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 :

Affine varieties define a notion of 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:

We associate to such a ring the affine variety:
For example, we associate to the ring . Each point defines a ring homomorphism:
called the 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 .


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 .


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.


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.


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

and define an induced morphism between the associated affine varieties and to be the restriction of the polynomial map , where represents . This gives a well defined morphism when restricted to . Then, if and we have but so and . It is straight forward to check that and so that isomorphisms in one category induce isomorphisms in the other and vice versa.


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

Let , . The morphism , induces the following homomorphism on the rings of functions ,
This is clearly not onto and so this is not an isomorphism of varieties. (In fact, there is no surjective homomorphism . Suppose and with for some . Then, differentiating, so and are coprime and in particular and cannot have any multiple zeros in common. But we also have so and must have the same roots but neither can have any single roots. This is a contradiction.)

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

and see that if then Hence,

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

for some generators of and . Formally substituting and then multiplying through by a sufficiently higher power of we get an expression of the form
and conclude that

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?