2024-10-03
Last time
We considered the surjection
for an ideal
We also calculated that
It follows that to show
we just need that
Definition (Homogeneous ideals)
An ideal
if and only if the th graded component for all . is generated by homogeneous elements.
Examples 1
- Let
be a graded ring homomorphism, i.e., . Then is homogeneous. - (Homogenisation) Consider the ideal
. The ideal is not homogeneous, while is graded by total degree. But let’s think of these as not being graded: is certainly not graded. We now add an extra variable and map , . Then we get . Clearing denominators, we find , which is homogeneous! It turns out we can even recover what we started with. To do this, we localise at : . Here, is a -graded ring, and we pick off the degree- part. In fact, , and for our ideal, , which is what we started with. - (Prime ideals) A homogeneous ideal
is prime if and only if for all homogeneous, implies or . (We only have to check this for homogeneous elements.) - If
is a homogeneous ideal, then where . The quotient is again graded with .
Definition (Proj)
We define the set
(Recall that
Examples 2
. We know thatThe homogeneous ideals are just and , and contains . So . We have “cut down” the dimension by . is meant to be “lines through the origin”. . We haveThe homogeneous ones are , , and where is homogeneous and irreducible in . But if we have a homogeneous polynomial in two variables over , it must factor: we can turn into a polynomial in one variable by dividing out by a power of one of the variables. More precisely, we havefor some . Now factors as over , and therefore so does :By irreducibility of , for not both zero. SoThe ideals are precisely the lines with , .
Definition (Vanishing locus)
If
These define the closed subsets for a topology on
Lemma
A base of opens for a topology on
Key construction
Let
where
This means that
Another property
We also have (actual equalities)
Example 3
Let
These are the charts that appeared when we did projective space. Also, for gluing, we have
Lemma (Distinguished opens)
Let
Moreover, if
where
Remark
The purpose of this lemma is to show that the map
Lemma
Let
Structure sheaf on
Let
where
Theorem
Let
Quasi-coherent sheaves on Proj
Let
with an action of
Examples 4
- If
and is a graded -module, then we can make a “twist”: - So we have the graded
-modules for all .
Quasi-coherent sheaf of modules
Just as in the affine case, we can define for each
Examples 5
Let
The converse doesn’t always hold. It holds for projective space.
Definition (Serre twisting sheaf)
Let
Example 6
If
Actually,
Another thing
We have an isomorphism
However, this is not an isomorphism in general.