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Algebraic Geometry Lecture 7
- Uniformisers
A few months ago in Dan’s lecture we defined the order function that tells you whether a function has a zero or pole at a prime divisor (irreducible subvariety of codimension 1), and of what order. Given a prime divisor P of our variety and a rational function f ∈ k(V ) we defined
ordP (f ) =
n > 0 if f has a zero of order n at P , 0 if f 6 = 0, ∞ at P , −n < 0 if f has a pole of order n at P.
Theorem 1. The function ordP : k(V ) → Z ∪ {∞} is surjective.
Defn. Given a prime divisor P of V (k) we say a rational function t ∈ k(C) is a uniformiser of P if ordP (t) = 1.
Uniformisers exist by the theorem, since the ordP map is surjective there must be a rational function of order 1. The theorem does not ensure uniqueness though and in general there are many choices we can take. Geometrically in the case of curves it is safe to think of uniformisers as being lines through the point P that are not tangential to the curve at that point.
Example Take the variety V : y^2 − x^3 − 3 x^2 − 2 x = 0 over Q.
Consider the two points P 1 = (1,
- and P 2 = (− 2 , 0) in V (Q). For a uniformiser at P 1 we may take t 1 : x − 1. 1
This has a simple zero at P 1 so it is a uniformiser there.
At P 2 we may be tempted to try t 2 : x + 2.
This does have a zero at the point, so ordP 2 (t 2 ) > 1, but we should note that
y^2 − x^3 − 3 x^2 − 2 x = 0 ⇒ x(x + 1)(x + 2) = y^2
⇒ x + 2 =
y^2 x(x + 1)
So
ordP 2 (x + 2) = ordP 2
y^2 x(x + 1)
= ordP 2 (y^2 ) − ordP 2 (x) − ordP 2 (x + 1) = 2 − 0 − 0 = 2.
So t 2 is not a uniformiser. But if we use t′ 2 : y then we find that this rational function has a simple zero at P 2 so is a uniformiser there.
- Divisors
Recall from Dan’s lecture that, given a variety V over a field k, a divisor is just a formal sum of irreducible subvarieties of codimension 1 in V (k) with coefficients in Z. So in particular if V is a curve then divisors will be formal sums of points. The group of divisors is denoted Div(V ).
Example If V : y^2 − x^3 − 3 x^2 − 2 x = 0 over Q then
(− 2 , 0) + 4(1,
- ∈ Div(V ).
If we go through each subvariety in a divisor and replace the coordinates with their conjugates then we would expect to get a different divisor. If we in fact get the same divisor then the divisor is called k-rational. The subgroup of k-rational divisors is denoted Div(V ).
Example Suppose that
(2, 0 , 1) + 4(1,
- ∈ Div(V )
for some variety V. Then if we apply the embedding
7 then we get (2, 0 , 1) + 4(1, −
But this is our original point, so this point is Q-rational.
Recall: Using the order function we may associate a divisor to each rational function f by div(f ) =
P ⊂V (k) P prime divisor
ordP (f )P.
These divisors are called principal divisors.
Using this product we can define the space of r-dimensional differentials, ΩrV as the space of products dx 1 ∧ dx 2 ∧ · · · ∧ dxr
where dxi are 1-dimensional differentials. If V has dimension n then ΩrV has di-
mension
n r
over k(V ).
If we have an n-dimensional differential ω on an n-dimensional variety then we can define its order and a corresponding divisor. It hinges on the fact that the
dimension of ΩnV will be
n n
= 1, so we can write ω in the form
ω = f du 1 ∧ · · · ∧ dun
for some f ∈ k(V ). We then define
ordP (ω) = ordP (f ).
This is well defined despite the apparent ambiguity in choosing a basis vector. Using this notion we can go on to define the divisor of a differential as
div(ω) =
Prime divisors P
ordP (ω)P.
These divisors are called the canonical divisors. It turns out that these divisors are invariant under conjugation, so they are in Div(V ). Moreover, the divisors of any two differentials differ by a principal divisor. Recall that we defined the divisor class group to be the group of k-rational divisors modulo the principal divisors. So all the canonical divisors lie in the same equivalence class of the divisor class group, known as the canonical class.
Defn. We say a differential ω ∈ ΩV is:
(1) regular (or holomorphic) if ordP (ω) > 0 for every prime divisor P ; (2) non-vanishing if ordP (ω) 6 0 for every prime divisor P.
In the case of curves things simplify somewhat and any one-dimensional differ- ential ω can be written in the form
ω = f dt
for a rational function f and a uniformiser t.
Example Calculate ordP (dx) at P = (− 2 , 0) for
C : y^2 − x^3 − 3 x^2 − 2 x = 0.
Recall we found that y was a uniformiser for C at P. So if we can write dx in the form dx = f dy then ordP (dx) = ordP (f ). But:
x^3 + 3x^2 + 2x = y^2 (3x^2 + 6x + 2)dx = 2ydy
dx =
2 y 3 x^2 + 6x + 2
dy.
So
ordP (dx) = ordP
2 y 3 x^2 + 6x + 2
= ordP (2) + ordP (y) − ordP (3x^2 + 6x + 2) = 0 + 1 − 0 = 1.
