§3. Product of Preschemes

3.1. Sums of preschemes

(3.1.1) Let $(X_{\alpha })$ be an arbitrary family of preschemes, and let $X$ be a topological space that is the sum of the underlying spaces $X_{\alpha }$; thus $X$ is the union of pairwise disjoint open subspaces $X_{\alpha }^{\prime }$, and for each $\alpha$ there is a homeomorphism ${\phi }_{\alpha }:X_{\alpha }\to X_{\alpha }^{\prime }$. If each $X_{\alpha }^{\prime }$ is equipped with the sheaf $({\phi }_{\alpha }{)}_{\ast }({\mathcal{O}}_{X_{\alpha }})$, it is clear that $X$ becomes a prescheme, called the sum of the family $(X_{\alpha })$ and written ${\coprod }_{\alpha }{X}_{\alpha }$. If $Y$ is a prescheme, the map $f\mapsto (f\circ {\phi }_{\alpha })$ is a bijection of $\mathrm{Hom}(X,Y)$ onto the product set ${\prod }_{\alpha }\mathrm{Hom}(X_{\alpha },Y)$. In particular, if the $X_{\alpha }$ are $S$-preschemes with structure morphisms ${\psi }_{\alpha }$, then $X$ is an $S$-prescheme for the unique morphism $\psi :X\to S$ such that $\psi \circ {\phi }_{\alpha }={\psi }_{\alpha }$ for every $\alpha$; that is, $X$ is the coproduct in the category of $S$-preschemes. The sum of two preschemes $X$, $Y$ is written $X⨿Y$. It is immediate that if $X=\mathrm{Spec}(A)$ and $Y=\mathrm{Spec}(B)$, then $X⨿Y$ is canonically identified with $\mathrm{Spec}(A\times B)$.

3.2. Products of preschemes

Definition (3.2.1). Given two $S$-preschemes $X$, $Y$, a triple $(Z,p_1,p_2)$ formed of an $S$-prescheme $Z$ and two $S$-morphisms $p_1:Z\to X$, $p_2:Z\to Y$ is called a product of the $S$-preschemes $X$ and $Y$ if, for every $S$-prescheme $T$, the map $f\mapsto (p_1\circ f,p_2\circ f)$ is a bijection of the set of $S$-morphisms $T\to Z$ onto the set of pairs formed of an $S$-morphism $T\to X$ and an $S$-morphism $T\to Y$; in other words, a bijection

$${\mathrm{Hom}}_S(T,Z)\stackrel{\sim }{\to }{\mathrm{Hom}}_S(T,X)\times {\mathrm{Hom}}_S(T,Y).$$

This is thus the general notion of the product of two objects of a category (T, I, 1.1) applied to the category of $S$-preschemes; in particular, a product of two $S$-preschemes is unique up to a unique $S$-isomorphism. Because of this uniqueness, one usually denotes a product of two $S$-preschemes $X$, $Y$ by $X{\times }_{S}Y$ (or simply $X\times Y$ when no confusion can arise), the projections $p_1$, $p_2$ — called the canonical projections of $X{\times }_{S}Y$ onto $X$ and $Y$ respectively — being suppressed from the notation. If $g:T\to X$, $h:T\to Y$ are two $S$-morphisms, we write $(g,h{)}_S$, or simply $(g,h)$, for the $S$-morphism $f:T\to X{\times }_{S}Y$ such that $p_1\circ f=g$, $p_2\circ f=h$. If $X^{\prime }$, $Y^{\prime }$ are two $S$-preschemes with canonical projections $p_1^{\prime }$, $p_2^{\prime }$ of $X^{\prime }{\times }_S{Y}^{\prime }$ (assumed to exist), and $u:X^{\prime }\to X$, $v:Y^{\prime }\to Y$ are $S$-morphisms, we write $u{\times }_{S}v$ (or simply $u\times v$) for the $S$-morphism $(u\circ p_1^{\prime },v\circ p_2^{\prime }{)}_S$ of $X^{\prime }{\times }_S{Y}^{\prime }$ into $X{\times }_{S}Y$.

When $S$ is an affine scheme with ring $A$, one often replaces $S$ by $A$ in the preceding notation.

Proposition (3.2.2). Let $X$, $Y$, $S$ be three affine schemes, with respective rings $B$, $C$, $A$. Let $Z=\mathrm{Spec}(B{\otimes }_{A}C)$, and let $p_1$, $p_2$ be the $S$-morphisms corresponding (2.2.4) to the canonical $A$-homomorphisms $u:b\mapsto b\otimes 1$ and $v:c\mapsto 1\otimes c$ of $B$ and $C$ into $B{\otimes }_{A}C$. Then $(Z,p_1,p_2)$ is a product of $X$ and $Y$.

Proof. By (2.2.4), everything reduces to checking that, by associating to each $A$-homomorphism $f:B{\otimes }_{A}C\to L$ (where $L$ is an $A$-algebra) the pair $(f\circ u,f\circ v)$, one defines a bijection ${\mathrm{Hom}}_A(B{\otimes }_{A}C,L)\stackrel{\sim }{\to }{\mathrm{Hom}}_A(B,L)\times {\mathrm{Hom}}_A(C,L)$¹; this is immediate from the definitions and the relation $b\otimes c=(b\otimes 1)(1\otimes c)$.

Corollary (3.2.3). Let $T$ be an affine scheme with ring $D$, and let $\xi =({}^a\rho ,\dots )$ (resp. $\zeta =({}^a\sigma ,\dots )$) be an $S$-morphism $T\to X$ (resp. $T\to Y$), where $\rho$ (resp. $\sigma$) is an $A$-homomorphism of $B$ (resp. $C$) into $D$. Then $(\xi ,\zeta {)}_S={}^a\tau$, where $\tau :B{\otimes }_{A}C\to D$ is the homomorphism such that $\tau (b\otimes c)=\rho (b)\sigma (c)$.

Proposition (3.2.4). Let $f:S^{\prime }\to S$ be a monomorphism of preschemes (T, I, 1.1), and let $X$, $Y$ be two $S^{\prime }$-preschemes, regarded also as $S$-preschemes by means of $f$. Then every product of the $S$-preschemes $X$, $Y$ is a product of the $S^{\prime }$-preschemes $X$, $Y$, and conversely.

