We defined

We defined the maps

The fibre over the origin is the exceptional divisor . In particular, is closed, and therefore is a closed subscheme of . We would like to compute its sheaf of ideals.

Let’s look at this over one chart , that is to say, in :

We conclude that in , is locally principal, i.e., effective Cartier.

Let be a ring and be an ideal. Recall the Rees algebra

We define the blow-up of in to be

The map is the structure map.

Let and . Then recall from 6.6. and 7.7. from these examplesthese examples that

Therefore .

If , then we have an isomorphism

If instead , then let us pick an element . Then since is a unit in , we have

Therefore we have an isomorphism :

Hence, if then .

Now consider the irrelevant ideal , which is by generated by the collection of all (in degree ). Therefore,

We have

Claim: In fact, we have an isomorphism (of graded -modules)

Proof. For injectivity, let us consider

Therefore for some . This means that in . For surjectivity, let . We can always factor in the following way:

Therefore is a non-zero-divisor and generates . □

We conclude from the above that we have the graded exact sequence (of modules)

Moreover, we have on the associated structure sheaves

We have proven above that is a line bundle, and therefore is effective Cartier. All squares are Cartesian in the following diagram:

where . We call the exceptional divisor.

In our example with and , we have

And then we take Proj. Note here that is the homogeneous ideal of elements of degree .

Let be an algebraically closed field. A curve over is an algebraic -variety of Krull dimension .

If is a curve, it is an integral scheme (reduced and irreducible). So there exists a unique generic point . We define the function field of to be

This is a field. This is because if is open and non-empty, then and actually , the generic point of (since is a domain). So . More generally, for any , (because is a domain). It is a fact that

Now if is a non-constant function, then is transcendental: for otherwise, would be algebraic over , but is assumed to be algebraically closed, which would force . In fact, the extension is finite and algebraic: this is because has transcendence degree over , which forces to be algebraic, and finiteness follows from the fact that is the function field of the finitely-generated -algebra .

If is a non-constant map of curves, then . To see this, because is a curve (Krull dimension ) it only has closed points and generic points. But now if and is closed (and not generic) then is closed and contains , which forces , i.e., is constant. This means that we have an extension

If then we have with ( cannot be in as is transcendental). Therefore a map a rational function, .

Consider and the map

induced by . Let . If this element gave us a map , then we would have a diagram

This is impossible as the composition corresponds to the open immersion , which means that is an immersion (a composition where is a closed immersion and is an open immersion), as the diagonal of is an immersion. But since is finite, it must be a closed immersion, which is impossible as is not surjective.

Let denote the set of closed points of . A divisor on a curve is an element of the free abelian group

So a divisor is a linear combination

We write or say that is effective if the . One can always write

for , effective.

There is a degree homomorphism

i.e., .

Let . Then since sends closed points to closed points, we get a morphism of abelian groups

and therefore .

For open, we also get

We will denote or . Note that is trivially split surjective, and the kernel is generated by the points (they are all closed points since must contain the generic point).

Let be a local domain. Then the following are equivalent:

We call satisfying any of the above equivalent conditions a discrete valuation ring.