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 \input{preamble} % OK, start here. % \begin{document} \title{Morphisms of Schemes} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we introduce some types of morphisms of schemes. A basic reference is \cite{EGA}. \section{Closed immersions} \label{section-closed-immersions} \noindent In this section we elucidate some of the results obtained previously on closed immersions of schemes. Recall that a morphism of schemes $i : Z \to X$ is defined to be a closed immersion if (a) $i$ induces a homeomorphism onto a closed subset of $X$, (b) $i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z$ is surjective, and (c) the kernel of $i^\sharp$ is locally generated by sections, see Schemes, Definitions \ref{schemes-definition-immersion} and \ref{schemes-definition-closed-immersion-locally-ringed-spaces}. It turns out that, given that $Z$ and $X$ are schemes, there are many different ways of characterizing a closed immersion. \begin{lemma} \label{lemma-closed-immersion} Let $i : Z \to X$ be a morphism of schemes. The following are equivalent: \begin{enumerate} \item The morphism $i$ is a closed immersion. \item For every affine open $\Spec(R) = U \subset X$, there exists an ideal $I \subset R$ such that $i^{-1}(U) = \Spec(R/I)$ as schemes over $U = \Spec(R)$. \item There exists an affine open covering $X = \bigcup_{j \in J} U_j$, $U_j = \Spec(R_j)$ and for every $j \in J$ there exists an ideal $I_j \subset R_j$ such that $i^{-1}(U_j) = \Spec(R_j/I_j)$ as schemes over $U_j = \Spec(R_j)$. \item The morphism $i$ induces a homeomorphism of $Z$ with a closed subset of $X$ and $i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z$ is surjective. \item The morphism $i$ induces a homeomorphism of $Z$ with a closed subset of $X$, the map $i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z$ is surjective, and the kernel $\Ker(i^\sharp)\subset \mathcal{O}_X$ is a quasi-coherent sheaf of ideals. \item The morphism $i$ induces a homeomorphism of $Z$ with a closed subset of $X$, the map $i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z$ is surjective, and the kernel $\Ker(i^\sharp)\subset \mathcal{O}_X$ is a sheaf of ideals which is locally generated by sections. \end{enumerate} \end{lemma} \begin{proof} Condition (6) is our definition of a closed immersion, see Schemes, Definitions \ref{schemes-definition-closed-immersion-locally-ringed-spaces} and \ref{schemes-definition-immersion}. So (6) $\Leftrightarrow$ (1). We have (1) $\Rightarrow$ (2) by Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme}. Trivially (2) $\Rightarrow$ (3). \medskip\noindent Assume (3). Each of the morphisms $\Spec(R_j/I_j) \to \Spec(R_j)$ is a closed immersion, see Schemes, Example \ref{schemes-example-closed-immersion-affines}. Hence $i^{-1}(U_j) \to U_j$ is a homeomorphism onto its image and $i^\sharp|_{U_j}$ is surjective. Hence $i$ is a homeomorphism onto its image and $i^\sharp$ is surjective since this may be checked locally. We conclude that (3) $\Rightarrow$ (4). \medskip\noindent The implication (4) $\Rightarrow$ (1) is Schemes, Lemma \ref{schemes-lemma-characterize-closed-immersions}. The implication (5) $\Rightarrow$ (6) is trivial. And the implication (6) $\Rightarrow$ (5) follows from Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme}. \end{proof} \begin{lemma} \label{lemma-closed-immersion-ideals} Let $X$ be a scheme. Let $i : Z \to X$ and $i' : Z' \to X$ be closed immersions and consider the ideal sheaves $\mathcal{I} = \Ker(i^\sharp)$ and $\mathcal{I}' = \Ker((i')^\sharp)$ of $\mathcal{O}_X$. \begin{enumerate} \item The morphism $i : Z \to X$ factors as $Z \to Z' \to X$ for some $a : Z \to Z'$ if and only if $\mathcal{I}' \subset \mathcal{I}$. If this happens, then $a$ is a closed immersion. \item We have $Z \cong Z'$ over $X$ if and only if $\mathcal{I} = \mathcal{I}'$. \end{enumerate} \end{lemma} \begin{proof} This follows from our discussion of closed subspaces in Schemes, Section \ref{schemes-section-closed-immersion} especially Schemes, Lemmas \ref{schemes-lemma-closed-immersion} and \ref{schemes-lemma-characterize-closed-subspace}. It also follows in a straightforward way from characterization (3) in Lemma \ref{lemma-closed-immersion} above. \end{proof} \begin{lemma} \label{lemma-closed-immersion-bijection-ideals} Let $X$ be a scheme. Let $\mathcal{I} \subset \mathcal{O}_X$ be a sheaf of ideals. The following are equivalent: \begin{enumerate} \item $\mathcal{I}$ is locally generated by sections as a sheaf of $\mathcal{O}_X$-modules, \item $\mathcal{I}$ is quasi-coherent as a sheaf of $\mathcal{O}_X$-modules, and \item there exists a closed immersion $i : Z \to X$ of schemes whose corresponding sheaf of ideals $\Ker(i^\sharp)$ is equal to $\mathcal{I}$. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (2) is immediate from Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme}. If (1) holds, then there is a closed subspace $i : Z \to X$ with $\mathcal{I} = \Ker(i^\sharp)$ by Schemes, Definition \ref{schemes-definition-closed-subspace} and Example \ref{schemes-example-closed-subspace}. By Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} this is a closed immersion of schemes and (3) holds. Conversely, if (3) holds, then (2) holds by Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} (which applies because a closed immersion of schemes is a fortiori a closed immersion of locally ringed spaces). \end{proof} \begin{lemma} \label{lemma-base-change-closed-immersion} The base change of a closed immersion is a closed immersion. \end{lemma} \begin{proof} See Schemes, Lemma \ref{schemes-lemma-base-change-immersion}. \end{proof} \begin{lemma} \label{lemma-composition-closed-immersion} A composition of closed immersions is a closed immersion. \end{lemma} \begin{proof} We have seen this in Schemes, Lemma \ref{schemes-lemma-composition-immersion}, but here is another proof. Namely, it follows from the characterization (3) of closed immersions in Lemma \ref{lemma-closed-immersion}. Since if $I \subset R$ is an ideal, and $\overline{J} \subset R/I$ is an ideal, then $\overline{J} = J/I$ for some ideal $J \subset R$ which contains $I$ and $(R/I)/\overline{J} = R/J$. \end{proof} \begin{lemma} \label{lemma-closed-immersion-quasi-compact} A closed immersion is quasi-compact. \end{lemma} \begin{proof} This lemma is a duplicate of Schemes, Lemma \ref{schemes-lemma-closed-immersion-quasi-compact}. \end{proof} \begin{lemma} \label{lemma-closed-immersion-separated} A closed immersion is separated. \end{lemma} \begin{proof} This lemma is a special case of Schemes, Lemma \ref{schemes-lemma-immersions-monomorphisms}. \end{proof} \section{Immersions} \label{section-immersions} \noindent In this section we collect some facts on immersions. \begin{lemma} \label{lemma-immersion-permanence} Let $Z \to Y \to X$ be morphisms of schemes. \begin{enumerate} \item If $Z \to X$ is an immersion, then $Z \to Y$ is an immersion. \item If $Z \to X$ is a quasi-compact immersion and $Z \to Y$ is quasi-separated, then $Z \to Y$ is a quasi-compact immersion. \item If $Z \to X$ is a closed immersion and $Y \to X$ is separated, then $Z \to Y$ is a closed immersion. \end{enumerate} \end{lemma} \begin{proof} In each case the proof is to contemplate the commutative diagram $$\xymatrix{ Z \ar[r] \ar[rd] & Y \times_X Z \ar[r] \ar[d] & Z \ar[d] \\ & Y \ar[r] & X }$$ where the composition of the top horizontal arrows is the identity. Let us prove (1). The first horizontal arrow is a section of $Y \times_X Z \to Z$, whence an immersion by Schemes, Lemma \ref{schemes-lemma-section-immersion}. The arrow $Y \times_X Z \to Y$ is a base change of $Z \to X$ hence an immersion (Schemes, Lemma \ref{schemes-lemma-base-change-immersion}). Finally, a composition of immersions is an immersion (Schemes, Lemma \ref{schemes-lemma-composition-immersion}). This proves (1). The other two results are proved in exactly the same manner. \end{proof} \begin{lemma} \label{lemma-factor-quasi-compact-immersion} Let $h : Z \to X$ be an immersion. If $h$ is quasi-compact, then we can factor $h = i \circ j$ with $j : Z \to \overline{Z}$ an open immersion and $i : \overline{Z} \to X$ a closed immersion. \end{lemma} \begin{proof} Note that $h$ is quasi-compact and quasi-separated (see Schemes, Lemma \ref{schemes-lemma-immersions-monomorphisms}). Hence $h_*\mathcal{O}_Z$ is a quasi-coherent sheaf of $\mathcal{O}_X$-modules by Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}. This implies that $\mathcal{I} = \Ker(\mathcal{O}_X \to h_*\mathcal{O}_Z)$ is a quasi-coherent sheaf of ideals, see Schemes, Section \ref{schemes-section-quasi-coherent}. Let $\overline{Z} \subset X$ be the closed subscheme corresponding to $\mathcal{I}$, see Lemma \ref{lemma-closed-immersion-bijection-ideals}. By Schemes, Lemma \ref{schemes-lemma-characterize-closed-subspace} the morphism $h$ factors as $h = i \circ j$ where $i : \overline{Z} \to X$ is the inclusion morphism. To see that $j$ is an open immersion, choose an open subscheme $U \subset X$ such that $h$ induces a closed immersion of $Z$ into $U$. Then it is clear that $\mathcal{I}|_U$ is the sheaf of ideals corresponding to the closed immersion $Z \to U$. Hence we see that $Z = \overline{Z} \cap U$. \end{proof} \begin{lemma} \label{lemma-factor-reduced-immersion} Let $h : Z \to X$ be an immersion. If $Z$ is reduced, then we can factor $h = i \circ j$ with $j : Z \to \overline{Z}$ an open immersion and $i : \overline{Z} \to X$ a closed immersion. \end{lemma} \begin{proof} Let $\overline{Z} \subset X$ be the closure of $h(Z)$ with the reduced induced closed subscheme structure, see Schemes, Definition \ref{schemes-definition-reduced-induced-scheme}. By Schemes, Lemma \ref{schemes-lemma-map-into-reduction} the morphism $h$ factors as $h = i \circ j$ with $i : \overline{Z} \to X$ the inclusion morphism and $j : Z \to \overline{Z}$. From the definition of an immersion we see there exists an open subscheme $U \subset X$ such that $h$ factors through a closed immersion into $U$. Hence $\overline{Z} \cap U$ and $h(Z)$ are reduced closed subschemes of $U$ with the same underlying closed set. Hence by the uniqueness in Schemes, Lemma \ref{schemes-lemma-reduced-closed-subscheme} we see that $h(Z) \cong \overline{Z} \cap U$. So $j$ induces an isomorphism of $Z$ with $\overline{Z} \cap U$. In other words $j$ is an open immersion. \end{proof} \begin{example} \label{example-thibaut} Here is an example of an immersion which is not a composition of an open immersion followed by a closed immersion. Let $k$ be a field. Let $X = \Spec(k[x_1, x_2, x_3, \ldots])$. Let $U = \bigcup_{n = 1}^{\infty} D(x_n)$. Then $U \to X$ is an open immersion. Consider the ideals $$I_n = (x_1^n, x_2^n, \ldots, x_{n - 1}^n, x_n - 1, x_{n + 1}, x_{n + 2}, \ldots) \subset k[x_1, x_2, x_3, \ldots][1/x_n].$$ Note that $I_n k[x_1, x_2, x_3, \ldots][1/x_nx_m] = (1)$ for any $m \not = n$. Hence the quasi-coherent ideals $\widetilde I_n$ on $D(x_n)$ agree on $D(x_nx_m)$, namely $\widetilde I_n|_{D(x_nx_m)} = \mathcal{O}_{D(x_n x_m)}$ if $n \not = m$. Hence these ideals glue to a quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_U$. Let $Z \subset U$ be the closed subscheme corresponding to $\mathcal{I}$. Thus $Z \to X$ is an immersion. \medskip\noindent We claim that we cannot factor $Z \to X$ as $Z \to \overline{Z} \to X$, where $\overline{Z} \to X$ is closed and $Z \to \overline{Z}$ is open. Namely, $\overline{Z}$ would have to be defined by an ideal $I \subset k[x_1, x_2, x_3, \ldots]$ such that $I_n = I k[x_1, x_2, x_3, \ldots][1/x_n]$. But the only element $f \in k[x_1, x_2, x_3, \ldots]$ which ends up in all $I_n$ is $0$! Hence $I$ does not exist. \end{example} \section{Closed immersions and quasi-coherent sheaves} \label{section-closed-immersions-quasi-coherent} \noindent The following lemma finally does for quasi-coherent sheaves on schemes what Modules, Lemma \ref{modules-lemma-i-star-exact} does for abelian sheaves. See also the discussion in Modules, Section \ref{modules-section-closed-immersion}. \begin{lemma} \label{lemma-i-star-equivalence} Let $i : Z \to X$ be a closed immersion of schemes. Let $\mathcal{I} \subset \mathcal{O}_X$ be the quasi-coherent sheaf of ideals cutting out $Z$. The functor $$i_* : \QCoh(\mathcal{O}_Z) \longrightarrow \QCoh(\mathcal{O}_X)$$ is exact, fully faithful, with essential image those quasi-coherent $\mathcal{O}_X$-modules $\mathcal{G}$ such that $\mathcal{I}\mathcal{G} = 0$. \end{lemma} \begin{proof} A closed immersion is quasi-compact and separated, see Lemmas \ref{lemma-closed-immersion-quasi-compact} and \ref{lemma-closed-immersion-separated}. Hence Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent} applies and the pushforward of a quasi-coherent sheaf on $Z$ is indeed a quasi-coherent sheaf on $X$. \medskip\noindent By Modules, Lemma \ref{modules-lemma-i-star-equivalence} the functor $i_*$ is fully faithful. \medskip\noindent Now we turn to the description of the essential image of the functor $i_*$. It is clear that $\mathcal{I}(i_*\mathcal{F}) = 0$ for any quasi-coherent $\mathcal{O}_Z$-module, for example by our local description above. Next, suppose that $\mathcal{G}$ is any quasi-coherent $\mathcal{O}_X$-module such that $\mathcal{I}\mathcal{G} = 0$. It suffices to show that the canonical map $$\mathcal{G} \longrightarrow i_* i^*\mathcal{G}$$ is an isomorphism. By exactly the same arguments as above we see that it suffices to prove the following algebraic statement: Given a ring $R$, an ideal $I$ and an $R$-module $N$ such that $IN = 0$ the canonical map $$N \longrightarrow N \otimes_R R/I,\quad n \longmapsto n \otimes 1$$ is an isomorphism of $R$-modules. Proof of this easy algebra fact is omitted. \end{proof} \noindent Let $i : Z \to X$ be a closed immersion. Because of the lemma above we often, by abuse of notation, denote $\mathcal{F}$ the sheaf $i_*\mathcal{F}$ on $X$. \begin{lemma} \label{lemma-largest-quasi-coherent-subsheaf} Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module. Let $\mathcal{G} \subset \mathcal{F}$ be a $\mathcal{O}_X$-submodule. There exists a unique quasi-coherent $\mathcal{O}_X$-submodule $\mathcal{G}' \subset \mathcal{G}$ with the following property: For every quasi-coherent $\mathcal{O}_X$-module $\mathcal{H}$ the map $$\Hom_{\mathcal{O}_X}(\mathcal{H}, \mathcal{G}') \longrightarrow \Hom_{\mathcal{O}_X}(\mathcal{H}, \mathcal{G})$$ is bijective. In particular $\mathcal{G}'$ is the largest quasi-coherent $\mathcal{O}_X$-submodule of $\mathcal{F}$ contained in $\mathcal{G}$. \end{lemma} \begin{proof} Let $\mathcal{G}_a$, $a \in A$ be the set of quasi-coherent $\mathcal{O}_X$-submodules contained in $\mathcal{G}$. Then the image $\mathcal{G}'$ of $$\bigoplus\nolimits_{a \in A} \mathcal{G}_a \longrightarrow \mathcal{F}$$ is quasi-coherent as the image of a map of quasi-coherent sheaves on $X$ is quasi-coherent and since a direct sum of quasi-coherent sheaves is quasi-coherent, see Schemes, Section \ref{schemes-section-quasi-coherent}. The module $\mathcal{G}'$ is contained in $\mathcal{G}$. Hence this is the largest quasi-coherent $\mathcal{O}_X$-module contained in $\mathcal{G}$. \medskip\noindent To prove the formula, let $\mathcal{H}$ be a quasi-coherent $\mathcal{O}_X$-module and let $\alpha : \mathcal{H} \to \mathcal{G}$ be an $\mathcal{O}_X$-module map. The image of the composition $\mathcal{H} \to \mathcal{G} \to \mathcal{F}$ is quasi-coherent as the image of a map of quasi-coherent sheaves. Hence it is contained in $\mathcal{G}'$. Hence $\alpha$ factors through $\mathcal{G}'$ as desired. \end{proof} \begin{lemma} \label{lemma-i-upper-shriek} Let $i : Z \to X$ be a closed immersion of schemes. There is a functor\footnote{This is likely nonstandard notation.