Proof. Let $\phi :X\to S^{\prime }$, $\psi :Y\to S^{\prime }$ be the structure morphisms. If $T$ is an $S$-prescheme and $u:T\to X$, $v:T\to Y$ are two $S$-morphisms, then by definition $f\circ \phi \circ u=f\circ \psi \circ v=\theta$, the structure morphism of $T$; the hypothesis on $f$ gives $\phi \circ u=\psi \circ v={\theta }^{\prime }$, and one sees that $T$ may be regarded as an $S^{\prime }$-prescheme with structure morphism ${\theta }^{\prime }$, and $u$, $v$ as $S^{\prime }$-morphisms. The conclusion follows at once, taking (3.2.1) into account.

Corollary (3.2.5). Let $X$, $Y$ be two $S$-preschemes with structure morphisms $\phi :X\to S$, $\psi :Y\to S$, and let $S^{\prime }$ be an open subset of $S$ such that $\phi (X)\subset S^{\prime }$, $\psi (Y)\subset S^{\prime }$. Then every product of the $S$-preschemes $X$, $Y$ is also a product of the $S^{\prime }$-preschemes $X$, $Y$, and conversely.

Proof. It suffices to apply (3.2.4) to the canonical injection $S^{\prime }\to S$.

Theorem (3.2.6). Given two $S$-preschemes $X$, $Y$, a product $X{\times }_{S}Y$ exists.

We proceed in several steps.

Lemma (3.2.6.1). Let $(Z,p,q)$ be a product of $X$ and $Y$, and let $U$, $V$ be open subsets of $X$, $Y$ respectively. If $W=p^{-1}(U)\cap q^{-1}(V)$, then the triple formed of $W$ and the restrictions of $p$ and $q$ to $W$ (regarded as morphisms $W\to U$, $W\to V$) is a product of $U$ and $V$.

Proof. If $T$ is an $S$-prescheme, the $S$-morphisms $T\to W$ may be identified with the $S$-morphisms $T\to Z$ whose image lies in $W$. If $g:T\to U$, $h:T\to V$ are two $S$-morphisms, they may be regarded as $S$-morphisms of $T$ into $X$ and $Y$ respectively, so by hypothesis there is a unique $S$-morphism $f:T\to Z$ with $g=p\circ f$, $h=q\circ f$. Since $p(f(T))\subset U$ and $q(f(T))\subset V$, we have $f(T)\subset p^{-1}(U)\cap q^{-1}(V)=W$, whence the assertion.

Lemma (3.2.6.2). Let $Z$ be an $S$-prescheme, $p:Z\to X$, $q:Z\to Y$ two $S$-morphisms, $(U_{\alpha })$ an open cover of $X$, and $(V_{\lambda })$ an open cover of $Y$. Suppose that for every pair $(\alpha ,\lambda )$ the $S$-prescheme $W_{\alpha \lambda }=p^{-1}(U_{\alpha })\cap q^{-1}(V_{\lambda })$, with the restrictions of $p$ and $q$, constitutes a product of $U_{\alpha }$ and $V_{\lambda }$. Then $(Z,p,q)$ is a product of $X$ and $Y$.

Proof. We first show that if $f_1$, $f_2$ are two $S$-morphisms $T\to Z$, then $p\circ f_1=p\circ f_2$ and $q\circ f_1=q\circ f_2$ imply $f_1=f_2$. Indeed, $Z$ is the union of the $W_{\alpha \lambda }$, so the $f_1^{-1}(W_{\alpha \lambda })$ form an open cover of $T$, as do the $f_2^{-1}(W_{\alpha \lambda })$; moreover $f_1^{-1}(W_{\alpha \lambda })=f_2^{-1}(W_{\alpha \lambda })$ by the hypotheses, so it suffices to see that the restrictions of $f_1$ and $f_2$ to this open set agree for each pair of indices. But these restrictions may be regarded as $S$-morphisms into $W_{\alpha \lambda }$, so the assertion follows from the hypotheses and (3.2.1). Now suppose given two $S$-morphisms $g:T\to X$, $h:T\to Y$. Set $T_{\alpha \lambda }=g^{-1}(U_{\alpha })\cap h^{-1}(V_{\lambda })$; the $T_{\alpha \lambda }$ form an open cover of $T$. By hypothesis there is an $S$-morphism $f_{\alpha \lambda }$ such that $p\circ f_{\alpha \lambda }$ and $q\circ f_{\alpha \lambda }$ are the restrictions of $g$ and $h$ to $T_{\alpha \lambda }$. The restrictions of $f_{\alpha \lambda }$ and $f_{\beta \mu }$ to $T_{\alpha \lambda }\cap T_{\beta \mu }$ coincide: their images lie in $W_{\alpha \lambda }\cap W_{\beta \mu }$, and since by (3.2.6.1) this open set is a product of $U_{\alpha }\cap U_{\beta }$ and $V_{\lambda }\cap V_{\mu }$, the equality of $p\circ f_{\alpha \lambda }$ with $p\circ f_{\beta \mu }$ and of $q\circ f_{\alpha \lambda }$ with $q\circ f_{\beta \mu }$ there forces the two restrictions to agree. The $f_{\alpha \lambda }$ therefore glue to an $S$-morphism $f:T\to Z$ with $g=p\circ f$, $h=q\circ f$, which proves (3.2.6.2).

(3.2.6.3) Let $(U_{\alpha })$ be an open cover of $X$ and $(V_{\lambda })$ an open cover of $Y$, and suppose that for every pair $(\alpha ,\lambda )$ a product of $U_{\alpha }$ and $V_{\lambda }$ exists. Then a product of $X$ and $Y$ exists.

Proof. Applying (3.2.6.1) to the open sets $U_{\alpha }\cap U_{\beta }$ and $V_{\lambda }\cap V_{\mu }$, one sees that a product of the induced $S$-preschemes exists; the uniqueness of the product yields canonical gluing isomorphisms $f_{ij}$ between the overlaps (writing $i=(\alpha ,\lambda )$, $j=(\beta ,\mu )$) satisfying the cocycle condition $f_{ik}=f_{jk}\circ f_{ij}$, by (3.2.6.1) applied to triple intersections. By the gluing of preschemes (2.3.1) there is a prescheme $Z$, an open cover $(Z_i)$, and isomorphisms $g_i$ of the induced prescheme $Z_i$ onto $U_{\alpha }{\times }_S{V}_{\lambda }$ with $f_{ij}=g_i\circ g_j^{-1}$ and $g_i(Z_i\cap Z_j)=W_{ij}$. The projections and structure morphisms of the $U_{\alpha }{\times }_S{V}_{\lambda }$ glue to morphisms $p:Z\to X$, $q:Z\to Y$, $\theta :Z\to S$ coinciding with $p_i\circ g_i$, $q_i\circ g_i$, ${\theta }_i\circ g_i$ on each $Z_i$; this makes $Z$ an $S$-prescheme. One checks (using (3.2.6.1) and uniqueness of products) that $p^{-1}(U_{\alpha })\cap q^{-1}(V_{\lambda })=Z_i$, so by (3.2.6.2) $(Z,p,q)$ is a product of $X$ and $Y$.

