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properties.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Properties of Schemes}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we introduce some absolute properties of schemes.
A foundational reference is \cite{EGA}.
\section{Constructible sets}
\label{section-constructible}
\noindent
Constructible and locally constructible sets are introduced in
Topology, Section \ref{topology-section-constructible}.
We may characterize locally constructible subsets of schemes as
follows.
\begin{lemma}
\label{lemma-locally-constructible}
Let $X$ be a scheme.
A subset $E$ of $X$ is locally constructible in $X$ if and only if
$E \cap U$ is constructible in $U$ for every affine open $U$ of $X$.
\end{lemma}
\begin{proof}
Assume $E$ is locally constructible. Then there exists an open covering
$X = \bigcup U_i$ such that $E \cap U_i$ is constructible in $U_i$
for each $i$. Let $V \subset X$ be any affine open. We can find a finite
open affine covering $V = V_1 \cup \ldots \cup V_m$ such that for each $j$
we have $V_j \subset U_i$ for some $i = i(j)$. By
Topology, Lemma \ref{topology-lemma-open-immersion-constructible-inverse-image}
we see that each $E \cap V_j$ is constructible in $V_j$. Since the inclusions
$V_j \to V$ are quasi-compact (see
Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine})
we conclude that $E \cap V$ is constructible in $V$ by
Topology, Lemma \ref{topology-lemma-collate-constructible}.
The converse implication is immediate.
\end{proof}
\begin{lemma}
\label{lemma-quasi-separated-quasi-compact-open-retrocompact}
Let $X$ be a quasi-separated scheme. The intersection of any two
quasi-compact opens of $X$ is a quasi-compact open of $X$.
Every quasi-compact open of $X$ is retrocompact in $X$.
\end{lemma}
\begin{proof}
If $U$ and $V$ are quasi-compact open then
$U \cap V = \Delta^{-1}(U \times V)$, where $\Delta : X \to X \times X$
is the diagonal. As $X$ is quasi-separated we see that $\Delta$ is
quasi-compact. Hence we see that $U \cap V$ is quasi-compact as
$U \times V$ is quasi-compact (details omitted; use
Schemes, Lemma \ref{schemes-lemma-affine-covering-fibre-product}
to see $U \times V$ is a finite union of affines).
The other assertions follow from the first and
Topology, Lemma \ref{topology-lemma-topology-quasi-separated-scheme}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-quasi-separated-spectral}
Let $X$ be a quasi-compact and quasi-separated scheme.
Then the underlying topological space of $X$ is a spectral space.
\end{lemma}
\begin{proof}
By Topology, Definition \ref{topology-definition-spectral-space}
we have to check that $X$ is sober, quasi-compact, has a basis
of quasi-compact opens, and the intersection of any two
quasi-compact opens is quasi-compact. This follows from
Schemes, Lemma \ref{schemes-lemma-scheme-sober} and
\ref{schemes-lemma-basis-affine-opens}
and
Lemma \ref{lemma-quasi-separated-quasi-compact-open-retrocompact}
above.
\end{proof}
\begin{lemma}
\label{lemma-constructible-quasi-compact-quasi-separated}
Let $X$ be a quasi-compact and quasi-separated scheme.
Any locally constructible subset of $X$ is constructible.
\end{lemma}
\begin{proof}
As $X$ is quasi-compact we can choose a finite affine open covering
$X = V_1 \cup \ldots \cup V_m$. As $X$ is quasi-separated each $V_i$ is
retrocompact in $X$ by
Lemma \ref{lemma-quasi-separated-quasi-compact-open-retrocompact}.
Hence by
Topology, Lemma \ref{topology-lemma-collate-constructible}
we see that $E \subset X$ is constructible in $X$ if and only if
$E \cap V_j$ is constructible in $V_j$. Thus we win by
Lemma \ref{lemma-locally-constructible}.
\end{proof}
\begin{lemma}
\label{lemma-retrocompact}
Let $X$ be a scheme. A subset $Z$ of $X$ is retrocompact in $X$ if and only if
$E \cap U$ is quasi-compact for every affine open $U$ of $X$.
\end{lemma}
\begin{proof}
Immediate from the fact that every quasi-compact open of $X$ is a finite
union of affine opens.