} $i^! : \QCoh(\mathcal{O}_X) \to \QCoh(\mathcal{O}_Z)$ which is a right adjoint to $i_*$. (Compare Modules, Lemma \ref{modules-lemma-i-star-right-adjoint}.) \end{lemma} \begin{proof} Given quasi-coherent $\mathcal{O}_X$-module $\mathcal{G}$ we consider the subsheaf $\mathcal{H}_Z(\mathcal{G})$ of $\mathcal{G}$ of local sections annihilated by $\mathcal{I}$. By Lemma \ref{lemma-largest-quasi-coherent-subsheaf} there is a canonical largest quasi-coherent $\mathcal{O}_X$-submodule $\mathcal{H}_Z(\mathcal{G})'$. By construction we have $$\Hom_{\mathcal{O}_X}(i_*\mathcal{F}, \mathcal{H}_Z(\mathcal{G})') = \Hom_{\mathcal{O}_X}(i_*\mathcal{F}, \mathcal{G})$$ for any quasi-coherent $\mathcal{O}_Z$-module $\mathcal{F}$. Hence we can set $i^!\mathcal{G} = i^*(\mathcal{H}_Z(\mathcal{G})')$. Details omitted. \end{proof} \noindent Using the $1$-to-$1$ corresponding between quasi-coherent sheaves of ideals and closed subschemes (see Lemma \ref{lemma-closed-immersion-bijection-ideals}) we can define scheme theoretic intersections and unions of closed subschemes. \begin{definition} \label{definition-scheme-theoretic-intersection-union} Let $X$ be a scheme. Let $Z, Y \subset X$ be closed subschemes corresponding to quasi-coherent ideal sheaves $\mathcal{I}, \mathcal{J} \subset \mathcal{O}_X$. The {\it scheme theoretic intersection} of $Z$ and $Y$ is the closed subscheme of $X$ cut out by $\mathcal{I} + \mathcal{J}$. Then {\it scheme theoretic union} of $Z$ and $Y$ is the closed subscheme of $X$ cut out by $\mathcal{I} \cap \mathcal{J}$. \end{definition} \begin{lemma} \label{lemma-scheme-theoretic-intersection} Let $X$ be a scheme. Let $Z, Y \subset X$ be closed subschemes. Let $Z \cap Y$ be the scheme theoretic intersection of $Z$ and $Y$. Then $Z \cap Y \to Z$ and $Z \cap Y \to Y$ are closed immersions and $$\xymatrix{ Z \cap Y \ar[r] \ar[d] & Z \ar[d] \\ Y \ar[r] & X }$$ is a cartesian diagram of schemes, i.e., $Z \cap Y = Z \times_X Y$. \end{lemma} \begin{proof} The morphisms $Z \cap Y \to Z$ and $Z \cap Y \to Y$ are closed immersions by Lemma \ref{lemma-closed-immersion-ideals}. Let $U = \Spec(A)$ be an affine open of $X$ and let $Z \cap U$ and $Y \cap U$ correspond to the ideals $I \subset A$ and $J \subset A$. Then $Z \cap Y \cap U$ corresponds to $I + J \subset A$. Since $A/I \otimes_A A/J = A/(I + J)$ we see that the diagram is cartesian by our description of fibre products of schemes in Schemes, Section \ref{schemes-section-fibre-products}. \end{proof} \begin{lemma} \label{lemma-scheme-theoretic-union} Let $S$ be a scheme. Let $X, Y \subset S$ be closed subschemes. Let $X \cup Y$ be the scheme theoretic union of $X$ and $Y$. Let $X \cap Y$ be the scheme theoretic intersection of $X$ and $Y$. Then $X \to X \cup Y$ and $Y \to X \cup Y$ are closed immersions, there is a short exact sequence $$0 \to \mathcal{O}_{X \cup Y} \to \mathcal{O}_X \times \mathcal{O}_Y \to \mathcal{O}_{X \cap Y} \to 0$$ of $\mathcal{O}_S$-modules, and the diagram $$\xymatrix{ X \cap Y \ar[r] \ar[d] & X \ar[d] \\ Y \ar[r] & X \cup Y }$$ is cocartesian in the category of schemes, i.e., $X \cup Y = X \amalg_{X \cap Y} Y$. \end{lemma} \begin{proof} The morphisms $X \to X \cup Y$ and $Y \to X \cup Y$ are closed immersions by Lemma \ref{lemma-closed-immersion-ideals}. In the short exact sequence we use the equivalence of Lemma \ref{lemma-i-star-equivalence} to think of quasi-coherent modules on closed subschemes of $S$ as quasi-coherent modules on $S$. For the first map in the sequence we use the canonical maps $\mathcal{O}_{X \cup Y} \to \mathcal{O}_X$ and $\mathcal{O}_{X \cup Y} \to \mathcal{O}_Y$ and for the second map we use the canonical map $\mathcal{O}_X \to \mathcal{O}_{X \cap Y}$ and the negative of the canonical map $\mathcal{O}_Y \to \mathcal{O}_{X \cap Y}$. Then to check exactness we may work affine locally. Let $U = \Spec(A)$ be an affine open of $S$ and let $X \cap U$ and $Y \cap U$ correspond to the ideals $I \subset A$ and $J \subset A$. Then $(X \cup Y) \cap U$ corresponds to $I \cap J \subset A$ and $X \cap Y \cap U$ corresponds to $I + J \subset A$. Thus exactness follows from the exactness of $$0 \to A/I \cap J \to A/I \times A/J \to A/(I + J) \to 0$$ To show the diagram is cocartesian, suppose we are given a scheme $T$ and morphisms of schemes $f : X \to T$, $g : Y \to T$ agreeing as morphisms $X \cap Y \to T$. Goal: Show there exists a unique morphism $h : X \cup Y \to T$ agreeing with $f$ and $g$. To construct $h$ we may work affine locally on $X \cup Y$, see Schemes, Section \ref{schemes-section-glueing-schemes}. If $s \in X$, $s \not \in Y$, then $X \to X \cup Y$ is an isomorphism in a neighbourhood of $s$ and it is clear how to construct $h$. Similarly for $s \in Y$, $s \not \in X$. For $s \in X \cap Y$ we can pick an affine open $V = \Spec(B) \subset T$ containing $f(s) = g(s)$. Then we can choose an affine open $U = \Spec(A) \subset S$ containing $s$ such that $f(X \cap U)$ and $g(Y \cap U)$ are contained in $V$. The morphisms $f|_{X \cap U}$ and $g|_{Y \cap V}$ into $V$ correspond to ring maps $$B \to A/I \quad\text{and}\quad B \to A/J$$ which agree as maps into $A/(I + J)$. By the short exact sequence displayed above there is a unique lift of these ring homomorphism to a ring map $B \to A/I \cap J$ as desired. \end{proof} \section{Supports of modules} \label{section-support} \noindent In this section we collect some elementary results on supports of quasi-coherent modules on schemes. Recall that the support of a sheaf of modules has been defined in Modules, Section \ref{modules-section-support}. On the other hand, the support of a module was defined in Algebra, Section \ref{algebra-section-support}. These match. \begin{lemma} \label{lemma-support-affine-open} Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $\Spec(A) = U \subset X$ be an affine open, and set $M = \Gamma(U, \mathcal{F})$. Let $x \in U$, and let $\mathfrak p \subset A$ be the corresponding prime. The following are equivalent \begin{enumerate} \item $\mathfrak p$ is in the support of $M$, and \item $x$ is in the support of $\mathcal{F}$. \end{enumerate} \end{lemma} \begin{proof} This follows from the equality $\mathcal{F}_x = M_{\mathfrak p}$, see Schemes, Lemma \ref{schemes-lemma-spec-sheaves} and the definitions. \end{proof} \begin{lemma} \label{lemma-support-closed-specialization} Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. The support of $\mathcal{F}$ is closed under specialization. \end{lemma} \begin{proof} If $x' \leadsto x$ is a specialization and $\mathcal{F}_x = 0$ then $\mathcal{F}_{x'}$ is zero, as $\mathcal{F}_{x'}$ is a localization of the module $\mathcal{F}_x$. Hence the complement of $\text{Supp}(\mathcal{F})$ is closed under generalization. \end{proof} \noindent For finite type quasi-coherent modules the support is closed, can be checked on fibres, and commutes with base change. \begin{lemma} \label{lemma-support-finite-type} Let $\mathcal{F}$ be a finite type quasi-coherent module on a scheme $X$. Then \begin{enumerate} \item The support of $\mathcal{F}$ is closed. \item For $x \in X$ we have $$x \in \text{Supp}(\mathcal{F}) \Leftrightarrow \mathcal{F}_x \not = 0 \Leftrightarrow \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) \not = 0.$$ \item For any morphism of schemes $f : Y \to X$ the pullback $f^*\mathcal{F}$ is of finite type as well and we have $\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$. \end{enumerate} \end{lemma} \begin{proof} Part (1) is a reformulation of Modules, Lemma \ref{modules-lemma-support-finite-type-closed}. You can also combine Lemma \ref{lemma-support-affine-open}, Properties, Lemma \ref{properties-lemma-finite-type-module}, and Algebra, Lemma \ref{algebra-lemma-support-closed} to see this. The first equivalence in (2) is the definition of support, and the second equivalence follows from Nakayama's lemma, see Algebra, Lemma \ref{algebra-lemma-NAK}. Let $f : Y \to X$ be a morphism of schemes. Note that $f^*\mathcal{F}$ is of finite type by Modules, Lemma \ref{modules-lemma-pullback-finite-type}. For the final assertion, let $y \in Y$ with image $x \in X$. Recall that $$(f^*\mathcal{F})_y = \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y},$$ see Sheaves, Lemma \ref{sheaves-lemma-stalk-pullback-modules}. Hence $(f^*\mathcal{F})_y \otimes \kappa(y)$ is nonzero if and only if $\mathcal{F}_x \otimes \kappa(x)$ is nonzero. By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of assertion (3). \end{proof} \begin{lemma} \label{lemma-scheme-theoretic-support} Let $\mathcal{F}$ be a finite type quasi-coherent module on a scheme $X$. There exists a smallest closed subscheme $i : Z \to X$ such that there exists a quasi-coherent $\mathcal{O}_Z$-module $\mathcal{G}$ with $i_*\mathcal{G} \cong \mathcal{F}$. Moreover: \begin{enumerate} \item If $\Spec(A) \subset X$ is any affine open, and $\mathcal{F}|_{\Spec(A)} = \widetilde{M}$ then $Z \cap \Spec(A) = \Spec(A/I)$ where $I = \text{Ann}_A(M)$. \item The quasi-coherent sheaf $\mathcal{G}$ is unique up to unique isomorphism. \item The quasi-coherent sheaf $\mathcal{G}$ is of finite type. \item The support of $\mathcal{G}$ and of $\mathcal{F}$ is $Z$. \end{enumerate} \end{lemma} \begin{proof} Suppose that $i' : Z' \to X$ is a closed subscheme which satisfies the description on open affines from the lemma. Then by Lemma \ref{lemma-i-star-equivalence} we see that $\mathcal{F} \cong i'_*\mathcal{G}'$ for some unique quasi-coherent sheaf $\mathcal{G}'$ on $Z'$. Furthermore, it is clear that $Z'$ is the smallest closed subscheme with this property (by the same lemma). Finally, using Properties, Lemma \ref{properties-lemma-finite-type-module} and Algebra, Lemma \ref{algebra-lemma-finite-over-subring} it follows that $\mathcal{G}'$ is of finite type. We have $\text{Supp}(\mathcal{G}') = Z$ by Algebra, Lemma \ref{algebra-lemma-support-closed}. Hence, in order to prove the lemma it suffices to show that the characterization in (1) actually does define a closed subscheme. And, in order to do this it suffices to prove that the given rule produces a quasi-coherent sheaf of ideals, see Lemma \ref{lemma-closed-immersion-bijection-ideals}. This comes down to the following algebra fact: If $A$ is a ring, $f \in A$, and $M$ is a finite $A$-module, then $\text{Ann}_A(M)_f = \text{Ann}_{A_f}(M_f)$. We omit the proof. \end{proof} \begin{definition} \label{definition-scheme-theoretic-support} Let $X$ be a scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module of finite type. The {\it scheme theoretic support of $\mathcal{F}$} is the closed subscheme $Z \subset X$ constructed in Lemma \ref{lemma-scheme-theoretic-support}. \end{definition} \noindent In this situation we often think of $\mathcal{F}$ as a quasi-coherent sheaf of finite type on $Z$ (via the equivalence of categories of Lemma \ref{lemma-i-star-equivalence}). \section{Scheme theoretic image} \label{section-scheme-theoretic-image} \noindent Caution: Some of the material in this section is ultra-general and behaves differently from what you might expect. \begin{lemma} \label{lemma-scheme-theoretic-image} Let $f : X \to Y$ be a morphism of schemes. There exists a closed subscheme $Z \subset Y$ such that $f$ factors through $Z$ and such that for any other closed subscheme $Z' \subset Y$ such that $f$ factors through $Z'$ we have $Z \subset Z'$. \end{lemma} \begin{proof} Let $\mathcal{I} = \Ker(\mathcal{O}_Y \to f_*\mathcal{O}_X)$. If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the closed subscheme determined by $\mathcal{I}$, see Lemma \ref{lemma-closed-immersion-bijection-ideals}. This works by Schemes, Lemma \ref{schemes-lemma-characterize-closed-subspace}. In general the same lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in $\mathcal{I}$. This follows from Lemma \ref{lemma-largest-quasi-coherent-subsheaf}. \end{proof} \begin{definition} \label{definition-scheme-theoretic-image} Let $f : X \to Y$ be a morphism of schemes. The {\it scheme theoretic image} of $f$ is the smallest closed subscheme $Z \subset Y$ through which $f$ factors, see Lemma \ref{lemma-scheme-theoretic-image} above. \end{definition} \noindent For a morphism $f : X \to Y$ of schemes with scheme theoretic image $Z$ we often denote $f : X \to Z$ the factorization of $f$ through its scheme theoretic image. If the morphism $f$ is not quasi-compact, then (in general) \begin{enumerate} \item the set theoretic inclusion $\overline{f(X)} \subset Z$ is not an equality, i.e., $f(X) \subset Z$ is not a dense subset, and \item the construction of the scheme theoretic image does not commute with restriction to open subschemes to $Y$. \end{enumerate} Namely, the immersion of Example \ref{example-thibaut} gives an example for both phenomena. (If $Z \to U \to X$ is as in Example \ref{example-thibaut}, then the scheme theoretic image of $Z \to X$ is $X$ and $Z$ is not topologically dense in $X$. Also, the scheme theoretic image of $Z = Z \cap U \to U$ is just $Z$ which is not equal to $U \cap X = U$.) However, the next lemma shows that both disasters are avoided when the morphism is quasi-compact. \begin{lemma} \label{lemma-quasi-compact-scheme-theoretic-image} Let $f : X \to Y$ be a morphism of schemes. Let $Z \subset Y$ be the scheme theoretic image of $f$. If $f$ is quasi-compact then \begin{enumerate} \item the sheaf of ideals $\mathcal{I} = \Ker(\mathcal{O}_Y \to f_*\mathcal{O}_X)$ is quasi-coherent, \item the scheme theoretic image $Z$ is the closed subscheme determined by $\mathcal{I}$, \item for any open $U \subset Y$ the scheme theoretic image of $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is equal to $Z \cap U$, and \item the image $f(X) \subset Z$ is a dense subset of $Z$, in other words the morphism $X \to Z$ is dominant (see Definition \ref{definition-dominant}). \end{enumerate} \end{lemma} \begin{proof} Part (4) follows from part (3). To show (3) it suffices to prove (1) since the formation of $\mathcal{I}$ commutes with restriction to open subschemes of $Y$. And if (1) holds then in the proof of Lemma \ref{lemma-scheme-theoretic-image} we showed (2). Thus it suffices to prove that $\mathcal{I}$ is quasi-coherent. Since the property of being quasi-coherent is local we may assume $Y$ is affine. As $f$ is quasi-compact, we can find a finite affine open covering $X = \bigcup_{i = 1, \ldots, n} U_i$. Denote $f'$ the composition $$X' = \coprod U_i \longrightarrow X \longrightarrow Y.$$ Then $f_*\mathcal{O}_X$ is a subsheaf of $f'_*\mathcal{O}_{X'}$, and hence $\mathcal{I} = \Ker(\mathcal{O}_Y \to \mathcal{O}_{X'})$. By Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent} the sheaf $f'_*\mathcal{O}_{X'}$ is quasi-coherent on $Y$. Hence we win. \end{proof} \begin{example} \label{example-scheme-theoretic-image} If $A \to B$ is a ring map with kernel $I$, then the scheme theoretic image of $\Spec(B) \to \Spec(A)$ is the closed subscheme $\Spec(A/I)$ of $\Spec(A)$. This follows from Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}. \end{example} \noindent If the morphism is quasi-compact, then the scheme theoretic image only adds points which are specializations of points in the image. \begin{lemma} \label{lemma-reach-points-scheme-theoretic-image} Let $f : X \to Y$ be a quasi-compact morphism. Let $Z$ be the scheme theoretic image of $f$. Let $z \in Z$. