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g/h " Fq(C). The divisor of f is defined to be div(f) := P -
Q, where P is the intersection divisor C )" Cg and Q is the
intersection divisor C )" Ch.
Let f = g/h be a rational function on C. Then intuitively, the
points where C and the curve defined by g intersect are the zeros
of f and the points where C and the curve defined by h intersect
are the poles of f, so we think of div(f) as being the zeros of f
minus the poles of f . Since deg(C )" Cg) = deg(C )" Ch) = de, we
have deg div(f) = 0. Intuitively, f has the same number of zeros
34 5. Points, Functions, and Divisors on Curves
as poles. Note that if P appears in both C )" Cg and C )" Ch, then
some cancellation will occur. In particular, P is only considered to
be a zero (resp., pole) of f if after the cancellation, P still appears in
div(f) with positive (resp., negative) coefficient. Notice also that the
divisor of a constant function f " Fq " Fq(C) is just 0.
Since rational functions are actually equivalence classes, we need
to be sure that our definition of div(f) is independent of the choice
of representative for the equivalence class of f. It is, but the proof
is messy. Instead, we ll just illustrate this in our example. On our
2
curve C0 over F3 defined by Y Z - X3 - 2XZ2 - 2Z3 = 0, we need
to compute the intersection divisor of C0 with the curves defined by
2
each of the following equations: X2 = 0, Z2 = 0, Y + Z2 + XZ = 0,
and XZ = 0. Any point (X0 : Y0 : Z0) of intersection between the
2 2
line X = 0 and the curve C0 must satisfy X0 = 0 and Z0(Y0 - 2Z0 ).
Writing F9 = F3[t]/(t2 + 1) and letting denote the element of F9
corresponding to t, we have that 2 = -1 = 2, so the polynomial
2
(Y0 - 2Z0)2 factors as (Y - Z)(Y + Z). This means that our point
(X0 : Y0 : Z0) must satisfy X0 = 0 and one of the following three
conditions: Z0 = 0, Y0 = Z0, or Y0 = 2Z0. Thus our three points
of intersection in P2(F9) are P ", (0 : : 1) and (0 : 2 : 1). Since
{(0 : : 1), (0 : 2 : 1)} is our point Q1 from before, we have that the
intersection divisor of the line X = 0 with C0 is P " + Q1. Therefore,
the intersection divisor of the double line X2 = 0 and the curve
C0 is 2P " + 2Q1. Notice that this divisor does indeed have degree
6 = 2 3.
Exercise 5.11. Show that the intersection divisor of C0 with the
curve defined by Z2 = 0 is 6P ". Show that the intersection divisor
of C0 with the curve defined by XZ = 0 is 4P " + Q1.
2
The intersection of C0 with the curve defined by Y +Z2+XZ = 0
is a little trickier to compute since this latter curve is not just the
union of two lines. However, the only point at infinity on the latter
curve is (1 : 0 : 0) and the only point at infinity on C0 is P " = (0 :
1 : 0), so the two curves do not intersect at infinity. Thus we may
assume Z = 0, divide through by Z2, and set x = X/Z, y = Y/Z to
get the affine portion of C0 defined by y2 - x3 - 2x - 2 = 0 and the
other curve defined by y2 + 1 + x = 0. We still don t have a product
5. Points, Functions, and Divisors on Curves 35
of two lines, but we can write x = -(1+y2) from the second equation
and substitute that in. We have 0 = y2 + (1 + y2)3 + 2(1 + y2) + 2 =
y6 + 1 = (y2 + 1)3 = (y - )3(y + )3. Thus these two curves intersect
with multiplicity 3 at Q1, so the intersection divisor is 3Q1.
Putting the results of the last two paragraphs and the exercise in
between them together, we have div(X2/Z2) = (2P " +2Q1)-6P " =
2
2Q1 - 4P " and div((Y + XZ + Z2)/XZ) = 3Q1 - (4P " + Q1) =
2Q1 - 4P ", so the two divisors do indeed agree.
Now that we know what divisors, rational functions, and divisors
of rational functions are, we are ready for our next definition.
Definition 5.12. Let D be a divisor on the nonsingular projective
plane curve C defined over the field Fq. Then the space of rational
functions associated to D is
L(D) := {f " Fq(C) | div(f) + D e" 0} *" {0}.
A few comments are in order. First, it s easy to see that L(D) is a
vector space over Fq. In fact, it s finite dimensional, but this is harder.
By collecting positive and negative coefficients appearing in the divi-
sor D, we can write D = Dpos -Dneg, where Dpos and Dneg are effec-
tive divisors. Also, we can write div(f) as a difference of two effective
divisors by saying div(f) = (zeros of f) - (poles of f) . Therefore,
we have div(f)+D = (Dpos - (poles of f) )+( (zeros of f) -Dneg).
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