varieties.tex

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\begin{document}

\title{Varieties}


\maketitle

\phantomsection
\label{section-phantom}

\tableofcontents

\section{Introduction}
\label{section-introduction}

\noindent
In this chapter we start studying varieties and more generally
schemes over a field. A fundamental reference is \cite{EGA}.








\section{Notation}
\label{section-notation}

\noindent
Throughout this chapter we use the letter $k$ to denote the ground field.










\section{Varieties}
\label{section-varieties}

\noindent
In the stacks project we will use the following as our definition
of a variety.

\begin{definition}
\label{definition-variety}
Let $k$ be a field. A {\it variety} is a scheme $X$ over $k$
such that $X$ is integral and the structure morphism
$X \to \Spec(k)$ is separated and of finite type.
\end{definition}

\noindent
This definition has the following drawback. Suppose that
$k \subset k'$ is an extension of fields. Suppose that $X$
is a variety over $k$. Then the base change
$X_{k'} = X \times_{\Spec(k)} \Spec(k')$ is
not necessarily a variety over $k'$. This phenomenon (in greater
generality) will be discussed in detail in the following sections.
The product of two varieties need not be a variety
(this is really the same phenomenon). Here is an example.

\begin{example}
\label{example-product-not-a-variety}
Let $k = \mathbf{Q}$. Let $X = \Spec(\mathbf{Q}(i))$
and $Y = \Spec(\mathbf{Q}(i))$. Then the product
$X \times_{\Spec(k)} Y$ of the varieties $X$ and $Y$
is not a variety, since it is reducible. (It is isomorphic
to the disjoint union of two copies of $X$.)
\end{example}

\noindent
If the ground field is algebraically closed however, then the
product of varieties is a variety. This follows from the results
in the algebra chapter, but there we treat much more general situations.
There is also a simple direct proof of it which we present here.

\begin{lemma}
\label{lemma-product-varieties}
Let $k$ be an algebraically closed field.
Let $X$, $Y$ be varieties over $k$.
Then $X \times_{\Spec(k)} Y$ is a variety over $k$.
\end{lemma}

\begin{proof}
The morphism $X \times_{\Spec(k)} Y \to \Spec(k)$ is of
finite type and separated because it is the composition of the
morphisms $X \times_{\Spec(k)} Y \to Y \to \Spec(k)$
which are separated and of finite type, see
Morphisms, Lemmas \ref{morphisms-lemma-base-change-finite-type} and
\ref{morphisms-lemma-composition-finite-type}
and
Schemes, Lemma \ref{schemes-lemma-separated-permanence}.
To finish the proof it suffices to show that $X \times_{\Spec(k)} Y$
is integral.
Let $X = \bigcup_{i = 1, \ldots, n} U_i$,
$Y = \bigcup_{j = 1, \ldots, m} V_j$ be finite affine open coverings.
If we can show that each $U_i \times_{\Spec(k)} V_j$ is integral,
then we are done by
Properties, Lemmas \ref{properties-lemma-characterize-reduced},
\ref{properties-lemma-characterize-irreducible}, and
\ref{properties-lemma-characterize-integral}.
This reduces us to the affine case.

\medskip\noindent
The affine case translates into the following algebra statement: Suppose
that $A$, $B$ are integral domains and finitely generated $k$-algebras.
Then $A \otimes_k B$ is an integral domain. To get a contradiction suppose that
$$
(\sum\nolimits_{i = 1, \ldots, n} a_i \otimes b_i)
(\sum\nolimits_{j = 1, \ldots, m} c_j \otimes d_j) = 0
$$
in $A \otimes_k B$ with both factors nonzero in $A \otimes_k B$.
We may assume that $b_1, \ldots, b_n$ are $k$-linearly
independent in $B$, and that $d_1, \ldots, d_m$ are $k$-linearly independent
in $B$. Of course we may also assume that $a_1$ and $c_1$ are nonzero
in $A$. Hence $D(a_1c_1) \subset \Spec(A)$ is nonempty. By the
Hilbert Nullstellensatz
(Algebra, Theorem \ref{algebra-theorem-nullstellensatz})
we can find a maximal ideal $\mathfrak m \subset A$ contained in
$D(a_1c_1)$ and $A/\mathfrak m = k$ as $k$ is algebraically closed.
Denote $\overline{a}_i, \overline{c}_j$ the residue classes of
$a_i, c_j$ in $A/\mathfrak m = k$. Then equation above becomes
$$
(\sum\nolimits_{i = 1, \ldots, n} \overline{a}_i b_i)
(\sum\nolimits_{j = 1, \ldots, m} \overline{c}_j d_j) = 0
$$
which is a contradiction with
$\mathfrak m \in D(a_1c_1)$, the linear independence of
$b_1, \ldots, b_n$ and $d_1, \ldots, d_m$, and the fact that $B$ is a domain.
\end{proof}






\section{Geometrically reduced schemes}
\label{section-geometrically-reduced}

\noindent
If $X$ is a reduced scheme over a field, then it can happen that $X$
becomes nonreduced after extending the ground field. This does not happen
for geometrically reduced schemes.

\begin{definition}
\label{definition-geometrically-reduced}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $x \in X$ be a point.
\begin{enumerate}
\item Let $x \in X$ be a point.
We say $X$ is {\it geometrically reduced at $x$}
if for any field extension $k \subset k'$
and any point $x' \in X_{k'}$ lying over $x$
the local ring $\mathcal{O}_{X_{k'}, x'}$ is reduced.
\item We say $X$ is {\it geometrically reduced} over $k$
if $X$ is geometrically reduced at every point of $X$.
\end{enumerate}
\end{definition}

\noindent
This may seem a little mysterious at first, but it is
really the same thing as the notion discussed in the algebra chapter.
Here are some basic results explaining the connection.

\begin{lemma}
\label{lemma-geometrically-reduced-at-point}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $x \in X$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically reduced at $x$, and
\item the ring $\mathcal{O}_{X, x}$ is geometrically
reduced over $k$ (see
Algebra, Definition \ref{algebra-definition-geometrically-reduced}).
\end{enumerate}
\end{lemma}

\begin{proof}
Assume (1). This in particular implies that $\mathcal{O}_{X, x}$
is reduced. Let $k \subset k'$ be a finite purely inseparable field
extension. Consider the ring $\mathcal{O}_{X, x} \otimes_k k'$.
By Algebra, Lemma \ref{algebra-lemma-p-ring-map}
its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$.
Hence it is a local ring also
(Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}).
Therefore there is a unique point $x' \in X_{k'}$ lying over $x$
and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$.
By assumption this is a reduced ring. Hence we deduce (2) by
Algebra, Lemma
\ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}.

\medskip\noindent
Assume (2). Let $k \subset k'$ be a field extension. Since
$\Spec(k') \to \Spec(k)$ is surjective, also
$X_{k'} \to X$ is surjective
(Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}).
Let $x' \in X_{k'}$ be any point lying over $x$.
The local ring $\mathcal{O}_{X_{k'}, x'}$
is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$.
Hence it is reduced by assumption and (1) is proved.
\end{proof}

\noindent
The notion isn't interesting in characteristic zero.

\begin{lemma}
\label{lemma-perfect-reduced}
Let $X$ be a scheme over a perfect field $k$ (e.g.\ $k$ has
characteristic zero). Let $x \in X$. If $\mathcal{O}_{X, x}$ is
reduced, then $X$ is geometrically reduced at $x$.
If $X$ is reduced, then $X$ is geometrically reduced over $k$.
\end{lemma}

\begin{proof}
The first statement follows from
Lemma \ref{lemma-geometrically-reduced-at-point} and
Algebra, Lemma \ref{algebra-lemma-separable-extension-preserves-reducedness}
and the definition of a perfect field
(Algebra, Definition \ref{algebra-definition-perfect}).
The second statement follows from the first.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-reduced}
Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically reduced,
\item $X_{k'}$ is reduced for every field extension $k \subset k'$,
\item $X_{k'}$ is reduced for every finite purely inseparable field extension
$k \subset k'$,
\item $X_{k^{1/p}}$ is reduced,
\item $X_{k^{perf}}$ is reduced,
\item $X_{\bar k}$ is reduced,
\item for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is geometrically reduced (see
Algebra, Definition \ref{algebra-definition-geometrically-reduced}).
\end{enumerate}
\end{lemma}

\begin{proof}
Assume (1). Then for every field extension $k \subset k'$ and
every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$
is reduced. In other words $X_{k'}$ is reduced. Hence (2).

\medskip\noindent
Assume (2). Let $U \subset X$ be an affine open. Then for
every field extension $k \subset k'$ the scheme $X_{k'}$ is reduced, hence
$U_{k'} = \Spec(\mathcal{O}(U)\otimes_k k')$ is reduced,
hence $\mathcal{O}(U)\otimes_k k'$ is reduced (see Properties,
Section \ref{properties-section-integral}). In other words
$\mathcal{O}(U)$ is geometrically reduced, so (7) holds.

\medskip\noindent
Assume (7). For any field extension $k \subset k'$ the base
change $X_{k'}$ is gotten by gluing the spectra of the
rings $\mathcal{O}_X(U) \otimes_k k'$ where $U$ is affine open
in $X$ (see Schemes, Section \ref{schemes-section-fibre-products}).
Hence $X_{k'}$ is reduced. So (1) holds.

\medskip\noindent
This proves that (1), (2), and (7) are equivalent. These are equivalent
to (3), (4), (5), and (6) because we can apply Algebra, Lemma
\ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}
to $\mathcal{O}_X(U)$ for $U \subset X$ affine open.
\end{proof}

\begin{lemma}
\label{lemma-check-only-finite-inseparable-extensions}
Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$.
Let $x \in X$. The following are equivalent
\begin{enumerate}
\item $X$ is geometrically reduced at $x$,
\item $\mathcal{O}_{X_{k'}, x'}$ is reduced for every
finite purely inseparable field extension $k'$ of $k$ and
$x' \in X_{k'}$ the unique point lying over $x$,
\item $\mathcal{O}_{X_{k^{1/p}}, x'}$ is reduced for
$x' \in X_{k'}$ the unique point lying over $x$, and
\item $\mathcal{O}_{X_{k^{perf}}, x'}$ is reduced for
$x' \in X_{k^{perf}}$ the unique point lying over $x$.
\end{enumerate}
\end{lemma}

\begin{proof}
Note that if $k \subset k'$ is purely inseparable, then
$X_{k'} \to X$ induces a homeomorphism on underlying topological
spaces, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}.
Whence the uniqueness of $x'$ lying over $x$ mentioned in the
statement. Moreover, in this case
$\mathcal{O}_{X_{k'}, x'} = \mathcal{O}_{X, x} \otimes_k k'$.
Hence the lemma follows from Lemma \ref{lemma-geometrically-reduced-at-point}
above and Algebra, Lemma
\ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-reduced-upstairs}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $k'/k$ be a field extension.
Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically reduced at $x$,
\item $X_{k'}$ is geometrically reduced at $x'$.
\end{enumerate}
In particular, $X$ is geometrically reduced over $k$ if and only if
$X_{k'}$ is geometrically reduced over $k'$.
\end{lemma}

\begin{proof}
It is clear that (1) implies (2). Assume (2).
Let $k \subset k''$ be a finite purely inseparable field extension
and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is
unique). We can find a common field extension $k \subset k'''$
(i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$) and a point
$x''' \in X_{k'''}$ lying over both $x'$ and $x''$.
Consider the map of local rings
$$
\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.
$$
This is a flat local ring homomorphism and hence faithfully flat.
By (2) we see that the local ring on the right is reduced.
Thus by Algebra, Lemma \ref{algebra-lemma-descent-reduced}
we conclude that $\mathcal{O}_{X_{k''}, x''}$ is reduced.
Thus by Lemma \ref{lemma-check-only-finite-inseparable-extensions}
we conclude that $X$ is geometrically reduced at $x$.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-reduced-any-base-change}
Let $k$ be a field.
Let $X$, $Y$ be schemes over $k$.
\begin{enumerate}
\item If $X$ is geometrically reduced at $x$, and $Y$ reduced,
then $X \times_k Y$ is reduced at every point lying over $x$.
\item If $X$ geometrically reduced over $k$ and $Y$ reduced.
Then $X \times_k Y$ is reduced.
\end{enumerate}
\end{lemma}

\begin{proof}
Combine, Lemmas \ref{lemma-geometrically-reduced-at-point}
and \ref{lemma-geometrically-reduced} and Algebra,
Lemma \ref{algebra-lemma-geometrically-reduced-any-reduced-base-change}.
\end{proof}

\begin{lemma}
\label{lemma-generic-points-geometrically-reduced}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
\begin{enumerate}
\item If $x' \leadsto x$ is a specialization and $X$ is geometrically
reduced at $x$, then $X$ is geometrically reduced at $x'$.
\item If $x \in X$ such that (a) $\mathcal{O}_{X, x}$
is reduced, and (b) for each specialization $x' \leadsto x$ where
$x'$ is a generic point of an irreducible component of $X$ the
scheme $X$ is geometrically reduced at $x'$, then $X$ is geometrically
reduced at $x$.
\item If $X$ is reduced and geometrically reduced at all generic
points of irreducible components of $X$, then $X$ is geometrically
reduced.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) follows from
Lemma \ref{lemma-geometrically-reduced-at-point}
and the fact that if $A$ is a geometrically reduced
$k$-algebra, then $S^{-1}A$ is a geometrically reduced $k$-algebra for
any multiplicative subset $S$ of $A$, see
Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-permanence}.

\medskip\noindent
Let $A = \mathcal{O}_{X, x}$. The assumptions (a) and (b) of (2) imply
that $A$ is reduced, and that $A_{\mathfrak q}$ is geometrically
reduced over $k$ for every minimal prime $\mathfrak q$ of $A$.
Hence $A$ is geometrically reduced over $k$, see
Algebra, Lemma \ref{algebra-lemma-generic-points-geometrically-reduced}.
Thus $X$ is geometrically reduced at $x$, see
Lemma \ref{lemma-geometrically-reduced-at-point}.

\medskip\noindent
Part (3) follows trivially from part (2).
\end{proof}

\begin{lemma}
\label{lemma-Noetherian-geometrically-reduced-at-point}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $x \in X$.
Assume $X$ locally Noetherian and geometrically reduced at $x$.
Then there exists an open neighbourhood $U \subset X$ of $x$
which is geometrically reduced over $k$.
\end{lemma}

\begin{proof}
Let $R$ be a Noetherian $k$-algebra.
Let $\mathfrak p \subset R$ be a prime.
Let $I = \text{Ker}(R \to R_{\mathfrak p}$.
Since $IR_{\mathfrak p} = R_{\mathfrak p}$ and $I$ is finitely generated
there exists an $f \in R$, $f \not \in \mathfrak p$ such that $fI = 0$.
Hence $R_f \subset R_{\mathfrak p}$.

\medskip\noindent
Assume $X$ locally Noetherian and geometrically reduced at $x$.
If we apply the above to $R = \mathcal{O}_X(U)$ for some affine
open neighbourhood of $x$, and $\mathfrak p \subset R$ the prime
corresponding to $x$, then we see that after shrinking $U$ we may
assume $R \subset R_{\mathfrak p}$. By
Lemma \ref{lemma-geometrically-reduced-at-point} the assumption
means that $R_{\mathfrak p}$ is geometrically reduced over $k$.
By Algebra, Lemma \ref{algebra-lemma-subalgebra-separable}
this implies that $R$ is geometrically reduced over $k$, which
in turn implies that $U$ is geometrically reduced.
\end{proof}

\begin{example}
\label{example-not-geometrically-reduced}
Let $k = \mathbf{F}_p(s, t)$, i.e., a purely transcendental extension
of the prime field. Consider the variety
$X = \Spec(k[x, y]/(1 + sx^p + ty^p))$.
Let $k \subset k'$ be any extension such that
both $s$ and $t$ have a $p$th root in $k'$.
Then the base change $X_{k'}$ is not reduced.
Namely, the ring $k'[x, y]/(1 + s x^p + ty^p)$ contains the element
$1 + s^{1/p}x + t^{1/p}y$ whose $p$th power is zero but
which is not zero (since the ideal $(1 + sx^p + ty^p)$ certainly
does not contain any nonzero element of degree $< p$).
\end{example}

\begin{lemma}
\label{lemma-finite-extension-geometrically-reduced}
Let $k$ be a field.
Let $X \to \Spec(k)$ be locally of finite type.
Assume $X$ has finitely many irreducible components.
Then there exists a finite purely inseparable extension $k \subset k'$
such that $(X_{k'})_{red}$ is geometrically reduced over $k'$.
\end{lemma}

\begin{proof}
To prove this lemma we may replace $X$ by its reduction $X_{red}$.
Hence we may assume that $X$ is reduced and locally of finite type
over $k$.
Let $x_1, \ldots, x_n \in X$ be the generic points of the irreducible
components of $X$.
Note that for every purely inseparable algebraic extension $k \subset k'$
the morphism $(X_{k'})_{red} \to X$ is a homeomorphism, see
Algebra, Lemma \ref{algebra-lemma-p-ring-map}. Hence the points
$x'_1, \ldots, x'_n$ lying over $x_1, \ldots, x_n$ are the generic
points of the irreducible components of $(X_{k'})_{red}$.
As $X$ is reduced the local rings $K_i = \mathcal{O}_{X, x_i}$ are fields, see
Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}.
As $X$ is locally of finite type over $k$ the field extensions
$k \subset K_i$ are finitely generated field extensions.
Finally, the local rings $\mathcal{O}_{X_{k'}, x'_i}$ are the
fields $(K_i \otimes_k k')_{red}$. By
Algebra, Lemma \ref{algebra-lemma-make-separable}
we can find a finite purely inseparable extension $k \subset k'$
such that $(K_i \otimes_k k')_{red}$ are separable field
extensions of $k'$. In particular each $(K_i \otimes_k k')_{red}$
is geometrically reduced over $k'$ by
Algebra, Lemma \ref{algebra-lemma-characterize-separable-field-extensions}.
At this point
Lemma \ref{lemma-generic-points-geometrically-reduced} part (3)
implies that $(X_{k'})_{red}$ is geometrically reduced.
\end{proof}






\section{Geometrically connected schemes}
\label{section-geometrically-connected}

\noindent
If $X$ is a connected scheme over a field, then it can happen that $X$
becomes disconnected after extending the ground field. This does not happen
for geometrically connected schemes.

\begin{definition}
\label{definition-geometrically-connected}
Let $X$ be a scheme over the field $k$. We say $X$ is
{\it geometrically connected} over $k$ if the scheme $X_{k'}$ is connected
for every field extension $k'$ of $k$.
\end{definition}

\noindent
By convention a connected topological space is nonempty; hence a fortiori
geometrically connected schemes are nonempty.
Here is an example of a variety which is not geometrically connected.

\begin{example}
\label{example-not-geometrically-irreducible}
Let $k = \mathbf{Q}$. The scheme
$X = \Spec(\mathbf{Q}(i))$ is a variety over $\Spec(\mathbf{Q})$.
But the base change $X_{\mathbf{C}}$ is the spectrum of
$\mathbf{C} \otimes_{\mathbf{Q}} \mathbf{Q}(i) \cong
\mathbf{C} \times \mathbf{C}$ which is the disjoint union of
two copies of $\Spec(\mathbf{C})$. So in fact, this is an
example of a non-geometrically connected variety.
\end{example}

\begin{lemma}
\label{lemma-geometrically-connected-check-after-extension}
Let $X$ be a scheme over the field $k$.
Let $k \subset k'$ be a field extension.
Then $X$ is geometrically connected over $k$ if and only if
$X_{k'}$ is geometrically connected over $k'$.
\end{lemma}

\begin{proof}
If $X$ is geometrically connected over $k$, then it is clear that
$X_{k'}$ is geometrically connected over $k'$. For the converse, note
that for any field extension $k \subset k''$ there exists a common
field extension $k' \subset k'''$ and $k'' \subset k'''$. As the
morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of
a surjective morphism between spectra of fields) we see that the
connectedness of $X_{k'''}$ implies the connectedness of $X_{k''}$.
Thus if $X_{k'}$ is geometrically connected over $k'$ then
$X$ is geometrically connected over $k$.
\end{proof}

\begin{lemma}
\label{lemma-bijection-connected-components}
Let $k$ be a field.
Let $X$, $Y$ be schemes over $k$.
Assume $X$ is geometrically connected over $k$.
Then the projection morphism
$$
p : X \times_k Y \longrightarrow Y
$$
induces a bijection between connected components.
\end{lemma}

\begin{proof}
The scheme theoretic fibres of $p$ are connected, since they
are base changes of the geometrically connected scheme $X$ by
field extensions. Moreover the scheme theoretic fibres are
homeomorphic to the set theoretic fibres, see
Schemes, Lemma \ref{schemes-lemma-fibre-topological}.
By
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}
the map $p$ is open.
Thus we may apply Topology,
Lemma \ref{topology-lemma-connected-fibres-connected-components}
to conclude.
\end{proof}

\begin{lemma}
\label{lemma-affine-geometrically-connected}
Let $k$ be a field.
Let $A$ be a $k$-algebra.
Then $X = \Spec(A)$ is geometrically connected over $k$
if and only if $A$ is geometrically connected over $k$ (see
Algebra, Definition \ref{algebra-definition-geometrically-connected}).
\end{lemma}

\begin{proof}
Immediate from the definitions.
\end{proof}

\begin{lemma}
\label{lemma-separably-closed-field-connected-components}
Let $k \subset k'$ be an extension of fields.
Let $X$ be a scheme over $k$.
Assume $k$ separably algebraically closed.
Then the morphism $X_{k'} \to X$ induces a bijection of connected
components. In particular, $X$ is geometrically connected over $k$
if and only if $X$ is connected.
\end{lemma}

\begin{proof}
Since $k$ is separably algebraically closed we see that
$k'$ is geometrically connected over $k$, see
Algebra,
Lemma \ref{algebra-lemma-separably-closed-connected-implies-geometric}.
Hence $Z = \Spec(k')$ is geometrically connected over $k$ by
Lemma \ref{lemma-affine-geometrically-connected}
above. Since $X_{k'} = Z \times_k X$ the result is a special case of
Lemma \ref{lemma-bijection-connected-components}.
\end{proof}

\begin{lemma}
\label{lemma-characterize-geometrically-connected}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $\overline{k}$ be a separable algebraic closure of $k$.
Then $X$ is geometrically connected if and only if the base change
$X_{\overline{k}}$ is connected.
\end{lemma}

\begin{proof}
Assume $X_{\overline{k}}$ is connected.
Let $k \subset k'$ be a field extension.
There exists a field extension $\overline{k} \subset \overline{k}'$
such that $k'$ embeds into $\overline{k}'$ as an extension of $k$.
By Lemma \ref{lemma-separably-closed-field-connected-components}
we see that $X_{\overline{k}'}$ is connected.
Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude
that $X_{k'}$ is connected as desired.
\end{proof}

\begin{lemma}
\label{lemma-descend-open}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $A$ be a $k$-algebra.
Let $V \subset X_A$ be a quasi-compact open.
Then there exists a finitely generated $k$-subalgebra $A' \subset A$
and a quasi-compact open $V' \subset X_{A'}$
such that $V = V'_A$.
\end{lemma}

\begin{proof}
We remark that if $X$ is also quasi-separated this follows from
Limits, Lemma \ref{limits-lemma-descend-opens}. Let
$U_1, \ldots, U_n$ be finitely many affine opens of $X$
such that $V \subset \bigcup U_{i, A}$. Say $U_i = \Spec(R_i)$.
Since $V$ is quasi-compact we can find finitely many
$f_{ij} \in R_i \otimes_k A$, $j = 1, \ldots, n_i$
such that $V = \bigcup_i \bigcup_{j = 1, \ldots, n_i} D(f_{ij})$
where $D(f_{ij}) \subset U_{i, A}$ is the corresponding standard
open. (We do not claim that $V \cap U_{i, A}$ is the union
of the $D(f_{ij})$, $j = 1, \ldots, n_i$.)
It is clear that we can find a finitely generated $k$-subalgebra
$A' \subset A$ such that $f_{ij}$ is the image of some
$f_{ij}' \in R_i \otimes_k A'$.
Set $V' = \bigcup D(f_{ij}')$ which is a quasi-compact open of $X_{A'}$.
Denote $\pi : X_A \to X_{A'}$ the canonical morphism.
We have $\pi(V) \subset V'$ as $\pi(D(f_{ij})) \subset D(f_{ij}')$.
If $x \in X_A$ with $\pi(x) \in V'$, then $\pi(x) \in D(f_{ij}')$
for some $i, j$ and we see that $x \in D(f_{ij})$ as $f_{ij}'$
maps to $f_{ij}$. Thus we see that $V = \pi^{-1}(V')$ as desired.
\end{proof}

