2024-10-01

Theorem (Evaluation criterion for separatedness)

Let be a quasi-separated morphism. Then is separated if and only if every diagram

has at most completion where is a valuation ring.

If we can take a limit, it is unique.

Proof

Suppose that there are two completions . Consider the diagram

Because is quasi-separated, is quasi-compact (Corollary 3). By assumption, , and therefore there exists a lift . By the evaluative criterion for universal closedness, is universally closed. But is an immersion, and therefore it is a closed immersion. So is separated.

Definition (Proper)

We say that is proper if it is finite-type, separated, and universally closed.

Examples 1

  1. Closed and finite morphisms are proper.
  2. Proper morphisms are stable under base change and composition.
  3. (In the problem sheets) In the following commutative diagram, if is proper and is separated

then this implies that is proper. There is also a “converse”.

Analogue: Compact topological space mapping into a Hausdorff topological space, then the image is compact.

Theorem (Evaluative criterion for properness)

Let be a finite-type and quasi-separated morphism. Then is proper if and only if there exists a unique completion

where is a valuation ring.

If is locally Noetherian, then every finite-type map is quasi-separated.

Combination of evaluative criterion for universal closedness and properness: universal closedness gives existence and properness gives uniqueness.

Proposition (Fundamental theorem of elimination theory)

Let be a ring and a projective -scheme. Then is proper.

Proof

By definition of projective, we know that there exists a closed immersion :

Because closed immersions are proper (they are finite-type, separated, and universally closed), it suffices to show that is proper. But is just the base change of :

is a fibre diagram. Hence, it suffices to show that is proper (since properness is stable under base change). We have seen that

The are all finite-type (finite-presentation even) over , and therefore is finite-type (finite-presentation) over and so is quasi-separated as well. We now just need to apply the evaluative criterion: we need to find a unique completion

A point corresponds to a quotient , i.e., . We need to complete this uniquely to

This is done in the problem sheets.

Called the “Fundamental theorem of elimination theory”.

Chow’s Lemma

If is proper then there exists an such that there exists where is birational:

Proj

was a way to take a ring and spit out a scheme: a “global” construction. is a way to take a graded ring and spit out (usually projective) schemes, also a “global” construction.

Definition (Graded ring)

is an -graded ring if:

  1. is a ring.
  2. are -modules.
  3. We have maps (-linear) such that makes an -algebra.

The irrelevant ideal is defined to be

We say that is homogeneous of degree if . Note that can be taken to be any value. If , then we can write where is the th graded piece/component.

Can change out the grading to be -graded or in general by some commutative monoid.

Examples 2

  1. and . We can give the different degrees.
  2. Given and , we can set , i.e., .
  3. is Noetherian if and only if is Noetherian and is a finitely-generated -algebra.
  4. Definition (Reese algebra). Let be a ring and an ideal. Then we define the Reese algebra of to be the graded ring .
  5. Let be a ring and an -module. We consider the symmetric algebra (this is , where is the tensor algebra). For an -algebra , we have
    It is the left adjoint to the forgetful functor . For example. we have (each indeterminate corresponds to a basis element of ). The adjunction also gives us a way to compute the symmetric algebra. First, we choose a presentation
    Since is the cokernel of , we can think of this as a coequaliser diagram

where is the coequaliser . Now since is a left adjoint, it preserves all colimits:

We can calculate the coequaliser here: it’s the quotient by the difference. We have and . Let be the standard basis for and be the standard basis for . Then we can write

Therefore

  1. We want to calculate for the ideal . We have a presentation

Since , we have where we think of as as . Therefore,

This calculation can be extended to polynomials in any number of variables.
7. How do we calculate the Reese algebra? We have a surjection . To see this, consider the diagram

and the bottom row is surjective. Let us calculate the kernel:

If then we have . Actually, because is a unit in . Therefore

i.e., we have an isomorphism on the localisations. Therefore, for all , which implies that is -torsion (every element of is annihilated by some power of for all ). Now, if , then is finitely-generated (as a graded subring of ), and so there exists such that for all .

In the case that is a Noetherian domain, is the -torsion ideal of . Therefore

if and only if is a domain.