Functor of points

Remarks

By the Yoneda lemma, the rule $X \mapsto h_X$ is fully faithful, so a scheme is determined up to unique isomorphism by its functor of points. It suffices to know $X(A)$ for all commutative rings $A$. Example: for $X = \mathbb{A}^n_\Z = \operatorname{Spec} \Z[x_1, \ldots, x_n]$, a morphism $\operatorname{Spec} A \to X$ is a ring map $\Z[x_1, \ldots, x_n] \to A$. Such a map is given by the images of the $x_i$, so $X(A) = A^n$. Non-example: the constant functor $F$ with $F(T) = \set{0, 1}$ for all $T$ is not of the form $h_X$. For the empty scheme $\emptyset$, the set $h_X(\emptyset)$ has exactly one element for every $X$, while $F(\emptyset)$ has two. A functor isomorphic to some $h_X$ is called representable, and $X$ then represents it.