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram $$\xymatrix{ \Spec(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\ \Spec(A) \ar[r] & Z \ar[r] & Y }$$ such that the closed point of $\Spec(A)$ maps to $z$. In particular any point of $Z$ is the specialization of a point of $f(X)$. \end{lemma} \begin{proof} Let $z \in \Spec(R) = V \subset Y$ be an affine open neighbourhood of $z$. By Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} the intersection $Z \cap V$ is the scheme theoretic image of $f^{-1}(V) \to V$. Hence we may replace $Y$ by $V$ and assume $Y = \Spec(R)$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Say $X = U_1 \cup \ldots \cup U_n$ is a finite affine open covering. Write $U_i = \Spec(A_i)$. Let $I = \Ker(R \to A_1 \times \ldots \times A_n)$. By Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} again we see that $Z$ corresponds to the closed subscheme $\Spec(R/I)$ of $Y$. If $\mathfrak p \subset R$ is the prime corresponding to $z$, then we see that $I_{\mathfrak p} \subset R_{\mathfrak p}$ is not an equality. Hence (as localization is exact, see Algebra, Proposition \ref{algebra-proposition-localization-exact}) we see that $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_1)_{\mathfrak p}$ is not zero. Hence one of the rings $(A_i)_{\mathfrak p}$ is not zero. Hence there exists an $i$ and a prime $\mathfrak q_i \subset A_i$ lying over a prime $\mathfrak p_i \subset \mathfrak p$. By Algebra, Lemma \ref{algebra-lemma-dominate} we can choose a valuation ring $A \subset K = f.f.(A_i/\mathfrak q_i)$ dominating the local ring $R_{\mathfrak p}/\mathfrak p_iR_{\mathfrak p} \subset f.f.(A_i/\mathfrak q_i)$. This gives the desired diagram. Some details omitted. \end{proof} \begin{lemma} \label{lemma-factor-factor} Let $$\xymatrix{ X_1 \ar[d] \ar[r]_{f_1} & Y_1 \ar[d] \\ X_2 \ar[r]^{f_2} & Y_2 }$$ be a commutative diagram of schemes. Let $Z_i \subset Y_i$, $i = 1, 2$ be the scheme theoretic image of $f_i$. Then the morphism $Y_1 \to Y_2$ induces a morphism $Z_1 \to Z_2$ and a commutative diagram $$\xymatrix{ X_1 \ar[r] \ar[d] & Z_1 \ar[d] \ar[r] & Y_1 \ar[d] \\ X_2 \ar[r] & Z_2 \ar[r] & Y_2 }$$ \end{lemma} \begin{proof} The scheme theoretic inverse image of $Z_2$ in $Y_1$ is a closed subscheme of $Y_1$ through which $f_1$ factors. Hence $Z_1$ is contained in this. This proves the lemma. \end{proof} \begin{lemma} \label{lemma-scheme-theoretic-image-reduced} Let $f : X \to Y$ be a morphism of schemes. If $X$ is reduced, then the scheme theoretic image of $f$ is the reduced induced scheme structure on $\overline{f(X)}$. \end{lemma} \begin{proof} This is true because the reduced induced scheme structure on $\overline{f(X)}$ is clearly the smallest closed subscheme of $Y$ through which $f$ factors, see Schemes, Lemma \ref{schemes-lemma-map-into-reduction}. \end{proof} \begin{lemma} \label{lemma-scheme-theoretic-image-of-partial-section} Let $f : X \to Y$ be a separated morphism of schemes. Let $V \subset Y$ be a restrocompact open. Let $s : V \to X$ be a morphism such that $f \circ s = \text{id}_V$. Let $Y'$ be the scheme theoretic image of $s$. Then $Y' \to Y$ is an isomorphism over $V$. \end{lemma} \begin{proof} The assumption that $V$ is retrocompact in $Y$ (Topology, Definition \ref{topology-definition-quasi-compact}) means that $V \to Y$ is a quasi-compact morphism. By Schemes, Lemma \ref{schemes-lemma-quasi-compact-permanence} the morphism $s : V \to Y$ is quasi-compact. Hence the construction of the scheme theoretic image $Y'$ of $s$ commutes with restriction to opens by Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}. In particular, we see that $Y' \cap f^{-1}(V)$ is the scheme theoretic image of a section of the separated morphism $f^{-1}(V) \to V$. Since a section of a separated morphism is a closed immersion (Schemes, Lemma \ref{schemes-lemma-section-immersion}), we conclude that $Y' \cap f^{-1}(V) \to V$ is an isomorphism as desired. \end{proof} \section{Scheme theoretic closure and density} \label{section-scheme-theoretic-closure} \noindent We take the following definition from \cite[IV, Definition 11.10.2]{EGA}. \begin{definition} \label{definition-scheme-theoretically-dense} Let $X$ be a scheme. Let $U \subset X$ be an open subscheme. \begin{enumerate} \item The scheme theoretic image of the morphism $U \to X$ is called the {\it scheme theoretic closure of $U$ in $X$}. \item We say $U$ is {\it scheme theoretically dense in $X$} if for every open $V \subset X$ the scheme theoretic closure of $U \cap V$ in $V$ is equal to $V$. \end{enumerate} \end{definition} \noindent With this definition it is {\bf not} the case that $U$ is scheme theoretically dense in $X$ if and only if the scheme theoretic closure of $U$ is $X$, see Example \ref{example-scheme-theretically-dense-not-dense}. This is somewhat inelegant; but see Lemmas \ref{lemma-scheme-theoretically-dense-quasi-compact} and \ref{lemma-reduced-scheme-theoretically-dense} below. On the other hand, with this definition $U$ is scheme theoretically dense in $X$ if and only if for every $V \subset X$ open the ring map $\mathcal{O}_X(V) \to \mathcal{O}_X(U \cap V)$ is injective, see Lemma \ref{lemma-characterize-scheme-theoretically-dense} below. In particular we see that scheme theoretically dense implies dense which is pleasing. \begin{example} \label{example-scheme-theretically-dense-not-dense} Here is an example where scheme theoretic closure being $X$ does not imply dense for the underlying topological spaces. Let $k$ be a field. Set $A = k[x, z_1, z_2, \ldots]/(x^n z_n)$ Set $I = (z_1, z_2, \ldots) \subset A$. Consider the affine scheme $X = \Spec(A)$ and the open subscheme $U = X \setminus V(I)$. Since $A \to \prod_n A_{z_n}$ is injective we see that the scheme theoretic closure of $U$ is $X$. Consider the morphism $X \to \Spec(k[x])$. This morphism is surjective (set all $z_n = 0$ to see this). But the restriction of this morphism to $U$ is not surjective because it maps to the point $x = 0$. Hence $U$ cannot be topologically dense in $X$. \end{example} \begin{lemma} \label{lemma-scheme-theoretically-dense-quasi-compact} Let $X$ be a scheme. Let $U \subset X$ be an open subscheme. If the inclusion morphism $U \to X$ is quasi-compact, then $U$ is scheme theoretically dense in $X$ if and only if the scheme theoretic closure of $U$ in $X$ is $X$. \end{lemma} \begin{proof} Follows from Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} part (3). \end{proof} \begin{example} \label{example-scheme-theoretic-closure} Let $A$ be a ring and $X = \Spec(A)$. Let $f_1, \ldots, f_n \in A$ and let $U = D(f_1) \cup \ldots \cup D(f_n)$. Let $I = \Ker(A \to \prod A_{f_i})$. Then the scheme theoretic closure of $U$ in $X$ is the closed subscheme $\Spec(A/I)$ of $X$. Note that $U \to X$ is quasi-compact. Hence by Lemma \ref{lemma-scheme-theoretically-dense-quasi-compact} we see $U$ is scheme theoretically dense in $X$ if and only if $I = 0$. \end{example} \begin{lemma} \label{lemma-characterize-scheme-theoretically-dense} Let $j : U \to X$ be an open immersion of schemes. Then $U$ is scheme theoretically dense in $X$ if and only if $\mathcal{O}_X \to j_*\mathcal{O}_U$ is injective. \end{lemma} \begin{proof} If $\mathcal{O}_X \to j_*\mathcal{O}_U$ is injective, then the same is true when restricted to any open $V$ of $X$. Hence the scheme theoretic closure of $U \cap V$ in $V$ is equal to $V$, see proof of Lemma \ref{lemma-scheme-theoretic-image}. Conversely, suppose that the scheme theoretic closure of $U \cap V$ is equal to $V$ for all opens $V$. Suppose that $\mathcal{O}_X \to j_*\mathcal{O}_U$ is not injective. Then we can find an affine open, say $\Spec(A) = V \subset X$ and a nonzero element $f \in A$ such that $f$ maps to zero in $\Gamma(V \cap U, \mathcal{O}_X)$. In this case the scheme theoretic closure of $V \cap U$ in $V$ is clearly contained in $\Spec(A/(f))$ a contradiction. \end{proof} \begin{lemma} \label{lemma-intersection-scheme-theoretically-dense} Let $X$ be a scheme. If $U$, $V$ are scheme theoretically dense open subschemes of $X$, then so is $U \cap V$. \end{lemma} \begin{proof} Let $W \subset X$ be any open. Consider the map $\mathcal{O}_X(W) \to \mathcal{O}_X(W \cap V) \to \mathcal{O}_X(W \cap V \cap U)$. By Lemma \ref{lemma-characterize-scheme-theoretically-dense} both maps are injective. Hence the composite is injective. Hence by Lemma \ref{lemma-characterize-scheme-theoretically-dense} $U \cap V$ is scheme theoretically dense in $X$. \end{proof} \begin{lemma} \label{lemma-quasi-compact-immersion} Let $h : Z \to X$ be an immersion. Assume either $h$ is quasi-compact or $Z$ is reduced. Let $\overline{Z} \subset X$ be the scheme theoretic image of $h$. Then the morphism $Z \to \overline{Z}$ is an open immersion which identifies $Z$ with a scheme theoretically dense open subscheme of $\overline{Z}$. Moreover, $Z$ is topologically dense in $\overline{Z}$. \end{lemma} \begin{proof} By Lemma \ref{lemma-factor-quasi-compact-immersion} or Lemma \ref{lemma-factor-reduced-immersion} we can factor $Z \to X$ as $Z \to \overline{Z}_1 \to X$ with $Z \to \overline{Z}_1$ open and $\overline{Z}_1 \to X$ closed. On the other hand, let $Z \to \overline{Z} \subset X$ be the scheme theoretic closure of $Z \to X$. We conclude that $\overline{Z} \subset \overline{Z}_1$. Since $Z$ is an open subscheme of $\overline{Z}_1$ it follows that $Z$ is an open subscheme of $\overline{Z}$ as well. In the case that $Z$ is reduced we know that $Z \subset \overline{Z}_1$ is topologically dense by the construction of $\overline{Z}_1$ in the proof of Lemma \ref{lemma-factor-reduced-immersion}. Hence $\overline{Z}_1$ and $\overline{Z}$ have the same underlying topological spaces. Thus $\overline{Z} \subset \overline{Z}_1$ is a closed immersion into a reduced scheme which induces a bijection on underlying topological spaces, and hence it is an isomorphism. In the case that $Z \to X$ is quasi-compact we argue as follows: The assertion that $Z$ is scheme theoretically dense in $\overline{Z}$ follows from Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} part (3). The last assertion follows from Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} part (4). \end{proof} \begin{lemma} \label{lemma-reduced-scheme-theoretically-dense} Let $X$ be a reduced scheme and let $U \subset X$ be an open subscheme. Then the following are equivalent \begin{enumerate} \item $U$ is topologically dense in $X$, \item the scheme theoretic closure of $U$ in $X$ is $X$, and \item $U$ is scheme theoretically dense in $X$. \end{enumerate} \end{lemma} \begin{proof} This follows from Lemma \ref{lemma-quasi-compact-immersion} and the fact that a closed subscheme $Z$ of $X$ whose underlying topological space equals $X$ must be equal to $X$ as a scheme. \end{proof} \begin{lemma} \label{lemma-reduced-subscheme-closure} Let $X$ be a scheme and let $U \subset X$ be a reduced open subscheme. Then the following are equivalent \begin{enumerate} \item the scheme theoretic closure of $U$ in $X$ is $X$, and \item $U$ is scheme theoretically dense in $X$. \end{enumerate} If this holds then $X$ is a reduced scheme. \end{lemma} \begin{proof} This follows from Lemma \ref{lemma-quasi-compact-immersion} and the fact that the scheme theoretic closure of $U$ in $X$ is reduced by Lemma \ref{lemma-scheme-theoretic-image-reduced}. \end{proof} \begin{lemma} \label{lemma-equality-of-morphisms} Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. Let $f, g : X \to Y$ be morphisms of schemes over $S$. Let $U \subset X$ be an open subscheme such that $f|_U = g|_U$. If the scheme theoretic closure of $U$ in $X$ is $X$ and $Y \to S$ is separated, then $f = g$. \end{lemma} \begin{proof} Follows from the definitions and Schemes, Lemma \ref{schemes-lemma-where-are-they-equal}. \end{proof} \section{Dominant morphisms} \label{section-dominant} \noindent The definition of a morphism of schemes being dominant is a little different from what you might expect if you are used to the notion of a dominant morphism of varieties. \begin{definition} \label{definition-dominant} A morphism $f : X \to S$ of schemes is called {\it dominant} if the image of $f$ is a dense subset of $S$. \end{definition} \noindent So for example, if $k$ is an infinite field and $\lambda_1, \lambda_2, \ldots$ is a countable collection of elements of $k$, then the morphism $$\coprod\nolimits_{i = 1, 2, \ldots } \Spec(k) \longrightarrow \Spec(k[x])$$ with $i$th factor mapping to the point $x = \lambda_i$ is dominant. \begin{lemma} \label{lemma-generic-points-in-image-dominant} Let $f : X \to S$ be a morphism of schemes. If every generic point of every irreducible component of $S$ is in the image of $f$, then $f$ is dominant. \end{lemma} \begin{proof} This is a topological fact which follows directly from the fact that the topological space underlying a scheme is sober, see Schemes, Lemma \ref{schemes-lemma-scheme-sober}, and that every point of $S$ is contained in an irreducible component of $S$, see Topology, Lemma \ref{topology-lemma-irreducible}. \end{proof} \noindent The expectation that morphisms are dominant only if generic points of the target are in the image does hold if the morphism is quasi-compact. \begin{lemma} \label{lemma-quasi-compact-dominant} \begin{slogan} Morphisms whose image contains the generic points are dominant \end{slogan} Let $f : X \to S$ be a quasi-compact morphism of schemes. Then $f$ is dominant (if and) only if for every irreducible component $Z \subset S$ the generic point of $Z$ is in the image of $f$. \end{lemma} \begin{proof} Let $V \subset S$ be an affine open. Because $f$ is quasi-compact we may choose finitely many affine opens $U_i \subset f^{-1}(V)$, $i = 1, \ldots, n$ covering $f^{-1}(V)$. Consider the morphism of affines $$f' : \coprod\nolimits_{i = 1, \ldots, n} U_i \longrightarrow V.$$ A disjoint union of affines is affine, see Schemes, Lemma \ref{schemes-lemma-disjoint-union-affines}. Generic points of irreducible components of $V$ are exactly the generic points of the irreducible components of $S$ that meet $V$. Also, $f$ is dominant if and only $f'$ is dominant no matter what choices of $V, n, U_i$ we make above. Thus we have reduced the lemma to the case of a morphism of affine schemes. The affine case is Algebra, Lemma \ref{algebra-lemma-image-dense-generic-points}. \end{proof} \noindent Here is a slightly more useful variant of the lemma above. \begin{lemma} \label{lemma-quasi-compact-generic-point-not-in-image} Let $f : X \to S$ be a quasi-compact morphism of schemes. Let $\eta \in S$ be a generic point of an irreducible component of $S$. If $\eta \not \in f(X)$ then there exists an open neighbourhood $V \subset S$ of $\eta$ such that $f^{-1}(V) = \emptyset$. \end{lemma} \begin{proof} Let $Z \subset S$ be the scheme theoretic image of $f$. We have to show that $\eta \not \in Z$. This follows from Lemma \ref{lemma-reach-points-scheme-theoretic-image} but can also be seen as follows. By Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} the morphism $X \to Z$ is dominant, which by Lemma \ref{lemma-quasi-compact-dominant} means all the generic points of all irreducible components of $Z$ are in the image of $X \to Z$. By assumption we see that $\eta \not \in Z$ since $\eta$ would be the generic point of some irreducible component of $Z$ if it were in $Z$. \end{proof} \noindent There is another case where dominant is the same as having all generic points of irreducible components in the image. \begin{lemma} \label{lemma-dominant-finite-number-irreducible-components} Let $f : X \to S$ be a morphism of schemes. Suppose that $X$ has finitely many irreducible components. Then $f$ is dominant (if and) only if for every irreducible component $Z \subset S$ the generic point of $Z$ is in the image of $f$. If so, then $S$ has finitely many irreducible components as well. \end{lemma} \begin{proof} Assume $f$ is dominant. Say $X = Z_1 \cup Z_2 \cup \ldots \cup Z_n$ is the decomposition of $X$ into irreducible components. Let $\xi_i \in Z_i$ be its generic point, so $Z_i = \overline{\{\xi_i\}}$. Note that $f(Z_i)$ is an irreducible subset of $S$. Hence $$S = \overline{f(X)} = \bigcup \overline{f(Z_i)} = \bigcup \overline{\{f(\xi_i)\}}$$ is a finite union of irreducible subsets whose generic points are in the image of $f$. The lemma follows. \end{proof} \begin{lemma} \label{lemma-dominant-between-integral} Let $f : X \to Y$ be a morphism of integral schemes. The following are equivalent \begin{enumerate} \item $f$ is dominant, \item $f$ maps the generic point of $X$ to the generic point of $Y$, \item for some nonempty affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$ is injective, \item for all nonempty affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$ is injective, \item for some $x \in X$ with image $y = f(x) \in Y$ the local ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective, and \item for all $x \in X$ with image $y = f(x) \in Y$ the local ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (2) follows from Lemma \ref{lemma-dominant-finite-number-irreducible-components}. Let $U \subset X$ and $V \subset Y$ be nonempty affine opens with $f(U) \subset V$. Recall that the rings $A = \mathcal{O}_X(U)$ and $B = \mathcal{O}_Y(V)$ are integral domains. The morphism $f|_U : U \to V$ corresponds to a ring map $\varphi : B \to A$. The generic points of $X$ and $Y$ correspond to the prime ideals $(0) \subset A$ and $(0) \subset B$. Thus (2) is equivalent to the condition $(0) = \varphi^{-1}((0))$, i.e., to the condition that $\varphi$ is injective. In this way we see that (2), (3), and (4) are equivalent. Similarly, given $x$ and $y$ as in (5) the local rings $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ are domains and the prime ideals $(0) \subset \mathcal{O}_{X, x}$ and $(0) \subset \mathcal{O}_{Y, y}$ correspond to the generic points of $X$ and $Y$ (via the identification of the spectrum of the local ring at $x$ with the set of points specializing to $x$, see Schemes, Lemma \ref{schemes-lemma-specialize-points}). Thus we can argue in the exact same manner as above to see that (2), (5), and (6) are equivalent. \end{proof} \section{Surjective morphisms} \label{section-surjective} \begin{definition} \label{definition-surjective} A morphism of schemes is said to be {\it surjective} if it is surjective on underlying topological spaces. \end{definition} \begin{lemma} \label{lemma-composition-surjective} The composition of surjective morphisms is surjective. