In this section, we want to sort out the various meanings of the word "point," as used by algebraic geometers. This word is used in at least three distinct ways. As a result, beginners in the subject often find themselves confused.

Let k be a field, and let S be an algebraic variety over k. (An algebraic variety is a scheme of finite type over a field. This usage differs from that of other authors, who reserve the word variety for an irreducible scheme of finite type. We want to work over arbitrary fields, and the property of being irreducible is not stable under the operation of extending the base field.) The most primitive notion of a point is also the rarest usage among algebraic geometers. We call any element of the underlying topological space of S a point. When using the term this way, we often make special mention of the closed points. This term emphasizes the fact that the underlying topological space is not Hausdorff, and many of its points are, in fact, not closed in S.

Let's assume for the moment that the variety in question can be embedded in a projective space Pⁿ_k. Then we can give an elementary definition of a k-rational point of S. A closed point s∈ S⊂Pⁿ is called a k-rational point if it has the form s = (s₀:…:s_n) with all s_i∈ k. We write S(k) for the set of all k-rational points of the scheme S.

If the field k is not algebraically closed, then the k-rational points usually form a very small subset of the collection of all closed points of S. We can reinterpret the definition of a k-rational point in a coordinate-free manner that will allow us to generalize the above definition in several ways. Notice that to give a k-rational point of S (in the sense described above) is equivalent to specifying a k-morphism Spec(k)→ S. The first generalization, then, is to drop the assumption that S can be embedded in projective space. Next, suppose that s∈ S is a closed point. Then the residue field k(s) = O_S,s/m_s is an extension field of k of finite degree. The inclusion of this closed point in S corresponds (loosely) to a k-morphism Spec(k(s)) → S. The looseness of the correspondence comes from the fact that we can compose this morphism with any k-automorphism of k(s) without detecting any difference in the closed point s∈ S. To elaborate, let L/k be any field extension of finite degree. Let us define an L-rational point of the k-scheme S to be a k-morphism Spec(L)→ S, and denote the set of all L-rational points of S by S(L). In general, the set S(L) is strictly larger than the set of closed points of S whose coefficients lie in L.

Example Let k=R be the field of real numbers, let L=C be the field of complex numbers, and let S=Spec(R). Now S is an R-scheme with one closed point, and S(R) is a one-point set. After all, there is only one R-algebra map from the reals into itself. Moreover, the set S(C) also consists of exactly one point, because any R-algebra homomorphism from R to C must take the number 1 to itself.

Example Now consider S=Spec(C) as a scheme over the field R. This is another scheme that has only one closed point. However, S(R) is the empty set (since there are no homomorphisms C→R) and S(C) is a two-point set (since there are two R-algebra maps from C to itself, depending on whether i|→ i or i|→ -i).

Example Now take S=Spec(R[t]) = A¹_R as a scheme over R. The set of closed points of S can be naturally identified with the complex upper half plane, the set S(R) with the real axis, and the set S(C) with the entire complex plane.

We are now ready to define yet another kind of point on a k-scheme S. There is no reason to restrict the previous definition to finite extension fields of k. If we let k^- denote the algebraic closure of k, then we can consider the set

Remark One nice aspect of the definition of rational points with values in extension fields is that it behaves well under base extension. If S is a k-scheme and if L/k is any extension field, we write S_L = Spec(L)×_Spec(k) S for the L-scheme obtained by base extension. It is easy to use the universal properties of fibre products to show that there is a natural bijection S_L(L) = S(L). (You should convince yourself that this is true in the examples given above.)

Example Take S=Spec(k). Since

S(T) = Mor_k(T,S) = Hom_k(k, Γ(T, O_T))

contains exactly one element for each k-scheme T, the scheme S corresponds to the constant functor T |→ {*}.

Example Take S=A¹_k = Spec(k[x]). What functor is defined by the affine line? Since this is an affine scheme, we have

S(T) = Mor_k(T,S) = Hom_k(k[x], Γ(T, O_T)) = Γ(T, O_T).

Thus, the affine line over k is equivalent to the functor T |→ Γ(T, O_T). In this case, the functor takes values not just in the category of sets, but in the category of k-algebras; this observation suggests that there is more structure to the affine line than we have so far used.

Example Take S= Spec(k[x, x^-1]). This is another affine scheme, so we have

S(T) = Mor_k(T,S) = Hom_k(k[x, x^-1], Γ(T, O_T)) = Γ(T, O_T)^*.

Thus, S is equivalent to the functor that maps a scheme T to the multiplicative group Γ(T, O_T)^*. For this reason, this scheme is often denoted G_m.

Let us denote the functor of points of S by the notation h_S, so that h_S(T) = S(T) = Hom_k(T, S). In this way, we can distinguish the scheme S from the functor h_S. Now we realize that there is another functor staring us in the face. The assignment S |→ h_S defines a functor

Proof: Let's recall what this statement means. For the functor h to be full, we need to show that for any two schemes S and T, any natural transformation η : h_S → h_T comes from a map of schemes S→ T. But this is obvious: take the identity map of S as an element of h_S(S) and apply η to get an element η_S(1_S) ∈ h_T(S) = Hom(S, T). It is easy to use the naturality of η to show that it must be given by composing with the morphism η_S(1_S) : S → T.

For the functor h to be faithful, we must show that if two morphisms of schemes f, g : S → T induce the same natural transformation h_S → h_T, then they must have started out as the same morphism. But this is also easy, since the natural transformation is defined by composing with the morphism, and we can test it on an identity morphism.

Proof: Left to the careful and industrious reader. One should note that you need to glue morphisms together on affine patches to show that h⁰ is full.

Remark The fully faithful functors h and h⁰ allow us to identify the category of schemes with a subcategory of a functor category. That may sound like abstract nonsense, but it has practical implications. In particular, we have shown that the functor of points determines the scheme. In other words, if we know enough about the points on a scheme, then we know everything about the scheme in question.

Pointed Remarks