(3.2.6.4) Let $\phi :X\to S$, $\psi :Y\to S$ be the structure morphisms of $X$ and $Y$, let $(S_i)$ be an open cover of $S$, and set $X_i={\phi }^{-1}(S_i)$, $Y_i={\psi }^{-1}(S_i)$. If each product $X_i{\times }_S{Y}_i$ exists, then $X{\times }_{S}Y$ exists.

Proof. By (3.2.6.3) it suffices to prove that the products $X_i{\times }_S{Y}_j$ exist for all $i$, $j$. Set $X_{ij}=X_i\cap X_j={\phi }^{-1}(S_i\cap S_j)$, $Y_{ij}=Y_i\cap Y_j={\psi }^{-1}(S_i\cap S_j)$; by (3.2.6.1) the product $Z_{ij}=X_{ij}{\times }_S{Y}_{ij}$ exists. If $T$ is an $S$-prescheme and $g:T\to X_i$, $h:T\to Y_j$ are $S$-morphisms, then necessarily $\phi (g(T))=\psi (h(T))\subset S_i\cap S_j$ by the definition of an $S$-morphism, so $g(T)\subset X_{ij}$ and $h(T)\subset Y_{ij}$; it is then immediate that $Z_{ij}$ is a product of $X_i$ and $Y_j$.

(3.2.6.5) We can now complete the proof of (3.2.6). If $S$ is an affine scheme, there are covers $(U_{\alpha })$, $(V_{\lambda })$ of $X$ and $Y$ by affine opens; since $U_{\alpha }{\times }_S{V}_{\lambda }$ exists by (3.2.2), so does $X{\times }_{S}Y$ by (3.2.6.3). If $S$ is an arbitrary prescheme, take a cover $(S_i)$ of $S$ by affine opens; with $X_i={\phi }^{-1}(S_i)$, $Y_i={\psi }^{-1}(S_i)$, the products $X_i{\times }_{S_i}{Y}_i$ exist by the foregoing, and then $X_i{\times }_S{Y}_i$ exists too by (3.2.5), so $X{\times }_{S}Y$ exists by (3.2.6.4). $\square$

Corollary (3.2.7). Let $Z=X{\times }_{S}Y$ be the product of two $S$-preschemes, $p$, $q$ its projections, and $\phi$ (resp. $\psi$) the structure morphism of $X$ (resp. $Y$). Let $S^{\prime }$ be an open subset of $S$, and $U$ (resp. $V$) an open subset of $X$ (resp. $Y$) contained in ${\phi }^{-1}(S^{\prime })$ (resp. ${\psi }^{-1}(S^{\prime })$). Then the product $U{\times }_{S^{\prime }}V$ is canonically identified with the prescheme induced by $Z$ on $p^{-1}(U)\cap q^{-1}(V)$ (regarded as an $S^{\prime }$-prescheme). Moreover, if $f:T\to X$, $g:T\to Y$ are $S$-morphisms with $f(T)\subset U$, $g(T)\subset V$, then the $S^{\prime }$-morphism $(f,g{)}_{S^{\prime }}$ is identified with the restriction of $(f,g{)}_S$. This follows from (3.2.5) and (3.2.6.1).

(3.2.8) Let $(X_{\alpha })$, $(Y_{\lambda })$ be two families of $S$-preschemes, and $X$ (resp. $Y$) the sum of the family $(X_{\alpha })$ (resp. $(Y_{\lambda })$) (3.1). Then $X{\times }_{S}Y$ is identified with the sum of the family $(X_{\alpha }{\times }_S{Y}_{\lambda })$; this follows at once from (3.2.6.3).

3.3. Formal properties of the product; change of base prescheme

(3.3.1) The reader will note that all the properties stated in this section, except (3.3.13) and (3.3.15), hold without change in any category, whenever the products that occur in the statements exist (for it is clear that the notions of $S$-object and $S$-morphism can be defined exactly as in (2.5) for any object $S$ of the category).

(3.3.2) First, $X{\times }_{S}Y$ is a covariant bifunctor in $X$ and $Y$ on the category of $S$-preschemes: it suffices to remark that the diagram

$$\begin{array}{ccc}X{\times }_{S}Y& \to & X^{\prime }{\times }_{S}Y\to X^{\prime \prime }{\times }_{S}Y\\ \downarrow & & \downarrow \\ X& \to & X^{\prime }\to X^{\prime \prime }\end{array}$$

is commutative.

Proposition (3.3.3). For every $S$-prescheme $X$, the first (resp. second) projection of $X{\times }_{S}S$ (resp. $S{\times }_{S}X$) is a functorial isomorphism of $X{\times }_{S}S$ (resp. $S{\times }_{S}X$) onto $X$, whose inverse is $(1_X,\phi {)}_S$ (resp. $(\phi ,1_X{)}_S$), where $\phi$ denotes the structure morphism $X\to S$. One may therefore write, up to a canonical isomorphism, $$X{\times }_{S}S=S{\times }_{S}X=X.$$

Proof. It suffices to prove that the triple $(X,1_X,\phi )$ is a product of $X$ and $S$. If $T$ is an $S$-prescheme, the only $S$-morphism of $T$ into $S$ is necessarily the structure morphism $\theta :T\to S$. If $f$ is an $S$-morphism of $T$ into $X$, then necessarily $\theta =\phi \circ f$, whence the assertion.