\noindent
Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite)
Galois extension. For example $\overline{k}$ could be the
separable algebraic closure of $k$.
For any $\sigma \in \text{Gal}(\overline{k}/k)$ we get a corresponding
automorphism
$
\Spec(\sigma) :
\Spec(\overline{k})
\longrightarrow
\Spec(\overline{k})
$.
Note that
$\Spec(\sigma) \circ \Spec(\tau) = \Spec(\tau \circ \sigma)$.
Hence we get an action
$$
\text{Gal}(\overline{k}/k)^{opp} \times \Spec(\overline{k})
\longrightarrow
\Spec(\overline{k})
$$
of the opposite group on the scheme $\Spec(\overline{k})$.
Let $X$ be a scheme over $k$. Since
$X_{\overline{k}} =
\Spec(\overline{k}) \times_{\Spec(k)} X$
by definition we see that the action above induces a canonical action
\begin{equation}
\label{equation-galois-action-base-change-kbar}
\text{Gal}(\overline{k}/k)^{opp} \times X_{\overline{k}}
\longrightarrow
X_{\overline{k}}.
\end{equation}

\begin{lemma}
\label{lemma-Galois-action-quasi-compact-open}
Let $k$ be a field. Let $X$ be a scheme over $k$.
Let $\overline{k}$ be a (possibly infinite) Galois extension of $k$.
Let $V \subset X_{\overline{k}}$ be a quasi-compact open.
Then
\begin{enumerate}
\item there exists a finite subextension $k \subset k' \subset \overline{k}$
and a quasi-compact open $V' \subset X_{k'}$ such that
$V = (V')_{\overline{k}}$,
\item there exists an open subgroup $H \subset \text{Gal}(\overline{k}/k)$
such that $\sigma(V) = V$ for all $\sigma \in H$.
\end{enumerate}
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-descend-open} there exists a finite subextension
$k \subset k' \subset \overline{k}$ and an open $V' \subset X_{k'}$
which pulls back to $V$. This proves (1). Since $\text{Gal}(\overline{k}/k')$
is open in $\text{Gal}(\overline{k}/k)$ part (2) is clear as well.
\end{proof}

\begin{lemma}
\label{lemma-closed-fixed-by-Galois}
Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite)
Galois extension. Let $X$ be a scheme over $k$. Let
$\overline{T} \subset X_{\overline{k}}$ have the following properties
\begin{enumerate}
\item $\overline{T}$ is a closed subset of $X_{\overline{k}}$,
\item for every $\sigma \in \text{Gal}(\overline{k}/k)$
we have $\sigma(\overline{T}) = \overline{T}$.
\end{enumerate}
Then there exists a closed subset $T \subset X$ whose inverse image
in $X_{k'}$ is $\overline{T}$.
\end{lemma}

\begin{proof}
This lemma immediately reduces to the case where $X = \Spec(A)$
is affine. In this case, let $\overline{I} \subset A \otimes_k \overline{k}$
be the radical ideal corresponding to $\overline{T}$.
Assumption (2) implies that $\sigma(\overline{I}) = \overline{I}$
for all $\sigma \in \text{Gal}(\overline{k}/k)$.
Pick $x \in \overline{I}$. There exists a finite Galois extension
$k \subset k'$ contained in $\overline{k}$ such that $x \in A \otimes_k k'$.
Set $G = \text{Gal}(k'/k)$. Set
$$
P(T) = \prod\nolimits_{\sigma \in G} (T - \sigma(x)) \in (A \otimes_k k')[T]
$$
It is clear that $P(T)$ is monic and is actually an element of
$(A \otimes_k k')^G[T] = A[T]$ (by basic Galois theory).
Moreover, if we write $P(T) = T^d + a_1T^{d - 1} + \ldots + a_0$
the we see that $a_i \in I := A \cap \overline{I}$. By
Algebra, Lemma \ref{algebra-lemma-polynomials-divide}
we see that $x$ is contained in the radical of $I(A \otimes_k \overline{k})$.
Hence $\overline{I}$ is the radical of $I(A \otimes_k \overline{k})$ and
setting $T = V(I)$ is a solution.
\end{proof}

\begin{lemma}
\label{lemma-characterize-geometrically-disconnected}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically connected,
\item for every finite separable field extension $k \subset k'$
the scheme $X_{k'}$ is connected.
\end{enumerate}
\end{lemma}

\begin{proof}
It follows immediately from the definition that (1) implies (2).
Assume that $X$ is not geometrically connected.
Let $k \subset \overline{k}$ be a separable algebraic
closure of $k$. By
Lemma \ref{lemma-characterize-geometrically-connected}
it follows that $X_{\overline{k}}$ is disconnected.
Say $X_{\overline{k}} = \overline{U} \amalg \overline{V}$
with $\overline{U}$ and $\overline{V}$ open, closed, and nonempty.

\medskip\noindent
Suppose that $W \subset X$ is any quasi-compact open.
Then $W_{\overline{k}} \cap \overline{U}$ and
$W_{\overline{k}} \cap \overline{V}$ are open and closed in
$W_{\overline{k}}$. In particular $W_{\overline{k}} \cap \overline{U}$ and
$W_{\overline{k}} \cap \overline{V}$ are quasi-compact, and by
Lemma \ref{lemma-Galois-action-quasi-compact-open}
both $W_{\overline{k}} \cap \overline{U}$ and
$W_{\overline{k}} \cap \overline{V}$
are defined over a finite subextension and invariant under an
open subgroup of $\text{Gal}(\overline{k}/k)$.
We will use this without further mention in the following.

\medskip\noindent
Pick $W_0 \subset X$ quasi-compact open such that both
$W_{0, \overline{k}} \cap \overline{U}$ and
$W_{0, \overline{k}} \cap \overline{V}$ are nonempty.
Choose a finite subextension $k \subset k' \subset \overline{k}$
and a decomposition $W_{0, k'} = U_0' \amalg V_0'$ into open and closed
subsets such that
$W_{0, \overline{k}} \cap \overline{U} = (U'_0)_{\overline{k}}$ and
$W_{0, \overline{k}} \cap \overline{V} = (V'_0)_{\overline{k}}$.
Let $H = \text{Gal}(\overline{k}/k') \subset \text{Gal}(\overline{k}/k)$.
In particular
$\sigma(W_{0, \overline{k}} \cap \overline{U}) =
W_{0, \overline{k}} \cap \overline{U}$ and similarly for
$\overline{V}$.

\medskip\noindent
Having chosen $W_0$, $k'$ as above, for every quasi-compact open
$W \subset X$ we set
$$
U_W =
\bigcap\nolimits_{\sigma \in H} \sigma(W_{\overline{k}} \cap \overline{U}),
\quad
V_W =
\bigcup\nolimits_{\sigma \in H} \sigma(W_{\overline{k}} \cap \overline{V}).
$$
Now, since $W_{\overline{k}} \cap \overline{U}$ and
$W_{\overline{k}} \cap \overline{V}$ are fixed by an open subgroup of
$\text{Gal}(\overline{k}/k)$ we see that the union and intersection
above are finite. Hence $U_W$ and $V_W$ are both open and closed.
Also, by construction $W_{\bar k} = U_W \amalg V_W$.

\medskip\noindent
We claim that if $W \subset W' \subset X$ are quasi-compact
open, then $W_{\overline{k}} \cap U_{W'} = U_W$ and
$W_{\overline{k}} \cap V_{W'} = V_W$. Verification omitted.
Hence we see that upon defining $U = \bigcup_{W \subset X} U_W$
and $V = \bigcup_{W \subset X} V_W$ we obtain
$X_{\overline{k}} = U \amalg V$ is a disjoint union of open
and closed subsets.
It is clear that $V$ is nonempty as it is constructed by taking
unions (locally). On the other hand, $U$ is nonempty since it contains
$W_0 \cap \overline{U}$ by construction. Finally, $U, V \subset X_{\bar k}$
are closed and $H$-invariant by construction. Hence by
Lemma \ref{lemma-closed-fixed-by-Galois}
we have $U = (U')_{\bar k}$, and $V = (V')_{\bar k}$ for some
closed $U', V' \subset X_{k'}$. Clearly $X_{k'} = U' \amalg V'$
and we see that $X_{k'}$ is disconnected as desired.
\end{proof}

\begin{lemma}
\label{lemma-tricky}
Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite)
Galois extension. Let $f : T \to X$ be a morphism of schemes over $k$.
Assume $T_{\overline{k}}$ connected and $X_{\overline{k}}$
disconnected. Then $X$ is disconnected.
\end{lemma}

\begin{proof}
Write $X_{\overline{k}} = \overline{U} \coprod \overline{V}$
with $\overline{U}$ and $\overline{V}$ open and closed.
Denote $\overline{f} : T_{\overline{k}} \to X_{\overline{k}}$ the base
change of $f$. Since $T_{\overline{k}}$ is connected we see that
$T_{\overline{k}}$ is contained in either $\overline{f}^{-1}(\overline{U})$
or $\overline{f}^{-1}(\overline{V})$.
Say $T_{\overline{k}} \subset \overline{f}^{-1}(\overline{U})$.

\medskip\noindent
Fix a quasi-compact open $W \subset X$. There exists a
finite Galois subextension $k \subset k' \subset \overline{k}$
such that $\overline{U} \cap W_{\overline{k}}$ and
$\overline{V} \cap W_{\overline{k}}$ come from quasi-compact
opens $U', V' \subset W_{k'}$. Then also $W_{k'} = U' \coprod V'$.
Consider
$$
U'' = \bigcap\nolimits_{\sigma \in \text{Gal}(k'/k)} \sigma(U'),
\quad
V'' = \bigcup\nolimits_{\sigma \in \text{Gal}(k'/k)} \sigma(V').
$$
These are Galois invariant, open and closed, and
$W_{k'} = U'' \coprod V''$.
By Lemma \ref{lemma-closed-fixed-by-Galois} we get open and closed subsets
$U_W, V_W \subset W$ such that
$U'' = (U_W)_{k'}$, $V'' = (V_W)_{k'}$ and
$W = U_W \coprod V_W$.

\medskip\noindent
We claim that if $W \subset W' \subset X$ are quasi-compact
open, then $W \cap U_{W'} = U_W$ and $W \cap V_{W'} = V_W$.
Verification omitted.
Hence we see that upon defining $U = \bigcup_{W \subset X} U_W$
and $V = \bigcup_{W \subset X} V_W$ we obtain $X = U \coprod V$.
It is clear that $V$ is nonempty as it is constructed by taking
unions (locally). On the other hand, $U$ is nonempty since it contains
$f(T)$ by construction.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-connected-criterion}
Let $k$ be a field. Let $T \to X$ be a morphism of schemes over $k$.
Assume $T$ is geometrically connected and $X$ connected.
Then $X$ is geometrically connected.
\end{lemma}

\begin{proof}
This is a reformulation of
Lemma \ref{lemma-tricky}.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-connected-if-connected-and-point}
Let $k$ be a field. Let $X$ be a scheme over $k$.
Assume $X$ is connected and has a point $x$ such that
$k$ is algebraically closed in $\kappa(x)$.
Then $X$ is geometrically connected.
In particular, if $X$ has a $k$-rational point and $X$ is connected,
then $X$ is geometrically connected.
\end{lemma}

\begin{proof}
Set $T = \Spec(\kappa(x))$. Let $k \subset \overline{k}$ be a
separable algebraic closure of $k$. The assumption on $k \subset \kappa(x)$
implies that $T_{\overline{k}}$ is irreducible, see
Algebra, Lemma \ref{algebra-lemma-field-extension-geometrically-irreducible}.
Hence by
Lemma \ref{lemma-geometrically-connected-criterion}
we see that $X_{\overline{k}}$ is connected. By
Lemma \ref{lemma-characterize-geometrically-connected}
we conclude that $X$ is geometrically connected.
\end{proof}

\begin{lemma}
\label{lemma-inverse-image-connected-component}
Let $k \subset K$ be an extension of fields.
Let $X$ be a scheme over $k$.
For every connected component $T$ of $X$ the inverse image
$T_K \subset X_K$ is a union of connected components of $X_K$.
\end{lemma}

\begin{proof}
This is a purely topological statement.
Denote $p : X_K \to X$ the projection morphism.
Let $T \subset X$ be a connected component of $X$.
Let $t \in T_K = p^{-1}(T)$. Let $C \subset X_K$ be a connected component
containing $t$. Then $p(C)$ is a connected subset of $X$
which meets $T$, hence $p(C) \subset T$. Hence $C \subset T_K$.
\end{proof}

\begin{lemma}
\label{lemma-image-connected-component-finite-extension}
Let $k \subset K$ be a finite extension of fields and let $X$ be a scheme over 
$k$. Denote by $p : X_K \to X$ the projection morphism. For every connected 
component $T$ of $X_K$ the image $p(T)$ is a connected component of 
$X$.
\end{lemma}

\begin{proof}
The image $p(T)$ is contained in some connected component $X'$ of $X$. Consider
$X'$ as a closed subscheme of $X$ in any way. Then $T$ is also a connected
component of $X'_K = p^{-1}(X')$ and we may therefore assume that $X$ is
connected. The morphism $p$ is open
(Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}), 
closed
(Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed})
and the fibers of $p$ are finite sets
(Morphisms, Lemma \ref{morphisms-lemma-finite-quasi-finite}).
Thus we may apply
Topology, Lemma \ref{topology-lemma-finite-fibre-connected-components}
to conclude.
\end{proof}

\begin{remark}
\label{remark-image-connected-component}
Let $k \subset K$ be an extension of fields.
Let $X$ be a scheme over $k$.
Denote $p : X_K \to X$ the projection morphism.
Let $\overline{T} \subset X_K$ be a connected component.
Is it true that $p(\overline{T})$ is a connected component of $X$?
When $k \subset K$ is finite
Lemma \ref{lemma-image-connected-component-finite-extension}
tells us the answer is ``yes''.
In general we do not know the answer.
If you do, or if you have a reference, please email
\href{mailto:stacks.project@gmail.com}{stacks.project@gmail.com}.
\end{remark}

\noindent
Let $X$ be a scheme. We denote $\pi_0(X)$ the set of connected
components of $X$.

\begin{lemma}
\label{lemma-galois-action-connected-components}
Let $k$ be a field, with separable algebraic closure $\overline{k}$.
Let $X$ be a scheme over $k$.
There is an action
$$
\text{Gal}(\overline{k}/k)^{opp} \times \pi_0(X_{\overline{k}})
\longrightarrow
\pi_0(X_{\overline{k}})
$$
with the following properties:
\begin{enumerate}
\item An element $\overline{T} \in \pi_0(X_{\overline{k}})$
is fixed by the action if and only if there exists a connected component
$T \subset X$, which is geometrically connected over $k$,
such that $T_{\overline{k}} = \overline{T}$.
\item For any field extension $k \subset k'$ with separable
algebraic closure $\overline{k}'$ the diagram
$$
\xymatrix{
\text{Gal}(\overline{k}'/k') \times \pi_0(X_{\overline{k}'})
\ar[r] \ar[d] &
\pi_0(X_{\overline{k}'}) \ar[d] \\
\text{Gal}(\overline{k}/k) \times \pi_0(X_{\overline{k}})
\ar[r] &
\pi_0(X_{\overline{k}})
}
$$
is commutative (where the right vertical arrow is a bijection
according to Lemma \ref{lemma-separably-closed-field-connected-components}).
\end{enumerate}
\end{lemma}

\begin{proof}
The action (\ref{equation-galois-action-base-change-kbar})
of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$
induces an action on its connected components.
Connected components are always closed
(Topology, Lemma \ref{topology-lemma-connected-components}).
Hence if $\overline{T}$ is as in (1), then by
Lemma \ref{lemma-closed-fixed-by-Galois} there exists a closed
subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$.
Note that $T$ is geometrically connected over $k$, see
Lemma \ref{lemma-characterize-geometrically-connected}.
To see that $T$ is a connected component of $X$, suppose that
$T \subset T'$, $T \not = T'$ where $T'$ is a connected component of $X$.
In this case $T'_{k'}$ strictly contains $\overline{T}$ and hence is
disconnected. By Lemma \ref{lemma-tricky} this means that $T'$ is
disconnected! Contradiction.

