Let be a non-singular proper curve and let be a line bundle on . Let . Then the induced morphism is finite by this theoremthis theorem. Take the affine opens and (which are finite). Consider exact sequence

The modules could potentially be quite large, since are finite algebras over the polynomial ring , and so could be infinite-dimensional vector spaces over . However, surprisingly, and is independent of “choices”. This is the first cohomology group!

In general, let be a scheme and let

be an exact sequence of quasi-coherent sheaves of -modules. Then this leads to

a long exact sequence of -modules.

If is proper over a field , then the Betti numbers

if is coherent. In this case, we can consider the following:

If is a proper curve then

In particular, .

To see this, we have

because . This is because if then it induces a map . We have (by thisthis)

So the image of in is a proper closed subscheme of : these can only be finite sets of points because is not proper over (for other wise would be an open and closed embedding). But because is connected (it’s irreducible), then the image has to be connected, so it’s a single point. Because is reduced, its image is reduced, so is constant.

Let be irreducible with degree (planar curve). Let (a possibly singular projective). Then (degree-genus formula)

So it is easy to produce curves with arbitrarily large arithmetic genus. We can calculate this by using the defining exact sequence of :

Then use the fact that

If is a non-singular proper curve, then there exists a divisor (called the canonical divisor such that

for all . Actually, , the cotangent sheaf on . We let common value (recalling Example 1Example 1)

be the genus of .

Let be a non-singular proper curve, a divisor on , and the canonical divisor. Then

With the same assumptions as aboveabove,

With the same assumptions as aboveabove,

Let in Theorem (Riemann-Roch)Theorem (Riemann-Roch). Then

In particular, on , since .

If then . In particular, if and then .

So if is big enough, we know exactly how many sections has.

Refer to this lemmathis lemma from last time.

By Serre dualitySerre duality, the LHS is and the RHS is . Let be a closed point. Then the defining exact sequence for the residue field is

Tensoring with the line bundle (i.e., applying which is exact because is flat), we have

Therefore

Equivalently,

Now write where . Then

Let be non-singular. Then if and only if .

Let . Then by this lemmathis lemma,

So . The constants form a subspace of dimension , and there must exist a non-constant with . Therefore has . So . But we certainly have . So is birational, and so they’re isomorphic.

Let be non-singular of degree (c.f. Example 2Example 2). Then , and by the above. These are “elliptic curves”, which are genus curves plus a point.

Let be an elliptic curve, where is a closed point. The canonical divisor has and . Therefore by this lemmathis lemma. We can thus define an injective map

Specifically, the map is

This map is injective by this corollarythis corollary.

This is bijective. That is, the closed points of form an abelian group, with the element given by .

We prove surjectivity. Let be given, i.e., with . Then by Theorem (Riemann-Roch)Theorem (Riemann-Roch),

So . Therefore there exists a unique (up to scaling) such that

where and are closed points. But taking degrees, we have

so there exists a unique with and . So

that is, by this propositionthis proposition.

Let be a non-singular proper curve with genus . Then there exists a finite morphism with degree .

Choose . By Corollary (Riemann’s inequality)Corollary (Riemann’s inequality),

So there exists a non-constant such that . Take . Now

So .

If , we can actually do better:

Hence, there exists a non-constant such that , and this time

So the degree of is . Curves admitting a degree map to are called hyperelliptic.