Corollary (3.3.4). Let $X$, $Y$ be two $S$-preschemes with structure morphisms $\phi :X\to S$, $\psi :Y\to S$. If one identifies $X$ with $X{\times }_{S}S$ and $Y$ with $S{\times }_{S}Y$, the projections $X{\times }_{S}Y\to X$ and $X{\times }_{S}Y\to Y$ are identified with $1_X{\times }_S\psi$ and $\phi {\times }_S{1}_Y$ respectively. (The verification is immediate and left to the reader.)

(3.3.5) One may define, in the same way as in (3.2), the product of any finite number $n$ of $S$-preschemes; the existence of these products follows from (3.2.6) by induction on $n$, on remarking that $(X_1{\times }_S\cdots {\times }_S{X}_{n-1}){\times }_S{X}_n$ satisfies the definition of the product. Uniqueness of the product entails, as in any category, its commutativity and associativity properties. For example, if $p_1$, $p_2$, $p_3$ denote the projections of $X_1{\times }_S{X}_2{\times }_S{X}_3$, and one identifies this prescheme with $(X_1{\times }_S{X}_2){\times }_S{X}_3$, then the projection onto $X_1{\times }_S{X}_2$ is identified with $(p_1,p_2{)}_S$.

(3.3.6) Let $S$, $S^{\prime }$ be two preschemes and $\phi :S^{\prime }\to S$ a morphism, making $S^{\prime }$ an $S$-prescheme. For every $S$-prescheme $X$, consider the product $X{\times }_S{S}^{\prime }$, and let $p$ and ${\pi }^{\prime }$ be its projections onto $X$ and $S^{\prime }$ respectively. Equipped with ${\pi }^{\prime }$, this product is an $S^{\prime }$-prescheme; regarded as such, it is denoted $X_{(S^{\prime })}$ or $X_{(\phi )}$, and one says it is the prescheme obtained by extension of the base prescheme from $S$ to $S^{\prime }$, by means of $\phi$, or the inverse image of $X$ by $\phi$. Note that if $\pi$ is the structure morphism of $X$ and $\theta$ the structure morphism of $X{\times }_S{S}^{\prime }$ regarded as an $S$-prescheme, the corresponding square is commutative.

(3.3.7) With the notation of (3.3.6), for every $S$-morphism $f:X\to Y$ one writes $f_{(S^{\prime })}$ for the $S^{\prime }$-morphism $f{\times }_{S}1:X_{(S^{\prime })}\to Y_{(S^{\prime })}$, and one says $f_{(S^{\prime })}$ is the inverse image of the morphism $f$ by $\phi$. Thus $X_{(S^{\prime })}$ is a covariant functor in $X$, from the category of $S$-preschemes to that of $S^{\prime }$-preschemes.

(3.3.8) The prescheme $X_{(S^{\prime })}$ may also be regarded as the solution of a universal mapping problem: every $S^{\prime }$-prescheme $T$ is also an $S$-prescheme by means of $\phi$; every $S$-morphism $g:T\to X$ then factors uniquely as $g=p\circ f$, where $f$ is an $S^{\prime }$-morphism $T\to X_{(S^{\prime })}$, as follows from the definition of the product applied to the $S$-morphisms $f$ and $\theta :T\to S^{\prime }$ (the structure morphism of $T$).

Proposition (3.3.9) (transitivity of extension of the base prescheme). Let $S^{\prime \prime }$ be a prescheme and ${\phi }^{\prime }:S^{\prime \prime }\to S^{\prime }$ a morphism. For every $S$-prescheme $X$, there is a canonical functorial isomorphism of the $S^{\prime \prime }$-prescheme $(X_{(S^{\prime })}{)}_{(S^{\prime \prime })}$ onto the $S^{\prime \prime }$-prescheme $X_{(S^{\prime \prime })}$.

Proof. Let $T$ be an $S^{\prime \prime }$-prescheme with structure morphism $\zeta$, and $g$ an $S$-morphism of $T$ into $X$ ($T$ being regarded as an $S$-prescheme with structure morphism $\phi \circ {\phi }^{\prime }\circ \zeta$). Since $T$ is also an $S^{\prime }$-prescheme with structure morphism ${\phi }^{\prime }\circ \zeta$, one may write $g=p\circ g^{\prime }$, where $g^{\prime }$ is an $S^{\prime }$-morphism $T\to X_{(S^{\prime })}$, then $g^{\prime }=p^{\prime }\circ g^{\prime \prime }$, where $g^{\prime \prime }$ is an $S^{\prime \prime }$-morphism $T\to (X_{(S^{\prime })}{)}_{(S^{\prime \prime })}$. The proposition follows from the uniqueness of the solution of a universal mapping problem.

This result is also expressed by writing the equality (up to canonical isomorphism) $(X_{(S^{\prime })}{)}_{(S^{\prime \prime })}=X_{(S^{\prime \prime })}$, or again $$(X{\times }_S{S}^{\prime }){\times }_{S^{\prime }}{S}^{\prime \prime }=X{\times }_S{S}^{\prime \prime }.\text{\hspace{1em}(3.3.9.1)}$$ The functorial character of the isomorphism is expressed by the transitivity formula for inverse images of morphisms $$(f_{(S^{\prime })}{)}_{(S^{\prime \prime })}=f_{(S^{\prime \prime })}\text{\hspace{1em}(3.3.9.2)}$$ for every $S$-morphism $f:X\to Y$.

Corollary (3.3.10). If $X$ and $Y$ are two $S$-preschemes, there is a canonical functorial isomorphism of the $S^{\prime }$-prescheme $X_{(S^{\prime })}{\times }_{S^{\prime }}{Y}_{(S^{\prime })}$ onto the $S^{\prime }$-prescheme $(X{\times }_{S}Y{)}_{(S^{\prime })}$.

Proof. Up to canonical isomorphisms, $$(X{\times }_S{S}^{\prime }){\times }_{S^{\prime }}(Y{\times }_S{S}^{\prime })=X{\times }_S(Y{\times }_S{S}^{\prime })=(X{\times }_{S}Y){\times }_S{S}^{\prime }$$ by (3.3.9.1) and the associativity of products of $S$-preschemes. The functorial character is expressed by $$(u_{(S^{\prime })},v_{(S^{\prime })}{)}_{S^{\prime }}=((u,v{)}_S{)}_{(S^{\prime })}\text{\hspace{1em}(3.3.10.1)}$$ for every pair of $S$-morphisms $u:T\to X$, $v:T\to Y$. In other words, the inverse-image functor $X\mapsto X_{(S^{\prime })}$ commutes with the formation of products; it also commutes with the formation of sums (3.2.8).