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-when-point-maps-to-pair} Let $X$ and $Y$ be schemes over a base scheme $S$. Given points $x \in X$ and $y \in Y$, there is a point of $X \times_S Y$ mapping to $x$ and $y$ under the projections if and only if $x$ and $y$ lie above the same point of $S$. \end{lemma} \begin{proof} The condition is obviously necessary, and the converse follows from the proof of Schemes, Lemma \ref{schemes-lemma-points-fibre-product}. \end{proof} \begin{lemma} \label{lemma-base-change-surjective} The base change of a surjective morphism is surjective. \end{lemma} \begin{proof} Let $f: X \to Y$ be a morphism of schemes over a base scheme $S$. If $S' \to S$ is a morphism of schemes, let $p: X_{S'} \to X$ and $q: Y_{S'} \to Y$ be the canonical projections. The commutative square $$\xymatrix{ X_{S'} \ar[d]_{f_{S'}} \ar[r]_p & X \ar[d]^{f} \\ Y_{S'} \ar[r]^{q} & Y. }$$ identifies $X_{S'}$ as a fibre product of $X \to Y$ and $Y_{S'} \to Y$. Let $Z$ be a subset of the underlying topological space of $X$. Then $q^{-1}(f(Z)) = f_{S'}(p^{-1}(Z))$, because $y' \in q^{-1}(f(Z))$ if and only if $q(y') = f(x)$ for some $x \in Z$, if and only if, by Lemma \ref{lemma-when-point-maps-to-pair}, there exists $x' \in X_{S'}$ such that $f_{S'}(x') = y'$ and $p(x') = x$. In particular taking $Z = X$ we see that if $f$ is surjective so is the base change $f_{S'}: X_{S'} \to Y_{S'}$. \end{proof} \begin{example} \label{example-injective-not-preserved-base-change} Bijectivity is not stable under base change, and so neither is injectivity. For example consider the bijection $\Spec(\mathbf{C}) \to \Spec(\mathbf{R})$. The base change $\Spec(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C}) \to \Spec(\mathbf{C})$ is not injective, since there is an isomorphism $\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C} \cong \mathbf{C} \times \mathbf{C}$ (the decomposition comes from the idempotent $\frac{1 \otimes 1 + i \otimes i}{2}$) and hence $\Spec(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C})$ has two points. \end{example} \begin{lemma} \label{lemma-surjection-from-quasi-compact} Let $$\xymatrix{ X \ar[rr]_f \ar[rd]_p & & Y \ar[dl]^q \\ & Z }$$ be a commutative diagram of morphisms of schemes. If $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact. \end{lemma} \begin{proof} Let $W \subset Z$ be a quasi-compact open. By assumption $p^{-1}(W)$ is quasi-compact. Hence by Topology, Lemma \ref{topology-lemma-image-quasi-compact} the inverse image $q^{-1}(W) = f(p^{-1}(W))$ is quasi-compact too. This proves the lemma. \end{proof} \section{Radicial and universally injective morphisms} \label{section-radicial} \noindent In this section we define what it means for a morphism of schemes to be {\it radicial} and what it means for a morphism of schemes to be {\it universally injective}. We then show that these notions agree. The reason for introducing both is that in the case of algebraic spaces there are corresponding notions which may not always agree. \begin{definition} \label{definition-universally-injective} Let $f : X \to S$ be a morphism. \begin{enumerate} \item We say that $f$ is {\it universally injective} if and only if for any morphism of schemes $S' \to S$ the base change $f' : X_{S'} \to S'$ is injective (on underlying topological spaces). \item We say $f$ is {\it radicial} if $f$ is injective as a map of topological spaces, and for every $x \in X$ the field extension $\kappa(x) \supset \kappa(f(x))$ is purely inseparable. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-universally-injective} Let $f : X \to S$ be a morphism of schemes. The following are equivalent: \begin{enumerate} \item For every field $K$ the induced map $\Mor(\Spec(K), X) \to \Mor(\Spec(K), S)$ is injective. \item The morphism $f$ is universally injective. \item The morphism $f$ is radicial. \item The diagonal morphism $\Delta_{X/S} : X \longrightarrow X \times_S X$ is surjective. \end{enumerate} \end{lemma} \begin{proof} Let $K$ be a field, and let $s : \Spec(K) \to S$ be a morphism. Giving a morphism $x : \Spec(K) \to X$ such that $f \circ x = s$ is the same as giving a section of the projection $X_K = \Spec(K) \times_S X \to \Spec(K)$, which in turn is the same as giving a point $x \in X_K$ whose residue field is $K$. Hence we see that (2) implies (1). \medskip\noindent Conversely, suppose that (1) holds. Assume that $x, x' \in X_{S'}$ map to the same point $s' \in S'$. Choose a commutative diagram $$\xymatrix{ K & \kappa(x) \ar[l] \\ \kappa(x') \ar[u] & \kappa(s') \ar[l] \ar[u] }$$ of fields. By Schemes, Lemma \ref{schemes-lemma-characterize-points} we get two morphisms $a, a' : \Spec(K) \to X_{S'}$. One corresponding to the point $x$ and the embedding $\kappa(x) \subset K$ and the other corresponding to the point $x'$ and the embedding $\kappa(x') \subset K$. Also we have $f' \circ a = f' \circ a'$. Condition (1) now implies that the compositions of $a$ and $a'$ with $X_{S'} \to X$ are equal. Since $X_{S'}$ is the fibre product of $S'$ and $X$ over $S$ we see that $a = a'$. Hence $x = x'$. Thus (1) implies (2). \medskip\noindent If there are two different points $x, x' \in X$ mapping to the same point of $s$ then (2) is violated. If for some $s = f(x)$, $x \in X$ the field extension $\kappa(s) \subset \kappa(x)$ is not purely inseparable, then we may find a field extension $\kappa(s) \subset K$ such that $\kappa(x)$ has two $\kappa(s)$-homomorphisms into $K$. By Schemes, Lemma \ref{schemes-lemma-characterize-points} this implies that the map $\Mor(\Spec(K), X) \to \Mor(\Spec(K), S)$ is not injective, and hence (1) is violated. Thus we see that the equivalent conditions (1) and (2) imply $f$ is radicial, i.e., they imply (3). \medskip\noindent Assume (3). By Schemes, Lemma \ref{schemes-lemma-characterize-points} a morphism $\Spec(K) \to X$ is given by a pair $(x, \kappa(x) \to K)$. Property (3) says exactly that associating to the pair $(x, \kappa(x) \to K)$ the pair $(s, \kappa(s) \to \kappa(x) \to K)$ is injective. In other words (1) holds. At this point we know that (1), (2) and (3) are all equivalent. \medskip\noindent Finally, we prove the equivalence of (4) with (1), (2) and (3). A point of $X \times_S X$ is given by a quadruple $(x_1, x_2, s, \mathfrak p)$, where $x_1, x_2 \in X$, $f(x_1) = f(x_2) = s$ and $\mathfrak p \subset \kappa(x_1) \otimes_{\kappa(s)} \kappa(x_2)$ is a prime ideal, see Schemes, Lemma \ref{schemes-lemma-points-fibre-product}. If $f$ is universally injective, then by taking $S'=X$ in the definition of universally injective, $\Delta_{X/S}$ must be surjective since it is a section of the injective morphism $X \times_S X \longrightarrow X$. Conversely, if $\Delta_{X/S}$ is surjective, then always $x_1 = x_2 = x$ and there is exactly one such prime ideal $\mathfrak p$, which means that $\kappa(s) \subset \kappa(x)$ is purely inseparable. Hence $f$ is radicial. Alternatively, if $\Delta_{X/S}$ is surjective, then for any $S' \to S$ the base change $\Delta_{X_{S'}/S'}$ is surjective which implies that $f$ is universally injective. This finishes the proof of the lemma. \end{proof} \begin{lemma} \label{lemma-universally-injective-separated} A universally injective morphism is separated. \end{lemma} \begin{proof} Combine Lemma \ref{lemma-universally-injective} with the remark that $X \to S$ is separated if and only if the image of $\Delta_{X/S}$ is closed in $X \times_S X$, see Schemes, Definition \ref{schemes-definition-separated} and the discussion following it. \end{proof} \begin{lemma} \label{lemma-base-change-universally-injective} A base change of a universally injective morphism is universally injective. \end{lemma} \begin{proof} This is formal. \end{proof} \begin{lemma} \label{lemma-composition-universally-injective} A composition of radicial morphisms is radicial, and so the same holds for the equivalent condition of being universally injective. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Affine morphisms} \label{section-affine} \begin{definition} \label{definition-affine} A morphism of schemes $f : X \to S$ is called {\it affine} if the inverse image of every affine open of $S$ is an affine open of $X$. \end{definition} \begin{lemma} \label{lemma-affine-separated} An affine morphism is separated and quasi-compact. \end{lemma} \begin{proof} Let $f : X \to S$ be affine. Quasi-compactness is immediate from Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine}. We will show $f$ is separated using Schemes, Lemma \ref{schemes-lemma-characterize-separated}. Let $x_1, x_2 \in X$ be points of $X$ which map to the same point $s \in S$. Choose any affine open $W \subset S$ containing $s$. By assumption $f^{-1}(W)$ is affine. Apply the lemma cited with $U = V = f^{-1}(W)$. \end{proof} \begin{lemma} \label{lemma-characterize-affine} Let $f : X \to S$ be a morphism of schemes. The following are equivalent \begin{enumerate} \item The morphism $f$ is affine. \item There exists an affine open covering $S = \bigcup W_j$ such that each $f^{-1}(W_j)$ is affine. \item There exists a quasi-coherent sheaf of $\mathcal{O}_S$-algebras $\mathcal{A}$ and an isomorphism $X \cong \underline{\Spec}_S(\mathcal{A})$ of schemes over $S$. See Constructions, Section \ref{constructions-section-spec} for notation. \end{enumerate} Moreover, in this case $X = \underline{\Spec}_S(f_*\mathcal{O}_X)$. \end{lemma} \begin{proof} It is obvious that (1) implies (2). \medskip\noindent Assume $S = \bigcup_{j \in J} W_j$ is an affine open covering such that each $f^{-1}(W_j)$ is affine. By Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine} we see that $f$ is quasi-compact. By Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} we see the morphism $f$ is quasi-separated. Hence by Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent} the sheaf $\mathcal{A} = f_*\mathcal{O}_X$ is a quasi-coherent sheaf of $\mathcal{O}_X$-algebras. Thus we have the scheme $g : Y = \underline{\Spec}_S(\mathcal{A}) \to S$ over $S$. The identity map $\text{id} : \mathcal{A} = f_*\mathcal{O}_X \to f_*\mathcal{O}_X$ provides, via the definition of the relative spectrum, a morphism $can : X \to Y$ over $S$, see Constructions, Lemma \ref{constructions-lemma-canonical-morphism}. By assumption and the lemma just cited the restriction $can|_{f^{-1}(W_j)} : f^{-1}(W_j) \to g^{-1}(W_j)$ is an isomorphism. Thus $can$ is an isomorphism. We have shown that (2) implies (3). \medskip\noindent Assume (3). By Constructions, Lemma \ref{constructions-lemma-spec-properties} we see that the inverse image of every affine open is affine, and hence the morphism is affine by definition. \end{proof} \begin{remark} \label{remark-direct-argument} We can also argue directly that (2) implies (1) in Lemma \ref{lemma-characterize-affine} above as follows. Assume $S = \bigcup W_j$ is an affine open covering such that each $f^{-1}(W_j)$ is affine. First argue that $\mathcal{A} = f_*\mathcal{O}_X$ is quasi-coherent as in the proof above. Let $\Spec(R) = V \subset S$ be affine open. We have to show that $f^{-1}(V)$ is affine. Set $A = \mathcal{A}(V) = f_*\mathcal{O}_X(V) = \mathcal{O}_X(f^{-1}(V))$. By Schemes, Lemma \ref{schemes-lemma-morphism-into-affine} there is a canonical morphism $\psi : f^{-1}(V) \to \Spec(A)$ over $\Spec(R) = V$. By Schemes, Lemma \ref{schemes-lemma-good-subcover} there exists an integer $n \geq 0$, a standard open covering $V = \bigcup_{i = 1, \ldots, n} D(h_i)$, $h_i \in R$, and a map $a : \{1, \ldots, n\} \to J$ such that each $D(h_i)$ is also a standard open of the affine scheme $W_{a(i)}$. The inverse image of a standard open under a morphism of affine schemes is standard open, see Algebra, Lemma \ref{algebra-lemma-spec-functorial}. Hence we see that $f^{-1}(D(h_i))$ is a standard open of $f^{-1}(W_{a(i)})$, in particular that $f^{-1}(D(h_i))$ is affine. Because $\mathcal{A}$ is quasi-coherent we have $A_{h_i} = \mathcal{A}(D(h_i)) = \mathcal{O}_X(f^{-1}(D(h_i)))$, so $f^{-1}(D(h_i))$ is the spectrum of $A_{h_i}$. It follows that the morphism $\psi$ induces an isomorphism of the open $f^{-1}(D(h_i))$ with the open $\Spec(A_{h_i})$ of $\Spec(A)$. Since $f^{-1}(V) = \bigcup f^{-1}(D(h_i))$ and $\Spec(A) = \bigcup \Spec(A_{h_i})$ we win. \end{remark} \begin{lemma} \label{lemma-affine-equivalence-algebras} Let $S$ be a scheme. There is an anti-equivalence of categories $$\begin{matrix} \text{Schemes affine} \\ \text{over }S \end{matrix} \longleftrightarrow \begin{matrix} \text{quasi-coherent sheaves} \\ \text{of }\mathcal{O}_S\text{-algebras} \end{matrix}$$ which associates to $f : X \to S$ the sheaf $f_*\mathcal{O}_X$. Moreover, this equivalence if compatible with arbitrary base change. \end{lemma} \begin{proof} The functor from right to left is given by $\underline{\Spec}_S$. The two functors are mutually inverse by Lemma \ref{lemma-characterize-affine} and Constructions, Lemma \ref{constructions-lemma-spec-properties} part (3). The final statement is Constructions, Lemma \ref{constructions-lemma-spec-properties} part (2). \end{proof} \begin{lemma} \label{lemma-affine-equivalence-modules} Let $f : X \to S$ be an affine morphism of schemes. Let $\mathcal{A} = f_*\mathcal{O}_X$. The functor $\mathcal{F} \mapsto f_*\mathcal{F}$ induces an equivalence of categories $$\left\{ \begin{matrix} \text{category of quasi-coherent}\\ \mathcal{O}_X\text{-modules} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{category of quasi-coherent}\\ \mathcal{A}\text{-modules} \end{matrix} \right\}$$ Moreover, an $\mathcal{A}$-module is quasi-coherent as an $\mathcal{O}_S$-module if and only if it is quasi-coherent as an $\mathcal{A}$-module. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-composition-affine} The composition of affine morphisms is affine. \end{lemma} \begin{proof} Let $f : X \to Y$ and $g : Y \to Z$ be affine morphisms. Let $U \subset Z$ be affine open. Then $g^{-1}(U)$ is affine by assumption on $g$. Whereupon $f^{-1}(g^{-1}(U))$ is affine by assumption on $f$. Hence $(g \circ f)^{-1}(U)$ is affine. \end{proof} \begin{lemma} \label{lemma-base-change-affine} The base change of an affine morphism is affine. \end{lemma} \begin{proof} Let $f : X \to S$ be an affine morphism. Let $S' \to S$ be any morphism. Denote $f' : X_{S'} = S' \times_S X \to S'$ the base change of $f$. For every $s' \in S'$ there exists an open affine neighbourhood $s' \in V \subset S'$ which maps into some open affine $U \subset S$. By assumption $f^{-1}(U)$ is affine. By the material in Schemes, Section \ref{schemes-section-fibre-products} we see that $f^{-1}(U)_V = V \times_U f^{-1}(U)$ is affine and equal to $(f')^{-1}(V)$. This proves that $S'$ has an open covering by affines whose inverse image under $f'$ is affine. We conclude by Lemma \ref{lemma-characterize-affine} above. \end{proof} \begin{lemma} \label{lemma-closed-immersion-affine} A closed immersion is affine. \end{lemma} \begin{proof} The first indication of this is Schemes, Lemma \ref{schemes-lemma-closed-immersion-affine-case}. See Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} for a complete statement. \end{proof} \begin{lemma} \label{lemma-affine-s-open} Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module. Let $s \in \Gamma(X, \mathcal{L})$. The inclusion morphism $j : X_s \to X$ is affine. \end{lemma} \begin{proof} This follows from Properties, Lemma \ref{properties-lemma-affine-cap-s-open} and the definition. \end{proof} \begin{lemma} \label{lemma-affine-permanence} Suppose $g : X \to Y$ is a morphism of schemes over $S$. \begin{enumerate} \item If $X$ is affine over $S$ and $\Delta : Y \to Y \times_S Y$ is affine, then $g$ is affine. \item If $X$ is affine over $S$ and $Y$ is separated over $S$, then $g$ is affine. \item A morphism from an affine scheme to a scheme with affine diagonal is affine. \item A morphism from an affine scheme to a separated scheme is affine. \end{enumerate} \end{lemma} \begin{proof} Proof of (1). The base change $X \times_S Y \to Y$ is affine by Lemma \ref{lemma-base-change-affine}. The morphism $(1, g) : X \to X \times_S Y$ is the base change of $Y \to Y \times_S Y$ by the morphism $X \times_S Y \to Y \times_S Y$. Hence it is affine by Lemma \ref{lemma-base-change-affine}. The composition of affine morphisms is affine (see Lemma \ref{lemma-composition-affine}) and (1) follows. Part (2) follows from (1) as a closed immersion is affine (see Lemma \ref{lemma-closed-immersion-affine}) and $Y/S$ separated means $\Delta$ is a closed immersion. Parts (3) and (4) are special cases of (1) and (2). \end{proof} \begin{lemma} \label{lemma-morphism-affines-affine} A morphism between affine schemes is affine. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-affine-permanence} with $S = \Spec(\mathbf{Z})$. It also follows directly from the equivalence of (1) and (2) in Lemma \ref{lemma-characterize-affine}. \end{proof} \begin{lemma} \label{lemma-Artinian-affine} Let $S$ be a scheme. Let $A$ be an Artinian ring. Any morphism $\Spec(A) \to S$ is affine. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-get-affine} Let $j : Y \to X$ be an immersion of schemes. Assume there exists an open $U \subset X$ with complement $Z = X \setminus U$ such that \begin{enumerate} \item $U \to X$ is affine, \item $j^{-1}(U) \to U$ is affine, and \item $j(Y) \cap Z$ is closed. \end{enumerate} Then $j$ is affine. In particular, if $X$ is affine, so is $Y$. \end{lemma} \begin{proof} By Schemes, Definition \ref{schemes-definition-immersion} there exists an open subscheme $W \subset X$ such that $j$ factors as a closed immersion $i : Y \to W$ followed by the inclusion morphism $W \to X$. Since a closed immersion is affine (Lemma \ref{lemma-closed-immersion-affine}), we see see that for every $x \in W$ there is an affine open neighbourhood of $x$ in $X$ whose inverse image under $j$ is affine. If $x \in U$, then the same thing is true by assumption (2). Finally, assume $x \in Z$ and $x \not \in W$. Then $x \not \in j(Y) \cap Z$. By assumption (3) we can find an affine open neighbourhood $V \subset X$ of $x$ which does not meet $j(Y) \cap Z$. Then $j^{-1}(V) = j^{-1}(V \cap U)$ which is affine by assumptions (1) and (2). It follows that $j$ is affine by Lemma \ref{lemma-characterize-affine}. \end{proof} \section{Quasi-affine morphisms} \label{section-quasi-affine} \noindent Recall that a scheme $X$ is called {\it quasi-affine} if it is quasi-compact and isomorphic to an open subscheme of an affine scheme, see Properties, Definition \ref{properties-definition-quasi-affine}. \begin{definition} \label{definition-quasi-affine} A morphism of schemes $f : X \to S$ is called {\it quasi-affine} if the inverse image of every affine open of $S$ is a quasi-affine scheme. \end{definition} \begin{lemma} \label{lemma-quasi-affine-separated} A quasi-affine morphism is separated and quasi-compact. \end{lemma} \begin{proof} Let $f : X \to S$ be quasi-affine. Quasi-compactness is immediate from Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine}. We will show $f$ is separated using Schemes, Lemma \ref{schemes-lemma-characterize-separated}. Let $x_1, x_2 \in X$ be points of $X$ which map to the same point $s \in S$. Choose any affine open $W \subset S$ containing $s$. By assumption $f^{-1}(W)$ is isomorphic to an open subscheme of an affine scheme, say $f^{-1}(W) \to Y$ is such an open immersion. Choose affine open neighbourhoods $x_1 \in U \subset f^{-1}(W)$ and $x_2 \in V \subset f^{-1}(W)$. We may think of $U$ and $V$ as open subschemes of $Y$ and hence we see that $U \cap V$ is affine and that $\mathcal{O}(U) \otimes_{\mathbf{Z}} \mathcal{O}(V) \to \mathcal{O}(U \cap V)$ is surjective (by the lemma cited above applied to $U, V$ in $Y$). Hence by the lemma cited we conclude that $f$ is separated. \end{proof} \begin{lemma} \label{lemma-characterize-quasi-affine} Let $f : X \to S$ be a morphism of schemes. The following are equivalent \begin{enumerate} \item The morphism $f$ is quasi-affine. \item There exists an affine open covering $S = \bigcup W_j$ such that each $f^{-1}(W_j)$ is quasi-affine. \item There exists a quasi-coherent sheaf of $\mathcal{O}_S$-algebras $\mathcal{A}$ and a quasi-compact open immersion $$\xymatrix{ X \ar[rr] \ar[rd] & & \underline{\Spec}_S(\mathcal{A}) \ar[dl] \\ & S & }$$ over $S$. \item Same as in (3) but with $\mathcal{A} = f_*\mathcal{O}_X$ and the horizontal arrow the canonical morphism of Constructions, Lemma \ref{constructions-lemma-canonical-morphism}. \end{enumerate} \end{lemma} \begin{proof} It is obvious that (1) implies (2) and that (4) implies (3). \medskip\noindent Assume $S = \bigcup_{j \in J} W_j$ is an affine open covering such that each $f^{-1}(W_j)$ is quasi-affine. By Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine} we see that $f$ is quasi-compact. By Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} we see the morphism $f$ is quasi-separated. Hence by Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent} the sheaf $\mathcal{A} = f_*\mathcal{O}_X$ is a quasi-coherent sheaf of $\mathcal{O}_X$-algebras. Thus we have the scheme $g : Y = \underline{\Spec}_S(\mathcal{A}) \to S$ over $S$. The identity map $\text{id} : \mathcal{A} = f_*\mathcal{O}_X \to f_*\mathcal{O}_X$ provides, via the definition of the relative spectrum, a morphism $can : X \to Y$ over $S$, see Constructions, Lemma \ref{constructions-lemma-canonical-morphism}. By assumption, the lemma just cited, and Properties, Lemma \ref{properties-lemma-quasi-affine} the restriction $can|_{f^{-1}(W_j)} : f^{-1}(W_j) \to g^{-1}(W_j)$ is a quasi-compact open immersion. Thus $can$ is a quasi-compact open immersion. We have shown that (2) implies (4). \medskip\noindent Assume (3). Choose any affine open $U \subset S$. By Constructions, Lemma \ref{constructions-lemma-spec-properties} we see that the inverse image of $U$ in the relative spectrum is affine. Hence we conclude that $f^{-1}(U)$ is quasi-affine (note that quasi-compactness is encoded in (3) as well). Thus (3) implies (1). \end{proof} \begin{lemma} \label{lemma-composition-quasi-affine} The composition of quasi-affine morphisms is quasi-affine. \end{lemma} \begin{proof} Let $f : X \to Y$ and $g : Y \to Z$ be quasi-affine morphisms. Let $U \subset Z$ be affine open. Then $g^{-1}(U)$ is quasi-affine by assumption on $g$. Let $j : g^{-1}(U) \to V$ be a quasi-compact open immersion into an affine scheme $V$. By Lemma \ref{lemma-characterize-quasi-affine} above we see that $f^{-1}(g^{-1}(U))$ is a quasi-compact open subscheme of the relative spectrum $\underline{\Spec}_{g^{-1}(U)}(\mathcal{A})$ for some quasi-coherent sheaf of $\mathcal{O}_{g^{-1}(U)}$-algebras $\mathcal{A}$. By Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent} the sheaf $\mathcal{A}' = j_*\mathcal{A}$ is a quasi-coherent sheaf of $\mathcal{O}_V$-algebras with the property that $j^*\mathcal{A}' = \mathcal{A}$. Hence we get a commutative diagram $$\xymatrix{ f^{-1}(g^{-1}(U)) \ar[r] & \underline{\Spec}_{g^{-1}(U)}(\mathcal{A}) \ar[r] \ar[d] & \underline{\Spec}_V(\mathcal{A}') \ar[d] \\ & g^{-1}(U) \ar[r]^j & V }$$ with the square being a fibre square, see Constructions, Lemma \ref{constructions-lemma-spec-properties}. Note that the upper right corner is an affine scheme. Hence $(g \circ f)^{-1}(U)$ is quasi-affine. \end{proof} \begin{lemma} \label{lemma-base-change-quasi-affine} The base change of a quasi-affine morphism is quasi-affine. \end{lemma} \begin{proof} Let $f : X \to S$ be a quasi-affine morphism. By Lemma \ref{lemma-characterize-quasi-affine} above we can find a quasi-coherent sheaf of $\mathcal{O}_S$-algebras $\mathcal{A}$ and a quasi-compact open immersion $X \to \underline{\Spec}_S(\mathcal{A})$ over $S$. Let $g : S' \to S$ be any morphism. Denote $f' : X_{S'} = S' \times_S X \to S'$ the base change of $f$. Since the base change of a quasi-compact open immersion is a quasi-compact open immersion we see that $X_{S'} \to \underline{\Spec}_{S'}(g^*\mathcal{A})$ is a quasi-compact open immersion (we have used Schemes, Lemmas \ref{schemes-lemma-quasi-compact-preserved-base-change} and \ref{schemes-lemma-base-change-immersion} and Constructions, Lemma \ref{constructions-lemma-spec-properties}). By Lemma \ref{lemma-characterize-quasi-affine} again we conclude that $X_{S'} \to S'$ is quasi-affine. \end{proof} \begin{lemma} \label{lemma-quasi-compact-immersion-quasi-affine} A quasi-compact immersion is quasi-affine. \end{lemma} \begin{proof} Let $X \to S$ be a quasi-compact immersion. We have to show the inverse image of every affine open is quasi-affine. Hence, assuming $S$ is an affine scheme, we have to show $X$ is quasi-affine. By Lemma \ref{lemma-quasi-compact-immersion} the morphism $X \to S$ factors as $X \to Z \to S$ where $Z$ is a closed subscheme of $S$ and $X \subset Z$ is a quasi-compact open. Since $S$ is affine Lemma \ref{lemma-closed-immersion} implies $Z$ is affine. Hence we win. \end{proof} \begin{lemma} \label{lemma-affine-quasi-affine} Let $S$ be a scheme. Let $X$ be an affine scheme. A morphism $f : X \to S$ is quasi-affine if and only if it is quasi-compact. In particular any morphism from an affine scheme to a quasi-separated scheme is quasi-affine. \end{lemma} \begin{proof} Let $V \subset S$ be an affine open. Then $f^{-1}(V)$ is an open subscheme of the affine scheme $X$, hence quasi-affine if and only if it is quasi-compact. This proves the first assertion. The quasi-compactness of any $f : X \to S$ where $X$ is affine and $S$ quasi-separated follows from Schemes, Lemma \ref{schemes-lemma-quasi-compact-permanence} applied to $X \to S \to \Spec(\mathbf{Z})$. \end{proof} \begin{lemma} \label{lemma-quasi-affine-permanence} Suppose $g : X \to Y$ is a morphism of schemes over $S$. If $X$ is quasi-affine over $S$ and $Y$ is quasi-separated over $S$, then $g$ is quasi-affine. In particular, any morphism from a quasi-affine scheme to a quasi-separated scheme is quasi-affine. \end{lemma} \begin{proof} The base change $X \times_S Y \to Y$ is quasi-affine by Lemma \ref{lemma-base-change-quasi-affine}. The morphism $X \to X \times_S Y$ is a quasi-compact immersion as $Y \to S$ is quasi-separated, see Schemes, Lemma \ref{schemes-lemma-section-immersion}. A quasi-compact immersion is quasi-affine by Lemma \ref{lemma-quasi-compact-immersion-quasi-affine} and the composition of quasi-affine morphisms is quasi-affine (see Lemma \ref{lemma-composition-quasi-affine}). Thus we win. \end{proof} \section{Types of morphisms defined by properties of ring maps} \label{section-properties-ring-maps} \noindent In this section we study what properties of ring maps allow one to define local properties of morphisms of schemes. \begin{definition} \label{definition-property-local} Let $P$ be a property of ring maps. \begin{enumerate} \item We say that $P$ is {\it local} if the following hold: \begin{enumerate} \item For any ring map $R \to A$, and any $f \in R$ we have $P(R \to A) \Rightarrow P(R_f \to A_f)$. \item For any rings $R$, $A$, any $f \in R$, $a\in A$, and any ring map $R_f \to A$ we have $P(R_f \to A) \Rightarrow P(R \to A_a)$. \item For any ring map $R \to A$, and $a_i \in A$ such that $(a_1, \ldots, a_n) = A$ then $\forall i, P(R \to A_{a_i}) \Rightarrow P(R \to A)$. \end{enumerate} \item We say that $P$ is {\it stable under base change} if for any ring maps $R \to A$, $R \to R'$ we have $P(R \to A) \Rightarrow P(R' \to R' \otimes_R A)$. \item We say that $P$ is {\it stable under composition} if for any ring maps $A \to B$, $B \to C$ we have $P(A \to B) \wedge P(B \to C) \Rightarrow P(A \to C)$. \end{enumerate} \end{definition} \begin{definition} \label{definition-locally-P} Let $P$ be a property of ring maps. Let $f : X \to S$ be a morphism of schemes. We say $f$ is {\it locally of type $P$} if for any $x \in X$ there exists an affine open neighbourhood $U$ of $x$ in $X$ which maps into an affine open $V \subset S$ such that the induced ring map $\mathcal{O}_S(V) \to \mathcal{O}_X(U)$ has property $P$. \end{definition} \noindent This is not a good'' definition unless the property $P$ is a local property. Even if $P$ is a local property we will not automatically use this definition to say that a morphism is locally of type $P$'' unless we also explicitly state the definition elsewhere. \begin{lemma} \label{lemma-locally-P} Let $f : X \to S$ be a morphism of schemes. Let $P$ be a property of ring maps. Let $U$ be an affine open of $X$, and $V$ an affine open of $S$ such that $f(U) \subset V$. If $f$ is locally of type $P$ and $P$ is local, then $P(\mathcal{O}_S(V) \to \mathcal{O}_X(U))$ holds. \end{lemma} \begin{proof} As $f$ is locally of type $P$ for every $u \in U$ there exists an affine open $U_u \subset X$ mapping into an affine open $V_u \subset S$ such that $P(\mathcal{O}_S(V_u) \to \mathcal{O}_X(U_u))$ holds. Choose an open neighbourhood $U'_u \subset U \cap U_u$ of $u$ which is standard affine open in both $U$ and $U_u$, see Schemes, Lemma \ref{schemes-lemma-standard-open-two-affines}. By Definition \ref{definition-property-local} (1)(b) we see that $P(\mathcal{O}_S(V_u) \to \mathcal{O}_X(U'_u))$ holds. Hence we may assume that $U_u \subset U$ is a standard affine open. Choose an open neighbourhood $V'_u \subset V \cap V_u$ of $f(u)$ which is standard affine open in both $V$ and $V_u$, see Schemes, Lemma \ref{schemes-lemma-standard-open-two-affines}. Then $U'_u = f^{-1}(V'_u) \cap U_u$ is a standard affine open of $U_u$ (hence of $U$) and we have $P(\mathcal{O}_S(V'_u) \to \mathcal{O}_X(U'_u))$ by Definition \ref{definition-property-local} (1)(a). Hence we may assume both $U_u \subset U$ and $V_u \subset V$ are standard affine open. Applying Definition \ref{definition-property-local} (1)(b) one more time we conclude that $P(\mathcal{O}_S(V) \to \mathcal{O}_X(U_u))$ holds. Because $U$ is quasi-compact we may choose a finite number of points $u_1, \ldots, u_n \in U$ such that $$U = U_{u_1} \cup \ldots \cup U_{u_n}.