\medskip\noindent
We omit the proof of the functoriality in (2).
\end{proof}

\begin{lemma}
\label{lemma-galois-action-connected-components-continuous}
Let $k$ be a field, with separable algebraic closure $\overline{k}$.
Let $X$ be a scheme over $k$.
Assume
\begin{enumerate}
\item $X$ is quasi-compact, and
\item the connected components of $X_{\overline{k}}$ are open.
\end{enumerate}
Then
\begin{enumerate}
\item[(a)] $\pi_0(X_{\overline{k}})$ is finite, and
\item[(b)] the action of $\text{Gal}(\overline{k}/k)$ on
$\pi_0(X_{\overline{k}})$ is continuous.
\end{enumerate}
Moreover, assumptions (1) and (2) are satisfied when $X$ is
of finite type over $k$.
\end{lemma}

\begin{proof}
Since the connected components are open, cover $X_{\overline{k}}$
(Topology, Lemma \ref{topology-lemma-connected-components}) and
$X_{\overline{k}}$ is quasi-compact, we conclude that there are only
finitely many of them. Thus (a) holds.
By Lemma \ref{lemma-descend-open} these connected components
are each defined over a finite subextension of $k \subset \overline{k}$
and we get (b).
If $X$ is of finite type over $k$, then $X_{\overline{k}}$ is of finite
type over $\overline{k}$
(Morphisms, Lemma \ref{morphisms-lemma-base-change-finite-type}).
Hence $X_{\overline{k}}$ is a Noetherian scheme
(Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}) and has
an underlying Noetherian topological space
(Properties, Lemma \ref{properties-lemma-Noetherian-topology}).
Thus $X_{\overline{k}}$ has finitely many irreducible components
(Topology, Lemma \ref{topology-lemma-Noetherian})
and a fortiori finitely many connected components (which are
therefore open).
\end{proof}









\section{Geometrically irreducible schemes}
\label{section-geometrically-irreducible}

\noindent
If $X$ is an irreducible scheme over a field, then it can happen that $X$
becomes reducible after extending the ground field. This does not happen
for geometrically irreducible schemes.

\begin{definition}
\label{definition-geometrically-irreducible}
Let $X$ be a scheme over the field $k$.
We say $X$ is {\it geometrically irreducible} over $k$ if the scheme
$X_{k'}$ is irreducible\footnote{An irreducible space is nonempty.}
for any field extension $k'$ of $k$.
\end{definition}

\begin{lemma}
\label{lemma-geometrically-irreducible-check-after-extension}
Let $X$ be a scheme over the field $k$.
Let $k \subset k'$ be a field extension.
Then $X$ is geometrically irreducible over $k$ if and only if
$X_{k'}$ is geometrically irreducible over $k'$.
\end{lemma}

\begin{proof}
If $X$ is geometrically irreducible over $k$, then it is clear that
$X_{k'}$ is geometrically irreducible over $k'$. For the converse, note
that for any field extension $k \subset k''$ there exists a common
field extension $k' \subset k'''$ and $k'' \subset k'''$. As the
morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of
a surjective morphism between spectra of fields) we see that the
irreducibility of $X_{k'''}$ implies the irreducibility of $X_{k''}$.
Thus if $X_{k'}$ is geometrically irreducible over $k'$ then
$X$ is geometrically irreducible over $k$.
\end{proof}

\begin{lemma}
\label{lemma-separably-closed-irreducible}
Let $X$ be a scheme over a separably closed field $k$.
If $X$ is irreducible, then $X_K$ is irreducible for any
field extension $k \subset K$. I.e., $X$ is geometrically
irreducible over $k$.
\end{lemma}

\begin{proof}
Use Properties, Lemma \ref{properties-lemma-characterize-irreducible}
and Algebra, Lemma \ref{algebra-lemma-separably-closed-irreducible}.
\end{proof}

\begin{lemma}
\label{lemma-bijection-irreducible-components}
Let $k$ be a field.
Let $X$, $Y$ be schemes over $k$.
Assume $X$ is geometrically irreducible over $k$.
Then the projection morphism
$$
p : X \times_k Y \longrightarrow Y
$$
induces a bijection between irreducible components.
\end{lemma}

\begin{proof}
First, note that the scheme theoretic fibres of $p$ are irreducible,
since they are base changes of the geometrically irreducible scheme $X$
by field extensions. Moreover the scheme theoretic fibres are
homeomorphic to the set theoretic fibres, see
Schemes, Lemma \ref{schemes-lemma-fibre-topological}.
By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}
the map $p$ is open.
Thus we may apply Topology,
Lemma \ref{topology-lemma-irreducible-fibres-irreducible-components}
to conclude.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-irreducible-local}
Let $k$ be a field. Let $X$ be a scheme over $k$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically irreducible over $k$,
\item for every affine open $U$ the $k$-algebra $\mathcal{O}_X(U)$
is geometrically irreducible over $k$ (see
Algebra, Definition \ref{algebra-definition-geometrically-irreducible}),
\item $X$ is irreducible and there exists an affine open covering
$X = \bigcup U_i$ such that each $k$-algebra $\mathcal{O}_X(U_i)$ is
geometrically irreducible, and
\item there exists an open covering $X = \bigcup_{i \in I} X_i$ such
that $X_i$ is geometrically irreducible for each $i$ and such that
$X_i \cap X_j \not = \emptyset$ for all $i, j \in I$.
\end{enumerate}
Moreover, if $X$ is geometrically irreducible so is every
open subscheme of $X$.
\end{lemma}

\begin{proof}
An affine scheme $\Spec(A)$ over $k$ is geometrically
irreducible if and only if $A$ is geometrically irreducible over $k$;
this is immediate from the definitions.
Recall that if a scheme is irreducible so is every nonempty
open subscheme of $X$, any two nonempty open subsets have
a nonempty intersection. Also, if every affine open is irreducible
then the scheme is irreducible, see Properties,
Lemma \ref{properties-lemma-characterize-irreducible}.
Hence the final statement of the lemma
is clear, as well as the implications (1) $\Rightarrow$ (2),
(2) $\Rightarrow$ (3), and (3) $\Rightarrow$ (4). If (4) holds,
then for any field extension $k'/k$ the scheme $X_{k'}$
has a covering by irreducible opens which pairwise intersect.
Hence $X_{k'}$ is irreducible. Hence (4) implies (1).
\end{proof}

\begin{lemma}
\label{lemma-geometrically-irreducible-function-field}
Let $X$ be a geometrically irreducible scheme over the field $k$.
Let $\xi \in X$ be its generic point. Then $\kappa(\xi)$ is a
geometrically irreducible over $k$.
\end{lemma}

\begin{proof}
Combining
Lemma \ref{lemma-geometrically-irreducible-local}
and
Algebra, Lemma \ref{algebra-lemma-subalgebra-geometrically-irreducible}
we see that $\mathcal{O}_{X, \xi}$ is geometrically irreducible over $k$.
Since $\mathcal{O}_{X, \xi} \to \kappa(\xi)$ is a surjection with
locally nilpotent kernel (see
Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring})
it follows that $\kappa(\xi)$ is geometrically irreducible, see
Algebra, Lemma \ref{algebra-lemma-p-ring-map}.
\end{proof}

\begin{lemma}
\label{lemma-separably-closed-field-irreducible-components}
Let $k \subset k'$ be an extension of fields.
Let $X$ be a scheme over $k$. Set $X' = X_{k'}$.
Assume $k$ separably algebraically closed.
Then the morphism $X' \to X$ induces a bijection of irreducible components.
\end{lemma}

\begin{proof}
Since $k$ is separably algebraically closed we see that
$k'$ is geometrically irreducible over $k$, see Algebra,
Lemma \ref{algebra-lemma-separably-closed-irreducible-implies-geometric}.
Hence $Z = \Spec(k')$ is geometrically irreducible over $k$.
by Lemma \ref{lemma-geometrically-irreducible-local} above.
Since $X' = Z \times_k X$ the result is a special case
of Lemma \ref{lemma-bijection-irreducible-components}.
\end{proof}

\begin{lemma}
\label{lemma-characterize-geometrically-irreducible}
Let $k$ be a field. Let $X$ be a scheme over $k$.
Assume $X$ is quasi-compact. The following are equivalent:
\begin{enumerate}
\item $X$ is geometrically irreducible over $k$,
\item for every finite separable field extension $k \subset k'$
the scheme $X_{k'}$ is irreducible, and
\item $X_{\overline{k}}$ is irreducible, where $k \subset \overline{k}$
is a separable algebraic closure of $k$.
\end{enumerate}
\end{lemma}

\begin{proof}
Assume $X_{\overline{k}}$ is irreducible, i.e., assume (3).
Let $k \subset k'$ be a field extension.
There exists a field extension $\overline{k} \subset \overline{k}'$
such that $k'$ embeds into $\overline{k}'$ as an extension of $k$.
By Lemma \ref{lemma-separably-closed-field-irreducible-components}
we see that $X_{\overline{k}'}$ is irreducible.
Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude
that $X_{k'}$ is irreducible. Hence (1) holds.

\medskip\noindent
Let $k \subset \overline{k}$ be a separable algebraic closure of $k$.
Assume not (3), i.e., assume $X_{\overline{k}}$ is reducible.
Our goal is to show that also $X_{k'}$ is
reducible for some finite subextension
$k \subset k' \subset \overline{k}$.
Let $X = \bigcup_{i \in I} U_i$ be an affine open covering
with $U_i$ not empty. If for some $i$ the scheme
$U_i$ is reducible, or if for some pair $i \not = j$ the
intersection $U_i \cap U_j$ is empty, then $X$ is reducible
(Properties, Lemma \ref{properties-lemma-characterize-irreducible})
and we are done.
In particular we may assume that
$U_{i, \overline{k}} \cap U_{j, \overline{k}}$ for all $i, j \in I$
is nonempty and we conclude that $U_{i, \overline{k}}$ has
to be reducible for some $i$. According to
Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible}
this means that $U_{i, k'}$ is reducible for some
finite separable field extension $k \subset k'$.
Hence also $X_{k'}$ is reducible. Thus we see that
(2) implies (3).

\medskip\noindent
The implication (1) $\Rightarrow$ (2) is immediate.
This proves the lemma.
\end{proof}

\begin{lemma}
\label{lemma-inverse-image-irreducible}
Let $k \subset K$ be an extension of fields.
Let $X$ be a scheme over $k$.
For every irreducible component $T$ of $X$ the inverse image
$T_K \subset X_K$ is a union of irreducible components of $X_K$.
\end{lemma}

\begin{proof}
Let $T \subset X$ be an irreducible component of $X$.
The morphism $T_K \to T$ is flat, so generalizations lift
along $T_K \to T$. Hence every $\xi \in T_K$
which is a generic point of an irreducible component of $T_K$
maps to the generic point $\eta$ of $T$. If $\xi' \leadsto \xi$ is
a specialization in $X_K$ then $\xi'$ maps to $\eta$ since there
are no points specializing to $\eta$ in $X$. Hence $\xi' \in T_K$
and we conclude that $\xi = \xi'$. In other words $\xi$ is the
generic point of an irreducible component of $X_K$. This
means that the irreducible components of $T_K$ are all irreducible
components of $X_K$.
\end{proof}

\noindent
For a scheme $X$ we denote $\text{IrredComp}(X)$ the set of
irreducible components of $X$.

\begin{lemma}
\label{lemma-image-irreducible}
Let $k \subset K$ be an extension of fields.
Let $X$ be a scheme over $k$.
For every irreducible component $\overline{T} \subset X_K$
the image of $\overline{T}$ in $X$ is an irreducible component in $X$.
This defines a canonical map
$$
\text{IrredComp}(X_K)
\longrightarrow
\text{IrredComp}(X)
$$
which is surjective.
\end{lemma}

\begin{proof}
Consider the diagram
$$
\xymatrix{
X_K \ar[d] & X_{\overline{K}} \ar[d] \ar[l] \\
X & X_{\overline{k}} \ar[l]
}
$$
where $\overline{K}$ is the separable algebraic closure of $K$, and
where $\overline{k}$ is the separable algebraic closure of $k$. By
Lemma \ref{lemma-separably-closed-field-irreducible-components}
the morphism $X_{\overline{K}} \to X_{\overline{k}}$ induces
a bijection between irreducible components. Hence it suffices
to show the lemma for the morphisms
$X_{\overline{k}} \to X$ and $X_{\overline{K}} \to X_K$.
In other words we may assume that $K = \overline{k}$.

\medskip\noindent
The morphism $p : X_{\overline{k}} \to X$ is integral, flat and surjective.
Flatness implies that generalizations lift along $p$, see
Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}.
Hence generic points of irreducible components of $X_{\overline{k}}$
map to generic points of irreducible components of $X$.
Integrality implies that $p$ is universally closed, see
Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}.
Hence we conclude that the image $p(\overline{T})$ of an irreducible component
is a closed irreducible subset which contains a generic point of an
irreducible component of $X$, hence $p(\overline{T})$
is an irreducible component of $X$. This proves the first assertion.
If $T \subset X$ is an irreducible component, then $p^{-1}(T) =T_K$
is a nonempty union of irreducible components, see
Lemma \ref{lemma-inverse-image-irreducible}.
Each of these necessarily maps onto $T$ by the first part.
Hence the map is surjective.
\end{proof}

\begin{lemma}
\label{lemma-galois-action-irreducible-components}
Let $k$ be a field, with separable algebraic closure $\overline{k}$.
Let $X$ be a scheme over $k$.
There is an action
$$
\text{Gal}(\overline{k}/k)^{opp} \times \text{IrredComp}(X_{\overline{k}})
\longrightarrow
\text{IrredComp}(X_{\overline{k}})
$$
with the following properties:
\begin{enumerate}
\item An element $\overline{T} \in \text{IrredComp}(X_{\overline{k}})$
is fixed by the action if and only if there exists an irreducible
component $T \subset X$, which is geometrically irreducible over $k$,
such that $T_{\overline{k}} = \overline{T}$.
\item For any field extension $k \subset k'$ with separable
algebraic closure $\overline{k}'$ the diagram
$$
\xymatrix{
\text{Gal}(\overline{k}'/k') \times \text{IrredComp}(X_{\overline{k}'})
\ar[r] \ar[d] &
\text{IrredComp}(X_{\overline{k}'}) \ar[d] \\
\text{Gal}(\overline{k}/k) \times \text{IrredComp}(X_{\overline{k}})
\ar[r] &
\text{IrredComp}(X_{\overline{k}})
}
$$
is commutative (where the right vertical arrow is a bijection
according to Lemma \ref{lemma-separably-closed-field-irreducible-components}).
\end{enumerate}
\end{lemma}

\begin{proof}
The action (\ref{equation-galois-action-base-change-kbar})
of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$
induces an action on its irreducible components.
Irreducible components are always closed
(Topology, Lemma \ref{topology-lemma-connected-components}).
Hence if $\overline{T}$ is as in (1), then by
Lemma \ref{lemma-closed-fixed-by-Galois} there exists a closed
subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$.
Note that $T$ is geometrically irreducible over $k$, see
Lemma \ref{lemma-characterize-geometrically-irreducible}.
To see that $T$ is an irreducible component of $X$, suppose that
$T \subset T'$, $T \not = T'$ where $T'$ is an irreducible
component of $X$. Let $\overline{\eta}$ be the generic point of
$\overline{T}$. It maps to the generic point $\eta$ of $T$.
Then the generic point $\xi \in T'$ specializes to $\eta$.
As $X_{\overline{k}} \to X$ is flat there exists a point
$\overline{\xi} \in X_{\overline{k}}$ which maps to $\xi$ and
specializes to $\overline{\eta}$. It follows that
the closure of the singleton $\{\overline{\xi}\}$ is an
irreducible closed subset of $X_{\overline{\xi}}$ which
strictly contains $\overline{T}$. This is the desired contradiction.

\medskip\noindent
We omit the proof of the functoriality in (2).
\end{proof}

\begin{lemma}
\label{lemma-orbit-irreducible-components}
Let $k$ be a field, with separable algebraic closure $\overline{k}$.
Let $X$ be a scheme over $k$.
The fibres of the map
$$
\text{IrredComp}(X_{\overline{k}})
\longrightarrow
\text{IrredComp}(X)
$$
of
Lemma \ref{lemma-image-irreducible}
are exactly the orbits of $\text{Gal}(\overline{k}/k)$ under the action of
Lemma \ref{lemma-galois-action-irreducible-components}.
\end{lemma}

\begin{proof}
Let $T \subset X$ be an irreducible component of $X$.
Let $\eta \in T$ be its generic point. By
Lemmas \ref{lemma-inverse-image-irreducible} and
\ref{lemma-image-irreducible}
the generic points of irreducible components of $\overline{T}$
which map into $T$ map to $\eta$. By
Algebra, Lemma \ref{algebra-lemma-Galois-orbit}
the Galois group acts transitively on
all of the points of $X_{\overline{k}}$ mapping to $\eta$.
Hence the lemma follows.
\end{proof}

\begin{lemma}
\label{lemma-galois-action-irreducible-components-locally-finite-type}
Let $k$ be a field.
Assume $X \to \Spec(k)$ locally of finite type.
In this case
\begin{enumerate}
\item the action
$$
\text{Gal}(\overline{k}/k)^{opp} \times \text{IrredComp}(X_{\overline{k}})
\longrightarrow
\text{IrredComp}(X_{\overline{k}})
$$
is continuous if we give $\text{IrredComp}(X_{\overline{k}})$ the discrete
topology,
\item every irreducible component of $X_{\overline{k}}$
can be defined over a finite extension of $k$, and
\item given any irreducible component $T \subset X$ the scheme
$T_{\overline{k}}$ is a finite union of irreducible components of
$X_{\overline{k}}$ which are all in the same
$\text{Gal}(\overline{k}/k)$-orbit.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $\overline{T}$ be an irreducible component of $X_{\overline{k}}$.
We may choose an affine open $U \subset X$ such that
$\overline{T} \cap U_{\overline{k}}$ is not empty.
Write $U = \Spec(A)$, so $A$ is a finite type $k$-algebra, see
Morphisms, Lemma \ref{morphisms-lemma-locally-finite-type-characterize}.
Hence $A_{\overline{k}}$ is a finite type $\overline{k}$-algebra,
and in particular Noetherian. Let $\mathfrak p = (f_1, \ldots, f_n)$
be the prime ideal corresponding to $\overline{T} \cap U_{\overline{k}}$.
Since $A_{\overline{k}} = A \otimes_k \overline{k}$
we see that there exists a finite subextension
$k \subset k' \subset \overline{k}$ such that each $f_i \in A_{k'}$.
It is clear that $\text{Gal}(\overline{k}/k')$
fixes $\overline{T}$, which proves (1).

\medskip\noindent
Part (2) follows by applying
Lemma \ref{lemma-galois-action-irreducible-components} (1)
to the situation over $k'$ which implies the irreducible component
$\overline{T}$ is of the form $T'_{\overline{k}}$ for some irreducible
$T' \subset X_{k'}$.

\medskip\noindent
To prove (3), let $T \subset X$ be an irreducible component.
Choose an irreducible component $\overline{T} \subset X_{\overline{k}}$
which maps to $T$, see
Lemma \ref{lemma-image-irreducible}.
By the above the orbit of $\overline{T}$ is finite, say it is
$\overline{T}_1, \ldots, \overline{T}_n$. Then
$\overline{T}_1 \cup \ldots \cup \overline{T}_n$
is a $\text{Gal}(\overline{k}/k)$-invariant closed subset of $X_{\overline{k}}$
hence of the form $W_{\overline{k}}$ for some $W \subset X$ closed by
Lemma \ref{lemma-closed-fixed-by-Galois}.
Clearly $W = T$ and we win.
\end{proof}

\begin{lemma}
\label{lemma-finite-extension-geometrically-irreducible-components}
Let $k$ be a field.
Let $X \to \Spec(k)$ be locally of finite type.
Assume $X$ has finitely many irreducible components.
Then there exists a finite separable extension $k \subset k'$
such that every irreducible component of $X_{k'}$
is geometrically irreducible over $k'$.
\end{lemma}

\begin{proof}
Let $\overline{k}$ be a separable algebraic closure of $k$.
The assumption that $X$ has finitely many irreducible components
combined with
Lemma \ref{lemma-galois-action-irreducible-components-locally-finite-type} (3)
shows that $X_{\overline{k}}$ has finitely many irreducible components
$\overline{T}_1, \ldots, \overline{T}_n$. By
Lemma \ref{lemma-galois-action-irreducible-components-locally-finite-type} (2)
there exists a finite extension $k \subset k' \subset \overline{k}$ and
irreducible components $T_i \subset X_{k'}$ such that
$\overline{T}_i = T_{i, \overline{k}}$ and we win.
\end{proof}

\begin{lemma}
\label{lemma-irreducible-components-geometrically-irreducible}
Let $X$ be a scheme over the field $k$.
Assume $X$ has finitely many irreducible components which are
all geometrically irreducible.
Then $X$ has finitely many connected components each of which is
geometrically connected.
\end{lemma}

\begin{proof}
This is clear because a connected component is a union of irreducible
components. Details omitted.
\end{proof}







\section{Geometrically integral schemes}
\label{section-geometrically-integral}

\noindent
If $X$ is an irreducible scheme over a field, then it can happen that $X$
becomes reducible after extending the ground field. This does not happen
for geometrically irreducible schemes.

\begin{definition}
\label{definition-geometrically-integral}
Let $X$ be a scheme over the field $k$.
\begin{enumerate}
\item Let $x \in X$. We say $X$ is
{\it geometrically pointwise integral at $x$} if for every
field extension $k \subset k'$ and every $x' \in X_{k'}$ lying over $x$
the local ring $\mathcal{O}_{X_{k'}, x'}$ is integral.
\item We say $X$ is {\it geometrically pointwise integral} if $X$
is geometrically pointwise integral at every point.
\item We say $X$ is {\it geometrically integral} over $k$ if the scheme
$X_{k'}$ is integral for every field extension $k'$ of $k$.
\end{enumerate}
\end{definition}

\noindent
The distinction between notions (2) and (3) is necessary.
For example if $k = \mathbf{R}$ and $X = \Spec(\mathbf{C}[x])$,
then $X$ is geometrically pointwise integral over $\mathbf{R}$ but
of course not geometrically integral.

\begin{lemma}
\label{lemma-geometrically-integral}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Then $X$ is geometrically integral over $k$ if and only if
$X$ is both geometrically reduced and geometrically irreducible
over $k$.
\end{lemma}

\begin{proof}
See Properties, Lemma \ref{properties-lemma-characterize-integral}.
\end{proof}





\section{Geometrically normal schemes}
\label{section-geometrically-normal}

\noindent
In Properties, Definition \ref{properties-definition-normal}
we have defined the notion of a normal scheme.
This notion is defined even for non-Noetherian
schemes. Hence, contrary to our discussion of
``geometrically regular'' schemes we consider all
field extensions of the ground field.

\begin{definition}
\label{definition-geometrically-normal}
Let $X$ be a scheme over the field $k$.
\begin{enumerate}
\item Let $x \in X$. We say $X$ is
{\it geometrically normal at $x$} if for every
field extension $k \subset k'$ and every $x' \in X_{k'}$ lying over $x$
the local ring $\mathcal{O}_{X_{k'}, x'}$ is normal.
\item We say $X$ is {\it geometrically normal} over $k$ if $X$
is geometrically normal at every $x \in X$.
\end{enumerate}
\end{definition}

\begin{lemma}
\label{lemma-geometrically-normal-at-point}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $x \in X$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically normal at $x$,
\item for every finite purely inseparable field extension $k'$ of $k$
and $x' \in X_{k'}$ lying over over $x$ the local ring
$\mathcal{O}_{X_{k'}, x'}$ is normal, and
\item the ring $\mathcal{O}_{X, x}$ is geometrically
normal over $k$ (see
Algebra, Definition \ref{algebra-definition-geometrically-normal}).
\end{enumerate}
\end{lemma}

\begin{proof}
It is clear that (1) implies (2). Assume (2). Let $k \subset k'$ be a finite
purely inseparable field extension (for example $k = k'$). Consider the ring
$\mathcal{O}_{X, x} \otimes_k k'$.
By Algebra, Lemma \ref{algebra-lemma-p-ring-map}
its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$.
Hence it is a local ring also
(Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}).
Therefore there is a unique point $x' \in X_{k'}$ lying over $x$
and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$.
By assumption this is a normal ring. Hence we deduce (3) by
Algebra, Lemma
\ref{algebra-lemma-geometrically-normal}.