\end{proof}
\section{Integral, irreducible, and reduced schemes}
\label{section-integral}
\begin{definition}
\label{definition-integral}
Let $X$ be a scheme. We say $X$ is {\it integral} if it is nonempty and
for every nonempty affine open $\Spec(R) = U \subset X$ the ring $R$
is an integral domain.
\end{definition}
\begin{lemma}
\label{lemma-characterize-reduced}
Let $X$ be a scheme.
The following are equivalent.
\begin{enumerate}
\item The scheme $X$ is reduced, see
Schemes, Definition \ref{schemes-definition-reduced}.
\item There exists an affine open covering $X = \bigcup U_i$
such that each $\Gamma(U_i, \mathcal{O}_X)$ is reduced.
\item For every affine open $U \subset X$ the ring
$\mathcal{O}_X(U)$ is reduced.
\item For every open $U \subset X$ the ring $\mathcal{O}_X(U)$ is reduced.
\end{enumerate}
\end{lemma}
\begin{proof}
See Schemes, Lemmas \ref{schemes-lemma-reduced} and
\ref{schemes-lemma-affine-reduced}.
\end{proof}
\begin{lemma}
\label{lemma-characterize-irreducible}
Let $X$ be a scheme.
The following are equivalent.
\begin{enumerate}
\item The scheme $X$ is irreducible.
\item There exists an affine open covering $X = \bigcup_{i \in I} U_i$
such that $I$ is not empty, $U_i$ is irreducible for all $i \in I$, and
$U_i \cap U_j \not = \emptyset$ for all $i, j \in I$.
\item The scheme $X$ is nonempty and every nonempty affine open
$U \subset X$ is irreducible.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1). By Schemes, Lemma \ref{schemes-lemma-scheme-sober}
we see that $X$ has a unique generic point $\eta$. Then
$X = \overline{\{\eta\}}$. Hence $\eta$ is an element of
every nonempty affine open $U \subset X$. This implies
that $U = \overline{\{\eta\}}$ and that any two nonempty affines
meet. Thus (1) implies both (2) and (3).
\medskip\noindent
Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets.
For every $i$ we see that either $U_i \subset Z_1$ or $U_i \subset Z_2$.
Pick some $i \in I$ and assume $U_i \subset Z_1$ (possibly after renumbering
$Z_1$, $Z_2$). For any $j \in I$ the open subset $U_i \cap U_j$ is dense in
$U_j$ and contained in the closed subset $Z_1 \cap U_j$. We conclude that
also $U_j \subset Z_1$. Thus $X = Z_1$ as desired.
\medskip\noindent
Assume (3). Choose an affine open covering $X = \bigcup_{i \in I} U_i$.
We may assume that each $U_i$ is nonempty.
Since $X$ is nonempty we see that $I$ is not empty.
By assumption each $U_i$ is irreducible.
Suppose $U_i \cap U_j = \emptyset$ for some pair $i, j \in I$.
Then the open $U_i \coprod U_j = U_i \cup U_j$ is affine, see
Schemes, Lemma \ref{schemes-lemma-disjoint-union-affines}.
Hence it is irreducible by assumption which is absurd. We conclude that (3)
implies (2). The lemma is proved.
\end{proof}
\begin{lemma}
\label{lemma-characterize-integral}
A scheme $X$ is integral if and only if it is reduced and irreducible.
\end{lemma}
\begin{proof}
If $X$ is irreducible, then every affine open $\Spec(R) = U \subset X$
is irreducible. If $X$ is reduced, then $R$ is reduced, by
Lemma \ref{lemma-characterize-reduced} above. Hence $R$ is reduced
and $(0)$ is a prime ideal, i.e., $R$ is an integral domain.
\medskip\noindent
If $X$ is integral, then for every nonempty affine open
$\Spec(R) = U \subset X$ the ring $R$ is reduced
and hence $X$ is reduced by Lemma \ref{lemma-characterize-reduced}.
Moreover, every nonempty affine open is irreducible.
Hence $X$ is irreducible, see Lemma \ref{lemma-characterize-irreducible}.
\end{proof}
\noindent
In Examples, Section
\ref{examples-section-connected-locally-integral-not-integral}
we construct a connected affine scheme all of whose local rings are domains,
but which is not integral.