Corollary (3.3.11). Let $Y$ be an $S$-prescheme and $f:X\to Y$ a morphism making $X$ a $Y$-prescheme (and hence also an $S$-prescheme). Then $X_{(S^{\prime })}$ is identified with the product $X{\times }_Y{Y}_{(S^{\prime })}$, the projection $X{\times }_Y{Y}_{(S^{\prime })}\to Y_{(S^{\prime })}$ being identified with $f_{(S^{\prime })}$.

Proof. Let $\psi :Y\to S$ be the structure morphism of $Y$. Since $Y_{(S^{\prime })}$ is identified with $S_{(\dots )}^{\prime }$ and $X_{(S^{\prime })}$ with the corresponding base change, the assertion follows from (3.3.9) and (3.3.4) via the commutative diagram relating the morphisms $f$, $\psi$, and $\phi$.

(3.3.12) Let $f:X\to X^{\prime }$, $g:Y\to Y^{\prime }$ be two $S$-morphisms that are monomorphisms of preschemes (T, I, 1.1); then $f{\times }_{S}g$ is a monomorphism. Indeed, if $p$, $q$ are the projections of $X{\times }_{S}Y$, $p^{\prime }$, $q^{\prime }$ those of $X^{\prime }{\times }_S{Y}^{\prime }$, and $u$, $v$ two $S$-morphisms $T\to X{\times }_{S}Y$, the relation $(f{\times }_{S}g)\circ u=(f{\times }_{S}g)\circ v$ gives $p^{\prime }\circ (f{\times }_{S}g)\circ u=p^{\prime }\circ (f{\times }_{S}g)\circ v$, i.e. $f\circ p\circ u=f\circ p\circ v$, and since $f$ is a monomorphism, $p\circ u=p\circ v$; likewise $q\circ u=q\circ v$, whence $u=v$. It follows that for every extension $S^{\prime }\to S$ of the base prescheme, $f_{(S^{\prime })}:X_{(S^{\prime })}\to X_{(S^{\prime })}^{\prime }$ is a monomorphism.

(3.3.13) Let $S$, $S^{\prime }$ be two affine schemes with respective rings $A$, $A^{\prime }$; a morphism $S^{\prime }\to S$ thus corresponds to a ring homomorphism $A\to A^{\prime }$. If $X$ is an $S$-prescheme, one also writes $X_{(A^{\prime })}$ or $X{\otimes }_A{A}^{\prime }$ for the $S^{\prime }$-prescheme $X_{(S^{\prime })}$; when $X$ is itself affine with ring $B$, $X_{(A^{\prime })}$ is affine with ring $B_{(A^{\prime })}=B{\otimes }_A{A}^{\prime }$, obtained by extension to $A^{\prime }$ of the ring of scalars of the $A$-algebra $B$.

(3.3.14) With the notation of (3.3.6), for every $S$-morphism $f:S^{\prime }\to X$, $f^{\prime }=(f,1_{S^{\prime }}{)}_S$ is an $S$-morphism $S^{\prime }\to X^{\prime }=X_{(S^{\prime })}$ such that $p\circ f^{\prime }=f$ and ${\pi }^{\prime }\circ f^{\prime }=1_{S^{\prime }}$ — in other words, an $S^{\prime }$-section of $X^{\prime }$; and conversely, if $f^{\prime }$ is such an $S^{\prime }$-section, then $f=p\circ f^{\prime }$ is an $S$-morphism $S^{\prime }\to X$. One thus defines a canonical bijection $${\mathrm{Hom}}_S(S^{\prime },X)\stackrel{\sim }{\to }{\mathrm{Hom}}_{S^{\prime }}(S^{\prime },X^{\prime }).$$ One says $f^{\prime }$ is the graph morphism of $f$, denoted ${\Gamma }_f$.

(3.3.15) Given a prescheme $X$, which may always be regarded as a $\mathbf{Z}$-prescheme, it follows in particular from (3.3.14) that the $X$-sections of $X{\otimes }_{\mathbf{Z}}\mathbf{Z}[T]$ (where $T$ is an indeterminate) correspond bijectively to the morphisms $\mathbf{Z}[T]\to X$. We show these $X$-sections also correspond bijectively to the sections of the structure sheaf ${\mathcal{O}}_X$ over $X$. Let $(U_{\alpha })$ be a cover of $X$ by affine opens; let $u:X\to X{\otimes }_{\mathbf{Z}}\mathbf{Z}[T]$ be an $X$-morphism and $u_{\alpha }$ its restriction to $U_{\alpha }$. If $A_{\alpha }$ is the ring of the affine scheme $U_{\alpha }$, then $U_{\alpha }{\otimes }_{\mathbf{Z}}\mathbf{Z}[T]$ is affine with ring $A_{\alpha }[T]$ (3.2.2), and $u_{\alpha }$ corresponds canonically to an $A_{\alpha }$-homomorphism $A_{\alpha }[T]\to A_{\alpha }$ (1.7.3). Such a homomorphism is completely determined by the image $s_{\alpha }\in A_{\alpha }=\Gamma (U_{\alpha },{\mathcal{O}}_X)$ of $T$; the compatibility of $u_{\alpha }$ and $u_{\beta }$ on overlaps shows the $s_{\alpha }$ are the restrictions of a section $s$ of ${\mathcal{O}}_X$ over $X$, and conversely. This result will be generalized in (II, 1.7.12).

3.4. Points of a prescheme with values in a prescheme; geometric points

(3.4.1) Let $X$ be a prescheme. For every prescheme $T$, one writes $X(T)$ for the set $\mathrm{Hom}(T,X)$ of morphisms $T\to X$, and the elements of this set are called points of $X$ with values in $T$. Associating to each morphism $f:T\to T^{\prime }$ the map $u^{\prime }\mapsto u^{\prime }\circ f$ of $X(T^{\prime })$ into $X(T)$, one sees that, for fixed $X$, $X(T)$ is a contravariant functor in $T$ from preschemes to sets. Moreover, every morphism $g:X\to Y$ defines a functorial homomorphism $X(T)\to Y(T)$, sending $u\in X(T)$ to $g\circ u$.

