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Algebraic Geometry Lecture 16 – Geometry of Surfaces
Mike Harvey
Recap (Divisors).
A prime divisor of a variety X is an irreducible, closed subvariety of codimension one. A divisor is a finite formal sum of prime divisors:
D =
Prime divisors P
nP P, nP ∈ Z.
Let f ∈ k(X) be a rational function, then
Div(f ) =
Prime divisors P
ordP (f )P
where
ordP (f ) =
d if f has a zero of order d on P, −d if f has a pole of order d on P, 0 otherwise.
A principal divisor is a divisor D such that D = Div(f ) for some f ∈ k(X).
A canonical divisor of X is
D = Div(ω) =
Prime divisors P
ordP (ω)P
for a differential ω, where ordP (ω) = ordP (f ) when ω = f dt 1 ∧... ∧ dtn.
Two divisors are called linearly equivalent if their difference is a principal divisor:
D ∼ D′^ ⇔ D − D′^ = Div(f ) for some f ∈ k(X).
We define
Pic(X) = Cl(X) =
Div(X) PDiv(X)
where
Div(X) = Group of divisors PDiv(X) = Group of principal divisors.
1
2
Surfaces.
Proposition 1. Let X ⊂ P^3 be a surface, then any two plane sections (intersections of planes with X) are linearly equivalent, this gives the “hyperplane class” H in Pic(X).
Proof. The hyperplane sections will be of the form
D 1 = X ∩ {1 = 0} D 2 = X ∩ { 2 = 0}
for linear equations 1 , 2. Then
D 1 − D 2 = Div
` 1
` 2
Intersection Numbers.
Let D, D′^ be two prime divisors on a surface X that intersect transversely:
× rather than ⊃⊂
Then we define the intersection number of D and D′^ to be
D.D′^ := #{D ∩ D′}.
Properties
- Respects linear equivalence: if C, D, D′^ are divisors and D ∼ D′^ then D.C = D′.C;
- Symmetric: if C, D are divisors then D.C = C.D;
- Bilinear: If C, D 1 , D 2 are divisors then (D 1 + D 2 ).C = D 1 .C + D 2 .C.
“Example” We denote D.D by D^2 , but what is D^2? Find a divisor D′^ ∼ D such that D′^6 = D, then we define D^2 = D′.D.
Example Let X = P^2 , so Pic(X) ∼= Z. Let h be a generator of Pic(X), so h is a “line-class”. Any two lines are linearly equivalent and two distinct lines meet in one point, so h^2 = 1. Now let C, D be curves of degree m, n respectively. So
C ∼ mh D ∼ nh.
We then have
C.D = mh.nh = (mn)h^2 = mn.