\medskip\noindent
Assume (3). Let $k \subset k'$ be a field extension. Since
$\Spec(k') \to \Spec(k)$ is surjective, also
$X_{k'} \to X$ is surjective
(Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}).
Let $x' \in X_{k'}$ be any point lying over $x$.
The local ring $\mathcal{O}_{X_{k'}, x'}$
is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$.
Hence it is normal by assumption and (1) is proved.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-normal}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically normal,
\item $X_{k'}$ is a normal scheme for every field extension $k \subset k'$,
\item $X_{k'}$ is a normal scheme for every finitely generated field
extension $k \subset k'$,
\item $X_{k'}$ is a normal scheme for every finite purely inseparable
field extension $k \subset k'$, and
\item for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is geometrically normal (see
Algebra, Definition \ref{algebra-definition-geometrically-normal}).
\end{enumerate}
\end{lemma}

\begin{proof}
Assume (1). Then for every field extension $k \subset k'$ and
every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$
is normal. By definition this means that $X_{k'}$ is normal.
Hence (2).

\medskip\noindent
It is clear that (2) implies (3) implies (4).

\medskip\noindent
Assume (4) and let $U \subset X$ be an affine open subscheme.
Then $U_{k'}$ is a normal scheme for any finite purely inseparable
extension $k \subset k'$ (including $k = k'$). This means that
$k' \otimes_k \mathcal{O}(U)$ is a normal ring for all
finite purely inseparable extensions $k \subset k'$. Hence
$\mathcal{O}(U)$ is a geometrically normal $k$-algebra by definition.

\medskip\noindent
Assume (5). For any field extension $k \subset k'$ the base
change $X_{k'}$ is gotten by gluing the spectra of the
rings $\mathcal{O}_X(U) \otimes_k k'$ where $U$ is affine open
in $X$ (see Schemes, Section \ref{schemes-section-fibre-products}).
Hence $X_{k'}$ is normal. So (1) holds.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-normal-upstairs}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $k'/k$ be a field extension.
Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically normal at $x$,
\item $X_{k'}$ is geometrically normal at $x'$.
\end{enumerate}
In particular, $X$ is geometrically normal over $k$ if and only if
$X_{k'}$ is geometrically normal over $k'$.
\end{lemma}

\begin{proof}
It is clear that (1) implies (2). Assume (2).
Let $k \subset k''$ be a finite purely inseparable field extension
and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is
unique). We can find a common field extension $k \subset k'''$
(i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$) and a point
$x''' \in X_{k'''}$ lying over both $x'$ and $x''$.
Consider the map of local rings
$$
\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.
$$
This is a flat local ring homomorphism and hence faithfully flat.
By (2) we see that the local ring on the right is normal.
Thus by Algebra, Lemma \ref{algebra-lemma-descent-normal}
we conclude that $\mathcal{O}_{X_{k''}, x''}$ is normal.
By Lemma \ref{lemma-geometrically-normal-at-point} we see that $X$
is geometrically normal at $x$.
\end{proof}

\begin{lemma}
\label{lemma-fibre-product-normal}
Let $k$ be a field. Let $X$ be a geometrically normal scheme over $k$
and let $Y$ be a normal scheme over $k$. Then $X \times_k Y$ is a normal
scheme.
\end{lemma}

\begin{proof}
This reduces to
Algebra, Lemma \ref{algebra-lemma-geometrically-normal-tensor-normal}
by
Lemma \ref{lemma-geometrically-normal}.
\end{proof}










\section{Change of fields and locally Noetherian schemes}
\label{section-locally-Noetherian}

\noindent
Let $X$ a locally Noetherian scheme over a field $k$.
It is not always that case that $X_{k'}$ is locally Noetherian too.
For example if $X = \Spec(\overline{\mathbf{Q}})$ and
$k = \mathbf{Q}$, then $X_{\overline{\mathbf{Q}}}$ is the spectrum
of $\overline{\mathbf{Q}} \otimes_{\mathbf{Q}} \overline{\mathbf{Q}}$
which is not Noetherian. (Hint: It has too many idempotents).
But if we only base change using finitely generated field extensions
then the Noetherian property is preserved. (Or if $X$ is locally of finite
type over $k$, since this property is preserved under base change.)

\begin{lemma}
\label{lemma-locally-Noetherian-base-change}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $k \subset k'$ be a finitely generated field extension.
Then $X$ is locally Noetherian if and only if $X_{k'}$ is locally
Noetherian.
\end{lemma}

\begin{proof}
Using Properties, Lemma \ref{properties-lemma-locally-Noetherian}
we reduce to the case where $X$ is
affine, say $X = \Spec(A)$. In this case we have to prove that
$A$ is Noetherian if and only if $A_{k'}$ is Noetherian.
Since $A \to A_{k'} = k' \otimes_k A$ is faithfully flat, we see
that if $A_{k'}$ is Noetherian, then so is $A$, by
Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}.
Conversely, if $A$ is Noetherian then $A_{k'}$ is Noetherian by
Algebra, Lemma \ref{algebra-lemma-Noetherian-field-extension}.
\end{proof}







\section{Geometrically regular schemes}
\label{section-geometrically-regular}

\noindent
A geometrically regular scheme over a field $k$ is a locally Noetherian
scheme over $k$ which remains regular upon suitable changes of base field.
A finite type scheme over $k$ is geometrically regular if and only
if it is smooth over $k$ (see Lemma \ref{lemma-geometrically-regular-smooth}).
The notion of geometric regularity is most interesting in situations
where smoothness cannot be used such as formal fibres (insert future
reference here).

\medskip\noindent
In the following definition we restrict ourselves to locally Noetherian
schemes, since the property of being a regular local ring is only
defined for Noetherian local rings. By Lemma \ref{lemma-geometrically-normal}
above, if we restrict ourselves to finitely generated field extensions then
this property is preserved under change of base field. This comment will be
used without further reference in this section. In particular the following
definition makes sense.

\begin{definition}
\label{definition-geometrically-regular}
Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$.
\begin{enumerate}
\item Let $x \in X$. We say $X$ is {\it geometrically regular at $x$}
over $k$ if for every finitely generated field extension $k \subset k'$
and any $x' \in X_{k'}$ lying over $x$ the local ring
$\mathcal{O}_{X_{k'}, x'}$ is regular.
\item We say $X$ is {\it geometrically regular over $k$} if
$X$ is geometrically regular at all of its points.
\end{enumerate}
\end{definition}

\noindent
A similar definition works to define geometrically
Cohen-Macaulay, $(R_k)$, and $(S_k)$ schemes over a field.
We will add a section for these separately as needed.

\begin{lemma}
\label{lemma-geometrically-regular-at-point}
Let $k$ be a field.
Let $X$ be a locally Noetherian scheme over $k$.
Let $x \in X$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically regular at $x$,
\item for every finite purely inseparable field extension $k'$ of $k$
and $x' \in X_{k'}$ lying over over $x$ the local ring
$\mathcal{O}_{X_{k'}, x'}$ is regular, and
\item the ring $\mathcal{O}_{X, x}$ is geometrically
regular over $k$ (see
Algebra, Definition \ref{algebra-definition-geometrically-regular}).
\end{enumerate}
\end{lemma}

\begin{proof}
It is clear that (1) implies (2).
Assume (2). This in particular implies that $\mathcal{O}_{X, x}$
is a regular local ring. Let $k \subset k'$ be a finite purely inseparable
field extension. Consider the ring $\mathcal{O}_{X, x} \otimes_k k'$.
By Algebra, Lemma \ref{algebra-lemma-p-ring-map}
its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$.
Hence it is a local ring also
(Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}).
Therefore there is a unique point $x' \in X_{k'}$ lying over $x$
and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$.
By assumption this is a regular ring. Hence we deduce (3)
from the definition of a geometrically regular ring.

\medskip\noindent
Assume (3). Let $k \subset k'$ be a field extension. Since
$\Spec(k') \to \Spec(k)$ is surjective, also
$X_{k'} \to X$ is surjective
(Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}).
Let $x' \in X_{k'}$ be any point lying over $x$.
The local ring $\mathcal{O}_{X_{k'}, x'}$
is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$.
Hence it is regular by assumption and (1) is proved.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-regular}
Let $k$ be a field.
Let $X$ be a locally Noetherian scheme over $k$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically regular,
\item $X_{k'}$ is a regular scheme for every finitely generated field
extension $k \subset k'$,
\item $X_{k'}$ is a regular scheme for every finite purely inseparable
field extension $k \subset k'$,
\item for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is geometrically regular (see
Algebra, Definition \ref{algebra-definition-geometrically-regular}), and
\item there exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is geometrically regular over $k$.
\end{enumerate}
\end{lemma}

\begin{proof}
Assume (1). Then for every finitely generated field extension
$k \subset k'$ and
every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$
is regular. By Properties, Lemma \ref{properties-lemma-characterize-regular}
this means that $X_{k'}$ is regular. Hence (2).

\medskip\noindent
It is clear that (2) implies (3).

\medskip\noindent
Assume (3) and let $U \subset X$ be an affine open subscheme.
Then $U_{k'}$ is a regular scheme for any finite purely inseparable
extension $k \subset k'$ (including $k = k'$). This means that
$k' \otimes_k \mathcal{O}(U)$ is a regular ring for all
finite purely inseparable extensions $k \subset k'$. Hence
$\mathcal{O}(U)$ is a geometrically regular $k$-algebra
and we see that (4) holds.

\medskip\noindent
It is clear that (4) implies (5). Let $X = \bigcup U_i$ be an affine
open covering as in (5). For any field extension $k \subset k'$ the base
change $X_{k'}$ is gotten by gluing the spectra of the
rings $\mathcal{O}_X(U_i) \otimes_k k'$ (see
Schemes, Section \ref{schemes-section-fibre-products}).
Hence $X_{k'}$ is regular. So (1) holds.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-regular-upstairs}
Let $k$ be a field.
Let $X$ be a scheme over $k$.
Let $k'/k$ be a finitely generated field extension.
Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$.
The following are equivalent
\begin{enumerate}
\item $X$ is geometrically regular at $x$,
\item $X_{k'}$ is geometrically regular at $x'$.
\end{enumerate}
In particular, $X$ is geometrically regular over $k$ if and only if
$X_{k'}$ is geometrically regular over $k'$.
\end{lemma}

\begin{proof}
It is clear that (1) implies (2). Assume (2).
Let $k \subset k''$ be a finite purely inseparable field extension
and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is
unique). We can find a common, finitely generated, field extension
$k \subset k'''$ (i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$)
and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$.
Consider the map of local rings
$$
\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.
$$
This is a flat local ring homomorphism of Noetherian local rings
and hence faithfully flat.
By (2) we see that the local ring on the right is regular.
Thus by Algebra, Lemma \ref{algebra-lemma-flat-under-regular}
we conclude that $\mathcal{O}_{X_{k''}, x''}$ is regular.
By Lemma \ref{lemma-geometrically-regular-at-point} we see that $X$
is geometrically regular at $x$.
\end{proof}

\noindent
The following lemma is a geometric variant of
Algebra, Lemma \ref{algebra-lemma-geometrically-regular-descent}.

\begin{lemma}
\label{lemma-flat-under-geometrically-regular}
Let $k$ be a field.
Let $f : X \to Y$ be a morphism of locally Noetherian schemes over $k$.
Let $x \in X$ be a point and set $y = f(x)$.
If $X$ is geometrically regular at $x$ and
$f$ is flat at $x$ then $Y$ is geometrically regular at $y$.
In particular, if $X$ is geometrically regular over $k$ and
$f$ is flat and surjective, then $Y$ is geometrically regular over $k$.
\end{lemma}

\begin{proof}
Let $k'$ be finite purely inseparable extension of $k$.
Let $f' : X_{k'} \to Y_{k'}$ be the base change of $f$.
Let $x' \in X_{k'}$ be the unique point lying over $x$.
If we show that $Y_{k'}$ is regular at $y' = f'(x')$, then
$Y$ is geometrically regular over $k$ at $y'$, see
Lemma \ref{lemma-geometrically-regular}.
By
Morphisms, Lemma \ref{morphisms-lemma-base-change-module-flat}
the morphism $X_{k'} \to Y_{k'}$ is flat at $x'$.
Hence the ring map
$$
\mathcal{O}_{Y_{k'}, y'}
\longrightarrow
\mathcal{O}_{X_{k'}, x'}
$$
is a flat local homomorphism of local Noetherian rings with
right hand side regular by assumption. Hence the left hand side
is a regular local ring by
Algebra, Lemma \ref{algebra-lemma-flat-under-regular}.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-regular-smooth}
Let $k$ be a field.
Let $X$ be a scheme of finite type over $k$.
Let $x \in X$.
Then $X$ is geometrically regular at $x$ if and only if $X \to \Spec(k)$
is smooth at $x$ (Morphisms, Definition \ref{morphisms-definition-smooth}).
\end{lemma}

\begin{proof}
The question is local around $x$,
hence we may assume that $X = \Spec(A)$
for some finite type $k$-algebra.
Let $x$ correspond to the prime $\mathfrak p$.

\medskip\noindent
If $A$ is smooth over $k$ at $\mathfrak p$, then we may localize $A$
and assume that $A$ is smooth over $k$. In this case $k' \otimes_k A$
is smooth over $k'$ for all extension fields $k'/k$, and each of
these Noetherian rings is regular by
Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}.

\medskip\noindent
Assume $X$ is geometrically regular at $x$.
Consider the residue field $K := \kappa(x) = \kappa(\mathfrak p)$ of $x$.
It is a finitely generated extension of $k$.
By Algebra, Lemma \ref{algebra-lemma-make-separable}
there exists a finite purely inseparable
extension $k \subset k'$ such that the compositum
$k'K$ is a separable field extension of $k'$.
Let $\mathfrak p' \subset A' = k' \otimes_k A$ be a prime ideal
lying over $\mathfrak p$. It is the unique prime lying over $\mathfrak p$, see
Algebra, Lemma \ref{algebra-lemma-p-ring-map}.
Hence the residue field $K' := \kappa(\mathfrak p')$
is the compositum $k'K$. By assumption the local ring
$(A')_{\mathfrak p'}$ is regular. Hence by
Algebra, Lemma \ref{algebra-lemma-separable-smooth}
we see that $k' \to A'$ is smooth at $\mathfrak p'$.
This in turn implies that $k \to A$ is smooth at $\mathfrak p$ by
Algebra, Lemma \ref{algebra-lemma-smooth-field-change-local}.
The lemma is proved.
\end{proof}


\begin{example}
\label{example-geometrically-reduced-not-normal}
Let $k =\mathbf{F}_p(t)$. It is quite easy to give an example of a regular
variety $V$ over $k$ which is not geometrically reduced. For example we
can take $\Spec(k[x]/(x^p - t))$. In fact, there exists an
example of a regular variety $V$ which is geometrically reduced, but
not even geometrically normal. Namely, take for $p > 2$ the scheme
$V = \Spec(k[x, y]/(y^2 - x^p + t))$. This is a variety as the
polynomial $y^2 - x^p + t \in k[x, y]$ is irreducible.
The morphism $V \to \Spec(k)$ is smooth at all points
except at the point $v_0 \in V$ corresponding to the maximal ideal
$(y, x^p - t)$ (because $2y$ is invertible). In particular we see that
$V$ is (geometrically) regular at all points, except possibly $v_0$.
The local ring
$$
\mathcal{O}_{V, v_0} = \left(k[x, y]/(y^2 - x^p + t)\right)_{(y, x^p - t)}
$$
is a domain of dimension $1$. Its maximal ideal is generated by $1$ element,
namely $y$. Hence it is a discrete valuation ring and regular.
Let $k' = k[t^{1/p}]$. Denote $t' = t^{1/p} \in k'$,
$V' = V_{k'}$, $v'_0 \in V'$ the unique point lying over $v_0$.
Over $k'$ we can write $x^p - t = (x - t')^p$, but the polynomial
$y^2 - (x - t')^p$ is still irreducible and $V'$ is still a variety.
But the element
$$
\frac{y}{x - t'} \in f.f.(\mathcal{O}_{V', v'_0})
$$
is integral over $\mathcal{O}_{V', v'_0}$ (just compute its square)
and not contained in it, so $V'$ is not normal at $v'_0$. This concludes
the example.
\end{example}







\section{Change of fields and the Cohen-Macaulay property}
\label{section-CM}

\noindent
The following lemma says that it does not make sense to define
geometrically Cohen-Macaulay schemes, since these would be the
same as Cohen-Macaulay schemes.

\begin{lemma}
\label{lemma-CM-base-change}
Let $X$ be a locally Noetherian scheme over the field $k$.
Let $k \subset k'$ be a finitely generated field extension.
Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying
over $x$. Then we have
$$
\mathcal{O}_{X, x}\text{ is Cohen-Macaulay}
\Leftrightarrow
\mathcal{O}_{X_{k'}, x'}\text{ is Cohen-Macaulay}
$$
If $X$ is locally of finite type over $k$, the same holds for any
field extension $k \subset k'$.
\end{lemma}

\begin{proof}
The first case of the lemma follows from
Algebra, Lemma \ref{algebra-lemma-CM-geometrically-CM}.
The second case of the lemma is equivalent to
Algebra, Lemma \ref{algebra-lemma-extend-field-CM-locus}.
\end{proof}







\section{Change of fields and the Jacobson property}
\label{section-overfield}

\noindent
A scheme locally of finite type over a field has plenty of closed
points, namely it is Jacobson. Moreover, the residue fields are
finite extensions of the ground field.

\begin{lemma}
\label{lemma-locally-finite-type-Jacobson}
Let $X$ be a scheme which is locally of finite type over $k$.
Then
\begin{enumerate}
\item for any closed point $x \in X$ the extension $k \subset \kappa(x)$
is algebraic, and
\item $X$ is a Jacobson scheme
(Properties, Definition \ref{properties-definition-jacobson}).
\end{enumerate}
\end{lemma}

\begin{proof}
A scheme is Jacobson if and only if it has an affine open covering
by Jacobson schemes, see
Properties, Lemma \ref{properties-lemma-locally-jacobson}.
The property on residue fields at closed points is also local on $X$.
Hence we may assume that $X$ is affine. In this case the result
is a consequence of the Hilbert Nullstellensatz, see
Algebra, Theorem \ref{algebra-theorem-nullstellensatz}.
It also follows from a combination of
Morphisms, Lemmas \ref{morphisms-lemma-jacobson-finite-type-points},
\ref{morphisms-lemma-Jacobson-universally-Jacobson}, and
\ref{morphisms-lemma-ubiquity-Jacobson-schemes}.
\end{proof}

\noindent
It turns out that if $X$ is not locally of finite type, then we can
achieve the same result after making a suitably large base field extension.

\begin{lemma}
\label{lemma-make-Jacobson}
Let $X$ be a scheme over a field $k$.
For any field extension $k \subset K$ whose cardinality is large enough
we have
\begin{enumerate}
\item for any closed point $x \in X_K$ the extension $K \subset \kappa(x)$
is algebraic, and
\item $X_K$ is a Jacobson scheme
(Properties, Definition \ref{properties-definition-jacobson}).
\end{enumerate}
\end{lemma}

\begin{proof}
Choose an affine open covering $X = \bigcup U_i$.
By
Algebra, Lemma \ref{algebra-lemma-base-change-Jacobson}
and
Properties, Lemma \ref{properties-lemma-affine-jacobson}
there exist cardinals $\kappa_i$ such that $U_{i, K}$ has
the desired properties over $K$ if $\#(K) \geq \kappa_i$.
Set $\kappa = \max\{\kappa_i\}$. Then if the cardinality of
$K$ is larger than $\kappa$ we see that each $U_{i, K}$ satisfies
the conclusions of the lemma. Hence $X_K$ is Jacobson by
Properties, Lemma \ref{properties-lemma-locally-jacobson}.
The statement on residue fields at closed points of $X_K$
follows from the corresponding
statements for residue fields of closed points of the $U_{i, K}$.
\end{proof}





\section{Algebraic schemes}
\label{section-algebraic-schemes}

\noindent
The following definition is taken from
\cite[I Definition 6.4.1]{EGA}.

\begin{definition}
\label{definition-algebraic-scheme}
Let $k$ be a field. An {\it algebraic $k$-scheme} is a scheme $X$ over $k$
such that the structure morphism $X \to \Spec(k)$ is of
finite type. A {\it locally algebraic $k$-scheme} is a scheme $X$ over $k$
such that the structure morphism $X \to \Spec(k)$ is
locally of finite type.
\end{definition}

\noindent
Note that every (locally) algebraic $k$-scheme is (locally) Noetherian, see
Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}.
The category of algebraic $k$-schemes has all products and fibre products
(unlike the category of varieties over $k$). Similarly for the category
of locally algebraic $k$-schemes.

\begin{lemma}
\label{lemma-algebraic-scheme-dim-0}
Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme of
dimension $0$. Then $X$ is a disjoint union of spectra of local Artinian
$k$-algebras $A$ with $\dim_k(A) < \infty$. If $X$ is an algebraic $k$-scheme
of dimension $0$, then in addition $X$ is affine and the morphism
$X \to \Spec(k)$ is finite.
\end{lemma}

\begin{proof}
Let $X$ be a locally algebraic $k$-scheme of dimension $0$.
Let $U = \Spec(A) \subset X$ be an affine open subscheme.
Since $\dim(X) = 0$ we see that $\dim(A) = 0$.
By Noether normalization, see
Algebra, Lemma \ref{algebra-lemma-Noether-normalization}
we see that there exists a finite injection $k \to A$, i.e.,
$\dim_k(A) < \infty$. Hence $A$ is Artinian, see
Algebra, Lemma \ref{algebra-lemma-finite-dimensional-algebra}.
This implies that $A = A_1 \times \ldots \times A_r$ is a product
of finitely many Artinian local rings, see
Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}.
Of course $\dim_k(A_i) < \infty$ for each $i$ as the sum of
these dimensions equals $\dim_k(A)$.

\medskip\noindent
The arguments above show that $X$ has an open covering whose members are
finite discrete topological spaces. Hence $X$ is a discrete topological space.
It follows that $X$ is isomorphic to the disjoint union of its connected
components each of which is a singleton. Since a singleton scheme is affine
we conclude (by the results of the paragraph above) that each of these
singletons is the spectrum of a local Artinian $k$-algebra $A$ with
$\dim_k(A) < \infty$.

