one of them ; suppose it contains x. 
If (f> be not a factor of xfj, proceed as in the process for finding the common factor of 
highest dimensions in x of (f) and xfj; and if, at any step of the process, fractional 
quotients in which the denominators are functions of y, z are obtained, let the 
denominators be removed in the usual way by multiplication throughout by a factor. 
Then either the process will terminate, or there will at last be a remainder, which 
is a function of y, z only, not x. 
In the first case <f> and xfj will have a common factor, and (f> will be decomposable, 
which is contrary to the hypothesis. 
In the second case a relation of the form 
A xfi = B cf> -j- C 
exists, where A, B, C are rational integral functions of y, z only. In this case, since 
all the values of x, y, z which make <j> = 0, also make xfj = 0, therefore they make 
C — 0. But C is a function of y, z only, not x. Now, the values under consideration 
are values of x, y, z. This is impossible. Hence this alternative does not hold. 
Hence </> must be a factor of xfj. 
If (f> be decomposable, its indecomposable factors may be taken separately, and 
shown as above to be factors of xfj. 
As it is further supposed that <£ contains no repeated factors, it follows that xfj 
contains (f) as a factor. 
To prove (ii.). 
By the same argument as in (i.) it follows that xfj contains (f> as a factor. Let 
xp — Pup. 
Z 2 
