Chapter 6

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 3 June 2013

Preschemes

If $(X,{𝒪}_X)$ is a ringed space, an open subset $V$ of $X$ is said to be an affine open set if the ringed space $(V,{𝒪}_X|V)$ is isomorphic to some affine scheme.

Definition. A prescheme is a ringed space $(X,{𝒪}_X)$ such that every $x\in X$ has an affine open neighbourhood, i.e., it is a locally affine ringed space.

Let $(X,{𝒪}_X)$ be a prescheme.

Lemma (6.1).

(i)	The affine open sets form a basis of the topology of $X\text{.}$
(ii)	If $U$ is any open set in $X,$ the ringed space $(U,{𝒪}_X|U)$ is a prescheme, called the restriction of $(X,{𝒪}_X)$ to $U\text{.}$
(iii)	$X$ is a $T_0\text{-space.}$
(iv)	Every irreducible closed subset $F$ of $X$ has a unique generic point $x,$ and $x\mapsto \left\{\bar{x}\right\}$ is a one-one correspondence between the points of $X$ and the irreducible closed subsets of $X\text{.}$

		Proof.
	(i) Let $U$ be an open set in $X,$ and for each $x\in U$ let $V_x$ be an affine open neighbourhood of $x\text{;}$ then $U$ is the union of the sets $U_x=U\cap V_x\text{;}$ each $U_x$ is open in $V_x$ and is therefore a union of basic open sets contained in $V_x,$ by (3.3); and these basic open sets are affine by (5.1). Hence $U$ is a union of affine open sets. (ii) follows from (i). (iii) Let $x,y\in X,$ $x\ne y\text{.}$ If $x$ and $y$ are not in the same affine open set, it is clear that the $T_0$ condition is satisfied. If they are in the same affine open set, use the fact (Chapter 3) that an affine scheme is a $T_0\text{-space.}$ (iv) Let $y\in F$ and let $U$ be an affine open neighbourhood of $y$ in $X\text{.}$ Then $U\cap F$ is dense in $F$ (since $F$ is irreducible) and is itself irreducible, hence is the closure in $U$ of some $x\in U\text{.}$ Hence if $F\prime =\left\{\bar{x}\right\}$ is the closure of $\left\{x\right\}$ in $X,$ we have $F\prime \subseteq F$ (since $x\in F\text{);}$ but $U\cap F\prime =U\cap F,$ hence $U\cap (F\prime -F)=\varnothing ,$ hence $F\prime -F=\varnothing$ since $F\prime$ is irreducible. Hence $F=\left\{\bar{x}\right\}\text{.}$ The uniqueness of the generic point follows from (iii), for the $T_0\text{-axiom}$ is equivalent to the statement: $\left\{\bar{x}\right\}=\left\{\bar{u}\right\}\Rightarrow x=y\text{.}$ $\square$

Morphisms of preschemes

Let $(X,{𝒪}_X)$ and $(Y,{𝒪}_Y)$ be preschemes. A morphism of ringed spaces $(\psi ,\theta ):(X,{𝒪}_X)\to (Y,{𝒪}_Y)$ is a morphism of preschemes if, for each $x\in X,$ ${\theta }_x^{\#}$ is a local homomorphism ${𝒪}_{Y,\psi \left(x\right)}\to {𝒪}_{X,x}\text{.}$ Hence ${\theta }_x^{\#}$ defines a field monomorphism ${\theta }^x:k\left(\psi \left(x\right)\right)\to k\left(x\right),$ so that $k\left(x\right)$ is an extension of the field $k\left(\psi \left(x\right)\right)\text{.}$

Relative theory: S-preschemes

Let $S$ be a fixed prescheme. (Strictly speaking, we should write $(S,{𝒪}_S),$ but from now on we shall drop the structure sheaf from the notation.) An $S\text{-prescheme}$ is a pair $(X,f),$ where $X$ is a prescheme and $f:X\to S$ is a morphism of preschemes. If $S$ is the affine scheme of a ring $A,$ we speak of an $A\text{-prescheme}\text{.}$

If $(X,f)$ and $(Y,g)$ are $S\text{-preschemes,}$ an $S\text{-morphism}$ $\phi :X\to Y$ is a morphism of preschemes such that the diagram

$$\begin{array}{ccccc}X& \multicolumn{3}{c}{\stackrel{\phi }{⟶}}& Y\\ & f\searrow & & \swarrow g& \\ & & S\end{array}$$

is commutative.