\section{Types of schemes defined by properties of rings}
\label{section-properties-rings}
\noindent
In this section we study what properties of rings
allow one to define local properties of schemes.
\begin{definition}
\label{definition-property-local}
Let $P$ be a property of rings.
We say that $P$ is {\it local} if the following hold:
\begin{enumerate}
\item For any ring $R$, and any $f \in R$ we have
$P(R) \Rightarrow P(R_f)$.
\item For any ring $R$, and $f_i \in R$ such that
$(f_1, \ldots, f_n) = R$ then
$\forall i, P(R_{f_i}) \Rightarrow P(R)$.
\end{enumerate}
\end{definition}
\begin{definition}
\label{definition-locally-P}
Let $P$ be a property of rings. Let $X$ be a scheme.
We say $X$ is {\it locally $P$} if for any $x \in X$
there exists an affine open neighbourhood $U$ of $x$
in $X$ such that $\mathcal{O}_X(U)$ has property $P$.
\end{definition}
\noindent
This is only a good notion if the property is local.
Even if $P$ is a local property we will not
automatically use this definition to say that a scheme is
``locally $P$'' unless we also explicitly state the definition
elsewhere.
\begin{lemma}
\label{lemma-locally-P}
Let $X$ be a scheme. Let $P$ be a local property of rings.
The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is locally $P$.
\item For every affine open $U \subset X$ the property
$P(\mathcal{O}_X(U))$ holds.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ satisfies $P$.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is locally $P$.
\end{enumerate}
Moreover, if $X$ is locally $P$ then every open subscheme
is locally $P$.
\end{lemma}
\begin{proof}
Of course (1) $\Leftrightarrow$ (3) and (2) $\Rightarrow$ (1).
If (3) $\Rightarrow$ (2), then the final statement of the lemma
holds and it follows easily that (4) is also equivalent to (1).
Thus we show (3) $\Rightarrow$ (2).
\medskip\noindent
Let $X = \bigcup U_i$ be an affine open covering, say
$U_i = \Spec(R_i)$. Assume $P(R_i)$.
Let $\Spec(R) = U \subset X$ be an arbitrary affine open.
By Schemes, Lemma \ref{schemes-lemma-good-subcover}
there exists a standard covering of $U = \Spec(R)$ by
standard opens $D(f_j)$ such that each ring $R_{f_j}$ is a
principal localization of one of the rings $R_i$. By
Definition \ref{definition-property-local} (1) we get $P(R_{f_j})$.
Whereupon $P(R)$ by Definition \ref{definition-property-local} (2).
\end{proof}
\noindent
Here is a sample application.
\begin{lemma}
\label{lemma-reduced-is-locally-reduced}
Let $X$ be a scheme. Then $X$ is reduced if and only if $X$ is
``locally reduced'' in the sense of Definition \ref{definition-locally-P}.
\end{lemma}
\begin{proof}
This is clear from Lemma \ref{lemma-characterize-reduced}.
\end{proof}
\begin{lemma}
\label{lemma-properties-local}
The following properties of a ring $R$ are local.
\begin{enumerate}
\item (Cohen-Macaulay.)
The ring $R$ is Noetherian and CM, see
Algebra, Definition \ref{algebra-definition-ring-CM}.
\item (Regular.)
The ring $R$ is Noetherian and regular, see
Algebra, Definition \ref{algebra-definition-regular}.
\item (Absolutely Noetherian.)
The ring $R$ is of finite type over $Z$.
\item Add more here as needed.\footnote{But we only list those properties
here which we have not already dealt with separately somewhere else.}
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Noetherian schemes}
\label{section-noetherian}
\noindent
Recall that a ring $R$ is {\it Noetherian} if it satisfies the ascending
chain condition of ideals. Equivalently every ideal of $R$ is finitely
generated.
\begin{definition}
\label{definition-noetherian}
Let $X$ be a scheme.
\begin{enumerate}
\item We say $X$ is {\it locally Noetherian} if every
$x \in X$ has an affine open neighbourhood
$\Spec(R) = U \subset X$ such that the ring $R$ is Noetherian.
\item We say $X$ is {\it Noetherian} if $X$ is locally Noetherian
and quasi-compact.
\end{enumerate}
\end{definition}
\noindent
Here is the standard result characterizing locally Noetherian schemes.
\begin{lemma}
\label{lemma-locally-Noetherian}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is locally Noetherian.
