prime spectrum

Definition

Let be a commutative ring. Its prime spectrum is the set of prime ideals .

Relation to maximal spectrum

For a ring , if an ideal is maximal then it is prime, therefore the maximal spectrum is contained in the prime spectrum

Bijection of sets of and ideals of .

Given a subset , let

Note that if is the ideal generated by then

so we might as well only think about ideals of .

Let be a subset. Define

Then the following properties hold:

1. is a radical ideal.

2. .

3. (the closure in the Zariski topology (described below))

4. The maps

are mutually injective bijections.

Proof

todo Lecture 13.

Topology on .

Thus, the sets

form the closed sets of a topology on . It is analogous to (and called) the Zariski topology.

The sets

form an open topological basis for the Zariski topology on .

Closed points versus fat points

Let be a point. By the properties above,

Note that if is maximal, then , that is points are closed. If is not maximal, then this will be bigger.

So contains points that aren’t closed.

Also, we can take the point . This point is not closed:

So the point is dense in .

as a functor

defines a contravariant functor between the categories of commutative rings and topological spaces.

Proof

First, we must show that

is a well-defined continuous map.

Subproof: As is clearly an ideal, we need to check that is prime. Let , by definition . Since is prime, then

so is prime, and defines a set map.

Next, we need to check that it is continuous. We can look at the closed sets for ideals . We aim to show that the preimage of is closed.

is contained in if and only if . This gives

ans is the ideal generated by so the inverse image is closed and is continuous.

Next, we need to check composition. That is if , then

So is a functor.

Structure sheaf

We can construct a structure sheaf on (for this section denote for simplicity) such that we can think of can be thought of as continuous functions on . We define this on the open basis as follows

where denotes the localization of at the multiplicative subset .

Statement of theorem

There is a unique sheaf of rings on such that with restriction maps

being the localization of first, then (note this is canonical).

Well-defined

Let , then we must show that .

We know that for this open basis if and only if for some and . This is the same as saying that is a unit in since

as is a unit in .

In this case, by the universal property of localization of a ring there exists a map

such that the following diagram commutes

So if we have the case that then we get the following diagram

So when we get the case that , so using the chain above we get the unique isomorphism

Satisfies the sheaf condition

Suppose then it remains to check that

by the sheaf property on a topological basis.

We can show this by proving the following 2 facts:

1. If with for all , then
2. If satisfy , then there exists such that .

These two conditions are equivalent to saying that the homomorphism

that sends the element from to the various restrictions is both injective and surjective.

Proof

Since the opens set is compact, we may assume the union of is finite. We can consider the space where

Since , we have and we can write

Relation to partition of unity

This is like an algebraic geometry version of a partition of unity. We can see this as it as addition of things that are locally “supported” with “functions” that are multiplied to get 1.

First, we start by proving 1. Say is an element such that

we aim to show first that .

In the localization means that (because of the defining quotient by the equivalence relationship). Let . This gives for all . Therefore,

Next, for 2. if such that for all then we aim to prove that there exists such that .

For each we can write

since they are equivalent in the localization at .

By the assumption on we have that the images under the localizations are equal, that is

since . This means that

clearing fractions gives

Replace by and by so that

so the equation above simplifies to become . Now we define the element to be

where is the , -dependent elements chosen in the proof for part 1. Then

so

So the construction is a sheaf.