\medskip\noindent
Finally, if $X$ is an algebraic $k$-scheme of dimension $0$, then
$X$ is quasi-compact hence is a finite disjoint union
$X = \Spec(A_1) \amalg \ldots \amalg \Spec(A_r)$
hence affine (see
Schemes, Lemma \ref{schemes-lemma-disjoint-union-affines})
and we have seen the finiteness of $X \to \Spec(k)$ in the
first paragraph of the proof.
\end{proof}






\section{Closures of products}
\label{section-closure-of-products}

\noindent
Some results on the relation between closure and products.

\begin{lemma}
\label{lemma-closure-of-product}
Let $k$ be a field.
Let $X$, $Y$ be schemes over $k$, and let
$A \subset X$, $B \subset Y$ be subsets.
Set
$$
AB =
\{z \in X \times_k Y \mid
\text{pr}_X(\gamma) \in A, \ \text{pr}_Y(\gamma) \in B\}
\subset X \times_k Y
$$
Then set theoretically we have
$$
\overline{A} \times_k \overline{B} = \overline{AB}
$$
\end{lemma}

\begin{proof}
The inclusion $\overline{AB} \subset \overline{A} \times_k \overline{B}$
is immediate.
We may replace $X$ and $Y$ by the reduced closed subschemes $\overline{A}$
and $\overline{B}$.
Let $W \subset X \times_k Y$ be a nonempty open subset. By
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}
the subset $U = \text{pr}_X(W)$ is nonempty open in $X$.
Hence $A \cap U$ is nonempty. Pick $a \in A \cap U$.
Denote $Y_{\kappa(a)} = \{a\} \times_k Y$
the fibre of $\text{pr}_X : X \times_k Y \to X$ over $a$. By
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}
again the morphism $Y_a \to Y$ is open as
$\Spec(\kappa(a)) \to \Spec(k)$ is universally open.
Hence the nonempty open
subset $W_a = W \times_{X \times_k Y} Y_a$
maps to a nonempty open subset of $Y$.
We conclude there exists a $b \in B$ in the image.
Hence $AB \cap W \not = \emptyset$ as desired.
\end{proof}

\begin{lemma}
\label{lemma-closure-image-product-map}
Let $k$ be a field.
Let $f : A \to X$, $g : B \to Y$ be morphisms of schemes over $k$.
Then set theoretically we have
$$
\overline{f(A)} \times_k \overline{g(B)} =
\overline{(f \times g)(A \times_k B)}
$$
\end{lemma}

\begin{proof}
This follows from
Lemma \ref{lemma-closure-of-product}
as the image of $f \times g$ is $f(A)g(B)$
in the notation of that lemma.
\end{proof}

\begin{lemma}
\label{lemma-scheme-theoretic-image-product-map}
Let $k$ be a field.
Let $f : A \to X$, $g : B \to Y$ be quasi-compact morphisms of schemes
over $k$. Let $Z \subset X$ be the scheme theoretic image of $f$, see
Morphisms, Definition \ref{morphisms-definition-scheme-theoretic-image}.
Similarly, let $Z' \subset Y$ be the scheme theoretic image of $g$.
Then $Z \times_k Z'$ is the scheme theoretic image of $f \times g$.
\end{lemma}

\begin{proof}
Recall that $Z$ is the smallest closed subscheme of $X$ through which
$f$ factors. Similarly for $Z'$. Let $W \subset X \times_k Y$ be the
scheme theoretic image of $f \times g$. As $f \times g$ factors through
$Z \times_k Z'$ we see that $W \subset Z \times_k Z'$.

\medskip\noindent
To prove the other inclusion let $U \subset X$ and $V \subset Y$ be
affine opens. By
Morphisms, Lemma \ref{morphisms-lemma-quasi-compact-scheme-theoretic-image}
the scheme $Z \cap U$ is the scheme theoretic image of
$f|_{f^{-1}(U)} : f^{-1}(U) \to U$, and similarly for
$Z' \cap V$ and $W \cap U \times_k V$. Hence we may assume
$X$ and $Y$ affine. As $f$ and $g$ are quasi-compact this implies
that $A = \bigcup U_i$ is a finite union of affines and
$B = \bigcup V_j$ is a finite union of affines.
Then we may replace $A$ by $\coprod U_i$ and $B$ by
$\coprod V_j$, i.e., we may assume that $A$ and $B$ are affine as well.
In this case $Z$ is cut out by
$\text{Ker}(\Gamma(X, \mathcal{O}_X) \to \Gamma(A, \mathcal{O}_A))$
and similarly for $Z'$ and $W$. Hence the result follows from
the equality
$$
\Gamma(A \times_k B, \mathcal{O}_{A \times_k B})
=
\Gamma(A, \mathcal{O}_A) \otimes_k \Gamma(B, \mathcal{O}_B)
$$
which holds as $A$ and $B$ are affine. Details omitted.
\end{proof}





\section{Schemes smooth over fields}
\label{section-smooth}

\noindent
Here are two lemmas characterizing smooth schemes over fields.

\begin{lemma}
\label{lemma-char-zero-differentials-free-smooth}
Let $k$ be a field. Let $X$ be a scheme over $k$.
Assume
\begin{enumerate}
\item $X$ is locally of finite type over $k$,
\item $\Omega_{X/k}$ is locally free, and
\item $k$ has characteristic zero.
\end{enumerate}
Then the structure morphism $X \to \Spec(k)$ is smooth.
\end{lemma}

\begin{proof}
This follows from
Algebra, Lemma \ref{algebra-lemma-characteristic-zero-local-smooth}.
\end{proof}

\noindent
In positive characteristic there exist nonreduced schemes
of finite type whose sheaf of differentials is free, for example
$\Spec(\mathbf{F}_p[t]/(t^p))$ over $\Spec(\mathbf{F}_p)$.
If the ground field $k$ is nonperfect of characteristic $p$,
there exist reduced schemes $X/k$ with free $\Omega_{X/k}$ which are
nonsmooth, for example $\Spec(k[t]/(t^p-a)$ where $a \in k$
is not a $p$th power.

\begin{lemma}
\label{lemma-char-p-differentials-free-smooth}
Let $k$ be a field. Let $X$ be a scheme over $k$.
Assume
\begin{enumerate}
\item $X$ is locally of finite type over $k$,
\item $\Omega_{X/k}$ is locally free,
\item $X$ is reduced, and
\item $k$ is perfect.
\end{enumerate}
Then the structure morphism $X \to \Spec(k)$ is smooth.
\end{lemma}

\begin{proof}
Let $x \in X$ be a point. As $X$ is locally Noetherian (see
Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian})
there are finitely many irreducible components
$X_1, \ldots, X_n$ passing through $x$ (see
Properties, Lemma \ref{properties-lemma-Noetherian-topology} and
Topology, Lemma \ref{topology-lemma-Noetherian}).
Let $\eta_i \in X_i$ be the generic point. As $X$ is reduced we have
$\mathcal{O}_{X, \eta_i} = \kappa(\eta_i)$, see
Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}.
Moreover, $\kappa(\eta_i)$ is a finitely generated field extension
of the perfect field $k$ hence separably generated over $k$ (see
Algebra, Section \ref{algebra-section-separability}).
It follows that $\Omega_{X/k, \eta_i} = \Omega_{\kappa(\eta_i)/k}$
is free of rank the transcendence degree of $\kappa(\eta_i)$ over $k$. By
Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point}
we conclude that $\dim_{\eta_i}(X_i) = \text{rank}_{\eta_i}(\Omega_{X/k})$.
Since $x \in X_1 \cap \ldots \cap X_n$ we see that
$$
\text{rank}_x(\Omega_{X/k}) = \text{rank}_{\eta_i}(\Omega_{X/k}) = \dim(X_i).
$$
Therefore $\dim_x(X) = \text{rank}_x(\Omega_{X/k})$, see
Algebra, Lemma \ref{algebra-lemma-dimension-at-a-point-finite-type-over-field}.
It follows that $X \to \Spec(k)$ is smooth at $x$ for example by
Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}.
\end{proof}

\begin{lemma}
\label{lemma-smooth-regular}
Let $X \to \Spec(k)$ be a smooth morphism where $k$ is a field.
Then $X$ is a regular scheme.
\end{lemma}

\begin{proof}
(See also
Lemma \ref{lemma-geometrically-regular-smooth}.)
By
Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}
every local ring $\mathcal{O}_{X, x}$ is regular.
And because $X$ is locally of finite type over $k$ it is locally
Noetherian. Hence $X$ is regular by
Properties, Lemma \ref{properties-lemma-characterize-regular}.
\end{proof}

\begin{lemma}
\label{lemma-smooth-geometrically-normal}
Let $X \to \Spec(k)$ be a smooth morphism where $k$ is a field.
Then $X$ is geometrically regular, geometrically normal, and
geometrically reduced over $k$.
\end{lemma}

\begin{proof}
(See also
Lemma \ref{lemma-geometrically-regular-smooth}.)
Let $k'$ be a finite purely inseparable extension of $k$.
It suffices to prove that $X_{k'}$ is regular, normal, reduced, see
Lemmas \ref{lemma-geometrically-regular},
\ref{lemma-geometrically-normal}, and
\ref{lemma-check-only-finite-inseparable-extensions}.
By
Morphisms, Lemma \ref{morphisms-lemma-base-change-smooth}
the morphism $X_{k'} \to \Spec(k')$ is smooth too.
Hence it suffices to show that a scheme $X$ smooth over a field is regular,
normal, and reduced. We see that $X$ is regular by
Lemma \ref{lemma-smooth-regular}.
Hence
Properties, Lemma \ref{properties-lemma-regular-normal}
guarantees that $X$ is normal.
\end{proof}

\begin{lemma}
\label{lemma-affine-space-over-field}
Let $k$ be a field. Let $d \geq 0$. Let $W \subset \mathbf{A}^d_k$
be nonempty open. Then there exists a closed point $w \in W$ such that
$k \subset \kappa(w)$ is finite separable.
\end{lemma}

\begin{proof}
After possible shrinking $W$ we may assume that
$W = \mathbf{A}^d_k \setminus V(f)$ for some $f \in k[x_1, \ldots, x_n]$.
If the lemma is wrong then $f(a_1, \ldots, a_n) = 0$ for all
$(a_1, \ldots, a_n) \in (k^{sep})^n$. This is absurd as $k^{sep}$
is an infinite field.
\end{proof}

\begin{lemma}
\label{lemma-smooth-separable-closed-points-dense}
Let $k$ be a field. If $X$ is smooth over $\Spec(k)$ then
the set
$$
\{x \in X\text{ closed such that }k \subset \kappa(x)
\text{ is finite separable}\}
$$
is dense in $X$.
\end{lemma}

\begin{proof}
It suffices to show that given a nonempty smooth $X$ over $k$
there exists at least one closed point whose residue field is
finite separable over $k$. To see this, choose a diagram
$$
\xymatrix{
X & U \ar[l] \ar[r]^-\pi & \mathbf{A}^d_k
}
$$
with $\pi$ \'etale, see
Morphisms, Lemma \ref{morphisms-lemma-smooth-etale-over-affine-space}.
The morphism $\pi : U \to \mathbf{A}^d_k$ is open, see
Morphisms, Lemma \ref{morphisms-lemma-etale-open}.
By
Lemma \ref{lemma-affine-space-over-field}
we may choose a closed point $w \in \pi(V)$ whose residue field is
finite separable over $k$. Pick any $x \in V$ with $\pi(x) = w$. By
Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}
the field extension $\kappa(w) \subset \kappa(x)$ is finite separable.
Hence $k \subset \kappa(x)$ is finite separable. The point $x$ is a
closed point of $X$ by
Morphisms, Lemma
\ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}.
\end{proof}

\begin{lemma}
\label{lemma-geometrically-reduced-dense-smooth-open}
Let $X$ be a scheme over a field $k$.
If $X$ is locally of finite type and geometrically reduced over $k$
then $X$ contains a dense open which is smooth over $k$.
\end{lemma}

\begin{proof}
The problem is local on $X$, hence we may assume $X$ is quasi-compact.
Let $X = X_1 \cup \ldots \cup X_n$ be the irreducible components of $X$.
Then $Z = \bigcup_{i \not = j} X_i \cap X_j$ is nowhere dense in $X$.
Hence we may replace $X$ by $X \setminus Z$. As $X \setminus Z$ is a
disjoint union of irreducible schemes, this reduces us to the case
where $X$ is irreducible. As $X$ is irreducible and reduced, it is
integral, see
Properties, Lemma \ref{properties-lemma-characterize-integral}.
Let $\eta \in X$ be its generic point.
Then the function field $K = k(X) = \kappa(\eta)$ is geometrically
reduced over $k$, hence separable over $k$, see
Algebra, Lemma \ref{algebra-lemma-characterize-separable-field-extensions}.
Let $U = \Spec(A) \subset X$ be any nonempty affine open
so that $K = f.f.(A) = A_{(0)}$. Apply
Algebra, Lemma \ref{algebra-lemma-separable-smooth}
to conclude that $A$ is smooth at $(0)$ over $k$.
By definition this means that some principal localization
of $A$ is smooth over $k$ and we win.
\end{proof}

\begin{lemma}
\label{lemma-flat-under-smooth}
Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes locally
of finite type over $k$. Let $x \in X$ be a point and set $y = f(x)$.
If $X \to \Spec(k)$ is smooth at $x$ and $f$ is flat at $x$
then $Y \to \Spec(k)$ is smooth at $y$. In particular, if $X$ is
smooth over $k$ and $f$ is flat and surjective, then $Y$ is smooth over $k$.
\end{lemma}

\begin{proof}
It suffices to show that $Y$ is geometrically regular at $y$, see
Lemma \ref{lemma-geometrically-regular-smooth}.
This follows from
Lemma \ref{lemma-flat-under-geometrically-regular}
(and
Lemma \ref{lemma-geometrically-regular-smooth}
applied to $(X, x)$).
\end{proof}




\section{Types of varieties}
\label{section-types}

\noindent
Short section discussion some elementary global properties of varieties.

\begin{definition}
\label{definition-variety-type}
Let $k$ be a field. Let $X$ be a variety over $k$.
\begin{enumerate}
\item We say $X$ is an {\it affine variety} if $X$ is an affine scheme.
This is equivalent to requiring $X$ it be isomorphic to a closed
subscheme of $\mathbf{A}^n_k$ for some $n$.
\item We say $X$ is a {\it projective variety} if the
structure morphism $X \to \Spec(k)$ is projective. By
Morphisms, Lemma \ref{morphisms-lemma-characterize-locally-projective}
this is true if and only if $X$ is isomorphic to a closed
subscheme of $\mathbf{P}^n_k$ for some $n$.
\item We say $X$ is a {\it quasi-projective variety} if
the structure morphism $X \to \Spec(k)$ is quasi-projective. By
Morphisms, Lemma \ref{morphisms-lemma-characterize-locally-quasi-projective}
this is true if and only if $X$ is isomorphic to a
locally closed subscheme of $\mathbf{P}^n_k$ for some $n$.
\item A {\it proper variety} is a variety such that the
morphism $X \to \Spec(k)$ is proper.
\end{enumerate}
\end{definition}

\noindent
Note that a projective variety is a proper variety, see
Morphisms, Lemma \ref{morphisms-lemma-locally-projective-proper}.
Also, an affine variety is quasi-projective as $\mathbf{A}^n_k$
is isomorphic to an open subscheme of $\mathbf{P}^n_k$, see
Constructions,
Lemma \ref{constructions-lemma-standard-covering-projective-space}.

\begin{lemma}
\label{lemma-regular-functions-proper-variety}
Let $X$ be a proper variety over $k$.
Then $\Gamma(X, \mathcal{O}_X)$ is a field which is
a finite extension of the field $k$.
\end{lemma}

\begin{proof}
By
Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-pushforward-coherent}
we see that $\Gamma(X, \mathcal{O}_X)$ is a finite dimensional
$k$-vector space. It is also a $k$-algebra without zero-divisors.
Hence it is a field, see
Algebra, Lemma \ref{algebra-lemma-integral-over-field}.
\end{proof}




\section{Groups of invertible functions}
\label{section-units}

\noindent
It is often (but not always) the case that $\mathcal{O}^*(X)/k^*$
is a finitely generated abelian group if $X$ is a variety over $k$.
We show this by a series of lemmas.
Everything rests on the following special case.

\begin{lemma}
\label{lemma-open-in-normal-proper}
Let $k$ be an algebraically closed field.
Let $\overline{X}$ be a proper variety over $k$.
Let $X \subset \overline{X}$ be an open subscheme.
Assume $X$ is normal.
Then $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group.
\end{lemma}

\begin{proof}
We will use without further mention that for any affine open $U$ of
$\overline{X}$ the ring $\mathcal{O}(U)$ is a finitely generated
$k$-algebra, which is Noetherian, a domain and normal, see
Algebra, Lemma \ref{algebra-lemma-Noetherian-permanence},
Properties, Definition \ref{properties-definition-integral},
Properties, Lemmas \ref{properties-lemma-locally-Noetherian} and
\ref{properties-lemma-locally-normal},
Morphisms, Lemma \ref{morphisms-lemma-locally-finite-type-characterize}.

\medskip\noindent
Let $\xi_1, \ldots, \xi_r$ be the generic points of the complement of $X$
in $\overline{X}$. There are finitely many since $\overline{X}$ has a
Noetherian underlying topological space (see
Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian},
Properties, Lemma \ref{properties-lemma-Noetherian-topology}, and
Topology, Lemma \ref{topology-lemma-Noetherian}).
For each $i$ the local ring $\mathcal{O}_i = \mathcal{O}_{X, \xi_i}$
is a normal Noetherian local domain (as a localization of a
Noetherian normal domain). Let $J \subset \{1, \ldots, r\}$ be the set of
indices $i$ such that $\dim(\mathcal{O}_i) = 1$. For $j \in J$ the
local ring $\mathcal{O}_j$ is a discrete valuation ring, see
Algebra, Lemma \ref{algebra-lemma-characterize-dvr}.
Hence we obtain a valuation
$$
v_j : k(\overline{X})^* \longrightarrow \mathbf{Z}
$$
with the property that $v_j(f) \geq 0 \Leftrightarrow f \in \mathcal{O}_j$.

\medskip\noindent
Think of $\mathcal{O}(X)$ as a sub $k$-algebra of $k(X) = k(\overline{X})$.
We claim that the kernel of the map
$$
\mathcal{O}(X)^* \longrightarrow
\prod\nolimits_{j \in J} \mathbf{Z},
\quad
f \longmapsto \prod v_j(f)
$$
is $k^*$. It is clear that this claim proves the lemma.
Namely, suppose that $f \in \mathcal{O}(X)$ is an element of the kernel.
Let $U = \Spec(B) \subset \overline{X}$ be any affine open.
Then $B$ is a Noetherian normal domain.
For every height one prime $\mathfrak q \subset B$ with corresponding
point $\xi \in X$ we see that either $\xi = \xi_j$ for some $j \in J$
or that $\xi \in X$. The reason is that
$\text{codim}(\overline{\{\xi\}}, \overline{X}) = 1$ by
Properties, Lemma \ref{properties-lemma-codimension-local-ring}
and hence if $\xi \in \overline{X} \setminus X$ it must be a
generic point of $\overline{X} \setminus X$, hence equal to some
$\xi_j$, $j \in J$.
We conclude that $f \in \mathcal{O}_{X, \xi} = B_{\mathfrak q}$
in either case as $f$ is in the kernel of the map. Thus
$f \in \bigcap_{\text{ht}(\mathfrak q) = 1} B_{\mathfrak q} = B$, see
Algebra, Lemma
\ref{algebra-lemma-normal-domain-intersection-localizations-height-1}.
In other words, we see that
$f \in \Gamma(\overline{X}, \mathcal{O}_{\overline{X}})$.
But since $k$ is algebraically closed we conclude that
$f \in k$ by
Lemma \ref{lemma-regular-functions-proper-variety}.
\end{proof}

\noindent
Next, we generalize the case above by some elementary arguments, still
keeping the field algebraically closed.

\begin{lemma}
\label{lemma-units-integral-finite-type-algebraically-closed}
Let $k$ be an algebraically closed field.
Let $X$ be an integral scheme locally of finite type over $k$.
Then $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group.
\end{lemma}

\begin{proof}
As $X$ is integral the restriction mapping
$\mathcal{O}(X) \to \mathcal{O}(U)$ is injective for any
nonempty open subscheme $U \subset X$. Hence we may assume
that $X$ is affine. Choose a closed immersion
$X \to \mathbf{A}^n_k$
and denote $\overline{X}$ the closure of $X$ in $\mathbf{P}^n_k$
via the usual immersion $\mathbf{A}^n_k \to \mathbf{P}^n_k$.
Thus we may assume that $X$ is an affine open of a projective
variety $\overline{X}$.