$$ By Definition \ref{definition-property-local} (1)(c) we conclude that $P(\mathcal{O}_S(V) \to \mathcal{O}_X(U))$ holds. \end{proof} \begin{lemma} \label{lemma-locally-P-characterize} Let $P$ be a local property of ring maps. Let $f : X \to S$ be a morphism of schemes. The following are equivalent \begin{enumerate} \item The morphism $f$ is locally of type $P$. \item For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ we have $P(\mathcal{O}_S(V) \to \mathcal{O}_X(U))$. \item There exists an open covering $S = \bigcup_{j \in J} V_j$ and open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that each of the morphisms $U_i \to V_j$, $j\in J, i\in I_j$ is locally of type $P$. \item There exists an affine open covering $S = \bigcup_{j \in J} V_j$ and affine open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that $P(\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i))$ holds, for all $j\in J, i\in I_j$. \end{enumerate} Moreover, if $f$ is locally of type $P$ then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_U : U \to V$ is locally of type $P$. \end{lemma} \begin{proof} This follows from Lemma \ref{lemma-locally-P} above. \end{proof} \begin{lemma} \label{lemma-composition-type-P} Let $P$ be a property of ring maps. Assume $P$ is local and stable under composition. The composition of morphisms locally of type $P$ is locally of type $P$. \end{lemma} \begin{proof} Let $f : X \to Y$ and $g : Y \to Z$ be morphisms locally of type $P$. Let $x \in X$. Choose an affine open neighbourhood $W \subset Z$ of $g(f(x))$. Choose an affine open neighbourhood $V \subset g^{-1}(W)$ of $f(x)$. Choose an affine open neighbourhood $U \subset f^{-1}(V)$ of $x$. By Lemma \ref{lemma-locally-P-characterize} the ring maps $\mathcal{O}_Z(W) \to \mathcal{O}_Y(V)$ and $\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$ satisfy $P$. Hence $\mathcal{O}_Z(W) \to \mathcal{O}_X(U)$ satisfies $P$ as $P$ is assumed stable under composition. \end{proof} \begin{lemma} \label{lemma-base-change-type-P} Let $P$ be a property of ring maps. Assume $P$ is local and stable under base change. The base change of a morphism locally of type $P$ is locally of type $P$. \end{lemma} \begin{proof} Let $f : X \to S$ be a morphism locally of type $P$. Let $S' \to S$ be any morphism. Denote $f' : X_{S'} = S' \times_S X \to S'$ the base change of $f$. For every $s' \in S'$ there exists an open affine neighbourhood $s' \in V' \subset S'$ which maps into some open affine $V \subset S$. By Lemma \ref{lemma-locally-P-characterize} the open $f^{-1}(V)$ is a union of affines $U_i$ such that the ring maps $\mathcal{O}_S(V) \to \mathcal{O}_X(U_i)$ all satisfy $P$. By the material in Schemes, Section \ref{schemes-section-fibre-products} we see that $f^{-1}(U)_{V'} = V' \times_V f^{-1}(V)$ is the union of the affine opens $V' \times_V U_i$. Since $\mathcal{O}_{X_{S'}}(V' \times_V U_i) = \mathcal{O}_{S'}(V') \otimes_{\mathcal{O}_S(V)} \mathcal{O}_X(U_i)$ we see that the ring maps $\mathcal{O}_{S'}(V') \to \mathcal{O}_{X_{S'}}(V' \times_V U_i)$ satisfy $P$ as $P$ is assumed stable under base change. \end{proof} \begin{lemma} \label{lemma-properties-local} The following properties of a ring map $R \to A$ are local. \begin{enumerate} \item (Isomorphism on local rings.) For every prime $\mathfrak q$ of $A$ lying over $\mathfrak p \subset R$ the ring map $R \to A$ induces an isomorphism $R_{\mathfrak p} \to A_{\mathfrak q}$. \item (Open immersion.) For every prime $\mathfrak q$ of $A$ there exists an $f \in R$, $\varphi(f) \not \in \mathfrak q$ such that the ring map $\varphi : R \to A$ induces an isomorphism $R_f \to A_f$. \item (Reduced fibres.) For every prime $\mathfrak p$ of $R$ the fibre ring $A \otimes_R \kappa(\mathfrak p)$ is reduced. \item (Fibres of dimension at most $n$.) For every prime $\mathfrak p$ of $R$ the fibre ring $A \otimes_R \kappa(\mathfrak p)$ has Krull dimension at most $n$. \item (Locally Noetherian on the target.) The ring map $R \to A$ has the property that $A$ is Noetherian. \item Add more here as needed\footnote{But only those properties that are not already dealt with separately elsewhere.}. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-properties-base-change} The following properties of ring maps are stable under base change. \begin{enumerate} \item (Isomorphism on local rings.) For every prime $\mathfrak q$ of $A$ lying over $\mathfrak p \subset R$ the ring map $R \to A$ induces an isomorphism $R_{\mathfrak p} \to A_{\mathfrak q}$. \item (Open immersion.) For every prime $\mathfrak q$ of $A$ there exists an $f \in R$, $\varphi(f) \not \in \mathfrak q$ such that the ring map $\varphi : R \to A$ induces an isomorphism $R_f \to A_f$. \item Add more here as needed\footnote{But only those properties that are not already dealt with separately elsewhere.}. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-properties-composition} The following properties of ring maps are stable under composition. \begin{enumerate} \item (Isomorphism on local rings.) For every prime $\mathfrak q$ of $A$ lying over $\mathfrak p \subset R$ the ring map $R \to A$ induces an isomorphism $R_{\mathfrak p} \to A_{\mathfrak q}$. \item (Open immersion.) For every prime $\mathfrak q$ of $A$ there exists an $f \in R$, $\varphi(f) \not \in \mathfrak q$ such that the ring map $\varphi : R \to A$ induces an isomorphism $R_f \to A_f$. \item (Locally Noetherian on the target.) The ring map $R \to A$ has the property that $A$ is Noetherian. \item Add more here as needed\footnote{But only those properties that are not already dealt with separately elsewhere.}. \end{enumerate} \end{lemma} \begin{proof} Omitted. \end{proof} \section{Morphisms of finite type} \label{section-finite-type} \noindent Recall that a ring map $R \to A$ is said to be of finite type if $A$ is isomorphic to a quotient of $R[x_1, \ldots, x_n]$ as an $R$-algebra, see Algebra, Definition \ref{algebra-definition-finite-type}. \begin{definition} \label{definition-finite-type} Let $f : X \to S$ be a morphism of schemes. \begin{enumerate} \item We say that $f$ is of {\it finite type at $x \in X$} if there exists an affine open neighbourhood $\Spec(A) = U \subset X$ of $x$ and an affine open $\Spec(R) = V \subset S$ with $f(U) \subset V$ such that the induced ring map $R \to A$ is of finite type. \item We say that $f$ is {\it locally of finite type} if it is of finite type at every point of $X$. \item We say that $f$ is of {\it finite type} if it is locally of finite type and quasi-compact. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-locally-finite-type-characterize} Let $f : X \to S$ be a morphism of schemes. The following are equivalent \begin{enumerate} \item The morphism $f$ is locally of finite type. \item For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_S(V) \to \mathcal{O}_X(U)$ is of finite type. \item There exists an open covering $S = \bigcup_{j \in J} V_j$ and open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that each of the morphisms $U_i \to V_j$, $j\in J, i\in I_j$ is locally of finite type. \item There exists an affine open covering $S = \bigcup_{j \in J} V_j$ and affine open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that the ring map $\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)$ is of finite type, for all $j\in J, i\in I_j$. \end{enumerate} Moreover, if $f$ is locally of finite type then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_U : U \to V$ is locally of finite type. \end{lemma} \begin{proof} This follows from Lemma \ref{lemma-locally-P} if we show that the property $R \to A$ is of finite type'' is local. We check conditions (a), (b) and (c) of Definition \ref{definition-property-local}. By Algebra, Lemma \ref{algebra-lemma-base-change-finiteness} being of finite type is stable under base change and hence we conclude (a) holds. By the same lemma being of finite type is stable under composition and trivially for any ring $R$ the ring map $R \to R_f$ is of finite type. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma \ref{algebra-lemma-cover-upstairs}. \end{proof} \begin{lemma} \label{lemma-composition-finite-type} The composition of two morphisms which are locally of finite type is locally of finite type. The same is true for morphisms of finite type. \end{lemma} \begin{proof} In the proof of Lemma \ref{lemma-locally-finite-type-characterize} we saw that being of finite type is a local property of ring maps. Hence the first statement of the lemma follows from Lemma \ref{lemma-composition-type-P} combined with the fact that being of finite type is a property of ring maps that is stable under composition, see Algebra, Lemma \ref{algebra-lemma-compose-finite-type}. By the above and the fact that compositions of quasi-compact morphisms are quasi-compact, see Schemes, Lemma \ref{schemes-lemma-composition-quasi-compact} we see that the composition of morphisms of finite type is of finite type. \end{proof} \begin{lemma} \label{lemma-base-change-finite-type} The base change of a morphism which is locally of finite type is locally of finite type. The same is true for morphisms of finite type. \end{lemma} \begin{proof} In the proof of Lemma \ref{lemma-locally-finite-type-characterize} we saw that being of finite type is a local property of ring maps. Hence the first statement of the lemma follows from Lemma \ref{lemma-composition-type-P} combined with the fact that being of finite type is a property of ring maps that is stable under base change, see Algebra, Lemma \ref{algebra-lemma-base-change-finiteness}. By the above and the fact that a base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma \ref{schemes-lemma-quasi-compact-preserved-base-change} we see that the base change of a morphism of finite type is a morphism of finite type. \end{proof} \begin{lemma} \label{lemma-immersion-locally-finite-type} A closed immersion is of finite type. An immersion is locally of finite type. \end{lemma} \begin{proof} This is true because an open immersion is a local isomorphism, and a closed immersion is obviously of finite type. \end{proof} \begin{lemma} \label{lemma-finite-type-noetherian} Let $f : X \to S$ be a morphism. If $S$ is (locally) Noetherian and $f$ (locally) of finite type then $X$ is (locally) Noetherian. \end{lemma} \begin{proof} This follows immediately from the fact that a ring of finite type over a Noetherian ring is Noetherian, see Algebra, Lemma \ref{algebra-lemma-Noetherian-permanence}. (Also: use the fact that the source of a quasi-compact morphism with quasi-compact target is quasi-compact.) \end{proof} \begin{lemma} \label{lemma-finite-type-Noetherian-quasi-separated} Let $f : X \to S$ be locally of finite type with $S$ locally Noetherian. Then $f$ is quasi-separated. \end{lemma} \begin{proof} In fact, it is true that $X$ is quasi-separated, see Properties, Lemma \ref{properties-lemma-locally-Noetherian-quasi-separated} and Lemma \ref{lemma-finite-type-noetherian} above. Then apply Schemes, Lemma \ref{schemes-lemma-compose-after-separated} to conclude that $f$ is quasi-separated. \end{proof} \begin{lemma} \label{lemma-permanence-finite-type} Let $X \to Y$ be a morphism of schemes over a base scheme $S$. If $X$ is locally of finite type over $S$, then $X \to Y$ is locally of finite type. \end{lemma} \begin{proof} Via Lemma \ref{lemma-locally-finite-type-characterize} this translates into the following algebra fact: Given ring maps $A \to B \to C$ such that $A \to C$ is of finite type, then $B \to C$ is of finite type. (See Algebra, Lemma \ref{algebra-lemma-compose-finite-type}). \end{proof} \section{Points of finite type and Jacobson schemes} \label{section-points-finite-type} \noindent Let $S$ be a scheme. A finite type point $s$ of $S$ is a point such that the morphism $\Spec(\kappa(s)) \to S$ is of finite type. The reason for studying this is that finite type points can replace closed points in a certain sense and in certain situations. There are always enough of them for example. Moreover, a scheme is Jacobson if and only if all finite type points are closed points. \begin{lemma} \label{lemma-point-finite-type} Let $S$ be a scheme. Let $k$ be a field. Let $f : \Spec(k) \to S$ be a morphism. The following are equivalent: \begin{enumerate} \item The morphism $f$ is of finite type. \item The morphism $f$ is locally of finite type. \item There exists an affine open $U = \Spec(R)$ of $S$ such that $f$ corresponds to a finite ring map $R \to k$. \item There exists an affine open $U = \Spec(R)$ of $S$ such that the image of $f$ consists of a closed point $u$ in $U$ and the field extension $\kappa(u) \subset k$ is finite. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (2) is obvious as $\Spec(k)$ is a singleton and hence any morphism from it is quasi-compact. \medskip\noindent Suppose $f$ is locally of finite type. Choose any affine open $\Spec(R) = U \subset S$ such that the image of $f$ is contained in $U$, and the ring map $R \to k$ is of finite type. Let $\mathfrak p \subset R$ be the kernel. Then $R/\mathfrak p \subset k$ is of finite type. By Algebra, Lemma \ref{algebra-lemma-field-finite-type-over-domain} there exist a $\overline{f} \in R/\mathfrak p$ such that $(R/\mathfrak p)_{\overline{f}}$ is a field and $(R/\mathfrak p)_{\overline{f}} \to k$ is a finite field extension. If $f \in R$ is a lift of $\overline{f}$, then we see that $k$ is a finite $R_f$-module. Thus (2) $\Rightarrow$ (3). \medskip\noindent Suppose that $\Spec(R) = U \subset S$ is an affine open such that $f$ corresponds to a finite ring map $R \to k$. Then $f$ is locally of finite type by Lemma \ref{lemma-locally-finite-type-characterize}. Thus (3) $\Rightarrow$ (2). \medskip\noindent Suppose $R \to k$ is finite. The image of $R \to k$ is a field over which $k$ is finite by Algebra, Lemma \ref{algebra-lemma-integral-under-field}. Hence the kernel of $R \to k$ is a maximal ideal. Thus (3) $\Rightarrow$ (4). \medskip\noindent The implication (4) $\Rightarrow$ (3) is immediate. \end{proof} \begin{lemma} \label{lemma-artinian-finite-type} Let $S$ be a scheme. Let $A$ be an Artinian local ring with residue field $\kappa$. Let $f : \Spec(A) \to S$ be a morphism of schemes. Then $f$ is of finite type if and only if the composition $\Spec(\kappa) \to \Spec(A) \to S$ is of finite type. \end{lemma} \begin{proof} Since the morphism $\Spec(\kappa) \to \Spec(A)$ is of finite type it is clear that if $f$ is of finite type so is the composition $\Spec(\kappa) \to S$ (see Lemma \ref{lemma-composition-finite-type}). For the converse, note that $\Spec(A) \to S$ maps into some affine open $U = \Spec(B)$ of $S$ as $\Spec(A)$ has only one point. To finish apply Algebra, Lemma \ref{algebra-lemma-essentially-of-finite-type-into-artinian-local} to $B \to A$. \end{proof} \noindent Recall that given a point $s$ of a scheme $S$ there is a canonical morphism $\Spec(\kappa(s)) \to S$, see Schemes, Section \ref{schemes-section-points}. \begin{definition} \label{definition-finite-type-point} Let $S$ be a scheme. Let us say that a point $s$ of $S$ is a {\it finite type point} if the canonical morphism $\Spec(\kappa(s)) \to S$ is of finite type. We denote $S_{\text{ft-pts}}$ the set of finite type points of $S$. \end{definition} \noindent We can describe the set of finite type points as follows. \begin{lemma} \label{lemma-identify-finite-type-points} Let $S$ be a scheme. We have $$S_{\text{ft-pts}} = \bigcup\nolimits_{U \subset S\text{ open}} U_0$$ where $U_0$ is the set of closed points of $U$. Here we may let $U$ range over all opens or over all affine opens of $S$. