Let be a non-singular curve. Let . Then there is an induced map

such that the map on function fields is

(This is the converse of this examplethis example.)

To do this, we view as

So we just need to define maps on each open:

These are both regular functions tautologically: has no poles in and has no poles in . These glue!

There is another way to define the above map. Pick an affine open , whence and we obtain a map

This restricts to a map from the generic point corresponding to the map , above. Now for a closed point , we look at the local ring and obtain the following diagram:

Thus, since is proper, there exists a unique map by the evaluative criteria for properness. This map actually extends to an affine neighbourhood of . Now recalling that we have , the complement of consists of finitely many points. Because is separated, if these maps agree on the intersection then they must agree everywhere: they glue to give a unique .

Let be a non-constant map of non-singular proper curves. Then is finite.

First, is proper. If is a closed point then is a closed subset of . Since has Krull dimension , the closed subsets of are finite sets of closed points or the whole of . By assumption, is non-constant, so is finite. Zariski’s Main Theorem says that a proper morphism with finite fibres is finite.

Let be a non-singular curve and . Let and in . Then

In particular, if is proper, then (using Theorem 1Theorem 1).

We have by definition

and

Now if then we also have by definitionby definition

If then is a unit in , so it is sent to a unit in . This implies that . Also note that , and so if then

as . This gives the first claim.

If is proper then is finite by TheoremTheorem. Now we may use the degree formula for pullbacksthe degree formula for pullbacks

If is proper and non-singular, then there is an induced degree map

which is surjective.

By Proposition 2Proposition 2, .

. Observe the following diagram:

Let be a non-singular projective curve. Then if and only if there exist distinct with , i.e., for some . In particular, on , we can take . Then . So .

Conversely, given , , and such that , then

is finite and . Therefore

But , which implies that , and is birational by this remarkthis remark.

The degree map

is an isomorphism if and only if . To see this, if is injective, then given distinct , in , we have which implies that by assumption.

Let be a divisor on a non-singular curve. We define in the following way. Let be an open set. Then

This is a sheaf of quasi-coherent -modules. Also,

where is the inclusion of the generic point. If , then .

Let

Now take local equations (uniformisers) that generated the maximal ideals . As we only have finitely many (closed) points, we can assume that if (by deleting the points we don’t need). We take the open

and therefore and cover . Now on each , we have an isomorphism

for if is open, then the restriction of to , , and

On , we have the isomorphism

We consider and as subsheaves . We have local equations and on a cover of . But now, will have local equations . Therefore will have local equations , which are local equations for .

It suffices to prove that if and only if , or equivalently, for some . If then there is a map

This map is an isomorphism because it is a map of line bundles, and is surjective. Conversely, if

then the image of is such that . If we take the dual of this isomorphism:

we get , i.e., . So , and therefore .

There is an isomorphism

The above Proposition 6Proposition 6 showed that is an injective group homomorphism. What about surjectivity? Let be a line bundle such that

( is a -dimensional vector space over .) Let be a trivialising cover for . Let . Then we get the compositions

where are the images of under .

Claim: If , then .
Proof. Consider the commutative diagram

We have the isomorphism on top . Following the maps around, we find that

and is a unit, so it doesn’t “interfere” with the valuation. □

Now, the following divisor is well-defined (the valuations agree on the intersections, so the divisors defined on each “glue”):

Then

Every line bundle on is of the form for some , i.e., there is an isomorphism

Let be a line bundle on . What is ? What about when ? We already know that (from herehere)

Let be a proper and non-singular curve. Let be a divisor.

If then . Therefore (recalling Proposition 2Proposition 2)

If further , then implies that is linearly equivalent to an effective divisor (take ) of . But an effective divisor is a non-negative sum of points. So this can only happen if it is a sum of zero points, i.e., .