\medskip\noindent
Let $\nu : \overline{X}^\nu \to \overline{X}$ be the normalization
morphism, see
Morphisms, Definition \ref{morphisms-definition-normalization}.
We know that $\nu$ is finite, dominant, and that $\overline{X}^\nu$
is a normal irreducible scheme, see
Morphisms, Lemmas \ref{morphisms-lemma-normalization-normal},
\ref{morphisms-lemma-Japanese-normalization}, and
\ref{morphisms-lemma-ubiquity-nagata}.
It follows that $\overline{X}^\nu$ is a proper variety,
because $\overline{X} \to \Spec(k)$ is proper as a composition
of a finite and a proper morphism (see results in
Morphisms, Sections \ref{morphisms-section-proper} and
\ref{morphisms-section-integral}).
It also follows that $\nu$ is a surjective morphism, because
the image of $\nu$ is closed and contains the generic point of $\overline{X}$.
Hence setting $X^\nu = \nu^{-1}(X)$ we see that it suffices to prove the
result for $X^\nu$. In other words, we may assume that $X$ is a nonempty
open of a normal proper variety $\overline{X}$. This case is handled by
Lemma \ref{lemma-open-in-normal-proper}.
\end{proof}

\noindent
The preceding lemma implies the following slight generalization.

\begin{lemma}
\label{lemma-units-general-algebraically-closed}
Let $k$ be an algebraically closed field.
Let $X$ be a connected reduced scheme which is locally of finite type
over $k$ with finitely many irreducible components.
Then $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group.
\end{lemma}

\begin{proof}
Let $X = \bigcup X_i$ be the irreducible components. By
Lemma \ref{lemma-units-integral-finite-type-algebraically-closed}
we see that $\mathcal{O}(X_i)^*/k^*$ is a finitely generated
abelian group. Let $f \in \mathcal{O}(X)^*$ be in the kernel
of the map
$$
\mathcal{O}(X)^* \longrightarrow \prod \mathcal{O}(X_i)^*/k^*.
$$
Then for each $i$ there exists an element $\lambda_i \in k$
such that $f|_{X_i} = \lambda_i$.
By restricting to $X_i \cap X_j$ we conclude that
$\lambda_i = \lambda_j$ if $X_i \cap X_j \not = \emptyset$.
Since $X$ is connected we conclude that all $\lambda_i$ agree
and hence that $f \in k^*$. This proves that
$$
\mathcal{O}(X)^*/k^* \subset \prod \mathcal{O}(X_i)^*/k^*
$$
and the lemma follows as on the right we have a product of finitely
many finitely generated abelian groups.
\end{proof}

\begin{lemma}
\label{lemma-integral-closure-ground-field}
Let $k$ be a field.
Let $X$ be a scheme over $k$ which is connected and reduced.
Then the integral closure of $k$ in $\Gamma(X, \mathcal{O}_X)$
is a field.
\end{lemma}

\begin{proof}
Let $k' \subset \Gamma(X, \mathcal{O}_X)$ be the integral closure of
$k$. Then $X \to \Spec(k)$ factors through $\Spec(k')$, see
Schemes, Lemma \ref{schemes-lemma-morphism-into-affine}.
As $X$ is reduced we see that $k'$ has no nonzero nilpotent elements.
As $k \to k'$ is integral we see that every prime ideal of $k'$ is
both a maximal ideal and a minimal prime,
and $\Spec(k')$ is totally disconnected, see
Algebra, Lemmas \ref{algebra-lemma-integral-no-inclusion} and
\ref{algebra-lemma-ring-with-only-minimal-primes}.
As $X$ is connected the morphism $X \to \Spec(k')$ is
constant, say with image the point corresponding to
$\mathfrak p \subset k'$. Then any $f \in k'$, $f \not \in \mathfrak p$
maps to an invertible element of $\mathcal{O}_X$. By definition
of $k'$ this then forces $f$ to be a unit of $k'$. Hence we see
that $k'$ is local with maximal ideal $\mathfrak p$, see
Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}.
Since we've already seen that $k'$ is reduced this implies that
$k'$ is a field, see
Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}.
\end{proof}

\begin{proposition}
\label{proposition-units-general}
Let $k$ be a field. Let $X$ be a scheme over $k$. Assume that $X$ is
locally of finite type over $k$, connected, reduced, and has finitely many
irreducible components. Then $\mathcal{O}(X)^*/k^*$ is a finitely generated
abelian group if in addition to the conditions above at least
one of the following conditions is satisfied:
\begin{enumerate}
\item the integral closure of $k$ in $\Gamma(X, \mathcal{O}_X)$ is $k$,
\item $X$ has a $k$-rational point, or
\item $X$ is geometrically integral.
\end{enumerate}
\end{proposition}

\begin{proof}
Let $\overline{k}$ be an algebraic closure of $k$.
Let $Y$ be a connected component of $(X_{\overline{k}})_{red}$.
Note that the canonical morphism $p : Y \to X$ is open (by
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open})
and closed (by
Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}).
Hence $p(Y) = X$ as $X$ was assumed connected. In particular, as
$X$ is reduced this implies $\mathcal{O}(X) \subset \mathcal{O}(Y)$. By
Lemma \ref{lemma-galois-action-irreducible-components-locally-finite-type}
we see that $Y$ has finitely many irreducible components.
Thus
Lemma \ref{lemma-units-general-algebraically-closed}
applies to $Y$. This implies that if $\mathcal{O}(X)^*/k^*$ is
not a finitely generated abelian group, then there exist elements
$f \in \mathcal{O}(X)$, $f \not \in k$ which map to an element of
$\overline{k}$ via the map $\mathcal{O}(X) \to \mathcal{O}(Y)$.
In this case $f$ is algebraic over $k$, hence
integral over $k$. Thus, if condition (1) holds, then this cannot happen.
To finish the proof we show that conditions (2) and (3) imply (1).

\medskip\noindent
Let $k \subset k' \subset \Gamma(X, \mathcal{O}_X)$ be the integral
closure of $k$ in $\Gamma(X, \mathcal{O}_X)$. By
Lemma \ref{lemma-integral-closure-ground-field}
we see that $k'$ is a field.
If $e : \Spec(k) \to X$ is a $k$-rational point, then
$e^\sharp : \Gamma(X, \mathcal{O}_X) \to k$ is a section to the
inclusion map $k \to \Gamma(X, \mathcal{O}_X)$. In particular the
restriction of $e^\sharp$ to $k'$ is a field map $k' \to k$ over $k$,
which clearly shows that (2) implies (1).

\medskip\noindent
If the integral closure $k'$ of $k$ in $\Gamma(X, \mathcal{O}_X)$
is not trivial, then we see that $X$ is either not geometrically connected
(if $k \subset k'$ is not purely inseparable) or that $X$ is not
geometrically reduced (if $k \subset k'$ is nontrivial purely inseparable).
Details omitted. Hence (3) implies (1).
\end{proof}

\begin{lemma}
\label{lemma-units-variety}
Let $k$ be a field.
Let $X$ be a variety over $k$.
The group $\mathcal{O}(X)^*/k^*$ is a finitely generated abelian group
provided at least one of the following conditions holds:
\begin{enumerate}
\item $k$ is integrally closed in $\Gamma(X, \mathcal{O}_X)$,
\item $k$ is algebraically closed in $k(X)$,
\item $X$ is geometrically integral over $k$, or
\item $k$ is the ``intersection'' of the field extensions
$k \subset \kappa(x)$ where $x$ runs over the closed points of $x$.
\end{enumerate}
\end{lemma}

\begin{proof}
We see that (1) is enough by
Proposition \ref{proposition-units-general}.
We omit the verification that each of (2), (3), (4) implies (1).
\end{proof}





\section{Uniqueness of base field}
\label{section-base-field}

\noindent
The phrase ``let $X$ be a scheme over $k$'' means that $X$ is a scheme
which comes equipped with a morphism $X \to \Spec(k)$. Now we can
ask whether the field $k$ is uniquely determined by the scheme $X$.
Of course this is not the case, since for example
$\mathbf{A}^1_{\mathbf{C}}$ which we ordinarily consider as a scheme
over the field $\mathbf{C}$ of complex numbers, could also be considered
as a scheme over $\mathbf{Q}$. But what if we ask that the morphism
$X \to \Spec(k)$ does not factor as
$X \to \Spec(k') \to \Spec(k)$ for any nontrivial field
extension $k \subset k'$? In other words we ask that $k$ is
somehow maximal such that $X$ lives over $k$.

\medskip\noindent
An example to show that this still does not guarantee uniqueness of $k$
is the scheme
$$
X =
\Spec\left(
\mathbf{Q}(x)[y]\left[\frac{1}{P(y)}, P \in \mathbf{Q}[y], P \not = 0\right]
\right)
$$
At first sight this seems to be a scheme over $\mathbf{Q}(x)$, but on a
second look it is clear that it is also a scheme over $\mathbf{Q}(y)$.
Moreover, the fields $\mathbf{Q}(x)$ and $\mathbf{Q}(y)$ are subfields
of $R = \Gamma(X, \mathcal{O}_X)$ which are maximal among the subfields
of $R$ (details omitted). In particular, both $\mathbf{Q}(x)$ and
$\mathbf{Q}(y)$ are maximal in the sense above. Note that both morphisms
$X \to \Spec(\mathbf{Q}(x))$
and $X \to \Spec(\mathbf{Q}(y))$ are ``essentially of finite type''
(i.e., the corresponding ring map is essentially of finite type).
Hence $X$ is a Noetherian scheme of finite dimension, i.e., it is
not completely pathological.

\medskip\noindent
Another issue that can prevent uniqueness is that the scheme $X$ may
be nonreduced. In that case there can be many different morphisms
from $X$ to the spectrum of a given field. As an explicit example consider
the dual numbers
$D = \mathbf{C}[y]/(y^2) = \mathbf{C} \oplus \epsilon \mathbf{C}$.
Given any derivation $\theta : \mathbf{C} \to \mathbf{C}$ over $\mathbf{Q}$
we get a ring map
$$
\mathbf{C} \longrightarrow D, \quad
c \longmapsto c + \epsilon \theta(c).
$$
The subfield of $\mathbf{C}$ on which all of these maps are the same
is the algebraic closure of $\mathbf{Q}$. This means that taking the
intersection of all the fields that $X$ can live over may end up
being a very small field if $X$ is nonreduced.

\medskip\noindent
One observation in this regard is the following: given a field
$k$ and two subfields $k_1, k_2$ of $k$ such that $k$ is finite
over $k_1$ and over $k_2$, then in general it is {\it not} the case
that $k$ is finite over $k_1 \cap k_2$. An example is the field
$k = \mathbf{Q}(t)$ and its subfields $k_1 = \mathbf{Q}(t^2)$ and
$\mathbf{Q}((t + 1)^2)$. Namely we have $k_1 \cap k_2 = \mathbf{Q}$
in this case. So in the following we have to be careful when taking
intersections of fields.

\medskip\noindent
Having said all of this we now show that if $X$ is locally of finite
type over a field, then some uniqueness holds. Here is the precise
result.

\begin{proposition}
\label{proposition-unique-base-field}
Let $X$ be a scheme. Let $a : X \to \Spec(k_1)$ and
$b : X \to \Spec(k_2)$ be morphisms from $X$ to spectra of fields.
Assume $a, b$ are locally of finite type, and
$X$ is reduced, and connected. Then we have
$k_1' = k_2'$, where $k_i' \subset \Gamma(X, \mathcal{O}_X)$ is
the integral closure of $k_i$ in $\Gamma(X, \mathcal{O}_X)$.
\end{proposition}

\begin{proof}
First, assume the lemma holds in case $X$ is quasi-compact (we will
do the quasi-compact case below).
As $X$ is locally of finite type over a field, it is locally Noetherian, see
Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}.
In particular this means that it is locally connected,
connected components of open subsets are open, and
intersections of quasi-compact opens are quasi-compact, see
Properties, Lemma \ref{properties-lemma-Noetherian-topology},
Topology, Lemma \ref{topology-lemma-locally-connected},
Topology, Section \ref{topology-section-noetherian}, and
Topology, Lemma \ref{topology-lemma-constructible-Noetherian-space}.
Pick an open covering $X = \bigcup_{i \in I} U_i$
such that each $U_i$ is quasi-compact and connected.
For each $i$ let $K_i \subset \mathcal{O}_X(U_i)$ be the integral
closure of $k_1$ and of $k_2$.
For each pair $i, j \in I$ we decompose
$$
U_i \cap U_j = \coprod U_{i, j, l}
$$
into its finitely many connected components. Write
$K_{i, j, l} \subset \mathcal{O}(U_{i, j, l})$
for the integral closure of $k_1$ and of $k_2$. By
Lemma \ref{lemma-integral-closure-ground-field}
the rings $K_i$ and $K_{i, j, l}$ are fields.
Now we claim that $k_1'$ and $k_2'$ both equal the kernel of the map
$$
\prod K_i \longrightarrow \prod K_{i, j, l}, \quad
(x_i)_i \longmapsto x_i|_{U_{i, j, l}} - x_j|_{U_{i, j, l}}
$$
which proves what we want.
Namely, it is clear that $k_1'$ is contained in this kernel.
On the other hand, suppose that $(x_i)_i$ is in the kernel.
By the sheaf condition $(x_i)_i$ corresponds to $f \in \mathcal{O}(X)$.
Pick some $i_0 \in I$ and let $P(T) \in k_1[T]$ be a monic polynomial
with $P(x_{i_0}) = 0$. Then we claim that $P(f) = 0$ which proves
that $f \in k_1$. To prove this we have to show that $P(x_i) = 0$
for all $i$. Pick $i \in I$. As $X$ is connected there exists a
sequence $i_0, i_1, \ldots, i_n = i \in I$ such that
$U_{i_t} \cap U_{i_{t + 1}} \not = \emptyset$. Now this means that
for each $t$ there exists an $l_t$ such that $x_{i_t}$ and
$x_{i_{t + 1}}$ map to the same element of the field $K_{i, j, l}$.
Hence if $P(x_{i_t}) = 0$, then $P(x_{i_{t + 1}}) = 0$. By
induction, starting with $P(x_{i_0}) = 0$ we deduce that
$P(x_i) = 0$ as desired.

\medskip\noindent
To finish the proof of the lemma we prove the lemma under the
additional hypothesis that $X$ is quasi-compact. By
Lemma \ref{lemma-integral-closure-ground-field}
after replacing $k_i$ by $k_i'$
we may assume that $k_i$ is integrally closed in $\Gamma(X, \mathcal{O}_X)$.
This implies that $\mathcal{O}(X)^*/k_i^*$ is a finitely generated
abelian group, see
Proposition \ref{proposition-units-general}.
Let $k_{12} = k_1 \cap k_2$ as a subring of $\mathcal{O}(X)$.
Note that $k_{12}$ is a field. Since
$$
k_1^*/k_{12}^* \longrightarrow \mathcal{O}(X)^*/k_2^*
$$
we see that $k_1^*/k_{12}^*$ is a finitely generated abelian group as well.
Hence there exist $\alpha_1, \ldots, \alpha_n \in k_1^*$ such that
every element $\lambda \in k_1$ has the form
$$
\lambda = c \alpha_1^{e_1} \ldots \alpha_n^{e_n}
$$
for some $e_i \in \mathbf{Z}$ and $c \in k_{12}$.
In particular, the ring map
$$
k_{12}[x_1, \ldots, x_n, \frac{1}{x_1 \ldots x_n}] \longrightarrow k_1, \quad
x_i \longmapsto \alpha_i
$$
is surjective. By the Hilbert Nullstellensatz,
Algebra, Theorem \ref{algebra-theorem-nullstellensatz}
we conclude that $k_1$ is a finite extension of $k_{12}$.
In the same way we conclude that $k_2$ is a finite extension of $k_{12}$.
In particular both $k_1$ and $k_2$ are contained in the integral closure
$k_{12}'$ of $k_{12}$ in $\Gamma(X, \mathcal{O}_X)$. But since $k_{12}'$
is a field by
Lemma \ref{lemma-integral-closure-ground-field}
and since we chose $k_i$ to be integrally closed in $\Gamma(X, \mathcal{O}_X)$
we conclude that $k_1 = k_{12} = k_2$ as desired.
\end{proof}






\section{Coherent sheaves on $\mathbf{P}^n$}
\label{section-coherent-Pn}

\noindent
In this section we prove some results on the cohomology of coherent
sheaves on $\mathbf{P}^n$ over a field which can be found in
\cite{Mum}. These will be useful later
when discussing Quot and Hilbert schemes.

\subsection{Preliminaries}
\label{subsection-preliminaries}

\noindent
Let $k$ be a field, $n \geq 1$, $d \geq 1$, and let
$s \in \Gamma(\mathbf{P}_k^n, \mathcal{O}(d))$
be a nonzero section. In this section we will write $\mathcal{O}(d)$
for the $d$th twist of the structure sheaf on projective space
(Constructions, Definitions \ref{constructions-definition-twist} and
\ref{constructions-definition-projective-space}).
Since $\mathbf{P}^n_k$ is a variety this section is regular, hence
$s$ is a regular section of $\mathcal{O}(d)$ and defines an effective
Cartier divisor $H = Z(s) \subset \mathbf{P}^n_k$, see
Divisors, Section \ref{divisors-section-effective-Cartier-divisors}.
Such a divisor $H$ is called a {\it hypersurface} and if $d = 1$
it is called a {\it hyperplane}.

\begin{lemma}
\label{lemma-hyperplane}
Let $k$ be a field. Let $n \geq 1$.
Let $i : H \to \mathbf{P}^n_k$ be a hyperplane.
Then there exists an isomorphism
$$
\varphi : \mathbf{P}^{n - 1}_k \longrightarrow H
$$
such that $i^*\mathcal{O}(1)$ pulls back to $\mathcal{O}(1)$.
\end{lemma}

\begin{proof}
We have $\mathbf{P}^n_k = \text{Proj}(k[T_0, \ldots, T_n])$.
The section $s$ corresponds to a homogeneous form in $T_0, \ldots, T_n$
of degree $1$, see
Cohomology of Schemes, Section
\ref{coherent-section-cohomology-projective-space}.
Say $s = \sum a_i T_i$.
Constructions, Lemma \ref{constructions-lemma-closed-in-projective-space}
gives that
$H = \text{Proj}(k[T_0, \ldots, T_n]/I)$ for the graded ideal $I$
defined by setting $I_d$ equal to the kernel of the map
$\Gamma(\mathbf{P}^n_k, \mathcal{O}(d)) \to \Gamma(H, i^*\mathcal{O}(d))$.
By our construction of $Z(s)$ in Divisors, 
Definition \ref{divisors-definition-zero-scheme-s}
we see that on $D_{+}(T_j)$ the ideal of $H$ is generated by
$\sum a_i T_i/T_j$ in the polynomial ring
$k[T_0/T_j, \ldots, T_n/T_j]$. Thus it is clear that $I$ is the ideal
generated by $\sum a_i T_i$. Note that
$$
k[T_0, \ldots, T_n]/I = k[T_0, \ldots, T_n]/(\sum a_i T_i) \cong
k[S_0, \ldots, S_{n - 1}]
$$
as graded rings. For example, if $a_n \not = 0$, then mapping
$S_i$ equal to the class of $T_i$ works. We obtain the desired isomorphism
by functoriality of $\text{Proj}$.
Equality of twists of structure sheaves follows for example from
Constructions, Lemma
\ref{constructions-lemma-surjective-graded-rings-generated-degree-1-map-proj}.
\end{proof}

\begin{lemma}
\label{lemma-exact-sequence-induction}
Let $k$ be an infinite field. Let $n \geq 1$.
Let $\mathcal{F}$ be a coherent module on $\mathbf{P}^n_k$.
Then there exist a nonzero section
$s \in \Gamma(\mathbf{P}^n_k, \mathcal{O}(1))$
and a short exact sequence
$$
0 \to \mathcal{F}(-1) \to \mathcal{F} \to i_*\mathcal{G} \to 0
$$
where $i : H \to \mathbf{P}^n_k$ is the hyperplane $H$ associated to $s$
and $\mathcal{G} = i^*\mathcal{F}$.
\end{lemma}

\begin{proof}
The map $\mathcal{F}(-1) \to \mathcal{F}$ comes from
Constructions, Equation (\ref{constructions-equation-multiply-on-sheaf})
with $n = 1$, $m = -1$ and the section $s$ of $\mathcal{O}(1)$.
Let's work out what this map looks like if we restrict it to
$D_{+}(T_0)$. Write $D_{+}(T_0) = \Spec(k[x_1, \ldots, x_n])$
with $x_i = T_i/T_0$. Identify $\mathcal{O}(1)|_{D_{+}(T_0)}$ with
$\mathcal{O}$ using the section $T_0$. Hence if
$s = \sum a_iT_i$ then $s|_{D_{+}(T_0)} = a_0 + \sum a_ix_i$
with the identification chosen above. Furthermore, suppose
$\mathcal{F}|_{D_{+}(T_0)}$ corresponds to the finite
$k[x_1, \ldots, x_n]$-module $M$. Via the identification
$\mathcal{F}(-1) = \mathcal{F} \otimes \mathcal{O}(-1)$
and our chosen trivialization of $\mathcal{O}(1)$ we see that
$\mathcal{F}(-1)$ corresponds to $M$ as well. Thus
restricting $\mathcal{F}(-1) \to \mathcal{F}$ to $D_{+}(T_0)$ gives
the map
$$
M \xrightarrow{a_0 + \sum a_ix_i} M
$$
To see that the arrow is injective, it suffices to pick
$a_0 + \sum a_ix_i$ outside any of the associated primes of $M$, see
Algebra, Lemma \ref{algebra-lemma-ass-zero-divisors}. By
Algebra, Lemma \ref{algebra-lemma-finite-ass}
the set $\text{Ass}(M)$ of associated primes of $M$ is finite.
Note that for $\mathfrak p \in \text{Ass}(M)$ the intersection
$\mathfrak p \cap \{a_0 + \sum a_i x_i\}$ is a proper $k$-subvector space.
We conclude that there is a finite family of proper sub vector spaces
$V_1, \ldots, V_m \subset \Gamma(\mathbf{P}^n_k, \mathcal{O}(1))$
such that if we take $s$ outside of $\bigcup V_i$, then multiplication
by $s$ is injective over $D_{+}(T_0)$.
Similarly for the restriction to $D_{+}(T_j)$ for $j = 1, \ldots, n$.
Since $k$ is infinite, a finite union of proper sub vector spaces
is never equal to the whole space, hence we may choose $s$ such
that the map is injective.
The cokernel of $\mathcal{F}(-1) \to \mathcal{F}$ is annihilated
by $\text{Im}(s : \mathcal{O}(-1) \to \mathcal{O})$ which is
the ideal sheaf of $H$ by
Divisors, Definition \ref{divisors-definition-zero-scheme-s}.
Hence we obtain $\mathcal{G}$ on $H$ using
Cohomology of Schemes, Lemma \ref{coherent-lemma-i-star-equivalence}.
\end{proof}

\begin{remark}
\label{remark-exact-sequence-induction}
Let $k$ be an infinite field. Let $n \geq 1$. Given a finite number
of coherent modules $\mathcal{F}_i$ on $\mathbf{P}^n_k$ we can choose
a single $s \in \Gamma(\mathbf{P}^n_k, \mathcal{O}(1))$ such
that the statement of Lemma \ref{lemma-exact-sequence-induction}
works for each of them.
To prove this, just apply the lemma to $\bigoplus \mathcal{F}_i$.
\end{remark}


