Page 62 - Geometric Modeling and Algebraic Geometry

4 Monoid Hypersurfaces 59
(ii) f is singular at p, g is singular at p,or ∇f(p) and ∇g(p) are nonzero and
parallel.
(iii) s∇f(p)+ t∇g(p)=0 for some (s, t) =(0, 0)

Proof. (ii) is equivalent to (iii) by a simple case study: f is singular at p if and
only if (iii) holds for (s, t)=(1, 0), g is singular at p if and only if (iii) holds for
(s, t)=(0, 1), and ∇f(p) and ∇g(p) are nonzero and parallel if and only if (iii)
holds for some s, t =0.
We can assume that p =(0:0:1),so I p (f, g)=lg S where
¯
S = k[x 1 ,x 2 ,x 3 ] (x 1 ,x 2 ) .
(f, g)
Furthermore, let d =deg f, e =deg g and write

d e
d−i e−i
f = f i x and g = g i x
3 3
i=1 i=1
where f i ,g i are homogeneous of degree i.
If f is singular at p, then f 1 =0. Choose
= ax 1 + bx 2 such that
is not a
multiple of g 1 . Then
will be a nonzero non-invertible element of S, so the length
of S is greater than 1.
We have ∇f(p)=(∇f 1 (p), 0) and ∇g(p)=(∇g 1 (p), 0). If they are parallel,
choose
= ax 0 + bx 1 such that
is not a multiple of f 1 (or g 1 ), and argue as above.
Finally assume that f and g intersect transversally at p. We may assume that
f 1 = x 1 and g 1 = x 2 . Then (f, g)=(x 1 ,x 2 ) as ideals in the local ring
¯ ¯
k[x 1 ,x 2 ,x 3 ] (x 1 ,x 2 ) . This means that S is isomorphic to the field k(x 3 ). The length
of any field is 1,so I p (f, g)=lg S =1.

Now we can say which are the lines L b , with b ∈ Z(f d−1 ,f d ), that contain a
singularity other than O:

Lemma 5. Let f d−1 and f d be as in Definition 2. The line L b contains a singular
point other than O if and only if Z(f d−1 ) is nonsingular at b and the intersection
multiplicity I b (f d−1 ,f d ) > 1.
Proof. Let b =(b 1 : b 2 : b 3 ) and assume that (b 0 : b 1 : b 2 : b 3 ) is a singu-
lar point of Z(F). Then, by Lemma 1, f d−1 (b 1 ,b 2 ,b 3 )= f d (b 1 ,b 2 ,b 3 )=0 and
b 0 ∇f d−1 (b 1 ,b 2 ,b 3 )+ ∇f d (b 1 ,b 2 ,b 3 )=0, which implies I b (f d−1 ,f d ) > 1. Fur-
thermore, if f d−1 is singular at b, then the gradient ∇f d−1 (b 1 ,b 2 ,b 3 )=0,so f d ,
too, is singular at b, contrary to our assumptions.
Now assume that Z(f d−1 ) is nonsingular at b =(b 1 : b 2 : b 3 ) and the intersection
multiplicity I b (f d−1 ,f d ) > 1. The second assumption implies f d−1 (b 1 ,b 2 ,b 3 )=
f d (b 1 ,b 2 ,b 3 )=0 and s∇f d−1 (b 1 ,b 2 ,b 3 )= t∇f d (b 1 ,b 2 ,b 3 ) for some (s, t) =
(0, 0). Since Z(f d−1 ) is nonsingular at b, we know that ∇f d−1 (b 1 ,b 2 ,b 3 ) =0,so
t =0.Now (−s/t : b 1 : b 2 : b 3 ) =(1:0:0:0) is a singular point of Z(F) on the
line L b .