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\input{preamble}
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\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.