# More Concepts in Algebraic Geometry

As before, *k* denotes an algebraically closed field.

Recall that we have a bijection between radical ideals of and closed subsets of .

The bijection reverses the inclusion so if and only if . Not too surprisingly, operations on ideals translate to operations on the corresponding closed subsets.

Proposition 1.Suppose closed subsets correspond to radical ideals .

- corresponds to .
- corresponds to .

**Proof**

For the first claim:

- Since for any
*j*, we have and thus . - Conversely, if , then any gives for some
*n*> 0 so is a finite sum of terms . Each such so and thus . We have shown: . - Finally, since is a radical ideal, we are done.

Second claim:

- First, we show .
- Since we have . Likewise and we have shown ⊆.
- Conversely, if lies outside
*S*and*T*then there exist and such that and thus . Since this gives . - Since intersection of radical ideals is radical, we have . And since from proposition 1 here, we are done. ♦

# Prime and Maximal Ideals

Next, we wish to find the closed subsets corresponding to maximal and prime ideals of *A*.

Proposition 2.Maximal ideals correspond to singleton subsets of .

**Proof**

For a singleton set {*P*}, consider the evaluation map:

which is a ring homomorphism. This is clearly surjective so its kernel is a maximal ideal. By definition .

Conversely, for a maximal ideal , since its corresponding subset by the Nullstellensatz. Thus it contains some *P*. From we get . By maximality of equality holds so *S* = {*P*}. ♦

**Exercise A**

Prove that if then .

Prime ideals are a little trickier, so we will have to introduce a new topological concept.

Definition.A topological space X is said to be

irreducibleif it is non-empty, and any non-empty open subset of X is dense in X. Otherwise we say it isreducible.

We are getting repetitive, but empty spaces are excluded from the class of irreducible spaces for the same reason 1 is not prime. Later, we will see that closed subsets can be uniquely “factored” as a union of irreducible closed subsets.

Lemma 1.The following are equivalent for any non-empty topological space X.

- X is irreducible.
- Any two non-empty open subsets of X must intersect.
- If closed subsets have union X, then C = X or C’ = X.

**Note**

All criteria for irreducibility are useful at some point of time so it pays to take heed. We will study such spaces in greater detail at a later time.

**Proof**

The equivalence between 2 and 3 is straightforward. For equivalence of 1 and 2, use the fact that a subset of a topological space is dense if and only if every non-empty *open* subset intersects it. ♦

Proposition 3.The prime ideals of A correspond to the irreducible closed subspaces of .

**Proof**

Suppose *V* is irreducible and . Then and are closed subsets of *V* and . Thus so we have .

Conversely suppose for a prime ideal . Let be closed subsets (of *V*, and hence of ) with union *V*. Write

for radical ideals

Thus corresponds to the ideal and we have . It remains to prove the following, which we will leave as an easy exercise. ♦

**Exercise B**

Suppose are ideals of any ring *A* with prime. If and , then or .

# Simple Example

We will work through a simple example step-by-step.

Let and consider the closed subset *V* cut out by and . Geometrically this gives the points of intersection between (a circle) and (a line). Clearly, we get two points (1, 1) and (-1, -1).

[Graph plotted by GeoGebra.]

Let us verify this algebraically.

Let . We have so it remains to show that is a radical ideal. We prove this by applying the following to the quotient ring .

Lemma 2.A ring A is said to be

reducedif (0) is a radical ideal in A. Then an ideal is radical if and only if is reduced.

**Proof**

Exercise. ♦

First note that for any ring *A* and , we have an evaluation map taking . The kernel of this map is precisely (*X* – *a*); indeed it clearly contains (*X* – *a*), conversely any can be written as with ; if we have *r* = 0. Hence we have where the isomorphism takes *X* to *a*.

With that in mind we get:

because the first isomorphism takes *X* to *Y*. This corresponds to a set of two points {-1, +1} in the affine line . Now by Chinese Remainder Theorem, we have

which is a reduced ring. Hence is a reduced ring and is a radical ideal.

The set of two points is reducible so its corresponding ideal is not prime. Indeed, we saw that , but .

## Slight Variation

Let us now consider the set cut out by and . Geometrically, we now have only one point of intersection.

[Graph plotted by GeoGebra.]

Let now. We have

This ring is non-reduced so is not a radical ideal. In fact, which corresponds to the geometric picture.

*But the story is not quite over!*

Note that upon taking the radical of , we are actually losing information on the multiplicity of intersection. Since satisfies , this suggests that the intersection multiplicity is 2 here. *Indeed we can construct a theory of intersection via considering general ideals instead of merely radical ideals. *

## Note

This seems like a tremendous amount of work for such simple geometric examples. But it looks tedious only because we took pains to explicitly justify every step. As you progress, the above computations will eventually seem too trivial to even contemplate.

In the following articles, we will consider a much harder example when we have more tools at our disposal.

It would be useful to mention that k is algebraically closed at the beginning of the post.

Good point. Done.

Could you explain how to obtain the radical of the ideal”a” in the secret example.

“Second example

By observation. Note that . Also is a maximal ideal so cannot contain anything more.

The first part of the proof of the statement “The prime ideals of A correspond to the irreducible closed subspaces of \mathbb A^n_k.” does not seem entirely right. Should there be a comma between f and g? like “Suppose V is irreducible and . Then the closed subsets …’

Thanks good catch!