(3.4.2) Given three sets $P$, $Q$, $R$ and two maps $\phi :P\to R$, $\psi :Q\to R$, the fiber product of $P$ and $Q$ over $R$ (relative to $\phi$ and $\psi$) is the subset of $P\times Q$ formed of the pairs $(p,q)$ with $\phi (p)=\psi (q)$; it is written $P{\times }_{R}Q$. Definition (3.2.1) of the product of $S$-preschemes may be reinterpreted, with the notation of (3.4.1), by the formula $$(X{\times }_{S}Y)(T)=X(T){\times }_{S(T)}Y(T),\text{\hspace{1em}(3.4.2.1)}$$ the maps $X(T)\to S(T)$ and $Y(T)\to S(T)$ corresponding to the structure morphisms $X\to S$ and $Y\to S$.

(3.4.3) If one fixes a prescheme $S$ and considers only $S$-preschemes and $S$-morphisms, one writes $X(T{)}_S$ for the set ${\mathrm{Hom}}_S(T,X)$ of $S$-morphisms $T\to X$ (suppressing the index $S$ when no confusion is possible); the elements of $X(T{)}_S$ are the $S$-points of the $S$-prescheme $X$ with values in $T$. In particular, an $S$-section of $X$ is nothing but a point of $X$ with values in $S$. Formula (3.4.2.1) then reads $$(X{\times }_{S}Y)(T{)}_S=X(T{)}_S\times Y(T{)}_S;\text{\hspace{1em}(3.4.3.1)}$$ more generally, if $Z$ is an $S$-prescheme and $X$, $Y$, $T$ are $Z$-preschemes (hence ipso facto $S$-preschemes), $$(X{\times }_{Z}Y)(T{)}_S=X(T{)}_S{\times }_{Z(T{)}_S}Y(T{)}_S.\text{\hspace{1em}(3.4.3.2)}$$ To prove that a triple $(W,r,s)$ is a product of $X$ and $Y$ over $Z$, it suffices to verify that for every $S$-prescheme $T$, the corresponding diagram makes $W(T{)}_S$ the fiber product of $X(T{)}_S$ and $Y(T{)}_S$ over $Z(T{)}_S$.

(3.4.4) When $T$ (resp. $S$) is an affine scheme with ring $B$ (resp. $A$), one replaces $T$ (resp. $S$) by $B$ (resp. $A$) in the preceding notation, and speaks of points of $X$ with values in the ring $B$, or of points of the $A$-prescheme $X$ with values in the $A$-algebra $B$, for the elements of $X(B)$ and $X(B{)}_A$ respectively. Note that $X(B)$ and $X(B{)}_A$ are now covariant functors in $B$. One likewise writes $X(T{)}_A$ for the set of points of the $A$-prescheme $X$ with values in the $A$-prescheme $T$.

(3.4.5) Consider the case where $T=\mathrm{Spec}(A)$ with $A$ a local ring; the elements of $X(A)$ then correspond bijectively to the local homomorphisms ${\mathcal{O}}_x\to A$ for $x\in X$ (2.4.4), and one says the point $x$ is the locality of the corresponding point of $X$ with values in $A$. More particularly, one calls geometric points of a prescheme $X$ the points of $X$ with values in a field $K$: giving such a point amounts to giving its locality $x$ in the underlying space of $X$ together with an extension $K$ of $\kappa (x)$. $K$ is called the value field of the geometric point, which is said to be localized at $x$. One thus defines a map $X(K)\to X$ sending a geometric point to its locality. If $S^{\prime }=\mathrm{Spec}(K)$ is an $S$-prescheme (i.e. $K$ is regarded as an extension of a residue field $\kappa (s)$, $s\in S$) and $X$ is an $S$-prescheme, an element of $X(K{)}_S$ — a geometric point of $X$ above $s$ with values in $K$ — consists of a $\kappa (s)$-monomorphism of a residue field $\kappa (x)$ into $K$, where $x$ is a point of $X$ above $s$. In particular, if $S=\mathrm{Spec}(K)=\{\xi \}$, the geometric points of $X$ with values in $K$ are identified with the points $x\in X$ such that $\kappa (x)=K$; these are called the points of the $K$-prescheme $X$ rational over $K$.

Lemma (3.4.6). Let $X_i$ ($1\le i\le n$) be $S$-preschemes, $s$ a point of $S$, and $x_i$ a point of $X_i$ above $s$. Then there exist an extension $K$ of $\kappa (s)$ and a geometric point of the product $Y=X_1{\times }_S\cdots {\times }_S{X}_n$, with values in $K$, whose projections are localized at the $x_i$.

Proof. There exist $\kappa (s)$-monomorphisms $\kappa (x_i)\to K$ into a common extension $K$ of $\kappa (s)$ (Bourbaki, Alg., ch. V, §4, prop. 2). The composites $\kappa (s)\to \kappa (x_i)\to K$ are all the same, so the morphisms $\mathrm{Spec}(K)\to X_i$ are $S$-morphisms and define a unique morphism $\mathrm{Spec}(K)\to Y$. The corresponding point $y$ clearly projects to $x_i$ in each $X_i$.

Proposition (3.4.7). Let $X_i$ ($1\le i\le n$) be $S$-preschemes, and for each $i$ let $x_i$ be a point of $X_i$. For there to exist a point $y$ of $Y=X_1{\times }_S\cdots {\times }_S{X}_n$ whose $i$-th projection is $x_i$ for $1\le i\le n$, it is necessary and sufficient that the $x_i$ lie above a common point $s$ of $S$.

Proof. The condition is clearly necessary; Lemma (3.4.6) shows it is sufficient. In other words, writing $(X)$ for the underlying set of $X$, there is a canonical surjective map $(X{\times }_{S}Y)\to (X){\times }_{(S)}(Y)$; this map is not injective in general — there may be several distinct points of $X{\times }_{S}Y$ with the same projections $x\in X$, $y\in Y$, as one already sees when $S$, $X$, $Y$ are prime spectra of fields $k$, $K$, $K^{\prime }$, since $K{\otimes }_k{K}^{\prime }$ generally has several distinct prime ideals (cf. 3.4.9).

