# Spec

In the introduction, we have explained the general idea of duality between spaces and rings of functions, and we have given two examples: one linear algebraic and the other topological. We present in what follows an algebro-geometric example. The functions considered will be *algebraic*. More precisely, the starting
point will be to take as our ring of functions the ring of *polynomials* .

*Algebraic geometry* is concerned with the geometry of spaces whose functions are of this kind. Using different starting rings, it is possible to develop in a parallel way different theories, such as *analytic geometry*, starting from the ring of power series converging on an open neighborhood of the origin, or *formal geometry*, starting from the ring of all formal power series.

An *affine (algebraic) variety* over is the vanishing locus in
of finitely many polynomials :

*space*in algebraic geometry.

We take as our notion of *ring of functions* the *finitely generated unital commutative algebras* over . Let be such an algebra. By hypothesis, there exists a finite number
of generators of , i.e. is the quotient of the polynomial algebra by some ideal of relations. By the Hilbert basis theorem, i.e. the fact that the ring is *Noetherian*, the ideal of relations is itself finitely generated. Let ,..., be some generators of the relations. We denote by the *ideal* generated by the 's, i.e. the set of all elements of the form , , i.e. the -submodule of generated by the 's.

Then we have an isomorphism:

*evaluation map at*. The

*spectrum*of the ring , denoted by , is the set of all ring homomorphisms (since all our rings are in fact -algebras,

*ring homomorphism*will always mean

*homomorphism of algebras*). The evaluation map induces a map defined by . It is possible to show that this map is a bijection, i.e. every ring homorphism is the evaluation for some . In other words, we have

**Exercise**: Check this is true by assuming that this is known for ,
i.e. that . The space seen as an affine variety
is called the *affine space of dimension *.

Conversely, starting from an affine variety , it is possible to construct
a ring of functions on by quotienting
by the ideal of functions vanishing on .
The resulting finitely generated
unital algebra is called the *coordinate ring of * and is denoted by
.

**Example**

We have an isomorphism . The ring is the coordinate ring of the -axis in . The isomorphism with corresponds to the obvious geometric fact that the -axis in is a copy of .

**Example**

We have an isomorphism given by . The ring is the coordinate ring of the parabola in . The isomorphism with corresponds to the geometric fact that the projection of the parabola on the -axis is an isomorphism of affine varieties.

**Example**

Let and . The ring is the coordinate ring of the graph in of for such that . The projection on the -axis gives an isomorphism between this affine variety and the -axis minus the points where . In other words, we have isomorphisms and .

The ring is generally denoted by and is called the *localization* of along the multiplicative set consisting of powers of .

We have explained how to associate an affine variety to a finitely generated unital commutative algebra , by taking its spectrum , and how to associate a finitely generated unital commutative algebra to an affine variety , by taking its ring of functions . But the constructions and are not in general the inverse of each other.

**Example**

Let and .

The equations and define the same subset of , so we have , but .

To solve this difficulty and have a better correspondence between spaces and rings of functions, there are two possible approaches:

1) Restrict the notion of ring. Rather than to consider rings of the form for a general ideal of relations, one can consider only rings of the form for an ideal which is *radical*, i.e. such that where . With this notion of rings, one obtains a one-to-one correspondence bewteen affine varieties
and rings. This results from the *Nullstellensatz* which states that .

2) Enlarge the notion of space. Rather than considering spaces as subsets of , one
could remember more information. For example, one would like to say that the space associated to the ring is different from the one associated to the ring , the former being a first order *infinitesimal thickening* of the latter. The notion of *scheme* enlarges the notion of space to take into account the possibility of *infinitesimal thickenings*. There is a one-to-one correspondence between affine schemes and rings of the form .

We will describe the first approach in more detail. Given an ideal we have the associated affine variety . To an affine variety we have associated the ideal of polynomials which vanish on it and have defined its ring of functions to be . It is straightforward to check that . But in general . For example, . One of the many forms of the Nullstellensatz states that for an affine variety we have . So the ring of functions on is just , that is . This gives a bijection between radical ideals and affine varieties which in turn sets up a contravariant equivalence between the category of affine varieties and the category of finitely-generated reduced algebras (i.e. nilpotent elements, it is easy to see that is reduced if and only if is radical).

For instance, given a morphism of varieties we have an induced homomorphism on the rings of functions given by .

Going the other way, given a homomorphism between finitely-generated reduced -algebras, , we can choose representations

**Example**

Let , . The morphism , has inverse morphism . This corresponds to the isomorphisms of rings of functions ,

**Example**

There is a table of dualities:

There is an alternative description of as the set of maximal ideals of ; we will now show that this is equivalent to our original definition. Firstly, if is a maximal ideal of then we let be the quotient morphism . It is a basic fact (see the accompanying exercises) that , so we have a morphism . In the reverse direction, given a ring homomorphism , we let . Then is an ideal of , and it is maximal because (since is an algebra morphism and ), and so is a field. It is straightforward to check that these constructions are mutually inverse.

**Exercise**: Show that , i.e. that for every maximal ideal of is of the form for some (unique) .

Let and consider the evaluation map defined by . It is clear that the ideal is contained in the kernel of . Now, given we can write it in the form

Conversely, given a maximal ideal we know by exercise 1 that the quotient is isomorphic to . If is identified with under this isomorphism then is contained in the natural projection since . By maximality, we conclude that it must be equal to .

**Exercise**: Using that , prove the
Nullstellensatz, i.e. for we have

Checking that is straightforward.

Let . Now introduce a new variable and consider the ideal . The corresponding variety must be empty. For if not, there exists some such that and yet since and . Now it must be the case that . Again, supose not, then lies in some maximal ideal but then we would have . So write

**Remark**:
What we call spectrum is what is generally called the *maximal spectrum*,
since it is the set of maximal ideals. What is generally called the spectrum is the set
of prime ideals. If we only consider varieties over , the maximal spectrum is enough.

**Exercise**: Why the name *spectrum*? Given consider the (commutative!) -algebra generated by . What is its Spec?