Let be a non-singular curve. We had

and we defined the pullback for a (non-constant) map to be given by

where is the uniformiser of the DVR , and is the induced map on stalks.

If is finite, then

We are going to need the following facts.

If is a finitely-generated and torsion free module over a DVR , then is free of rank

This follows because DVRs are PIDs: torsion-free modules over a PID are free.

If is a finite -algebra (finite-dimensional over ), then is finite (as a set), discrete topologised, and is an isomorphism.

Since is finite, then is a coherent -module. If is closed then . Let be the inclusion of the generic point. Then , and because pushforwards are left exact, we can localise at to get the inclusion in the bottom row:

where we get back because being finite implies it is non-constant, and so maps a generic point to a generic point (herehere). The three maps being injective means that the blue map is also injective. We find that is torsion free. Since it’s a -module and finitely-generated (coherent module), it is a free -module of rank (we can just check the generic rank).

We just need to check the formula on a single point , where . We need to show

Another way to calculate this: we have by Fact 1Fact 1 that

Now observe the diagram

But now, lets take an affine open such that . Because is finite, is still affine. So we have

So calculating the fibre over is the same as calculating the fibre over .

Continuing the calculation, we have

where the last equality follows by the fact that is a DVR. But now, because is a finite-dimensional -algebra. So by Fact 2Fact 2, this breaks up into a product

where these correspond precisely to such that . This identity tells us that

where the last equality follows because is precisely the valuation: is an ideal and so is generated by some power of , which is the valuation, since has a filtration whose length is the same as the size of the valuation.

Let be a finite map of non-singular curves. Then is an is an isomorphism of schemes if and only if .

If then

Since , then , and therefore for some unique . In particular, is injective on points in . Since is finite, and therefore universally closed and surjective, this implies that is a homeomorphism.

Also, is a free -module of rank , and the map is an isomorphism after quotienting by . Therefore we have an isomorphism of sheaves .

Let be a non-singular curve. Let . Then we can define an associated principal divisor .

Let . Then there exists only finitely many such that .

Let be an affine open. Then the complement is a closed subset not containing the generic point of , and therefore is finite () and we can disregard it. In other words, we may assume that . Therefore for and . Now, are also finite, so we can discard these too. We thus further replace with . Therefore and for all closed points .

Therefore we can define the homomorphism (well-defined by Lemma 2Lemma 2)

We let the principal divisors be . Note that for every , we can write

We call the zeroes of (the sum over those with ), and the poles of (the sum over those with ).

Let and a primitive cube root of . Let . We consider the function

So

where the last line is in coordinates (the closed points).

Consider now

Then the same function aboveabove gives us a rational function on :

Therefore using our calculation from before, we have the restriction to

where the last line is in homogeneous coordinates. But this isn’t the whole divisor: we are missing the point at . What happens at ? We can map . Then

The point at corresponds to the prime ideal . We find that . So

Define the divisor class group to be the quotient

We say that are linearly equivalent if they agree in . That is, if and only if for some .

If is finite, then (since sends principal divisors to principal divisors) there is an induced pullback map .

If then . We have and therefore

is surjective.

Let . Let , and write for the inclusion. Then the sequence

is exact.

Consider the diagram

The top row is exact: we know that the kernel of the restriction map is generated by . So we find the following diagram

where the map is an isomorphism because a rational function on extends to a rational function on . We thus conclude by the Snake Lemma that the blue map is surjective.

We will soon prove that

To show this, we consider the following diagram:

by Example $ Cl( A_k 1) = 0$Example: . This tells us that is surjective. It remains to show that the map is injective. To do this, we will show that the degree map factors through , giving us .