\subsection{Regularity}
\label{subsection-regularity}

\begin{definition}
\label{definition-regularity}
Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent
sheaf on $\mathbf{P}^n_k$. We say $\mathcal{F}$ is {\it $m$-regular}
if
$$
H^i(\mathbf{P}^n_k, \mathcal{F}(m - i)) = 0
$$
for $i = 1, \ldots, n$.
\end{definition}

\noindent
Note that $\mathcal{F} = \mathcal{O}(d)$ is $m$-regular if and only
if $d \geq m$. This follows from the computation of cohomology groups
in Cohomology of Schemes, Equation (\ref{coherent-equation-identify}).
Namely, we see that $H^n(\mathbf{P}^n_k, \mathcal{O}(d)) = 0$
if and only if $d \geq -n$.

\begin{lemma}
\label{lemma-m-regular-extend-base-field}
Let $k \subset k'$ be an extension of fields. Let $n \geq 0$.
Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^n_k$.
Let $\mathcal{F}'$ be the pullback of $\mathcal{F}$ to $\mathbf{P}^n_{k'}$.
Then $\mathcal{F}$ is $m$-regular if and only if $\mathcal{F}'$ is
$m$-regular.
\end{lemma}

\begin{proof}
This is true because
$$
H^i(\mathbf{P}^n_{k'}, \mathcal{F}') =
H^i(\mathbf{P}^n_k, \mathcal{F}) \otimes_k k'
$$
by flat base change, see
Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}.
\end{proof}

\begin{lemma}
\label{lemma-m-regular}
In the situation of Lemma \ref{lemma-exact-sequence-induction},
if $\mathcal{F}$ is $m$-regular, then $\mathcal{G}$ is $m$-regular
on $H \cong \mathbf{P}^{n - 1}_k$.
\end{lemma}

\begin{proof}
Recall that $H^i(\mathbf{P}^n_k, i_*\mathcal{G}) = H^i(H, \mathcal{G})$ by
Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}.
Hence we see that for $i \geq 1$ we get
$$
H^i(\mathbf{P}^n_k, \mathcal{F}(m - i)) \to
H^i(H, \mathcal{G}(m - i)) \to
H^{i + 1}(\mathbf{P}^n_k, \mathcal{F}(m - 1 - i))
$$
as part of the long exact sequence associated to the short exact sequence
$0 \to \mathcal{F}(m - 1 - i) \to \mathcal{F}(m - i)
\to i_*\mathcal{G}(m - i) \to 0$
we obtain from the exact sequence of
Lemma \ref{lemma-exact-sequence-induction} by tensoring with
the invertible sheaf $\mathcal{O}(m - i)$. The lemma follows.
\end{proof}

\begin{lemma}
\label{lemma-m-regular-up}
Let $k$ be a field. Let $n \geq 0$.
Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^n_k$.
If $\mathcal{F}$ is $m$-regular, then $\mathcal{F}$ is
$(m + 1)$-regular.
\end{lemma}

\begin{proof}
We prove this by induction on $n$. If $n = 0$ every sheaf is $m$-regular
for all $m$ and there is nothing to prove. By
Lemma \ref{lemma-m-regular-extend-base-field} we may replace $k$
by an infinite overfield and assume $k$ is infinite.
Thus we may apply Lemma \ref{lemma-exact-sequence-induction}.
By Lemma \ref{lemma-m-regular} we know that $\mathcal{G}$ is
$m$-regular. By induction on $n$ we see that $\mathcal{G}$ is
$(m + 1)$-regular. Considering the long exact cohomology sequence
associated to the sequence
$$
0 \to \mathcal{F}(m - i) \to \mathcal{F}(m + 1 - i)
\to i_*\mathcal{G}(m + 1 - i) \to 0
$$
the reader easily deduces for $i \geq 1$ the vanishing of
$H^i(\mathbf{P}^n_k, \mathcal{F}(m + 1 - i))$ from the (known) vanishing of
$H^i(\mathbf{P}^n_k, \mathcal{F}(m - i))$ and
$H^i(\mathbf{P}^n_k, \mathcal{G}(m + 1 - i))$.
\end{proof}

\begin{lemma}
\label{lemma-m-regular-multiply}
Let $k$ be a field. Let $n \geq 0$.
Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^n_k$.
If $\mathcal{F}$ is $m$-regular, then the multiplication map
$$
H^0(\mathbf{P}^n_k, \mathcal{F}(m)) \otimes_k
H^0(\mathbf{P}^n_k, \mathcal{O}(1))
\longrightarrow
H^0(\mathbf{P}^n_k, \mathcal{F}(m + 1))
$$
is surjective.
\end{lemma}

\begin{proof}
Let $k \subset k'$ be an extension of fields. Let $\mathcal{F}'$
be as in Lemma \ref{lemma-m-regular-extend-base-field}. By
Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}
the base change of the linear map of the lemma to $k'$ is the
same linear map for the sheaf $\mathcal{F}'$. Since $k \to k'$
is faithfully flat it suffices to prove the lemma over $k'$, i.e.,
we may assume $k$ is infinite.

\medskip\noindent
Assume $k$ is infinite. We prove the lemma by induction on $n$.
The case $n = 0$ is trivial as $\mathcal{O}(1) \cong \mathcal{O}$
is generated by $T_0$. For $n > 0$ apply
Lemma \ref{lemma-exact-sequence-induction}
and tensor the sequence by $\mathcal{O}(m + 1)$ to get
$$
0 \to \mathcal{F}(m) \xrightarrow{s} \mathcal{F}(m + 1) \to
i_*\mathcal{G}(m + 1) \to 0
$$
Let $t \in H^0(\mathbf{P}^n_k, \mathcal{F}(m + 1))$.
By induction the image $\overline{t} \in H^0(H, \mathcal{G}(m + 1))$
is the image of $\sum g_i \otimes \overline{s}_i$ with
$\overline{s}_i \in \Gamma(H, \mathcal{O}(1))$ and
$g_i \in H^0(H, \mathcal{G}(m))$. Since $\mathcal{F}$ is $m$-regular
we have $H^1(\mathbf{P}^n_k, \mathcal{F}(m - 1)) = 0$, hence long
exact cohomology sequence associated to the short exact sequence
$$
0 \to \mathcal{F}(m - 1) \xrightarrow{s} \mathcal{F}(m) \to
i_*\mathcal{G}(m) \to 0
$$
shows we can lift $g_i$ to $f_i \in H^0(\mathbf{P}^n_k, \mathcal{F}(m))$.
We can also lift $\overline{s}_i$ to
$s_i \in H^0(\mathbf{P}^n_k, \mathcal{O}(1))$ (see proof of
Lemma \ref{lemma-hyperplane} for example). After substracting the
image of $\sum f_i \otimes s_i$ from $t$ we see that we may assume
$\overline{t} = 0$. But this exactly means that $t$ is the
image of $f \otimes s$ for some $f \in H^0(\mathbf{P}^n_k, \mathcal{F}(m))$
as desired.
\end{proof}

\begin{lemma}
\label{lemma-m-regular-globally-generated}
Let $k$ be a field. Let $n \geq 0$.
Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^n_k$.
If $\mathcal{F}$ is $m$-regular, then $\mathcal{F}(m)$ is
globally generated.
\end{lemma}

\begin{proof}
For all $d \gg 0$ the sheaf $\mathcal{F}(d)$ is globally generated.
This follows for example from the first part of
Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-projective}.
Pick $d \geq m$ such that $\mathcal{F}(d)$ is globally generated.
Choose a basis $f_1, \ldots, f_r \in H^0(\mathbf{P}^n_k, \mathcal{F})$.
By Lemma \ref{lemma-m-regular-multiply} every element
$f \in H^0(\mathbf{P}^n_k, \mathcal{F}(d))$ can be written as
$f = \sum P_if_i$ for some $P_i \in k[T_0, \ldots, T_n]$ homogeneous
of degree $d - m$. Since the sections $f$ generate $\mathcal{F}(d)$
it follows that the sections $f_i$ generate $\mathcal{F}(m)$.
\end{proof}

\subsection{Hilbert polynomials}
\label{subsection-hilbert}

\noindent
Let $k$ be a field. Let $X$ be a proper scheme over $k$. Let $\mathcal{F}$
be a coherent $\mathcal{O}_X$-module. In this situation the
{\it Euler characteristic of $\mathcal{F}$} is the integer
$$
\chi(X, \mathcal{F}) = \sum (-1)^i \dim_k H^i(X, \mathcal{F}).
$$
Note that only a finite number of the vector spaces $H^i(X, \mathcal{F})$
are nonzero (Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherence-higher-direct-images})
and that each of these spaces is finite dimensional
(Cohomology of Schemes, Lemma
\ref{coherent-lemma-proper-over-affine-cohomology-finite}). Thus
$\chi(X, \mathcal{F}) \in \mathbf{Z}$ is well defined. Observe that
this definition depends on the field $k$ and not just on the pair
$(X, \mathcal{F})$.

\begin{lemma}
\label{lemma-euler-characteristic-additive}
Let $k$ be a field. Let $X$ be a proper scheme over $k$.
Let $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$
be a short exact sequence of coherent modules on $X$. Then
$$
\chi(X, \mathcal{F}_2) = \chi(X, \mathcal{F}_1) + \chi(X, \mathcal{F}_3)
$$
\end{lemma}

\begin{proof}
Consider the long exact sequence of cohomology
$$
0 \to H^0(X, \mathcal{F}_1) \to H^0(X, \mathcal{F}_2) \to
H^0(X, \mathcal{F}_3) \to H^1(X, \mathcal{F}_1) \to \ldots
$$
associated to the short exact sequence of the lemma. The rank-nullity theorem
in linear algebra shows that
$$
0 = \dim H^0(X, \mathcal{F}_1) - \dim H^0(X, \mathcal{F}_2)
+ \dim H^0(X, \mathcal{F}_3) - \dim H^1(X, \mathcal{F}_1) + \ldots
$$
This immediately implies the lemma.
\end{proof}

\begin{lemma}
\label{lemma-euler-characteristic-extend-base-field}
Let $k \subset k'$ be an extension of fields. Let $X$ be a proper scheme
over $k$. Let $\mathcal{F}$ be a coherent sheaf on $X$.
Let $\mathcal{F}'$ be the pullback of $\mathcal{F}$ to $X_{k'}$.
Then $\chi(X, \mathcal{F}) = \chi(X', \mathcal{F}')$.
\end{lemma}

\begin{proof}
This is true because
$$
H^i(X_{k'}, \mathcal{F}') = H^i(X, \mathcal{F}) \otimes_k k'
$$
by flat base change, see
Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}.
\end{proof}

\begin{lemma}
\label{lemma-hilbert-polynomial}
Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf
on $\mathbf{P}^n_k$. The function
$$
d \longmapsto \chi(\mathbf{P}^n_k, \mathcal{F}(d))
$$
is a polynomial.
\end{lemma}

\begin{proof}
We prove this by induction on $n$. If $n = 0$, then
$\mathbf{P}^n_k = \Spec(k)$ and $\mathcal{F}(d) = \mathcal{F}$.
Hence in this case the function is constant, i.e., a polynomial
of degree $0$. Assume $n > 0$. By
Lemma \ref{lemma-euler-characteristic-extend-base-field}
we may assume $k$ is infinite. Apply
Lemma \ref{lemma-exact-sequence-induction}.
Applying Lemma \ref{lemma-euler-characteristic-additive}
to the twisted sequences
$0 \to \mathcal{F}(d - 1) \to \mathcal{F}(d) \to i_*\mathcal{G}(d) \to 0$
we obtain
$$
\chi(\mathbf{P}^n_k, \mathcal{F}(d)) -
\chi(\mathbf{P}^n_k, \mathcal{F}(d - 1)) =
\chi(H, \mathcal{G}(d))
$$
(this also uses the identification of the cohomology of
$i_*\mathcal{G}$ with the cohomology of $\mathcal{G}$, see
Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}).
Since $H \cong \mathbf{P}^{n - 1}_k$ (Lemma \ref{lemma-hyperplane})
by induction the right hand side is a polynomial.
The lemma is finished by noting that any function
$f : \mathbf{Z} \to \mathbf{Z}$ with the property that the map
$d \mapsto f(d) - f(d - 1)$ is a polynomial, is itself a polynomial.
We omit the proof of this fact (hint: compare with
Algebra, Lemma \ref{algebra-lemma-numerical-polynomial}).
\end{proof}

\begin{definition}
\label{definition-hilbert-polynomial}
Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf
on $\mathbf{P}^n_k$. The function
$d \mapsto \chi(\mathbf{P}^n_k, \mathcal{F}(d))$ is called the
{\it Hilbert polynomial} of $\mathcal{F}$.
\end{definition}

\noindent
The Hilbert polynomial has coefficients in $\mathbf{Q}$ and not
in general in $\mathbf{Z}$. For example the Hilbert polynomial
of $\mathcal{O}_{\mathbf{P}^n_k}$ is
$$
d \longmapsto {{d + n}\choose{n}} = \frac{d^n}{n!} + \ldots
$$
This follows from the following lemma and the fact that
$$
H^0(\mathbf{P}^n_k, \mathcal{O}_{\mathbf{P}^n_k}) = k[T_0, \ldots, T_n]_d
$$
(degree $d$ part) whose dimension over $k$ is ${{d + n}\choose{n}}$.

\begin{lemma}
\label{lemma-hilbert-polynomial-H0}
Let $k$ be a field. Let $n \geq 0$. Let $\mathcal{F}$ be a coherent sheaf
on $\mathbf{P}^n_k$ with Hilbert polynomial $P \in \mathbf{Q}[t]$.
Then
$$
P(d) = \dim_k H^0(\mathbf{P}^n_k, \mathcal{F}(d))
$$
for all $d \gg 0$.
\end{lemma}

\begin{proof}
This follows from the vanishing of cohomology of high enough twists
of $\mathcal{F}$. See
Cohomology of Schemes,
Lemma \ref{coherent-lemma-coherent-projective}.
\end{proof}



\subsection{Boundedness of quotients}
\label{subsection-boundedness}

\noindent
In this subsection we bound the regularity of quotients of
a given coherent sheaf on $\mathbf{P}^n$ in terms of the
Hilbert polynomial.

\begin{lemma}
\label{lemma-bound-quotients-free}
Let $k$ be a field. Let $n \geq 0$. Let $r \geq 1$. Let $P \in \mathbf{Q}[t]$.
There exists an integer $m$ depending on $n$, $r$, and $P$
with the following property: if
$$
0 \to \mathcal{K} \to \mathcal{O}^{\oplus r} \to \mathcal{F} \to 0
$$
is a short exact sequence of coherent sheaves on $\mathbf{P}^n_k$
and $\mathcal{F}$ has Hilbert polynomial $P$, then
$\mathcal{K}$ is $m$-regular.
\end{lemma}

\begin{proof}
We prove this by induction on $n$. If $n = 0$, then
$\mathbf{P}^n_k = \Spec(k)$ and any coherent module is $0$-regular
and any surjective map is surjective on global sections.
Assume $n > 0$. Consider an exact sequence as in the lemma.
Let $P' \in \mathbf{Q}[t]$ be the polynomial
$P'(t) = P(t) - P(t - 1)$. Let $m'$ be the integer
which works for $n - 1$, $r$, and $P'$.
By Lemmas \ref{lemma-m-regular-extend-base-field} and
\ref{lemma-euler-characteristic-extend-base-field}
we may replace $k$ by a field extension, hence we may assume
$k$ is infinite. Apply 
Lemma \ref{lemma-exact-sequence-induction}
to the coherent sheaf $\mathcal{F}$.
The Hilbert polynomial of $\mathcal{F}' = i^*\mathcal{F}$
is $P'$ (see proof of Lemma \ref{lemma-hilbert-polynomial}).
Since $i^*$ is right exact we see that $\mathcal{F}'$ is
a quotient of $\mathcal{O}_H^{\oplus r} = i^*\mathcal{O}^{\oplus r}$.
Thus the induction hypothesis applies to $\mathcal{F}'$ on
$H \cong \mathbf{P}^{n - 1}_k$ (Lemma \ref{lemma-hyperplane}).
Note that the map $\mathcal{K}(-1) \to \mathcal{K}$ is injective
as $\mathcal{K} \subset \mathcal{O}^{\oplus r}$ and has
cokernel $i_*\mathcal{H}$ where $\mathcal{H} = i^*\mathcal{K}$.
By the snake lemma (Homology, Lemma \ref{homology-lemma-snake})
we obtain a commutative diagram with exact columns and rows
$$
\xymatrix{
& 0 \ar[d] & 0 \ar[d] & 0 \ar[d] \\
0 \ar[r] &
\mathcal{K}(-1) \ar[r] \ar[d] &
\mathcal{O}^{\oplus r}(-1) \ar[r] \ar[d] &
\mathcal{F}(-1) \ar[d] \ar[r] & 0 \\
0 \ar[r] &
\mathcal{K} \ar[r] \ar[d] &
\mathcal{O}^{\oplus r} \ar[r] \ar[d] &
\mathcal{F} \ar[d] \ar[r] & 0\\
0 \ar[r] &
i_*\mathcal{H} \ar[r] \ar[d] &
i_*\mathcal{O}_H^{\oplus r} \ar[r] \ar[d] &
i_*\mathcal{F}' \ar[r] \ar[d] & 0 \\
& 0 & 0 & 0
}
$$
Thus the induction hypothesis applies to the exact sequence
$0 \to \mathcal{H} \to \mathcal{O}_H^{\oplus r} \to \mathcal{F}' \to 0$
on $H \cong \mathbf{P}^{n - 1}_k$ (Lemma \ref{lemma-hyperplane})
and $\mathcal{H}$ is $m'$-regular. Recall that this implies that
$\mathcal{H}$ is $d$-regular for all $d \geq m'$
(Lemma \ref{lemma-m-regular-up}).

\medskip\noindent
Let $i \geq 2$ and $d \geq m'$. It follows from the long exact
cohomology sequence associated to the left column of the diagram
above and the vanishing of $H^{i - 1}(H, \mathcal{H}(d))$
that the map
$$
H^i(\mathbf{P}^n_k, \mathcal{K}(d - 1))
\longrightarrow
H^i(\mathbf{P}^n_k, \mathcal{K}(d))
$$
is injective. As these groups are zero for $d \gg 0$
(Cohomology of Schemes,
Lemma \ref{coherent-lemma-coherent-projective})
we conclude $H^i(\mathbf{P}^n_k, \mathcal{K}(d))$ are zero
for all $d \geq m'$ and $i \geq 2$.

\medskip\noindent
We still have to control $H^1$. First we observe that all the maps
$$
H^1(\mathbf{P}^n_k, \mathcal{K}(m' - 1)) \to
H^1(\mathbf{P}^n_k, \mathcal{K}(m')) \to
H^1(\mathbf{P}^n_k, \mathcal{K}(m' + 1)) \to \ldots
$$
are surjective by the vanishing of $H^1(H, \mathcal{H}(d))$ for $d \geq m'$.
Suppose $d > m'$ is such that
$$
H^1(\mathbf{P}^n_k, \mathcal{K}(d - 1))
\longrightarrow
H^1(\mathbf{P}^n_k, \mathcal{K}(d))
$$
is injective. Then
$H^0(\mathbf{P}^n_k, \mathcal{K}(d)) \to H^0(H, \mathcal{H}(d))$
is surjective. Consider the commutative diagram
$$
\xymatrix{
H^0(\mathbf{P}^n_k, \mathcal{K}(d)) \otimes_k
H^0(\mathbf{P}^n_k, \mathcal{O}(1))
\ar[r] \ar[d] &
H^0(\mathbf{P}^n_k, \mathcal{K}(d + 1)) \ar[d] \\
H^0(H, \mathcal{H}(d)) \otimes_k
H^0(H, \mathcal{O}_H(1))
\ar[r] &
H^0(H, \mathcal{H}(d + 1))
}
$$
By Lemma \ref{lemma-m-regular-multiply}
we see that the bottom horizontal arrow is surjective.
Hence the right vertical arrow is surjective. We conclude that
$$
H^1(\mathbf{P}^n_k, \mathcal{K}(d))
\longrightarrow
H^1(\mathbf{P}^n_k, \mathcal{K}(d + 1))
$$
is injective. By induction we see that
$$
H^1(\mathbf{P}^n_k, \mathcal{K}(d - 1)) \to
H^1(\mathbf{P}^n_k, \mathcal{K}(d)) \to
H^1(\mathbf{P}^n_k, \mathcal{K}(d + 1)) \to \ldots
$$
are all injective and we conclude that
$H^1(\mathbf{P}^n_k, \mathcal{K}(d - 1)) = 0$
because of the eventual vanishing of these groups. Thus the dimensions
of the groups $H^1(\mathbf{P}^n_k, \mathcal{K}(d))$ for $d \geq m'$
are strictly decreasing until they become zero. It follows that the
regularity of $\mathcal{K}$ is bounded by
$m' + \dim_k H^1(\mathbf{P}^n_k, \mathcal{K}(m'))$.
On the other hand, by the vanishing of the higher cohomology groups
we have
$$
\dim_k H^1(\mathbf{P}^n_k, \mathcal{K}(m')) = 
- \chi(\mathbf{P}^n_k, \mathcal{K}(m')) +
\dim_k H^0(\mathbf{P}^n_k, \mathcal{K}(m'))
$$
Note that the $H^0$ has dimension bounded by the dimension of
$H^0(\mathbf{P}^n_k, \mathcal{O}^{\oplus r}(m'))$ which is
at most $r{{n + m'}\choose{n}}$ if $m' > 0$ and zero if not.
Finally, the term $\chi(\mathbf{P}^n_k, \mathcal{K}(m'))$ is equal
to $r{{n + m'}\choose{n}} - P(m')$. This gives a bound of the
desired type finishing the proof of the lemma.
\end{proof}







\section{Glueing dimension one rings}
\label{section-glueing-dim-1}

\noindent
This section contains some algebraic preliminaries to proving that
a finite set of codimension $1$ points of a separated scheme is
contained in an affine open.

\begin{situation}
\label{situation-glue}
Here we are given a commutative diagram of rings
$$
\xymatrix{
A \ar[r] & K \\
R \ar[u] \ar[r] & B \ar[u]
}
$$
where $K$ is a field and $A$, $B$ are subrings of $K$ with fraction
field $K$. Finally, $R = A \times_K B = A \cap B$.
\end{situation}

\begin{lemma}
\label{lemma-glue-valuation-ring}
In Situation \ref{situation-glue} assume that $B$ is a valuation ring.
Then for every unit $u$ of $A$ either $u \in R$ or $u^{-1} \in R$.
\end{lemma}

\begin{proof}
Namely, if the image $c$ of $u$ in $K$ is in $B$, then $u \in R$.
Otherwise, $c^{-1} \in B$
(Algebra, Lemma \ref{algebra-lemma-valuation-ring-x-or-x-inverse})
and $u^{-1} \in R$.
\end{proof}

\noindent
The following lemma explains the meaning of the condition
``$A \otimes B \to K$ is surjective'' which comes up quite
a bit in the following.

\begin{lemma}
\label{lemma-glue-separated}
In Situation \ref{situation-glue} assume $A$ is a Noetherian
ring of dimension $1$. The following are equivalent
\begin{enumerate}
\item $A \otimes B \to K$ is not surjective,
\item there exists a discrete valuation ring $\mathcal{O} \subset K$
containing both $A$ and $B$.
\end{enumerate}
\end{lemma}

\begin{proof}
It is clear that (2) implies (1). On the other hand, if $A \otimes B \to K$
is not surjective, then the image $C \subset K$ is not a field hence
$C$ has a nonzero maximal ideal $\mathfrak m$. Choose a valuation ring
$\mathcal{O} \subset K$ dominating $C_\mathfrak m$. By
Algebra, Lemma \ref{algebra-lemma-krull-akizuki} applied to
$A \subset \mathcal{O}$ the ring $\mathcal{O}$ is Noetherian.
Hence $\mathcal{O}$ is a valuation ring by
Algebra, Lemma \ref{algebra-lemma-valuation-ring-Noetherian-discrete}.
\end{proof}

\begin{lemma}
\label{lemma-semi-local}
In Situation \ref{situation-glue} assume
\begin{enumerate}
\item $A$ is a Noetherian semi-local domain of dimension $1$,
\item $B$ is a discrete valuation ring,
\end{enumerate}
Then we have the following two possibilities
\begin{enumerate}
\item[(a)] If $A^*$ is not contained in $R$, then
$\Spec(A) \to \Spec(R)$ and $\Spec(B) \to \Spec(R)$
are open immersions and $K = A \otimes_R B$.
\item[(b)] If $A^*$ is contained in $R$, then $B$ dominates one of
the local rings of $A$ at a maximal ideal and $A \otimes B \to K$
is not surjective.
\end{enumerate}
\end{lemma}

\begin{proof}
Assumption (a) implies there is a unit of $A$ whose image in $K$ lies in
the maximal ideal of $B$. Then $u$ is a nonzerodivisor of $R$ and for
every $a \in A$ there exists an $n$ such that $u^n a \in R$. It follows
that $A = R_u$.

