Full text: Algebraic invariants

(001). For the remaining intersections, we may set £i = l 
and obtain from each root r of 
(1) P-\-Qbr 2 -\-4:ar — 2>b 2 = 0 
two intersections (1 ,r, ±^'3). For, if x' 3 = 0, then C = 0, so 
that (1) would have a multiple root, whence a 2 +46 3 = 0. But 
the three partial derivatives of F would then all vanish at 
{2b,—a, 0) or (1,0,0), according as h¿¿0 or ¿ = 0. Hence there 
are exactly nine distinct points of inflexion. 
For each of the four roots of (l), the three points of inflexion 
P and (1, r, ±x' 3 ) are collinear, being on x 2 =rx 1. Since we 
may proceed with any point of inflexion as we did with P, 
we see that there are 9-4/3 or 12 lines each joining three points 
of inflexion and such that four of the lines pass through any 
one of the nine points. The six points of inflexion not on a 
fixed one of these lines therefore lie by threes on two new 
lines; three such lines form an inflexion triangle. Thus there 
are |12 = 4 inflexion triangles. 
The fact that there are four inflexion triangles, one for 
each root r of (1), can also be seen as follows: 
TfH+rF = {rx 1 -x 2 ) \x 3 2 -rx 2 2 — {r 2 +3b)xix 2 - {r 3 +Qbr-\-3a)xi 2 \. 
The last factor equals 
X3 2 --\rx 2 +%{r 2 +Zb)x i} 2 , 
r 
and hence is the product of two linear functions. 
Corresponding results hold for any cubic curve /=0 without 
singular points. We have shown that / can be reduced to 
the special form F by a linear transformation of a certain 
determinant A. Follow this by the transformation which 
multiplies £3 by A and X\ by A~ 2 , and hence has the determin 
ant A -1 . Thus there is a transformation of determinant 
unity which replaces / by a form of type F, and hence replaces 
the Hessian h of / by the Hessian H of F. Hence there are 
exactly four values of r for which p=h-\-2\rJ has a linear factor 
and therefore three linear factors. These r’s are the roots 
of a quartic (l) in which a and b are functions of the coefficients 
§14] 
INFLEXION POINTS OF CUBIC
	        
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