The 'base prescheme' $S$ may be considered as a generalization of the ground field of algebraic geometry: if $A$ is the coordinate ring of an affine $k\text{-variety,}$ then $A$ is a $k\text{-algebra}$ with identity element, so we have a homomorphism $k\to A,$ hence $\text{Spec}\left(A\right)\to \text{Spec}\left(k\right)\text{.}$ Thus $\text{Spec}\left(A\right)$ is a $k\text{-prescheme.}$ (Of course, $\text{Spec}\left(k\right)$ consists of only one potnt, so the map $\text{Spec}\left(A\right)\to \text{Spec}\left(k\right)$ is trivial as a map of topological spaces; but a morphism of preschemes comprises also a map of the structure sheaves.)

Every prescheme may be considered canonically as a $\mathbb{Z}\text{-prescheme.}$ Namely the 'characteristic morphism' (Ex. 5, Chapter 3) is a morphism $\text{Spec}\left(A\right)\to \text{Spec}\left(\mathbb{Z}\right),$ and hence one defines a morphism of preschemes $X\to \text{Spec}\left(\mathbb{Z}\right)$ for any prescheme $X$ (do it on the affine open sets).

Products

Let $\text{C}$ be any category and for any two objects $X,T$ in $\text{C},$ let $X\left(T\right)$ denote the set of all morphisms $T\to X$ in $\text{C}\text{.}$ For fixed $X$ and variable $T,$ $T\mapsto X\left(T\right)$ is a contravariant functor $\text{C}\to \text{Sets}$ $\text{(}=$ category of all sets).

Let $F$ be any contravariant functor on $\text{C}$ with values in $\text{Sets}\text{.}$ $F$ is said to be representable if there exists an object $X$ in $\text{C}$ and a functorial isomorphism $F\left(T\right)\cong X\left(T\right)$ (for all $T\in \text{C}\text{).}$ If $X$ exists, it must be unique up to isomorphism.

A product of two objects $X,Y$ in $\text{C}$ is an object $X\times Y$ which (if it exists) represents the functor $T\mapsto X\left(T\right)\times Y\left(T\right)\text{:}$ in other words, there is a functorial isomorphism $(X\times Y)\left(T\right)\cong X\left(T\right)\times Y\left(T\right)$ for all objects $T$ in $\text{C}\text{.}$

Products exist in many categories: in the category of groups. ('direct products'), topological spaces, algebraic varieties, modules over a fixed ring (here the product is 'direct sum' $M\oplus N\text{),}$ etc. The dual concept is that of sum: in the category of groups, for example, sum is 'free product'; in the category of commutative $A\text{-algebras,}$ where $A$ is a fixed commutative ring, sum is tensor product over $A\text{.}$ Since the category of affine schemes over $A$ is dual to the category of $A\text{-algebras,}$ $\text{Spec}\left(B{\otimes }_{A}C\right)$ is a product of $\text{Spec}\left(B\right)$ and $\text{Spec}\left(C\right)$ in the category of affine schemes over $A$ (here $B,C$ are any two $A\text{-algebras).}$

Theorem (6.2). Let $S$ be a fixed prescheme, $X$ and $Y$ two $S\text{-preschemes.}$ Then the product $X{\times }_{S}Y$ exists in the category of $S\text{-preschemes.}$

The proof is tedious but not essentially difficult (EGA, I, 3.2.6), and we shall not reproduce it here. Locally, as we have just observed, it corresponds to the tensor product of rings, and it is a question of sticking things together so that it all fits.

This product of course has the usual associativity and commutativity properties, as in any category in which products exist.

The existence of products is fundamental, and arises in many contexts:

(1) Change of base. If $X$ is an $S\text{-prescheme}$ and if $S\prime \to S$ is a morphism of preschemes, then the product $X{\times }_{S}S\prime$ is denoted by $X_{\left(S\prime \right)}$ and is said to be obtained by extension of the base-prescheme from $S$ to $S\prime \text{.}$ We have a commutative diagram

$$\begin{array}{ccc}X& \leftarrow & X_{\left(S\prime \right)}\\ \downarrow & & \downarrow \\ S& \leftarrow & S\prime \end{array}$$

and $X_{\left(S\prime \right)}$ is to be regarded as an $S\prime \text{-prescheme.}$

Base extension is a transitive operation, i.e. if $S^{\prime \prime }\to S\prime \to S$ are morphisms of preschemes and $X$ is an $S\text{-prescheme,}$ then $\left(X{\times }_{S}S\prime \right){\times }_{S\prime }{S}^{\prime \prime }$ is canonically isomorphic to $X{\times }_S{S}^{\prime \prime }\text{.}$

This operation generalizes the notion of extension of the ground-field in algebraic geometry: if $X$ is a $k\text{-variety,}$ say affine with coordinate ring $A,$ and if $k\prime$ is an extension field of $k,$ then the embedding $k\to k\prime$ gives $\text{Spec}\left(k\prime \right)\to \text{Spec}\left(k\right),$ and $S{\otimes }_{k}k\prime$ gives rise to an affine variety $X_k,$ defined over $k\prime \text{.}$