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-point-finite-type}. \end{proof} \begin{lemma} \label{lemma-finite-type-points-morphism} Let $f : T \to S$ be a morphism of schemes. If $f$ is locally of finite type, then $f(T_{\text{ft-pts}}) \subset S_{\text{ft-pts}}$. \end{lemma} \begin{proof} If $T$ is the spectrum of a field this is Lemma \ref{lemma-point-finite-type}. In general it follows since the composition of morphisms locally of finite type is locally of finite type (Lemma \ref{lemma-composition-finite-type}). \end{proof} \begin{lemma} \label{lemma-finite-type-points-surjective-morphism} Let $f : T \to S$ be a morphism of schemes. If $f$ is locally of finite type and surjective, then $f(T_{\text{ft-pts}}) = S_{\text{ft-pts}}$. \end{lemma} \begin{proof} We have $f(T_{\text{ft-pts}}) \subset S_{\text{ft-pts}}$ by Lemma \ref{lemma-finite-type-points-morphism}. Let $s \in S$ be a finite type point. As $f$ is surjective the scheme $T_s = \Spec(\kappa(s)) \times_S T$ is nonempty, therefore has a finite type point $t \in T_s$ by Lemma \ref{lemma-identify-finite-type-points}. Now $T_s \to T$ is a morphism of finite type as a base change of $s \to S$ (Lemma \ref{lemma-base-change-finite-type}). Hence the image of $t$ in $T$ is a finite type point by Lemma \ref{lemma-finite-type-points-morphism} which maps to $s$ by construction. \end{proof} \begin{lemma} \label{lemma-enough-finite-type-points} Let $S$ be a scheme. For any locally closed subset $T \subset S$ we have $$T \not = \emptyset \Rightarrow T \cap S_{\text{ft-pts}} \not = \emptyset.$$ In particular, for any closed subset $T \subset S$ we see that $T \cap S_{\text{ft-pts}}$ is dense in $T$. \end{lemma} \begin{proof} Note that $T$ carries a scheme structure (see Schemes, Lemma \ref{schemes-lemma-reduced-closed-subscheme}) such that $T \to S$ is a locally closed immersion. Any locally closed immersion is locally of finite type, see Lemma \ref{lemma-immersion-locally-finite-type}. Hence by Lemma \ref{lemma-finite-type-points-morphism} we see $T_{\text{ft-pts}} \subset S_{\text{ft-pts}}$. Finally, any nonempty affine open of $T$ has at least one closed point which is a finite type point of $T$ by Lemma \ref{lemma-identify-finite-type-points}. \end{proof} \noindent It follows that most of the material from Topology, Section \ref{topology-section-space-jacobson} goes through with the set of closed points replaced by the set of points of finite type. In fact, if $S$ is Jacobson then we recover the closed points as the finite type points. \begin{lemma} \label{lemma-jacobson-finite-type-points} Let $S$ be a scheme. The following are equivalent: \begin{enumerate} \item the scheme $S$ is Jacobson, \item $S_{\text{ft-pts}}$ is the set of closed points of $S$, \item for all $T \to S$ locally of finite type closed points map to closed points, and \item for all $T \to S$ locally of finite type closed points $t \in T$ map to closed points $s \in S$ with $\kappa(s) \subset \kappa(t)$ finite. \end{enumerate} \end{lemma} \begin{proof} We have trivially (4) $\Rightarrow$ (3) $\Rightarrow$ (2). Lemma \ref{lemma-enough-finite-type-points} shows that (2) implies (1). Hence it suffices to show that (1) implies (4). Suppose that $T \to S$ is locally of finite type. Choose $t \in T$ closed and let $s \in S$ be the image. Choose affine open neighbourhoods $\Spec(R) = U \subset S$ of $s$ and $\Spec(A) = V \subset T$ of $t$ with $V$ mapping into $U$. The induced ring map $R \to A$ is of finite type (see Lemma \ref{lemma-locally-finite-type-characterize}) and $R$ is Jacobson by Properties, Lemma \ref{properties-lemma-locally-jacobson}. Thus the result follows from Algebra, Proposition \ref{algebra-proposition-Jacobson-permanence}. \end{proof} \begin{lemma} \label{lemma-Jacobson-universally-Jacobson} Let $S$ be a Jacobson scheme. Any scheme locally of finite type over $S$ is Jacobson. \end{lemma} \begin{proof} This is clear from Algebra, Proposition \ref{algebra-proposition-Jacobson-permanence} (and Properties, Lemma \ref{properties-lemma-locally-jacobson} and Lemma \ref{lemma-locally-finite-type-characterize}). \end{proof} \begin{lemma} \label{lemma-ubiquity-Jacobson-schemes} The following types of schemes are Jacobson. \begin{enumerate} \item Any scheme locally of finite type over a field. \item Any scheme locally of finite type over $\mathbf{Z}$. \item Any scheme locally of finite type over a $1$-dimensional Noetherian domain with infinitely many primes. \item A scheme of the form $\Spec(R) \setminus \{\mathfrak m\}$ where $(R, \mathfrak m)$ is a Noetherian local ring. Also any scheme locally of finite type over it. \end{enumerate} \end{lemma} \begin{proof} We will use Lemma \ref{lemma-Jacobson-universally-Jacobson} without mention. The spectrum of a field is clearly Jacobson. The spectrum of $\mathbf{Z}$ is Jacobson, see Algebra, Lemma \ref{algebra-lemma-pid-jacobson}. For (3) see Algebra, Lemma \ref{algebra-lemma-noetherian-dim-1-Jacobson}. For (4) see Properties, Lemma \ref{properties-lemma-complement-closed-point-Jacobson}. \end{proof} \section{Universally catenary schemes} \label{section-universally-catenary} \noindent Recall that a topological space $X$ is called {\it catenary} if for every pair of irreducible closed subsets $T \subset T'$ there exist a maximal chain of irreducible closed subsets $$T = T_0 \subset T_1 \subset \ldots \subset T_e = T'$$ and every such chain has the same length. See Topology, Definition \ref{topology-definition-catenary}. Recall that a scheme is catenary if its underlying topological space is catenary. See Properties, Definition \ref{properties-definition-catenary}. \begin{definition} \label{definition-universally-catenary} Let $S$ be a scheme. Assume $S$ is locally Noetherian. We say $S$ is {\it universally catenary} if for every morphism $X \to S$ locally of finite type the scheme $X$ is catenary. \end{definition} \noindent This is a better'' notion than catenary as there exist Noetherian schemes which are catenary but not universally catenary. See Examples, Section \ref{examples-section-non-catenary-Noetherian-local}. Many schemes are universally catenary, see Lemma \ref{lemma-ubiquity-uc} below. \medskip\noindent Recall that a ring $A$ is called {\it catenary} if for any pair of prime ideals $\mathfrak p \subset \mathfrak q$ there exists a maximal chain of primes $$\mathfrak p = \mathfrak p_0 \subset \ldots \subset \mathfrak p_e = \mathfrak q$$ and all of these have the same length. See Algebra, Definition \ref{algebra-definition-catenary}. We have seen the relationship between catenary schemes and catenary rings in Properties, Section \ref{properties-section-catenary}. Recall that a ring $A$ is called {\it universally catenary} if $A$ is Noetherian and for every finite type ring map $A \to B$ the ring $B$ is catenary. See Algebra, Definition \ref{algebra-definition-universally-catenary}. Many interesting rings which come up in algebraic geometry satisfy this property. \begin{lemma} \label{lemma-universally-catenary-local} Let $S$ be a locally Noetherian scheme. The following are equivalent \begin{enumerate} \item $S$ is universally catenary, \item there exists an open covering of $S$ all of whose members are universally catenary schemes, \item for every affine open $\Spec(R) = U \subset S$ the ring $R$ is universally catenary, and \item there exists an affine open covering $S = \bigcup U_i$ such that each $U_i$ is the spectrum of a universally catenary ring. \end{enumerate} Moreover, in this case any scheme locally of finite type over $S$ is universally catenary as well. \end{lemma} \begin{proof} By Lemma \ref{lemma-immersion-locally-finite-type} an open immersion is locally of finite type. A composition of morphisms locally of finite type is locally of finite type (Lemma \ref{lemma-composition-finite-type}). Thus it is clear that if $S$ is universally catenary then any open and any scheme locally of finite type over $S$ is universally catenary as well. This proves the final statement of the lemma and that (1) implies (2). \medskip\noindent If $\Spec(R)$ is a universally catenary scheme, then every scheme $\Spec(A)$ with $A$ a finite type $R$-algebra is catenary. Hence all these rings $A$ are catenary by Algebra, Lemma \ref{algebra-lemma-catenary}. Thus $R$ is universally catenary. Combined with the remarks above we conclude that (1) implies (3), and (2) implies (4). Of course (3) implies (4) trivially. \medskip\noindent To finish the proof we show that (4) implies (1). Assume (4) and let $X \to S$ be a morphism locally of finite type. We can find an affine open covering $X = \bigcup V_j$ such that each $V_j \to S$ maps into one of the $U_i$. By Lemma \ref{lemma-locally-finite-type-characterize} the induced ring map $\mathcal{O}(U_i) \to \mathcal{O}(V_j)$ is of finite type. Hence $\mathcal{O}(V_j)$ is catenary. Hence $X$ is catenary by Properties, Lemma \ref{properties-lemma-catenary-local}. \end{proof} \begin{lemma} \label{lemma-universally-catenary-local-rings-universally-catenary} Let $S$ be a locally Noetherian scheme. The following are equivalent: \begin{enumerate} \item $S$ is universally catenary, and \item all local rings $\mathcal{O}_{S, s}$ of $S$ are universally catenary. \end{enumerate} \end{lemma} \begin{proof} Assume that all local rings of $S$ are universally catenary. Let $f : X \to S$ be locally of finite type. We know that $X$ is catenary if and only if $\mathcal{O}_{X, x}$ is catenary for all $x \in X$. If $f(x) = s$, then $\mathcal{O}_{X, x}$ is essentially of finite type over $\mathcal{O}_{S, s}$. Hence $\mathcal{O}_{X, x}$ is catenary by the assumption that $\mathcal{O}_{S, s}$ is universally catenary. \medskip\noindent Conversely, assume that $S$ is universally catenary. Let $s \in S$. We may replace $S$ by an affine open neighbourhood of $s$ by Lemma \ref{lemma-universally-catenary-local}. Say $S = \Spec(R)$ and $s$ corresponds to the prime ideal $\mathfrak p$. Any finite type $R_{\mathfrak p}$-algebra $A'$ is of the form $A_{\mathfrak p}$ for some finite type $R$-algebra $A$. By assumption (and Lemma \ref{lemma-universally-catenary-local} if you like) the ring $A$ is catenary, and hence $A'$ (a localization of $A$) is catenary. Thus $R_{\mathfrak p}$ is universally catenary. \end{proof} \begin{lemma} \label{lemma-ubiquity-uc} The following types of schemes are universally catenary. \begin{enumerate} \item Any scheme locally of finite type over a field. \item Any scheme locally of finite type over a Cohen-Macaulay scheme. \item Any scheme locally of finite type over $\mathbf{Z}$. \item Any scheme locally of finite type over a $1$-dimensional Noetherian domain. \item And so on. \end{enumerate} \end{lemma} \begin{proof} All of these follow from the fact that a Cohen-Macaulay ring is universally catenary, see Algebra, Lemma \ref{algebra-lemma-CM-ring-catenary}. Also, use the last assertion of Lemma \ref{lemma-universally-catenary-local}. Some details omitted. \end{proof} \section{Nagata schemes, reprise} \label{section-nagata} \noindent See Properties, Section \ref{properties-section-nagata} for the definitions and basic properties of Nagata and universally Japanese schemes. \begin{lemma} \label{lemma-finite-type-nagata} Let $f : X \to S$ be a morphism. If $S$ is Nagata and $f$ locally of finite type then $X$ is Nagata. If $S$ is universally Japanese and $f$ locally of finite type then $X$ is universally Japanese. \end{lemma} \begin{proof} For universally Japanese'' this follows from Algebra, Lemma \ref{algebra-lemma-universally-japanese}. For Nagata'' this follows from Algebra, Proposition \ref{algebra-proposition-nagata-universally-japanese}. \end{proof} \begin{lemma} \label{lemma-ubiquity-nagata} The following types of schemes are Nagata. \begin{enumerate} \item Any scheme locally of finite type over a field. \item Any scheme locally of finite type over a Noetherian complete local ring. \item Any scheme locally of finite type over $\mathbf{Z}$. \item Any scheme locally of finite type over a Dedekind ring of characteristic zero. \item And so on. \end{enumerate} \end{lemma} \begin{proof} By Lemma \ref{lemma-finite-type-nagata} we only need to show that the rings mentioned above are Nagata rings. For this see Algebra, Proposition \ref{algebra-proposition-ubiquity-nagata}. \end{proof} \section{The singular locus, reprise} \label{section-singular-locus} \noindent We look for a criterion that implies openness of the regular locus for any scheme locally of finite type over the base. Here is the definition. \begin{definition} \label{definition-J} Let $X$ be a locally Noetherian scheme. We say $X$ is {\it J-2} if for every morphism $Y \to X$ which is locally of finite type the regular locus $\text{Reg}(Y)$ is open in $Y$. \end{definition} \noindent This is the analogue of the corresponding notion for Noetherian rings, see More on Algebra, Definition \ref{more-algebra-definition-J}. \begin{lemma} \label{lemma-J} Let $X$ be a locally Noetherian scheme. The following are equivalent \begin{enumerate} \item $X$ is J-2, \item there exists an open covering of $X$ all of whose members are J-2 schemes, \item for every affine open $\Spec(R) = U \subset X$ the ring $R$ is J-2, and \item there exists an affine open covering $S = \bigcup U_i$ such that each $\mathcal{O}(U_i)$ is J-2 for all $i$. \end{enumerate} Moreover, in this case any scheme locally of finite type over $X$ is J-2 as well. \end{lemma} \begin{proof} By Lemma \ref{lemma-immersion-locally-finite-type} an open immersion is locally of finite type. A composition of morphisms locally of finite type is locally of finite type (Lemma \ref{lemma-composition-finite-type}). Thus it is clear that if $X$ is J-2 then any open and any scheme locally of finite type over $X$ is J-2 as well. This proves the final statement of the lemma. \medskip\noindent If $\Spec(R)$ is J-2, then for every finite type $R$-algebra $A$ the regular locus of the scheme $\Spec(A)$ is open. Hence $R$ is J-2, by definition (see More on Algebra, Definition \ref{more-algebra-definition-J}). Combined with the remarks above we conclude that (1) implies (3), and (2) implies (4). Of course (1) $\Rightarrow$ (2) and (3) $\Rightarrow$ (4) trivially. \medskip\noindent To finish the proof we show that (4) implies (1). Assume (4) and let $Y \to X$ be a morphism locally of finite type. We can find an affine open covering $Y = \bigcup V_j$ such that each $V_j \to X$ maps into one of the $U_i$. By Lemma \ref{lemma-locally-finite-type-characterize} the induced ring map $\mathcal{O}(U_i) \to \mathcal{O}(V_j)$ is of finite type. Hence the regular locus of $V_j = \Spec(\mathcal{O}(V_j))$ is open. Since $\text{Reg}(Y) \cap V_j = \text{Reg}(V_j)$ we conclude that $\text{Reg}(Y)$ is open as desired. \end{proof} \begin{lemma} \label{lemma-ubiquity-J-2} The following types of schemes are J-2. \begin{enumerate} \item Any scheme locally of finite type over a field. \item Any scheme locally of finite type over a Noetherian complete local ring. \item Any scheme locally of finite type over $\mathbf{Z}$. \item Any scheme locally of finite type over a Noetherian local ring of dimension $1$. \item Any scheme locally of finite type over a Nagata ring of dimension $1$. \item Any scheme locally of finite type over a Dedekind ring of characteristic zero. \item And so on. \end{enumerate} \end{lemma} \begin{proof} By Lemma \ref{lemma-J} we only need to show that the rings mentioned above are J-2. For this see More on Algebra, Proposition \ref{more-algebra-proposition-ubiquity-J-2}. \end{proof} \section{Quasi-finite morphisms} \label{section-quasi-finite} \noindent A solid treatment of