Corollary (3.4.8). Let $f:X\to Y$ be an $S$-morphism, $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ the $S^{\prime }$-morphism deduced from $f$ by an extension $S^{\prime }\to S$ of the base prescheme. Let $p$ (resp. $q$) be the projection $X_{(S^{\prime })}\to X$ (resp. $Y_{(S^{\prime })}\to Y$); then for every subset $M$ of $X$, $$f_{(S^{\prime })}^{-1}(q^{-1}(M))=p^{-1}(f^{-1}(M)).$$

Proof. By (3.3.11), $X_{(S^{\prime })}$ is identified with the product $X{\times }_Y{Y}_{(S^{\prime })}$ via the commutative square. By (3.4.7), the relation $q(y^{\prime })=f(x)$ for $x\in M$, $y^{\prime }\in Y_{(S^{\prime })}$ is equivalent to the existence of $x^{\prime }\in X_{(S^{\prime })}$ with $p(x^{\prime })=x$ and $f_{(S^{\prime })}(x^{\prime })=y^{\prime }$, whence the corollary.

Proposition (3.4.9). Let $X$, $Y$ be two $S$-preschemes, $x$ a point of $X$ and $y$ a point of $Y$, above the same point $s\in S$. The set of points of $X{\times }_{S}Y$ with projections $x$ and $y$ is in canonical bijection with the set of types of composite extensions of $\kappa (x)$ and $\kappa (y)$, regarded as extensions of $\kappa (s)$ (Bourbaki, Alg., ch. VIII, §8, prop. 2).

Proof. Let $p$ (resp. $q$) be the projection of $X{\times }_{S}Y$ onto $X$ (resp. $Y$), and let $E=p^{-1}(x)\cap q^{-1}(y)$. The morphisms $\mathrm{Spec}(\kappa (x))\to S$ and $\mathrm{Spec}(\kappa (y))\to S$ factor through $\mathrm{Spec}(\kappa (s))\to S$; since the latter is a monomorphism (2.4.7), (3.2.4) gives $$P=\mathrm{Spec}(\kappa (x)){\times }_S\mathrm{Spec}(\kappa (y))=\mathrm{Spec}(\kappa (x){\otimes }_{\kappa (s)}\kappa (y)).$$ We define mutually inverse maps $\alpha :P_0\to E$, $\beta :E\to P_0$ ($P_0$ the underlying set of $P$). With $i:\mathrm{Spec}(\kappa (x))\to X$, $j:\mathrm{Spec}(\kappa (y))\to Y$ the canonical morphisms (2.4.5), $\alpha$ is the map of underlying spaces corresponding to $i{\times }_{S}j$. Conversely, every $z\in E$ defines two $\kappa (s)$-monomorphisms $\kappa (x)\to \kappa (z)$, $\kappa (y)\to \kappa (z)$, hence a $\kappa (s)$-monomorphism $\kappa (x){\otimes }_{\kappa (s)}\kappa (y)\to \kappa (z)$ and a morphism $\mathrm{Spec}(\kappa (z))\to P$; $\beta (z)$ is the image of $z$ under it. That $\alpha \circ \beta$ and $\beta \circ \alpha$ are the identity follows from (2.4.5) and (3.2.1). Finally, $P_0$ is in bijection with the set of types of composite extensions of $\kappa (x)$ and $\kappa (y)$ (Bourbaki, Alg., ch. VIII, §8, prop. 1).

3.5. Surjections and injections

(3.5.1) Consider in general a property $\mathrm{P}$ of morphisms of preschemes, and the two statements:

(i) If $f:X\to X^{\prime }$, $g:Y\to Y^{\prime }$ are two $S$-morphisms with property $\mathrm{P}$, then $f{\times }_{S}g$ has property $\mathrm{P}$.

(ii) If $f:X\to Y$ is an $S$-morphism with property $\mathrm{P}$, then every $S^{\prime }$-morphism $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ deduced from $f$ by extension of the base prescheme has property $\mathrm{P}$.

Since $f_{(S^{\prime })}=f{\times }_S{1}_{S^{\prime }}$, one sees that if the identity $1_X$ has property $\mathrm{P}$ for every prescheme $X$, then (i) implies (ii); and since $f{\times }_{S}g$ is the composite $X{\times }_{S}Y\stackrel{f\times 1}{\to }{X}^{\prime }{\times }_{S}Y\stackrel{1\times g}{\to }{X}^{\prime }{\times }_S{Y}^{\prime }$, if property $\mathrm{P}$ is stable under composition then (ii) implies (i).

Proposition (3.5.2). (i) If $f:X\to X^{\prime }$, $g:Y\to Y^{\prime }$ are surjective $S$-morphisms, then $f{\times }_{S}g$ is surjective. (ii) If $f:X\to Y$ is a surjective $S$-morphism, then $f_{(S^{\prime })}$ is surjective for every extension $S^{\prime }$ of the base prescheme.

Proof. Since a composite of surjections is a surjection, it suffices by (3.5.1) to prove (ii); this follows at once from (3.4.8) applied to $M=X$.

Proposition (3.5.3). For a morphism $f:X\to Y$ to be surjective, it is necessary and sufficient that for every field $K$ and every morphism $\mathrm{Spec}(K)\to Y$, there exist an extension $K^{\prime }$ of $K$ and a morphism $\mathrm{Spec}(K^{\prime })\to X$ making the diagram $$\begin{array}{ccc}X& \leftarrow & \mathrm{Spec}(K^{\prime })\\ \downarrow & & \downarrow \\ Y& \leftarrow & \mathrm{Spec}(K)\end{array}$$ commute.

Proof. The condition is sufficient: for any $y\in Y$, apply it to a morphism $\mathrm{Spec}(K)\to Y$ corresponding to a monomorphism $\kappa (y)\to K$ (2.4.6). Conversely, suppose $f$ surjective, and let $y\in Y$ be the image of the unique point of $\mathrm{Spec}(K)$; there is $x\in X$ with $f(x)=y$. Consider the corresponding monomorphism $\kappa (y)\to \kappa (x)$ (2.2.1); take $K^{\prime }$ an extension of $\kappa (y)$ admitting $\kappa (y)$-monomorphisms of $\kappa (x)$ and of $K$ (Bourbaki, Alg., ch. V, §4, prop. 2); the morphism $\mathrm{Spec}(K^{\prime })\to X$ corresponding to $\kappa (x)\to K^{\prime }$ answers the question. In the language of (3.4.5): every geometric point of $Y$ with values in $K$ comes from a geometric point of $X$ with values in an extension of $K$.

Definition (3.5.4). A morphism of preschemes $f:X\to Y$ is called universally injective, or a radicial morphism, if for every field $K$ the corresponding map $X(K)\to Y(K)$ is injective.

