SOLUTIONS PROBLEM SHEET 3
Exercise 1
(i) Let p1 , p2 , p3 , p4 distinct points of P2 such that no 3 are collinear. In particular p1 , p2 , p3 are
not collinear, therefore they are linearly independent. Changing the coordinates we can assume that
p1 = (1, 0, 0), p2 = (0, 1, 0), p3 = (0, 0, 1).
Let p4 = (a, b, c). Assume that a = 0. Then p4 = bp2 + cp3 . We obtain a contradiction because the
points are linearly independent. Therefore a 6= 0. Similarly b, c 6= 0. Making the change of coordinates
1
 
0 0
x
a
0 0 0
1



y ,
(x , y , z ) = 0 b 0
1
z
0 0 c
we obtain that p4 = (1, 1, 1). This change will leave p1 , p2 , p3 invariant because they belong to P2 .
(they are invariant under the multiplication with a nonzero real number)
(ii) Let C : a20 X 2 + a02 Y 2 + a11 XY + b10 X + b01 Y + c = 0. Writing the equation in the coordinates
from P2 we obtain
C : a20 X 2 + a02 Y 2 + a11 XY + b10 XZ + b01 Y Z + c1 Z 2 = 0
Assume that p1 , . . . , p4 , (a, b, c) ∈ C. We obtain that
a20 = a02 = c1 = 0,
a20 + a02 + a11 + b10 + b01 + c1 = 0,
2
a20 a + a02 b2 + a11 ab + b10 ac + b01 bc + c1 c2 = 0.
We obtain that
C : a11 XY + b10 XZ + b01 Y Z = 0,
where the coefficients a11 , b10 , b01 satisfy
a11 + b10 + b01 = 0,
a11 ab + b10 ac + b01 bc = 0.
Exercise 2
X
Let C = {y 2 = x3 } and P (X, Y ) =
am,n X m Y n . We observe that for n even, X m Y n = X k
m,n6=0
s
3
mod (X 3 − Y 2 ), where k ∈ N or, for n odd X m Y n = X
− Y 2 ), where s ∈ N. Therefore
XY mod (X X
it is enough to solve the problem when P (X, Y ) =
am,0 X m +
am,1 X m Y . Now we use the
m≥1
m≥1
parameterization X = t2 , Y = t3 . It follows that
X
P (X, Y ) |C ≡ 0 ⇐⇒
am,0 t2m + am,1 t2m+3 ≡ 0.
m≥1
We obtain that am,0 = am,1 = 0 for all m ≥ 1. It follows that X 3 − Y 2 divides P (X, Y ).
1
2
SOLUTIONS PROBLEM SHEET 3
Exercise 3
We can assume that a = b = 0. Then p = 0.
Let L be a line that contains the origin and that is tangent at 0 to V . The line L can be parameterized like {t(x, y); t ∈ C}. We write the Taylor expansion
f (tx, ty) = f1 (tx, ty) + f2 (tx, ty),
where f1 depends on the gradient of f . Then the intersection of L with V at 0 has multiplicity at
least 2 if and only if f1 (tx, ty) = 0. Here f1 (x, y) = fx (0)x + fy (0)y.
(ii) The equation of the secant trough q 6= 0 is
Sq,0 : (fx (0) + αq ) x + fy (0) + (αq ) y = 0,
where αq , βq 7→ 0, for q 7→ 0.
Exercise 4
Let C : y 2 = x(x − 1)(x + 1).
(i) Let x =
X
,
Z
y=
Y
.
Z
e ⊂ P2
Then projectivization of C is a cubic C
e : Y 2 Z = X (X − Z) (X + Z) .
C
e is obtained for Z = 0. Therefore the point at infinity for C
e is
(ii) The point at infinity of C
{(X, Y, Z) : X = Z = 0}.