\item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is Noetherian.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is Noetherian.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is locally Noetherian.
\end{enumerate}
Moreover, if $X$ is locally Noetherian then every open subscheme
is locally Noetherian.
\end{lemma}
\begin{proof}
To show this it suffices to show that being Noetherian is a local
property of rings, see Lemma \ref{lemma-locally-P}.
Any localization of a Noetherian ring is Noetherian, see
Algebra, Lemma \ref{algebra-lemma-Noetherian-permanence}.
By Algebra, Lemma \ref{algebra-lemma-cover} we see the second
property to Definition \ref{definition-property-local}.
\end{proof}
\begin{lemma}
\label{lemma-immersion-into-noetherian}
Any immersion $Z \to X$ with $X$ locally Noetherian is quasi-compact.
\end{lemma}
\begin{proof}
A closed immersion is clearly quasi-compact.
A composition of quasi-compact morphisms is quasi-compact,
see Topology, Lemma \ref{topology-lemma-composition-quasi-compact}.
Hence it suffices to show that an open immersion into
a locally Noetherian scheme is quasi-compact.
Using Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine}
we reduce to the case where $X$ is affine.
Any open subset of the spectrum of a Noetherian ring
is quasi-compact (for example
combine Algebra, Lemma \ref{algebra-lemma-Noetherian-topology} and
Topology, Lemmas \ref{topology-lemma-Noetherian} and
\ref{topology-lemma-Noetherian-quasi-compact}).
\end{proof}
\begin{lemma}
\label{lemma-locally-Noetherian-quasi-separated}
A locally Noetherian scheme is quasi-separated.
\end{lemma}
\begin{proof}
By Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated}
we have to show that the intersection $U \cap V$ of two
affine opens of $X$ is quasi-compact. This follows from
Lemma \ref{lemma-immersion-into-noetherian} above on
considering the open immersion $U \cap V \to U$ for example.
(But really it is just because any open of the spectrum of a
Noetherian ring is quasi-compact.)
\end{proof}
\begin{lemma}
\label{lemma-Noetherian-topology}
A (locally) Noetherian scheme has a (locally)
Noetherian underlying topological space,
see Topology, Definition \ref{topology-definition-noetherian}.
\end{lemma}
\begin{proof}
This is because a Noetherian scheme is a finite union of spectra
of Noetherian rings and
Algebra, Lemma \ref{algebra-lemma-Noetherian-topology} and
Topology, Lemma \ref{topology-lemma-finite-union-Noetherian}.
\end{proof}
\begin{lemma}
\label{lemma-morphism-Noetherian-schemes-quasi-compact}
Any morphism of schemes $f : X \to Y$ with $X$ Noetherian
is quasi-compact.
\end{lemma}
\begin{proof}
Use Lemma \ref{lemma-Noetherian-topology}
and use that any subset of a Noetherian topological
space is quasi-compact (see Topology, Lemmas
Lemmas \ref{topology-lemma-Noetherian} and
\ref{topology-lemma-Noetherian-quasi-compact}).
\end{proof}
\begin{lemma}
\label{lemma-locally-closed-in-Noetherian}
Any locally closed subscheme of a (locally) Noetherian
scheme is (locally) Noetherian.
\end{lemma}
\begin{proof}
Omitted. Hint: Any quotient, and any localization of a Noetherian
ring is Noetherian. For the Noetherian case use again
that any subset of a Noetherian space is a Noetherian space
(with induced topology).
\end{proof}
\noindent
Here is a fun lemma.
It says that every locally Noetherian scheme has plenty of
closed points (at least one in every closed subset).
\begin{lemma}
\label{lemma-locally-Noetherian-closed-point}
Any locally Noetherian scheme has a closed point.
Any closed subset of a locally Noetherian scheme has a closed point.
Equivalently, any point of a locally Noetherian scheme specializes
to a closed point.
\end{lemma}
\begin{proof}
The second assertion follows from the first (using
Schemes, Lemma \ref{schemes-lemma-reduced-closed-subscheme}
and Lemma \ref{lemma-locally-closed-in-Noetherian}).
Consider any nonempty affine open $U \subset X$.
