Definition Let XAn and YAm be algebraic sets. A regular mapping φ:X Y, also called a morphism, is defined to be any function on the underlying sets that can be represented by polynomials f1, , fm k[x1, , xn] such that

φ(x) = (f1(x), , fm(x))


g(f1(x), , fm(x)) = 0

whenever x X and g I(Y).

Definition The set of all morphisms from X to Y will be denoted Mor(X, Y).

Lemma Let XAn be an algebraic set. Then

Mor(X, A1) k[x1, , xn]/I(X) = A(X).

Proof: By definition, a morphism φ:X A1 is given by one polynomial in n variables. Two polynomials represent the same function on the underlying set of X if and only if their difference lies in the ideal I(X). The result follows.

Remark Suppose φ:X Y is a morphism of algebraic sets. Then φ induces a ring homorphism

φ# : A(Y) A(X)

defined by

φ#(f) = f°φ.

In this way, we can view A as a contravariant functor from the category ALGSET/k (whose objects are all algebraic sets with the definition of morphism just presented) taking values in the category k-ALG (whose objects are all commutative k-algebras with the evident definition of morphism).

Proposition Mor(X,Y) Homk-Alg(A(Y), A(X)).

Proof: As above, one can associate to any morphism φ of algebraic sets the k-algebra homomorphism φ#. On the other hand, suppose f:A(Y) A(X) is a homomorphism of k-algebras. Write

A(Y) = k[y1, , ym]/I(Y)


A(X) = k[x1, , xn]/I(X).

Define si = f(yi) A(X), and choose an arbitrary lifting s'i k[x1, , xn]. Then there is a morphism

s = (s'1, , s'm) : X Y

that is independent of the choice of lifting, and clearly satisfies s# = f.

Corollary The functor A:ALGSET/k k-ALG is a fully faithful embedding of categories. Moreover, the image of this functor allows us to identify algebraic sets with finitely generated k-algebras that contain no nilpotent elements.

Remark As we noted earlier, the algebra A(X) does not have enough information to allow us to recover X as a particular subset of a particular affine space An. Nevertheless, this result says that we can recover X up to isomorphism of algebraic sets. That means that any intrinsic geometric properties of X can be discovered by studying the algebra A(X); only the extrinsic properties that depend on the particular embedding are lost.

Definition Let k be an algebraically closed field. Let J k[x1, , xn] be any ideal. The structured algebraic set defined by J is the pair (Z(J), R(J)) consisting of the algebraic set Z(J) together with the coordinate ring R(J) = k[x1, , xn]/J.

By abuse of notation, if X=(Z(J), R(J)) is the structured algebraic set associated to the ideal J, we will write A(X)= R(J). When X is any (unstructured) algebraic set, unless otherwise specified, it can be identified in a natural way with the canonical structured algebraic set defined by its ideal I(X). With this convention, A(X) has the same meaning that it had previously. Finally, we can define

Mor(X, Y) = Homk(A(Y), A(X))

for any pair of structured algebraic sets. It follows easily from the definitions that there is a fully faithful contravariant functor


that identifies the category of structured algebraic sets with the category of finitely generated k-algebras. In particular, one has the following result.

Corollary Two (structured) algebraic sets are isomorphic if and only if their coordinate rings are isomorphic.

Remark One advantage of viewing all algebraic sets as structured algebraic sets is that the definition is coordinate-free, and all the geometric information can be extracted from the coordinate ring itself.

Theorem Let XAn and YAm be structured algebraic sets. Then their product exists in the category of structured algebraic sets over k.

Proof: Using the contravariant functor A to transfer the problem into the category of finitely generated k-algebras, we can try to construct a coproduct. It is straightforward to check that the tensor product is such a coproduct. Thus, the product X× Y of the structured algebraic sets is determined up to isomorphism by the formula

A(X× Y) = A(X)k A(Y).


In particular, we see that An× Am = An+m in the category of structured algebraic sets.

Definition Let X, YAn be structured algebraic sets. Define X Y to be the pullback in the category of structured algebraic sets of the diagram
X An

Lemma Let X, YAn be structured algebraic sets. Then X Y exists, and

I(X Y) = (I(X) I(Y)).

Proof: We again use the functor A to transport this problem into the category of finitely generated k-algebras, where one needs to complete the pushout diagram
k[x1,,xn] k[x1,,xn]/I(X)

It is easy to check that k[x1,,xn]/(I(X) I(Y)) satisfies the necessary universal property.


We need to work in the category of structured algebraic sets for this definition to be truly useful. Consider, for example, the intersection in A2 of the parabola X defined by y=x2 with the line Y defined by y=0. The intersection satisfies

I(X Y) = (y-x2, y) = (y, x2),

so the coordinate ring is

A(X Y) = k[x,y]/(y, x2) k[x]/(x2).

In particular, the intersection of algebraic sets can turn out to be a structured algebraic set; in the present case, this structured algebraic set conatins the extra geometric information that the two curves are tangent at the point of intersection.

Proposition Let XAn and YAm be structured algebraic sets. Then X× Y can be canonically identified with the structured algebraic set

(X×Am)(An× Y),

where the intersection is computed inside An×Am = An+m.