(2) Geometrical points. If $X$ is an affine $(k,K)\text{-variety}$ (Chapter 1) with coordinate ring $A,$ then the points of $X$ are in one-one correspondence with the $k\text{-homomorphisms}$ $A\to K,$ i.e. with the $k\text{-morphisms}$ $\text{Spec}\left(K\right)\to \text{Spec}\left(A\right)\text{.}$ This motivates the following definition: if $X,T$ are $S\text{-preschemes,}$ the $S\text{-morphisms}$ $T\to X$ are called points of the $S\text{-prescheme}$ $X$ with values in the $S\text{-prescheme}$ $T\text{.}$ Let $$ {X\left(T\right)}_S$$ denote the set of points of $X$ with values in $T,$ then the product of two $S\text{-preschemes}$ $X$ and $Y$ is characterized by the formula $\left(X{\times }_{X}Y\right){\left(T\right)}_S=X{\left(T\right)}_S\times Y{\left(T\right)}_S,$ for any $S\text{-prescheme}$ $T\text{.}$

In particular, a geometrical point of $X$ is a point of $X$ with values in an algebraically closed field $K,$ that is to say it is a morphism $\phi :\text{Spec}\left(K\right)\to X\text{.}$ $\text{Spec}\left(K\right)$ consists of a single point, whose image under $\phi$ is the locality of the geometrical point. Given the locality $x,$ the geometrical point $\phi$ is determined by an embedding of the residue field $k\left(x\right)$ in $K\text{.}$

Remark. The product $X{\times }_{S}Y$ is not the set-theoretic product of $X$ and $Y,$ nor even the fibre product of $X$ and $Y$ over $S\text{:}$ that is to say, if $\left(X\right)$ temporarily denotes the set underlying $X,$ then in general we have $\left(X{\times }_{S}Y\right)\ne \left(X\right){\times }_{\left(S\right)}\left(Y\right)\text{.}$ However, there is a surjective mapping $f:\left(X{\times }_{S}Y\right)\to \left(X\right){\times }_{\left(S\right)}\left(Y\right)\text{.}$ For if $x\in X$ and $y\in Y$ lie over the same point $s\in S,$ then $k\left(x\right)$ and $k\left(y\right)$ are extensions of $k\left(s\right),$ and can therefore both be embedded in an extension $K$ of $k\left(s\right)\text{;}$ hence we have $S\text{-morphisms}$ $\text{Spec}\left(K\right)\to X$ and $\text{Spec}\left(K\right)\to Y,$ localized at $x$ and $y$ respectively, and therefore an $S\text{-morphism}$ $\text{Spec}\left(K\right)\to X\times Y,$ localized at say $z\text{.}$ Clearly the projections of $z$ are $x$ and $y,$ i.e. $f\left(z\right)=(x,y)\text{.}$

To show that $f$ is not in general injective, it is enough to take $X,Y,S$ to be the spectra of fields $K,L,k$ respectively $\text{(}K,L$ being extensions of $k\text{);}$ then $K{\otimes }_{k}L$ in general has more than one prime ideal.

In fact it is not difficult to show that if $x\in X$ and $y\in Y$ lie over the same point $s\in S,$ then the points $z$ of $Z{\times }_{S}Y$ such that $f\left(z\right)=(x,y)$ are in one-one correspondence with the isomorphism types of composite extensions of $k\left(x\right)$ and $k\left(y\right)$ over $k\left(s\right)$ (E.G.A. I, 3.4.9).

(3) Fibres. Let $f:X\to Y$ be a morphism of preschemes and let $y$ be a point of $Y\text{.}$ Then the projection $p:X{\times }_Y\text{Spec}\left(k\left(y\right)\right)\to X$ is a homeomorphism of the space underlying $X{\times }_Y\text{Spec}\left(k\left(y\right)\right)$ onto the fibre $f^{-1}\left(y\right)$ (E.G.A, I, 3.6.1). Hence the fibre $f^{-1}\left(y\right)$ can be regarded as a prescheme over the field $k\left(y\right)\text{:}$ as such we denote it by $X_y\text{.}$ If $x\in f^{-1}\left(y\right)$ and $p\left(x\prime \right)=x,$ where $x\prime \in X{\times }_Y\text{Spec}\left(k\left(y\right)\right),$ it turns out that the residue fields $k\left(x\right)$ and $k\left(x\prime \right)$ are the same, i.e. the residue field $k\left(x\right)$ is the same whether $x$ is regarded as a point of the prescheme $X$ or as a point of the prescheme $X_y\text{.}$