Let $x \in U$ be a closed point. If $x$ is a closed point
of $X$ then we are done. If not, let $y \in \overline{\{x\}}$
be a specialization of $x$. Note that $y \in X \setminus U$.
Consider the local ring $R = \mathcal{O}_{X, y}$.
This is a Noetherian local ring. Denote $V \subset \Spec(R)$
the inverse image of $U$ in $\Spec(R)$ by the canonical morphism
$\Spec(R) \to X$ (see Schemes, Section \ref{schemes-section-points}.)
By construction $V$ is a singleton with unique point
corresponding to $x$ (use
Schemes, Lemma \ref{schemes-lemma-specialize-points}).
Say $V = \{\mathfrak q\}$. Consider the Noetherian
local domain $R/\mathfrak q$. By
Algebra, Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens}
we see that $\dim(R/\mathfrak q) = 1$.
In other words, we see that $y$ is an immediate specialization
of $x$ (see Topology, Definition \ref{topology-definition-dimension-function}).
In other words, any
point $y \not = x$ such that $x \leadsto y$ is an immediate
specialization of $x$. Clearly each of these points is a
closed point, and we win.
\end{proof}
\begin{lemma}
\label{lemma-locally-Noetherian-specialization-dvr}
Let $X$ be a locally Noetherian scheme.
Let $x' \leadsto x$ be a specialization of points of $X$.
Then
\begin{enumerate}
\item there exists a discrete valuation ring $R$ and a morphism
$f : \Spec(R) \to X$ such that the generic point $\eta$ of
$\Spec(R)$ maps to $x'$ and the special point maps to $x$, and
\item given a finitely generated field extension $\kappa(x') \subset K$
we may arrange it so that the extension $\kappa(x') \subset \kappa(\eta)$
induced by $f$ is isomorphic to the given one.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $x' \leadsto x$ be a specialization in $X$, and let
$\kappa(x') \subset K$ be a finitely generated extension of fields. By
Schemes, Lemma \ref{schemes-lemma-specialize-points}
and the discussion following
Schemes, Lemma \ref{schemes-lemma-characterize-points}
this leads to ring maps $\mathcal{O}_{X, x} \to \kappa(x') \to K$.
Let $R \subset K$ be any discrete valuation ring whose field of fractions is
$K$ and which dominates the image of $\mathcal{O}_{X, x} \to K$, see
Algebra, Lemma \ref{algebra-lemma-exists-dvr}.
The ring map $\mathcal{O}_{X, x} \to R$ induces the morphism
$f : \Spec(R) \to X$, see
Schemes, Lemma \ref{schemes-lemma-morphism-from-spec-local-ring}.
This morphism has all the desired properties by construction.
\end{proof}
\section{Jacobson schemes}
\label{section-jacobson}
\noindent
Recall that a space is said to be {\it Jacobson} if the closed points are
dense in every closed subset, see
Topology, Section \ref{topology-section-space-jacobson}.
\begin{definition}
\label{definition-jacobson}
A scheme $S$ is said to be {\it Jacobson} if its underlying topological
space is Jacobson.
\end{definition}
\noindent
Recall that a ring $R$ is Jacobson if every radical ideal of $R$
is the intersection of maximal ideals, see
Algebra, Definition \ref{algebra-definition-ring-jacobson}.
\begin{lemma}
\label{lemma-affine-jacobson}
An affine scheme $\Spec(R)$ is Jacobson if and only if
the ring $R$ is Jacobson.
\end{lemma}
\begin{proof}
This is Algebra, Lemma \ref{algebra-lemma-jacobson}.
\end{proof}
\noindent
Here is the standard result characterizing Jacobson schemes.
Intuitively it claims that Jacobson $\Leftrightarrow$ locally Jacobson.
\begin{lemma}
\label{lemma-locally-jacobson}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is Jacobson.
\item The scheme $X$ is ``locally Jacobson'' in the sense of
Definition \ref{definition-locally-P}.
\item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is Jacobson.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is Jacobson.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is Jacobson.
\end{enumerate}
Moreover, if $X$ is Jacobson then every open subscheme
is Jacobson.
\end{lemma}
\begin{proof}
The final assertion of the lemma holds by
Topology, Lemma \ref{topology-lemma-jacobson-inherited}.
The equivalence of (5) and (1) is
Topology, Lemma \ref{topology-lemma-jacobson-local}.
