From a graded ring , we defined . We also defined , the Serre twisting sheaf. (These need not be line bundles.)

We always have -module homomorphisms

Hence, by functoriality, we have maps

This is a -graded ring. There is a map of -graded rings:

Let , with grading by total degree. Then

is an isomorphism of graded rings. Let us consider a basic case: when , we have on the degree component

Write . By the sheaf condition, we have the exact sequence

But what are , ? They are (recalling this examplethis example)

Therefore corresponds to

For , this puts a restriction on the indices in the sums above: we need and . Therefore except and except . So

We have maps

This gives the -graded -module

Let be of homogeneous degree , and . Then

Moreover, if is quasi-coherent on , i.e., is a quasi-coherent -module and , then

In particular if is generated by and (uniformly bounded), e.g. if is quasi-compact, then is a line bundle for all and

for all .

Since is an affine scheme, it suffices to prove that as (ungraded, since everything is degree ) -modules. If , we can consider

Localising, becomes a unit and is an isomorphism. Taking degree components, we have . For the case , we map the other way to obtain the same result.

If is quasi-compact, then we have that

is an adjoint pair. Moreover,

. Recall that our definition of is that a morphism of -schemes is the same as a line bundle on and a surjection .

We show that satisfies the same universal property.

Corresponding to our global sections , we get a quotient map

So there is a map (by the universal property of ). In fact, because of this, any map will have a line bundle with sections, obtained via .

Conversely, let be an -scheme together with a surjection for a line bundle on . This corresponds to such that , where

open. This means that the map

is an isomorphism for each (a nowhere-vanishing section of a line bundle). We now try to map to our natural cover of .

Since is affine, a morphism is equivalent to a ring homomorphism

(By .) By the universal property of polynomial rings and quotients, this is equivalent to picking some elements , , such that . Now, if there is to be an isomorphism and (defined above), , then there must be a commutative diagram

Following this diagram around, we have

Therefore, must be the image of in under the homomorphism

It follows that is completely determined by the .

Now, suppose that are two morphisms of -schemes such that there are isomorphisms and with for all .

It suffices to check this on each . Now,

Similarly for . So it follows follows that . By the above calculations, we see that for all because any map on the opens are completely determined by the . So .

For existence of such an , for each we define a morphism

exactly as determined above: we take to be the appropriate image of . We need to show that they glue:

But a map is equivalent to a ring homomorphism . But the composition

sends for all . So the remaining question is whether in . But this boils down to the equation

which is indeed true in .