Proof: The natural inclusions

I(X) k[x1,,xn] k[x1,, xn, y1, , ym]


I(Y) k[y1,,ym] k[x1,, xn, y1, , ym]

clearly allow us to identify the images with

I1(X) = I(X×Am) and I2(Y) = I(An× Y).

By the previous lemma, we have

I((X×Am)(An× Y)) = (I1(X) I2(Y)).

Now one can verify that multiplication defines an isomorphism

A(X)k A(Y) k[x1,,xn,y1,,ym]/(I1(X) I2(Y)),

by the "formula"

f(x) g(y) f(x,0)g(0,y).


  1. Let * denote the algebraic set consisting of a single point. Then A(*) = k and * is a final object in the category of structured algebraic sets. In other words, given any structured algebraic set X, there is always a unique morphism X* corresponding to the structure map k A(X) that includes k as the ring of constant functions on X. More precisely, one has

    Mor(X, *) = Homk(k, A(X)),

    and it is the set of k-homomorphism that can easily be seen to consist of exactly one element.
  2. Consider again the one point algebraic set *. We have seen that Mor(X, *) always contains exactly one element. What is the set Mor(*, X)? Turning the question into one about algebras, we are asking about the set of k-algebra homomorphisms Homk(A(X), k). Since these must preserve the k-structure, and k is a field, such homomorphisms are necessarily onto. Each surjective homomorphism to the field k corresponds to a maximal ideal in A(X) (obtained as the kernel). So, using the Weak Nullstellensatz, one has

    Mor(*, X) = { maximal ideals in A(X)} = { points in X} = X(k).

  3. Let X be the (disjoint) union of two points, viewed as an algebraic set over k. Then

    A(X) = k× k.

    To see this, notice that we can take

    A(X) = k[x]/(x2-x) k[x]/(x) k[x]/(x-1).

    More generally, if X and Y are structured algebraic sets, then their disjoint union satisfies A(X Y) = A(X)× A(Y). In other words, A takes disjoint unions into Cartesian products of algebras.
  4. Let X be a structured algebraic set, and consider the diagonal map

    Δ:X X× X.

    This is actually a morphism of structured algebraic sets; it corresponds to the algebra homomorphism A(X) A(X) A(X) by multiplication.
  5. Let X be the affine plane curve defined by the equation y2=x3+x2. So, A(X) = k[x,y]/(y2-x3-x2). We can define a morphism from A1 X as follows. Let y=mx be a general line through the origin. This line intersects X at the points where

    x3 + x2(1-m2) = 0.

    Ignoring the obvious point at the origin, we get an additional point

    x = m2-1, y = m3-m.

    These equations define an algebra homomorphism

    k[x, y]/(y2-x3-x) k[m],

    since one can easily check that

    (m3-m)2 - (m2-1)3 - (m2-1)2 = 0.

  6. Let H be the hyperbola defined by the equation xy=1. So, the coordinate ring of H is A(H) = k[x,y]/(xy-1) k[x,x-1]. Now let X be any structured algebraic set. Then

    Mor(X,H) = Homk(k[x,x-1], A(X)) = A(X)*,

    the set of invertible elements in the ring A(X). It is interesting to note that, in this case, the set of morphisms always forms an abelian group. One interpretation of this remark is that there is a functor

    Mor(-,H) : STRALGSET AB.

    Another interpretation says that H is a group-object in the category of structured algebraic sets. Concretely, there are morphisms
    μ : H× H H μ#(x) = x x
    ε:* H ε#(x) = 1
    ι:H H ι#(x) = x-1.

    The first morphism is multiplication, the second is the inclusion of the identity element, and the third identifies inverses. These maps satisfy
    associativity: μ°(μ×1) = μ°(1×μ)
    existence of identity: μ(ε×1) = 1
    existence of inverse: μ(ι,1) = επ.

  7. Let T be the structured algebraic set with coordinate ring

    A(T) = k[x]/(x2) = k[ε].

    How should one interpret the set Mor(T,X)? By definition, every such morphism corresponds to an algebra homomorphism f:A(X) k[ε]. By composing with the canonical surjection k[ε] k (which takes ε to 0), one obtains a well-defined point x X, corresponding to the maximal ideal M that is the kernel of the composite. Now we can decompose the set Mor(T,X) into subsets

    Mor(T,X; x) = {φ:T X | φ(*) = x}.

    We now observe that f(M)(ε) and f(M2) (ε2) = 0. Therefore,

    Mor(T,X; x) = Homk-VS(M/M2, k).

    In a natural sense, this set of vector space homomorphism can be thought of as the tangent space to X at the point x. For example, let H be the hyperbola considered earlier, and look at the set Mor(T, H; (1,1)). Any such morphism is given by

    f(x) = 1+λε and f(y) = 1+με.

    Moreover, since xy=1, we have

    0 = (1+λε)(1+με) - 1 = (λ+μ)ε.

    Therefore, μ=-λ and the points x=1+λε, y = 1-λε all lie along the line x+y=2, the usual tangent line to the hyperbola at that point.

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210