Hence, using Lemma \ref{lemma-affine-jacobson},
we see that (1) $\Leftrightarrow$ (2).
To finish proving the lemma it suffices to show that ``Jacobson''
is a local property of rings, see Lemma \ref{lemma-locally-P}.
Any localization of a Jacobson ring at an element is Jacobson, see
Algebra, Lemma \ref{algebra-lemma-Jacobson-invert-element}.
Suppose $R$ is a ring, $f_1, \ldots, f_n \in R$ generate the unit
ideal and each $R_{f_i}$ is Jacobson. Then we see that
$\Spec(R) = \bigcup D(f_i)$ is a union of open subsets
which are all Jacobson, and hence $\Spec(R)$ is Jacobson
by Topology, Lemma \ref{topology-lemma-jacobson-local} again.
This proves the second property of Definition \ref{definition-property-local}.
\end{proof}
\noindent
Many schemes used commonly in algebraic geometry are Jacobson, see
Morphisms, Lemma \ref{morphisms-lemma-ubiquity-Jacobson-schemes}.
We mention here the following interesting case.
\begin{lemma}
\label{lemma-complement-closed-point-Jacobson}
Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$.
In this case the scheme $S = \Spec(R) \setminus \{\mathfrak m\}$
is Jacobson.
\end{lemma}
\begin{proof}
Since $\Spec(R)$ is a Noetherian scheme, hence
$S$ is a Noetherian scheme (Lemma \ref{lemma-locally-closed-in-Noetherian}).
Hence $S$ is a sober, Noetherian topological space (use
Schemes, Lemma \ref{schemes-lemma-scheme-sober}).
Assume $S$ is not Jacobson to
get a contradiction. By
Topology, Lemma \ref{topology-lemma-non-jacobson-Noetherian-characterize}
there exists some non-closed point $\xi \in S$
such that $\{\xi\}$ is locally closed. This corresponds
to a prime $\mathfrak p \subset R$ such that (1) there exists
a prime $\mathfrak q$, $\mathfrak p \subset \mathfrak q \subset \mathfrak m$
with both inclusions strict, and (2) $\{\mathfrak p\}$ is open in
$\Spec(R/\mathfrak p)$. This is impossible by Algebra,
Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens}.
\end{proof}
\section{Normal schemes}
\label{section-normal}
\noindent
Recall that a ring $R$ is said to be normal if all its local rings
are normal domains,
see Algebra, Definition \ref{algebra-definition-ring-normal}.
A normal domain is a domain which is integrally closed in its field
of fractions, see
Algebra, Definition \ref{algebra-definition-domain-normal}.
Thus it makes sense to define a normal scheme as follows.
\begin{definition}
\label{definition-normal}
A scheme $X$ is {\it normal} if and only if for all $x \in X$ the local ring
$\mathcal{O}_{X, x}$ is a normal domain.
\end{definition}
\noindent
This seems to be the definition used in EGA, see \cite[0, 4.1.4]{EGA}.
Suppose $X = \Spec(A)$, and $A$ is reduced. Then saying that $X$ is
normal is not equivalent to saying that $A$ is integrally closed in its
total ring of fractions. However, if $A$ is Noetherian then this is the case
(see Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}).
\begin{lemma}
\label{lemma-locally-normal}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is normal.
\item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is normal.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is normal.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is normal.
\end{enumerate}
Moreover, if $X$ is normal then every open subscheme
is normal.
\end{lemma}
\begin{proof}
This is clear from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-normal-reduced}
A normal scheme is reduced.
\end{lemma}
\begin{proof}
Immediate from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-integral-normal}
Let $X$ be an integral scheme.
Then $X$ is normal if and only if for every affine open
$U \subset X$ the ring $\mathcal{O}_X(U)$ is a normal domain.
\end{lemma}
\begin{proof}
This follows from
Algebra, Lemma \ref{algebra-lemma-normality-is-local}.
\end{proof}
\begin{lemma}
\label{lemma-normal-locally-finite-nr-irreducibles}
Let $X$ be a scheme with a finite number of irreducible components.
The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a finite disjoint union of normal integral schemes.
\end{enumerate}
\end{lemma}
\begin{proof}
It is immediate from the definitions that (2) implies (1).
Let $X$ be a normal scheme with a finite number of irreducible components.