It follows at once from the definitions that every monomorphism of preschemes (T, 1.1) is radicial.

(3.5.5) For a morphism $f:X\to Y$ to be radicial, it suffices that the condition of (3.5.4) hold for every algebraically closed field. Indeed, if $K$ is arbitrary and $K^{\prime }$ an algebraically closed extension of $K$, the square relating $X(K)\to Y(K)$ and $X(K^{\prime })\to Y(K^{\prime })$ is commutative, and since $X(K)\to X(K^{\prime })$ is injective and (by hypothesis) $X(K^{\prime })\to Y(K^{\prime })$ is injective, $X(K)\to Y(K)$ is necessarily injective.

Proposition (3.5.6). Let $f:X\to Y$, $g:Y\to Z$ be two morphisms of preschemes. (i) If $f$ and $g$ are radicial, so is $g\circ f$. (ii) Conversely, if $g\circ f$ is radicial, so is $f$.

Proof. By (3.5.4), the proposition reduces to the corresponding (evident) assertions for the maps $X(K)\to Y(K)\to Z(K)$.

Proposition (3.5.7). (i) If the $S$-morphisms $f:X\to X^{\prime }$, $g:Y\to Y^{\prime }$ are radicial, so is $f{\times }_{S}g$. (ii) If the $S$-morphism $f:X\to Y$ is radicial, so is $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ for every extension $S^{\prime }\to S$ of the base prescheme.

Proof. By (3.5.1) it suffices to prove (i). By (3.4.2.1), $(X{\times }_{S}Y)(K)=X(K){\times }_{S(K)}Y(K)$ and $(X^{\prime }{\times }_S{Y}^{\prime })(K)=X^{\prime }(K){\times }_{S(K)}{Y}^{\prime }(K)$; the map corresponding to $f{\times }_{S}g$ is then $(u,v)\mapsto (f\circ u,g\circ v)$, and the proposition follows.

Proposition (3.5.8). For a morphism $f=(\psi ,\theta ):X\to Y$ to be radicial, it is necessary and sufficient that $\psi$ be injective and that, for every $x\in X$, the monomorphism ${\theta }^x:\kappa (\psi (x))\to \kappa (x)$ make $\kappa (x)$ a radicial (purely inseparable) extension of $\kappa (\psi (x))$.

Proof. Suppose $f$ radicial. First, $\psi (x_1)=\psi (x_2)=y$ forces $x_1=x_2$: there is a field $K$, extension of $\kappa (y)$, with $\kappa (y)$-monomorphisms $\kappa (x_1)\to K$, $\kappa (x_2)\to K$ (Bourbaki, Alg., ch. V, §4, prop. 2); the corresponding morphisms $u_1,u_2:\mathrm{Spec}(K)\to X$ satisfy $f\circ u_1=f\circ u_2$, hence $u_1=u_2$, so $x_1=x_2$. Next, regarding $\kappa (x)$ as an extension of $\kappa (\psi (x))$ via ${\theta }^x$: if it were not radicial, there would be two distinct $\kappa (\psi (x))$-monomorphisms of $\kappa (x)$ into an algebraically closed extension $K$, and the two corresponding morphisms $\mathrm{Spec}(K)\to X$ would violate the hypothesis. Conversely, by (2.4.6), the stated conditions are immediately sufficient for $f$ to be radicial.

Corollary (3.5.9). If $A$ is a ring and $S$ a multiplicative subset of $A$, the canonical morphism $\mathrm{Spec}(S^{-1}A)\to \mathrm{Spec}(A)$ is radicial.

Proof. This morphism is a monomorphism (1.6.2).

Corollary (3.5.10). Let $f:X\to Y$ be a radicial morphism, $g:Y^{\prime }\to Y$ a morphism, and $X^{\prime }=X_{(Y^{\prime })}=X{\times }_Y{Y}^{\prime }$. Then the radicial morphism $f_{(Y^{\prime })}$ (3.5.7(ii)) is a bijection of the underlying space $X^{\prime }$ onto $g^{-1}(f(X))$; moreover, for every field $K$, the set $X^{\prime }(K)$ is identified with the subset of $Y^{\prime }(K)$ that is the inverse image, under $Y^{\prime }(K)\to Y(K)$ (corresponding to $g$), of the subset $X(K)$ of $Y(K)$.

Proof. The first assertion follows from (3.5.8) and (3.4.8); the second from the commutativity of the square relating $X^{\prime }(K)\to Y^{\prime }(K)$ and $X(K)\to Y(K)$.

Remark (3.5.11). We say a morphism $f=(\psi ,\theta )$ of preschemes is injective if the map $\psi$ is injective. For a morphism $f=(\psi ,\theta ):X\to Y$ to be radicial, it is necessary and sufficient that for every morphism $Y^{\prime }\to Y$ the morphism $f_{(Y^{\prime })}:X_{(Y^{\prime })}\to Y^{\prime }$ be injective (which justifies the terminology universally injective). The condition is necessary by (3.5.7(ii)) and (3.5.8). Conversely, it implies first that $\psi$ is injective; and if for some $x\in X$ the monomorphism ${\theta }^x:\kappa (\psi (x))\to \kappa (x)$ were not radicial, there would be an extension $K$ of $\kappa (\psi (x))$ and two distinct morphisms $\mathrm{Spec}(K)\to X$ over the same morphism $\mathrm{Spec}(K)\to Y$ (3.5.8); setting $Y^{\prime }=\mathrm{Spec}(K)$, there would be two distinct $Y^{\prime }$-sections of $X_{(Y^{\prime })}$ (3.3.14), contradicting the injectivity of $f_{(Y^{\prime })}$.

3.6. Fibers

Proposition (3.6.1). Let $f:X\to Y$ be a morphism, $y$ a point of $Y$, and ${\mathfrak{a}}_y$ an ideal of definition of ${\mathcal{O}}_y$ for the ${\mathfrak{m}}_y$-preadic topology. The projection $p:X{\times }_Y\mathrm{Spec}({\mathcal{O}}_y/{\mathfrak{a}}_y)\to X$ is a homeomorphism of the underlying space of $X{\times }_Y\mathrm{Spec}({\mathcal{O}}_y/{\mathfrak{a}}_y)$ onto the fiber $f^{-1}(y)$, equipped with the topology induced from that of $X$.