quasi-finite morphisms is the basis of many developments further down the road. It will lead to various versions of Zariski's Main Theorem, behaviour of dimensions of fibres, descent for \'etale morphisms, etc, etc. Before reading this section it may be a good idea to take a look at the algebra results in Algebra, Section \ref{algebra-section-quasi-finite}. \medskip\noindent Recall that a finite type ring map $R \to A$ is quasi-finite at a prime $\mathfrak q$ if $\mathfrak q$ defines an isolated point of its fibre, see Algebra, Definition \ref{algebra-definition-quasi-finite}. \begin{definition} \label{definition-quasi-finite} Let $f : X \to S$ be a morphism of schemes. \begin{enumerate} \item We say that $f$ is {\it quasi-finite at a point $x \in X$} if there exist an affine neighbourhood $\Spec(A) = U \subset X$ of $x$ and an affine open $\Spec(R) = V \subset S$ such that $f(U) \subset V$, the ring map $R \to A$ is of finite type, and $R \to A$ is quasi-finite at the prime of $A$ corresponding to $x$ (see above). \item We say $f$ is {\it locally quasi-finite} if $f$ is quasi-finite at every point $x$ of $X$. \item We say that $f$ is {\it quasi-finite} if $f$ is of finite type and every point $x$ is an isolated point of its fibre. \end{enumerate} \end{definition} \noindent Trivially, a locally quasi-finite morphism is locally of finite type. We will see below that a morphism $f$ which is locally of finite type is quasi-finite at $x$ if and only if $x$ is isolated in its fibre. Moreover, the set of points at which a morphism is quasi-finite is open; we will see this in Section \ref{section-Zariski} on Zariski's Main Theorem. \begin{lemma} \label{lemma-algebraic-residue-field-extension-closed-point-fibre} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. If $\kappa(x)/\kappa(s)$ is an algebraic field extension, then \begin{enumerate} \item $x$ is a closed point of its fibre, and \item if in addition $s$ is a closed point of $S$, then $x$ is a closed point of $X$. \end{enumerate} \end{lemma} \begin{proof} The second statement follows from the first by elementary topology. According to Schemes, Lemma \ref{schemes-lemma-fibre-topological} to prove the first statement we may replace $X$ by $X_s$ and $S$ by $\Spec(\kappa(s))$. Thus we may assume that $S = \Spec(k)$ is the spectrum of a field. In this case, let $\Spec(A) = U \subset X$ be any affine open containing $x$. The point $x$ corresponds to a prime ideal $\mathfrak q \subset A$ such that $k \subset \kappa(\mathfrak q)$ is an algebraic field extension. By Algebra, Lemma \ref{algebra-lemma-finite-residue-extension-closed} we see that $\mathfrak q$ is a maximal ideal, i.e., $x \in U$ is a closed point. Since the affine opens form a basis of the topology of $X$ we conclude that $\{x\}$ is closed. \end{proof} \noindent The following lemma is a version of the Hilbert Nullstellensatz. \begin{lemma} \label{lemma-closed-point-fibre-locally-finite-type} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. Assume $f$ is locally of finite type. Then $x$ is a closed point of its fibre if and only if $\kappa(s) \subset \kappa(x)$ is a finite field extension. \end{lemma} \begin{proof} If the extension is finite, then $x$ is a closed point of the fibre by Lemma \ref{lemma-algebraic-residue-field-extension-closed-point-fibre} above. For the converse, assume that $x$ is a closed point of its fibre. Choose affine opens $\Spec(A) = U \subset X$ and $\Spec(R) = V \subset S$ such that $f(U) \subset V$. By Lemma \ref{lemma-locally-finite-type-characterize} the ring map $R \to A$ is of finite type. Let $\mathfrak q \subset A$, resp.\ $\mathfrak p \subset R$ be the prime ideal corresponding to $x$, resp.\ $s$. Consider the fibre ring $\overline{A} = A \otimes_R \kappa(\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{A}$ corresponding to $\mathfrak q$. The assumption that $x$ is a closed point of its fibre implies that $\overline{\mathfrak q}$ is a maximal ideal of $\overline{A}$. Since $\overline{A}$ is an algebra of finite type over the field $\kappa(\mathfrak p)$ we see by the Hilbert Nullstellensatz, see Algebra, Theorem \ref{algebra-theorem-nullstellensatz}, that $\kappa(\overline{\mathfrak q})$ is a finite extension of $\kappa(\mathfrak p)$. Since $\kappa(s) = \kappa(\mathfrak p)$ and $\kappa(x) = \kappa(\mathfrak q) = \kappa(\overline{\mathfrak q})$ we win. \end{proof} \begin{lemma} \label{lemma-base-change-closed-point-fibre-locally-finite-type} Let $f : X \to S$ be a morphism of schemes which is locally of finite type. Let $g : S' \to S$ be any morphism. Denote $f' : X' \to S'$ the base change. If $x' \in X'$ maps to a point $x \in X$ which is closed in $X_{f(x)}$ then $x'$ is closed in $X'_{f'(x')}$. \end{lemma} \begin{proof} The residue field $\kappa(x')$ is a quotient of $\kappa(f'(x')) \otimes_{\kappa(f(x))} \kappa(x)$, see Schemes, Lemma \ref{schemes-lemma-points-fibre-product}. Hence it is a finite extension of $\kappa(f'(x'))$ as $\kappa(x)$ is a finite extension of $\kappa(f(x))$ by Lemma \ref{lemma-closed-point-fibre-locally-finite-type}. Thus we see that $x'$ is closed in its fibre by applying that lemma one more time. \end{proof} \begin{lemma} \label{lemma-residue-field-quasi-finite} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. If $f$ is quasi-finite at $x$, then the residue field extension $\kappa(s) \subset \kappa(x)$ is finite. \end{lemma} \begin{proof} This is clear from Algebra, Definition \ref{algebra-definition-quasi-finite}. \end{proof} \begin{lemma} \label{lemma-quasi-finite-at-point-characterize} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. Let $X_s$ be the fibre of $f$ at $s$. Assume $f$ is locally of finite type. The following are equivalent: \begin{enumerate} \item The morphism $f$ is quasi-finite at $x$. \item The point $x$ is isolated in $X_s$. \item The point $x$ is closed in $X_s$ and there is no point $x' \in X_s$, $x' \not = x$ which specializes to $x$. \item For any pair of affine opens $\Spec(A) = U \subset X$, $\Spec(R) = V \subset S$ with $f(U) \subset V$ and $x \in U$ corresponding to $\mathfrak q \subset A$ the ring map $R \to A$ is quasi-finite at $\mathfrak q$. \end{enumerate} \end{lemma} \begin{proof} Assume $f$ is quasi-finite at $x$. By assumption there exist opens $U \subset X$, $V \subset S$ such that $f(U) \subset V$, $x \in U$ and $x$ an isolated point of $U_s$. Hence $\{x\} \subset U_s$ is an open subset. Since $U_s = U \cap X_s \subset X_s$ is also open we conclude that $\{x\} \subset X_s$ is an open subset also. Thus we conclude that $x$ is an isolated point of $X_s$. \medskip\noindent Note that $X_s$ is a Jacobson scheme by Lemma \ref{lemma-ubiquity-Jacobson-schemes} (and Lemma \ref{lemma-base-change-finite-type}). If $x$ is isolated in $X_s$, i.e., $\{x\} \subset X_s$ is open, then $\{x\}$ contains a closed point (by the Jacobson property), hence $x$ is closed in $X_s$. It is clear that there is no point $x' \in X_s$, distinct from $x$, specializing to $x$. \medskip\noindent Assume that $x$ is closed in $X_s$ and that there is no point $x' \in X_s$, distinct from $x$, specializing to $x$. Consider a pair of affine opens $\Spec(A) = U \subset X$, $\Spec(R) = V \subset S$ with $f(U) \subset V$ and $x \in U$. Let $\mathfrak q \subset A$ correspond to $x$ and $\mathfrak p \subset R$ correspond to $s$. By Lemma \ref{lemma-locally-finite-type-characterize} the ring map $R \to A$ is of finite type. Consider the fibre ring $\overline{A} = A \otimes_R \kappa(\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{A}$ corresponding to $\mathfrak q$. Since $\Spec(\overline{A})$ is an open subscheme of the fibre $X_s$ we see that $\overline{q}$ is a maximal ideal of $\overline{A}$ and that there is no point of $\Spec(\overline{A})$ specializing to $\overline{\mathfrak q}$. This implies that $\dim(\overline{A}_{\overline{q}}) = 0$. Hence by Algebra, Definition \ref{algebra-definition-quasi-finite} we see that $R \to A$ is quasi-finite at $\mathfrak q$, i.e., $X \to S$ is quasi-finite at $x$ by definition. \medskip\noindent At this point we have shown conditions (1) -- (3) are all equivalent. It is clear that (4) implies (1). And it is also clear that (2) implies (4) since if $x$ is an isolated point of $X_s$ then it is also an isolated point of $U_s$ for any open $U$ which contains it. \end{proof} \begin{lemma} \label{lemma-finite-fibre} Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Assume that \begin{enumerate} \item $f$ is locally of finite type, and \item $f^{-1}(\{s\})$ is a finite set. \end{enumerate} Then $X_s$ is a finite discrete topological space, and $f$ is quasi-finite at each point of $X$ lying over $s$. \end{lemma} \begin{proof} Suppose $T$ is a scheme which (a) is locally of finite type over a field $k$, and (b) has finitely many points. Then Lemma \ref{lemma-ubiquity-Jacobson-schemes} shows $T$ is a Jacobson scheme. A finite Jacobson space is discrete, see Topology, Lemma \ref{topology-lemma-finite-jacobson}. Apply this remark to the fibre $X_s$ which is locally of finite type over $\Spec(\kappa(s))$ to see the first statement. Finally, apply Lemma \ref{lemma-quasi-finite-at-point-characterize} to see the second. \end{proof} \begin{lemma} \label{lemma-locally-quasi-finite-fibres} Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type. Then the following are equivalent \begin{enumerate} \item $f$ is locally quasi-finite, \item for every $s \in S$ the fibre $X_s$ is a discrete topological space, and \item for every morphism $\Spec(k) \to S$ where $k$ is a field the base change $X_k$ has an underlying discrete topological space. \end{enumerate} \end{lemma} \begin{proof} It is immediate that (3) implies (2). Lemma \ref{lemma-quasi-finite-at-point-characterize} shows that (2) is equivalent to (1). Assume (2) and let $\Spec(k) \to S$ be as in (3). Denote $s \in S$ the image of $\Spec(k) \to S$. Then $X_k$ is the base change of $X_s$ via $\Spec(k) \to \Spec(\kappa(s))$. Hence every point of $X_k$ is closed by Lemma \ref{lemma-base-change-closed-point-fibre-locally-finite-type}. As $X_k \to \Spec(k)$ is locally of finite type (by Lemma \ref{lemma-base-change-finite-type}), we may apply Lemma \ref{lemma-quasi-finite-at-point-characterize} to conclude that every point of $X_k$ is isolated, i.e., $X_k$ has a discrete underlying topological space. \end{proof} \begin{lemma} \label{lemma-quasi-finite-locally-quasi-compact} Let $f : X \to S$ be a morphism of schemes. Then $f$ is quasi-finite if and only if $f$ is locally quasi-finite and quasi-compact. \end{lemma} \begin{proof} Assume $f$ is quasi-finite. It is quasi-compact by Definition \ref{definition-finite-type}. Let $x \in X$. We see that $f$ is quasi-finite at $x$ by Lemma \ref{lemma-quasi-finite-at-point-characterize}. Hence $f$ is quasi-compact and locally quasi-finite. \medskip\noindent Assume $f$ is quasi-compact and locally quasi-finite. Then $f$ is of finite type. Let $x \in X$ be a point. By Lemma \ref{lemma-quasi-finite-at-point-characterize} we see that $x$ is an isolated point of its fibre. The lemma is proved. \end{proof} \begin{lemma} \label{lemma-quasi-finite} Let $f : X \to S$ be a morphism of schemes. The following are equivalent: \begin{enumerate} \item $f$ is quasi-finite, and \item $f$ is locally of finite type, quasi-compact, and has finite fibres. \end{enumerate} \end{lemma} \begin{proof} Assume $f$ is quasi-finite. In particular $f$ is locally of finite type and quasi-compact (since it is of finite type). Let $s \in S$. Since every $x \in X_s$ is isolated in $X_s$ we see that $X_s = \bigcup_{x \in X_s} \{x\}$ is an open covering. As $f$ is quasi-compact, the fibre $X_s$ is quasi-compact. Hence we see that $X_s$ is finite. \medskip\noindent Conversely, assume $f$ is locally of finite type, quasi-compact and has finite fibres. Then it is locally quasi-finite by Lemma \ref{lemma-finite-fibre}. Hence it is quasi-finite by Lemma \ref{lemma-quasi-finite-locally-quasi-compact}. \end{proof} \noindent Recall that a ring map $R \to A$ is quasi-finite if it is of finite type and quasi-finite at {\it all} primes of $A$, see Algebra, Definition \ref{algebra-definition-quasi-finite}. \begin{lemma} \label{lemma-locally-quasi-finite-characterize} Let $f : X \to S$ be a morphism of schemes. The following are equivalent \begin{enumerate} \item The morphism $f$ is locally quasi-finite. \item For every pair of affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_S(V) \to \mathcal{O}_X(U)$ is quasi-finite. \item There exists an open covering $S = \bigcup_{j \in J} V_j$ and open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that each of the morphisms $U_i \to V_j$, $j\in J, i\in I_j$ is locally quasi-finite. \item There exists an affine open covering $S = \bigcup_{j \in J} V_j$ and affine open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that the ring map $\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_i)$ is quasi-finite, for all $j\in J, i\in I_j$. \end{enumerate} Moreover, if $f$ is locally quasi-finite then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_U : U \to V$ is locally quasi-finite. \end{lemma} \begin{proof} For a ring map $R \to A$ let us define $P(R \to A)$ to mean $R \to A$ is quasi-finite'' (see remark above lemma). We claim that $P$ is a local property of ring maps. We check conditions (a), (b) and (c) of Definition \ref{definition-property-local}. In the proof of Lemma \ref{lemma-locally-finite-type-characterize} we have seen that (a), (b) and (c) hold for the property of being of finite type''. Note that, for a finite type ring map $R \to A$, the property $R \to A$ is quasi-finite at $\mathfrak q$ depends only on the local ring $A_{\mathfrak q}$ as an algebra over $R_{\mathfrak p}$ where $\mathfrak p = R \cap \mathfrak q$ (usual abuse of notation). Using these remarks (a), (b) and (c) of Definition \ref{definition-property-local} follow immediately. For example, suppose $R \to A$ is a ring map such that all of the ring maps $R \to A_{a_i}$ are quasi-finite for $a_1, \ldots, a_n \in A$ generating the unit ideal. We conclude that $R \to A$ is of finite type. Also, for any prime $\mathfrak q \subset A$ the local ring $A_{\mathfrak q}$ is isomorphic as an $R$-algebra to the local ring $(A_{a_i})_{\mathfrak q_i}$ for some $i$ and some $\mathfrak q_i \subset A_{a_i}$. Hence we conclude that $R \to A$ is quasi-finite at $\mathfrak q$. \medskip\noindent We conclude that Lemma \ref{lemma-locally-P} applies with $P$ as in the previous paragraph. Hence it suffices to prove that $f$ is locally quasi-finite is equivalent to $f$ is locally of type $P$. Since $P(R \to A)$ is $R \to A$ is quasi-finite'' which means $R \to A$ is quasi-finite at every prime of $A$, this follows from Lemma \ref{lemma-quasi-finite-at-point-characterize}. \end{proof} \begin{lemma} \label{lemma-composition-quasi-finite} The composition of two morphisms which are locally quasi-finite is locally quasi-finite. The same is true for quasi-finite morphisms. \end{lemma} \begin{proof} In the proof of Lemma \ref{lemma-locally-quasi-finite-characterize} we saw that $P =$quasi-finite'' is a local property of ring maps, and that a morphism of schemes is locally quasi-finite if and only if it is locally of type $P$ as in Definition \ref{definition-locally-P}. Hence the first statement of the lemma follows from Lemma \ref{lemma-composition-type-P} combined with the fact that being quasi-finite is a property of ring maps that is stable under composition, see Algebra, Lemma \ref{algebra-lemma-quasi-finite-composition}. By the above, Lemma \ref{lemma-quasi-finite-locally-quasi-compact} and the fact that compositions of quasi-compact morphisms are quasi-compact, see Schemes, Lemma \ref{schemes-lemma-composition-quasi-compact} we see that the composition of quasi-finite morphisms is quasi-finite.