If $X$ is affine then $X$ satisfies (2) by
Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}.
For a general $X$, let $X = \bigcup X_i$ be
an affine open covering. Note that also each $X_i$ has
but a finite number of irreducible components, and the lemma holds
for each $X_i$. Let $T \subset X$ be an irreducible component.
By the affine case each intersection $T \cap X_i$ is open in $X_i$
and an integral normal scheme.
Hence $T \subset X$ is open, and an integral normal scheme.
This proves that $X$ is the disjoint union of its irreducible components,
which are integral normal schemes. There are only finitely many
by assumption.
\end{proof}
\begin{lemma}
\label{lemma-normal-Noetherian}
Let $X$ be a Noetherian scheme.
The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a finite disjoint union of normal integral schemes.
\end{enumerate}
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-normal-locally-finite-nr-irreducibles} because a Noetherian
scheme has a Noetherian underlying topological space
(Lemma \ref{lemma-Noetherian-topology}
and
Topology, Lemma \ref{topology-lemma-Noetherian}.
\end{proof}
\begin{lemma}
\label{lemma-normal-locally-Noetherian}
Let $X$ be a locally Noetherian scheme.
The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a disjoint union of integral normal schemes.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Hint: This is purely topological from
Lemma \ref{lemma-normal-Noetherian}.
\end{proof}
\begin{remark}
\label{remark-normal-connected-irreducible}
Let $X$ be a normal scheme. If $X$ is locally Noetherian then we see that
$X$ is integral if and only if $X$ is connected, see
Lemma \ref{lemma-normal-locally-Noetherian}.
But there exists a connected affine scheme $X$ such that
$\mathcal{O}_{X, x}$ is a domain for all $x \in X$, but $X$ is not
irreducible, see Examples, Section
\ref{examples-section-connected-locally-integral-not-integral}.
This example is even a normal scheme (proof omitted), so beware!
\end{remark}
\begin{lemma}
\label{lemma-normal-integral-sections}
Let $X$ be an integral normal scheme.
Then $\Gamma(X, \mathcal{O}_X)$ is a normal domain.
\end{lemma}
\begin{proof}
Set $R = \Gamma(X, \mathcal{O}_X)$.
It is clear that $R$ is a domain.
Suppose $f = a/b$ is an element of its fraction field
which is integral over $R$. Say we have
$f^d + \sum_{i = 1, \ldots, d} a_i f^i = 0$ with
$a_i \in R$. Let $U \subset X$ be affine open.
Since $b \in R$ is not zero and since $X$ is integral we see
that also $b|_U \in \mathcal{O}_X(U)$ is not zero.
Hence $a/b$ is an element of the fraction field of
$\mathcal{O}_X(U)$ which is integral over $\mathcal{O}_X(U)$
(because we can use the same polynomial
$f^d + \sum_{i = 1, \ldots, d} a_i|_U f^i = 0$ on $U$).
Since $\mathcal{O}_X(U)$ is a normal domain
(Lemma \ref{lemma-locally-normal}), we see that
$f_U = (a|_U)/(b|_U) \in \mathcal{O}_X(U)$. It is easy to
see that $f_U|_V = f_V$ whenever $V \subset U \subset X$ are
affine open. Hence the local sections $f_U$ glue to a global
section $f$ as desired.
\end{proof}
\section{Cohen-Macaulay schemes}
\label{section-Cohen-Macaulay}
\noindent
Recall, see Algebra, Definition \ref{algebra-definition-local-ring-CM},
that a local Noetherian ring $(R, \mathfrak m)$ is
said to be Cohen-Macaulay if $\text{depth}_{\mathfrak m}(R) = \dim(R)$.
Recall that a Noetherian ring $R$ is said to be Cohen-Macaulay if
every local ring $R_{\mathfrak p}$ of $R$ is Cohen-Macaulay,
see Algebra, Definition \ref{algebra-definition-ring-CM}.
\begin{definition}
\label{definition-Cohen-Macaulay}
Let $X$ be a scheme. We say $X$ is {\it Cohen-Macaulay} if
for every $x \in X$ there exists an affine open neighbourhood
$U \subset X$ of $x$ such that the ring $\mathcal{O}_X(U)$ is
Noetherian and Cohen-Macaulay.
