Page 70 - Geometric Modeling and Algebraic Geometry

4 Monoid Hypersurfaces 67

Fig. 4.3. The monoids Z(x +y +5xyz−z (x+y)) and Z(x +y +5xyz−z (x−y)) both
3
3
3
3
3
3
have a T 3,3,5 singularity, but the singularities are not right equivalent over . (The pictures
are generated by the program [5].)
This polynomial will have roots at (0 : 1) and (1 : 0) if and only if f 4 (1, 0, 0) =
a 1 =0. When a 1 =0 we may (by symmetry) assume a 3 =0, so that (0 : 1) is a
simple root and (1 : 0) is a root of multiplicity m−1. Other roots of f 4 (θ) correspond
to intersections of Z(f 3 ) and Z(f 4 ) away from (1 :0:0). The multiplicity m i of
each root is equal to the corresponding intersection multiplicity, giving rise to an
A m i −1 singularity if m i > 0, as described by Proposition 6, or a line L a ⊂ Z(F)
with O as the only singular point if m i =1.
The polynomial f 4 (θ) defines a linear map from the coefficient space k 15 of f 4
to the space of homogeneous polynomials of degree 12 in s and t. By elementary
linear algebra, we see that the image of this map is the set of polynomials of the form

b 0 s 12 + b 1 s t + b 2 s t + ··· + b 12 t 12
10 2
11
where b 0 = b 12 . The kernel of the map corresponds to the set of polynomials of the
form
f 3 where
is a linear form. This means that f 4 (θ) ≡ 0 if and only if f 3 is a
factor in f 4 , making Z(F) reducible and not a monoid.
For every m =0, 2, 3, 4,... , 12 we can select r parameter points

p 1 ,...,p r ∈ P \{(0 : 1), (1 : 0)}
1
and positive multiplicities m 1 ,...,m r with m 1 + ··· + m r =12 − m and try to
describe the polynomials f 4 such that f 4 (θ) has a root of multiplicity m i at p i for
each i =1,...,r.
Still assuming a 3 =0 whenever a 1 =0, any such choice of parameter points
p 1 ,...,p r and multiplicities m 1 ,...,m r corresponds to a polynomial q = b 0 s 12 +
b 1 s t + ··· + b 12 t 12 that is, up to a nonzero constant, uniquely determined.
11
Now, q is equal to f 4 (θ) for some f 4 if and only if b 0 = b 12 .If m ≥ 2, then q
contains a factor st m−1 ,so b 0 = b 12 =0,giving q = f 4 (θ) for some f 4 . In fact,