(4) Separated morphisms. Schemes. Whenever the product $X\times X$ is defined in a category $\text{C}$ $\text{(}X$ being an object of $\text{C}\text{),}$ there is a well defined diagonal morphism ${\Delta }_X:X\to X\times X\text{.}$ ${\Delta }_X$ is the element of $(X\times X)\hspace{0.17em}\left(X\right)$ corresponding to $({\text{id}}_X,{\text{id}}_X)$ in $X\left(X\right)\times X\left(X\right)$ $\text{(}{\text{id}}_X=$ identity morphism of $X\text{).}$ Hence if $f:X\to X$ is a morphism of preschemes, we have a diagonal morphism ${\Delta }_{X|S}:X\to X{\times }_{S}X\text{.}$ If $X=\text{Spec}\left(A\right),$ $S=\text{Spec}\left(B\right)$ then $\Delta$ corresponds to the homomorphism $A{\otimes }_{B}A\to A$ which maps $x\otimes y$ to $xy\text{;}$ this homomorphism is surjective and therefore $\Delta$ is in this case a homeomorphism of $X$ onto a closed subset of the product.

If $X,S$ are arbitrary preschemes, let $p_1:X{\times }_{S}X\to X$ be the projection on the first factor; then $p_1\circ \Delta$ is the identity map of $X$ and therefore $\Delta$ is a homeomorphism of $X$ onto $\Delta \left(X\right)\text{.}$ If $\left(U_{\alpha }\right)$ is a covering of $X$ by affine open sets, then $\Delta \left(X\right)\cap (U_{\alpha }\times U_{\alpha })$ is the diagonal of $U_{\alpha }\times U_{\alpha },$ hence closed in $U_{\alpha }\times U_{\alpha }\text{;}$ and $\Delta \left(X\right)$ is contained in $\underset{\alpha }{\cup }(U_{\alpha }\times U_{\alpha }),$ hence $\Delta \left(X\right)$ is locally closed (Chapter 2) (but not necessarily closed) in $X\times X\text{.}$

The morphism $f:X\to S$ is said to be separated, or $X$ is separated over $S,$ if $\Delta \left(X\right)$ is closed in $X{\times }_{S}X\text{.}$

A prescheme $X$ is a scheme if it is separated over $\mathbb{Z},$ i.e. if the 'characteristic morphism' $X\to \text{Spec}\left(\mathbb{Z}\right)$ is separated. This is the formal analogue of Hausdorff's axiom, or of Serre's second axiom for algebraic varieties (see Chapter 1).

Remark. If $X=\text{Spec}\left(A\right),$ then $A{\otimes }_{\mathbb{Z}}A\to A$ is surjective, as we have remarked above, and therefore the characteristic morphism $X\to \text{Spec}\left(\mathbb{Z}\right)$ is separated. This justifies the terminology 'affine scheme' rather than 'affine prescheme'.

If $U,V$ are two affine open sets in a prescheme $X,$ then $U\cap V$ need not be affine. But if $X$ is a scheme, $U\cap V$ will be affine, for $U\cap V$ is isomorphic to $\Delta \left(X\right)\cap \left(U{\times }_{\mathbb{Z}}V\right),$ hence is closed in $U{\times }_{\mathbb{Z}}V$ and therefore affine.

(5) Proper morphisms. A morphism of preschemes $f:X\to S$ is of finite type if $S$ is a union of affine open sets $V_{\alpha }$ such that each $f^{-1}\left(V_{\alpha }\right)$ is a finite union of affine open sets $U_{i\alpha }$ with the property that each ring $A\left(U_{i\alpha }\right)$ is finitely generated as an algebra over $A\left(V_{\alpha }\right)$ (here, if $U$ is an affine scheme, $A\left(U\right)$ denotes the associated ring). If $X$ and $S$ are both affine, say $X=\text{Spec}\left(A\right),$ $S=\text{Spec}\left(B\right),$ then $f:X\to S$ is of finite type if and only if $A$ is finitely generated as a $B\text{-algebra}$ (E.G.A., I, 6.3.3).

A morphism $f:X\to S$ is proper if

(i)	$f$ is separated and of finite type;
(ii)	$f$ is universally closed, i.e. for every morphism $S\prime \to S$ the projection $X_{\left(S\prime \right)}=X{\times }_{S}S\prime \to S\prime$ is a closed mapping.

This is the generalization of the notion of completeness for an algebraic variety over a field (cf. Chapter I).

Notes and References

This is a typed excerpt of the book "Algebraic Geometry: Introduction to Schemes - I.G. Macdonald".

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