\end{definition}
\begin{lemma}
\label{lemma-characterize-Cohen-Macaulay}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item $X$ is Cohen-Macaulay,
\item $X$ is locally Noetherian and all of its local rings are Cohen-Macaulay,
and
\item $X$ is locally Noetherian and for any closed point $x \in X$
the local ring $\mathcal{O}_{X, x}$ is Cohen-Macaulay.
\end{enumerate}
\end{lemma}
\begin{proof}
Algebra, Lemma \ref{algebra-lemma-localize-CM} says that the localization of
a Cohen-Macaulay local ring is Cohen-Macaulay. The lemma follows
by combining this with Lemma \ref{lemma-locally-Noetherian},
with the existence of closed
points on locally Noetherian schemes
(Lemma \ref{lemma-locally-Noetherian-closed-point}), and
the definitions.
\end{proof}
\begin{lemma}
\label{lemma-locally-Cohen-Macaulay}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is Cohen-Macaulay.
\item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is Noetherian and Cohen-Macaulay.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is Noetherian and Cohen-Macaulay.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is Cohen-Macaulay.
\end{enumerate}
Moreover, if $X$ is Cohen-Macaulay then every open subscheme
is Cohen-Macaulay.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-locally-Noetherian}
and \ref{lemma-characterize-Cohen-Macaulay}.
\end{proof}
\noindent
More information on Cohen-Macaulay schemes and depth can be found in
Cohomology of Schemes, Section \ref{coherent-section-depth}.
\section{Regular schemes}
\label{section-regular}
\noindent
Recall, see Algebra, Definition \ref{algebra-definition-regular-local},
that a local Noetherian ring $(R, \mathfrak m)$ is
said to be {\it regular} if $\mathfrak m$ can be generated
by $\dim(R)$ elements.
Recall that a Noetherian ring $R$ is said to be {\it regular} if
every local ring $R_{\mathfrak p}$ of $R$ is regular,
see Algebra, Definition \ref{algebra-definition-regular}.
\begin{definition}
\label{definition-regular}
Let $X$ be a scheme. We say $X$ is {\it regular}, or {\it nonsingular} if
for every $x \in X$ there exists an affine open neighbourhood
$U \subset X$ of $x$ such that the ring $\mathcal{O}_X(U)$ is
Noetherian and regular.
\end{definition}
\begin{lemma}
\label{lemma-characterize-regular}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item $X$ is regular,
\item $X$ is locally Noetherian and all of its local rings are regular,
and
\item $X$ is locally Noetherian and for any closed point $x \in X$
the local ring $\mathcal{O}_{X, x}$ is regular.
\end{enumerate}
\end{lemma}
\begin{proof}
By the discussion in Algebra preceding Algebra, Definition
\ref{algebra-definition-regular} we know that the localization of
a regular local ring is regular. The lemma follows
by combining this with Lemma \ref{lemma-locally-Noetherian},
with the existence of closed
points on locally Noetherian schemes
(Lemma \ref{lemma-locally-Noetherian-closed-point}), and
the definitions.
\end{proof}
\begin{lemma}
\label{lemma-locally-regular}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is regular.
\item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is Noetherian and regular.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is Noetherian and regular.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is regular.
\end{enumerate}
Moreover, if $X$ is regular then every open subscheme is regular.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-locally-Noetherian}
and \ref{lemma-characterize-regular}.
\end{proof}
\begin{lemma}
\label{lemma-regular-normal}
A regular scheme is normal.
\end{lemma}
\begin{proof}
See
Algebra, Lemma \ref{algebra-lemma-regular-normal}.
\end{proof}
\section{Dimension}
\label{section-dimension}
\noindent
The dimension of a scheme is just the dimension of its underlying
topological space.
\begin{definition}
\label{definition-dimension}
Let $X$ be a scheme.
\begin{enumerate}
\item The {\it dimension} of $X$ is just the dimension of $X$
as a topological spaces, see
Topology, Definition \ref{topology-definition-Krull}.
\item For $x \in X$ we denote $\dim_x(X)$ the dimension of the underlying
topological space of $X$ at $x$ as in
Topology, Definition \ref{topology-definition-Krull}.
We say $\dim_x(X)$ is the {\it dimension of $X$ at $x$}.
\end{enumerate}
\end{definition}