\medskip\noindent
Let $\mathfrak m_A$ be the radical of $A$. Let $x \in \mathfrak m_A$
be a nonzero element. Since $\dim(A) = 1$ we see that $K = A_x$.
After replacing $x$ by $x^n u^m$ for some $n \geq 1$ and
$m \in \mathbf{Z}$ we may assume $x$ maps to a unit of $B$.
We see that for every $b \in B$ we have that $x^nb$
in the image of $R$ for some $n$. Thus $B = R_x$.

\medskip\noindent
Let $z \in R$. If $z \not \in \mathfrak m_A$ and $z$ does not map to
an element of $\mathfrak m_B$, then $z$ is invertible.
Thus $x + u$ is invertible in $R$. Hence $\Spec(R) = D(x) \cup D(u)$.
We have seen above that $D(u) = \Spec(A)$ and $D(x) = \Spec(B)$.

\medskip\noindent
Case (b). If $x \in \mathfrak m_A$, then $1 + x$ is a unit and
hence $1 + x \in R$, i.e, $x \in R$. Thus we see that
$\mathfrak m_A \subset R \subset A$. In fact, in this case
$A$ is integral over $R$. Namely, write
$A/\mathfrak m_A = \kappa_1 \times \ldots \times \kappa_n$
as a product of fields. Say $x = (c_1, \ldots, c_r, 0, \ldots, 0)$
is an element with $c_i \not = 0$. Then
$$
x^2 - x(c_1, \ldots, c_r, 1, \ldots, 1)  = 0
$$
Since $R$ contains all units we see that $A/\mathfrak m_A$ is
integral over the image of $R$ in it, and hence $A$ is integral over $R$.
It follows that $R \subset A \subset B$ as $B$ is integrally closed.
Moreover, if $x \in \mathfrak m_A$ is nonzero, then
$K = A_x = \bigcup x^{-n}A = \bigcup x^{-n}R$.
Hence $x^{-1} \not \in B$, i.e., $x \in \mathfrak m_B$.
We conclude $\mathfrak m_A \subset \mathfrak m_B$.
Thus $A \cap \mathfrak m_B$ is a maximal ideal of $A$ thereby
finishing the proof.
\end{proof}

\begin{lemma}
\label{lemma-semi-local-dimension-one-conductor}
Let $B$ be a semi-local Noetherian domain of dimension $1$.
Let $B'$ be the integral closure of $B$ in its fraction field.
Then $B'$ is a semi-local Dedekind domain.
Let $x$ be a nonzero element of the radical of $B'$.
Then for every $y \in B'$ there exists an $n$ such that
$x^n y \in B$.
\end{lemma}

\begin{proof}
Let $\mathfrak m_B$ be the radical of $B$. The structure of $B'$ results from
Algebra, Lemma \ref{algebra-lemma-integral-closure-Dedekind}.
Given $x, y \in B'$ as in the statement of the lemma consider
the subring $B \subset A \subset B'$ generated by $x$ and $y$.
Then $A$ is finite over $B$ (Algebra, Lemma
\ref{algebra-lemma-characterize-finite-in-terms-of-integral}).
Since the fraction fields of $B$ and $A$ are the same we see
that the finite module $A/B$ is supported on the set of
closed points of $B$. Thus $\mathfrak m_B^n A \subset B$ for
a suitable $n$. Moreover, $\Spec(B') \to \Spec(A)$ is
surjective (Algebra, Lemma \ref{algebra-lemma-integral-overring-surjective}),
hence $A$ is semi-local as well. It also follows that
$x$ is in the radical $\mathfrak m_A$ of $A$. Note that
$\mathfrak m_A = \sqrt{\mathfrak m_B A}$.
Thus $x^m y \in \mathfrak m_B A$ for some $m$.
Then $x^{nm} y \in B$.
\end{proof}

\begin{lemma}
\label{lemma-semi-local-both-side}
In Situation \ref{situation-glue} assume
\begin{enumerate}
\item $A$ is a Noetherian semi-local domain of dimension $1$,
\item $B$ is a Noetherian semi-local domain of dimension $1$,
\item $A \otimes B \to K$ is surjective.
\end{enumerate}
Then $\Spec(A) \to \Spec(R)$ and $\Spec(B) \to \Spec(R)$
are open immersions and $K = A \otimes_R B$.
\end{lemma}

\begin{proof}
Special case: $B$ is integrally closed in $K$. This means that
$B$ is a Dedekind domain
(Algebra, Lemma \ref{algebra-lemma-characterize-Dedekind})
whence all of its localizations at maximal ideals are discrete valuation rings.
Let $\mathfrak m_1, \ldots, \mathfrak m_r$ be the maximal ideals of
$B$. We set
$$
R_1 = A \times_K B_{\mathfrak m_1}
$$
Observing that $A \otimes_{R_1} B_{\mathfrak m_1} \to K$ is surjective
we conclude from Lemma \ref{lemma-semi-local} that $A$ and
$B_{\mathfrak m_1}$ define open subschemes covering $\Spec(R_1)$ and that
$K = A \otimes_{R_1} B_{\mathfrak m_1}$. In particular $R_1$ is a
semi-local Noetherian ring of dimension $1$. By induction we define
$$
R_{i + 1} = R_i \times_K B_{\mathfrak m_{i + 1}}
$$
for $i = 1, \ldots, r - 1$. Observe that $R = R_n$ because
$B = B_{\mathfrak m_1} \cap \ldots \cap B_{\mathfrak m_r}$ (see
Algebra, Lemma
\ref{algebra-lemma-normal-domain-intersection-localizations-height-1}).
It follows from the inductive procedure that $R \to A$ defines an
open immersion $\Spec(A) \to \Spec(R)$. On the other hand, the
maximal ideals $\mathfrak n_i$ of $R$ not in this open correspond
to the maximal ideals $\mathfrak m_i$ of $B$ and in fact
the ring map $R \to B$ defines an isomorphisms
$R_{\mathfrak n_i} \to B_{\mathfrak m_i}$ (details omitted; hint:
in each step we added exactly one maximal ideal to $\Spec(R_i)$).
It follows that $\Spec(B) \to \Spec(R)$ is an open immersion
as desired.

\medskip\noindent
General case. Let $B' \subset K$ be the integral closure of $B$.
See Lemma \ref{lemma-semi-local-dimension-one-conductor}.
Then the special case applies to $R' = A \times_K B'$.
Pick $x \in R'$ which is not contained in the maximal
ideals of $A$ and is contained in the maximal ideals of $B'$
(see Algebra, Lemma \ref{algebra-lemma-chinese-remainder}).
By Lemma \ref{lemma-semi-local-dimension-one-conductor}
there exists an integer $n$ such that $x^n \in R = A \times_K B$.
Replace $x$ by $x^n$ so $x \in R$. For every $y \in R'$ there exists
an integer $n$ such that $x^n y \in R$. On the other hand,
it is clear that $R'_x = A$. Thus $R_x = A$.
Exchanging the roles of $A$ and $B$ we also find an $y \in R$
such that $B = R_y$. Note that inverting both $x$ and $y$
leaves no primes except $(0)$. Thus $K = R_{xy} = R_x \otimes_R R_y$.
This finishes the proof.
\end{proof}

\begin{lemma}
\label{lemma-glue-a-bunch-of-local-rings}
Let $K$ be a field. Let $A_1, \ldots, A_r \subset K$ be Noetherian
semi-local rings of dimension $1$ with fraction field $K$. If
$A_i \otimes A_j \to K$ is surjective for all $i \not = j$, then
there exists a Noetherian semi-local domain $A \subset K$
of dimension $1$ containing $A_1, \ldots, A_r$ such that
\begin{enumerate}
\item $A \to A_i$ induces an open immersion $j_i : \Spec(A_i) \to \Spec(A)$,
\item $\Spec(A)$ is the union of the opens $j_i(\Spec(A_i))$,
\item each closed point of $\Spec(A)$ lies in exactly one of these
opens.
\end{enumerate}
\end{lemma}

\begin{proof}
Namely, we can take $A = A_1 \cap \ldots \cap A_r$. First we note that (3),
once (1) and (2) have been proven, follows from the assumption that
$A_i \otimes A_j \to K$ is surjective since if
$\mathfrak m \in j_i(\Spec(A_i)) \cap j_j(\Spec(A_j))$, then
$A_i \otimes A_j \to K$ ends up in $A_\mathfrak m$.
To prove (1) and (2) we argue by induction on $r$.
If $r > 1$ by induction we have the results (1) and (2) for
$B = A_2 \cap \ldots \cap A_r$. Then we apply
Lemma \ref{lemma-semi-local-both-side} to see they hold for
$A = A_1 \cap B$.
\end{proof}

\begin{lemma}
\label{lemma-create-globally-generated}
Let $A$ be a domain with fraction field $K$. Let $B_1, \ldots, B_r \subset K$
be Noetherian $1$-dimensional semi-local rings whose fraction
fields are $K$. If $A \otimes B_i \to K$ are surjective for $i = 1, \ldots, r$,
then there exists an $x \in A$ such that $x^{-1}$ is in the radical of
$B_i$ for $i = 1, \ldots, r$.
\end{lemma}

\begin{proof}
Let $B_i'$ be the integral closure of $B_i$ in $K$. Suppose we find a
nonzero $x \in A$ such that $x^{-1}$ is in the radical of $B'_i$ for
$i = 1, \ldots, r$. Then by
Lemma \ref{lemma-semi-local-dimension-one-conductor},
after replacing $x$ by a power we get $x^{-1} \in B_i$.
Since $\Spec(B'_i) \to \Spec(B_i)$ is surjective we
see that $x^{-1}$ is then also in the radical of $B_i$.
Thus we may assume that each $B_i$ is a semi-local Dedekind domain.

\medskip\noindent
If $B_i$ is not local, then remove $B_i$ from the list and
add back the finite collection of local rings $(B_i)_\mathfrak m$.
Thus we may assume that $B_i$ is a discrete valuation ring for
$i = 1, \ldots, r$.

\medskip\noindent
Let $v_i : K \to \mathbf{Z}$, $i = 1, \ldots, r$
be the corresponding discrete valuations (see
Algebra, Lemma \ref{algebra-lemma-characterize-Dedekind}).
We are looking for a nonzero $x \in A$ with $v_i(x) < 0$ for
$i = 1, \ldots, r$. We will prove this by induction on $r$.

\medskip\noindent
If $r = 1$ and the result is wrong, then $A \subset B$ and the map
$A \otimes B \to K$ is not surjective, contradiction.

\medskip\noindent
If $r > 1$, then by induction we can find a nonzero $x \in A$ such that
$v_i(x) < 0$ for $i = 1, \ldots, r - 1$. If $v_r(x) < 0$ then we are
done, so we may assume $v_r(x) \geq 0$. By the base case we can find
$y \in A$ nonzero such that $v_r(y) < 0$. After replacing $x$ by a
power we may assume that $v_i(x) < v_i(y)$ for $i = 1, \ldots, r - 1$.
Then $x + y$ is the element we are looking for.
\end{proof}

\begin{lemma}
\label{lemma-localization-semi-local}
Let $A$ be a ring. Let $\mathfrak p_1, \ldots, \mathfrak p_r$
be a finite set of a primes of $A$. Let
$S = A \setminus \bigcup \mathfrak p_i$. Then $S$ is a multiplicative
system and $S^{-1}A$ is a semi-local ring whose maximal ideals
correspond to the maximal elements of the set $\{\mathfrak p_i\}$.
\end{lemma}

\begin{proof}
If $a, b \in A$ and $a, b \in S$, then $a, b \not \in \mathfrak p_i$
hence $ab \not \in \mathfrak p_i$, hence $ab \in S$. Also $1 \in S$.
Thus $S$ is a multiplicative subset of $A$. By the description of
$\Spec(S^{-1}A)$ in
Algebra, Lemma \ref{algebra-lemma-spec-localization}
and by
Algebra, Lemma \ref{algebra-lemma-silly}
we see that the primes of $S^{-1}A$ correspond to the primes of
$A$ contained in one of the $\mathfrak p_i$.
Hence the maximal ideals of $S^{-1}A$ correspond one-to-one with the
maximal (w.r.t.\ inclusion) elements of the set
$\{\mathfrak p_1, \ldots, \mathfrak p_r\}$.
\end{proof}




\section{One dimensional Noetherian schemes}
\label{section-dimension-one}

\noindent
Some material leading up to a discussion of algebraic curves.

\begin{lemma}
\label{lemma-affine}
Let $X$ be a scheme all of whose local rings are Noetherian of dimension
$\leq 1$. Let $U \subset X$ be a retrocompact open. Denote
$j : U \to X$ the inclusion morphism. Then $R^pj_*\mathcal{F} = 0$, $p > 0$
for every quasi-coherent $\mathcal{O}_U$-module $\mathcal{F}$.
\end{lemma}

\begin{proof}
We may check the vanishing of $R^pj_*\mathcal{F}$ at stalks.
Formation of $R^qj_*$ commutes with flat base change, see
Cohomology of Schemes, Lemma
\ref{coherent-lemma-flat-base-change-cohomology}.
Thus we may assume that $X$ is the spectrum of a Noetherian local
ring of dimension $\leq 1$. In this case $X$ has a closed points
$x$ and finitely many other points $x_1, \ldots, x_n$ which specialize
to $x$ but not each other (see
Algebra, Lemma \ref{algebra-lemma-Noetherian-irreducible-components}).
If $x \in U$, then $U = X$ and the result is clear. If not, then
$U = \{x_1, \ldots, x_r\}$ for some $r$ after possibly renumbering
the points. Then $U$ is affine
(Schemes, Lemma \ref{schemes-lemma-scheme-finite-discrete-affine}).
Thus the result by Cohomology of Schemes, Lemma
\ref{coherent-lemma-relative-affine-vanishing}.
\end{proof}

\begin{lemma}
\label{lemma-open-in-affine-curve-affine}
Let $X$ be an affine scheme all of whose local rings are Noetherian
of dimension $\leq 1$. Then any quasi-compact open $U \subset X$ is affine.
\end{lemma}

\begin{proof}
Denote $j : U \to X$ the inclusion morphism. Let $\mathcal{F}$
be a quasi-coherent $\mathcal{O}_U$-module.
By Lemma \ref{lemma-affine} the higher direct images
$R^pj_*\mathcal{F}$ are zero. The $\mathcal{O}_X$-module $j_*\mathcal{F}$
is quasi-coherent
(Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}).
Hence it has vanishing higher cohomology groups by
Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}.
By the Leray spectral sequence
Cohomology, Lemma \ref{cohomology-lemma-apply-Leray}
we have $H^p(U, \mathcal{F}) = 0$ for all $p > 0$.
Thus $U$ is affine, for example by
Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-compact-h1-zero-covering}.
\end{proof}

\begin{lemma}
\label{lemma-complement-codim-1-closed-points}
Let $X$ be a scheme. Let $U \subset X$ be an open. Assume
\begin{enumerate}
\item $U$ is a retrocompact open of $X$,
\item $X \setminus U$ is discrete, and
\item for $x \in X \setminus U$ the local ring
$\mathcal{O}_{X, x}$ is Noetherian of dimension $\leq 1$.
\end{enumerate}
Then (1) there exists an invertible $\mathcal{O}_X$-module $\mathcal{L}$
and a section $s$ such that $U = X_s$ and (2) the map
$\text{Pic}(X) \to \text{Pic}(U)$ is surjective.
\end{lemma}

\begin{proof}
Let $X \setminus U = \{x_i; i \in I\}$.
Choose affine opens $U_i \subset X$ with $x_i \in X$ and
$x_j \not \in U_i$ for $j \not = i$. This is possible by condition (2).
Say $U_i = \Spec(A_i)$. Let $\mathfrak m_i \subset A_i$ be the maximal
ideal corresponding to $x_i$. By our assumption on the local rings
there are only a finite number of prime ideals
$\mathfrak q \subset \mathfrak m_i$,
$\mathfrak q \not = \mathfrak m_i$ (see
Algebra, Lemma \ref{algebra-lemma-Noetherian-irreducible-components}).
Thus by prime avoidance (Algebra, Lemma
\ref{algebra-lemma-silly}) we can find $f_i \in \mathfrak m_i$
not contained in any of those primes. Then
$V(f_i) = \{\mathfrak m_i\} \amalg Z_i$ for some closed subset
$Z_i \subset U_i$ because $Z_i$ is a retrocompact open subset of
$V(f_i)$ closed under specialization, see
Algebra, Lemma \ref{algebra-lemma-constructible-stable-specialization-closed}.
After shrinking $U_i$ we may assume $V(f_i) = \{x_i\}$. Then
$$
\mathcal{U} : X = U \cup \bigcup U_i
$$
is an open covering of $X$. Consider the $2$-cocycle with values
in $\mathcal{O}_X^*$ given by $f_i$ on $U \cap U_i$ and by
$f_i/f_j$ on $U_i \cap U_j$. This defines a line bundle
$\mathcal{L}$ such that the section $s$ defined by $1$ on $U$
and $f_i$ on $U_i$ is as in the statement of the lemma.

\medskip\noindent
Let $\mathcal{N}$ be an invertible $\mathcal{O}_U$-module.
Let $N_i$ be the invertible $(A_i)_{f_i}$ module such that
$\mathcal{N}|_{U \cap U_i}$ is equal to $\tilde N_i$.
Observe that $(A_{\mathfrak m_i})_{f_i}$ is an Artinian ring
(as a dimension zero Noetherian ring, see
Algebra, Lemma \ref{algebra-lemma-Noetherian-dimension-0}).
Thus it is a product of local rings
(Algebra, Lemma \ref{algebra-lemma-artinian-finite-length}) and
hence has trivial Picard group. Thus, after shrinking $U_i$
(i.e., after replacing $A_i$ by $(A_i)_g$ for some $g \in A_i$,
$g \not \in \mathfrak m_i$)
we can assume that $N_i = (A_i)_{f_i}$, i.e., that
$\mathcal{N}|_{U \cap U_i}$ is trivial. In this case it is
clear how to extend $\mathcal{N}$ to an invertible sheaf over $X$
(by extending it by a trivial invertible module over each $U_i$).
\end{proof}

\begin{lemma}
\label{lemma-find-globally-generated}
Let $X$ be an integral separated scheme. Let $U \subset X$ be a nonempty
affine open such that $X \setminus U$ is a finite set of points
$x_1, \ldots, x_r$ with $\mathcal{O}_{X, x_i}$ Noetherian of dimension $1$.
Then there exists a globally generated invertible $\mathcal{O}_X$-module
$\mathcal{L}$ and a section $s$ such that $U = X_s$.
\end{lemma}

\begin{proof}
Say $U = \Spec(A)$ and let $K$ be the fraction field of $X$.
Write $B_i = \mathcal{O}_{X, x_i}$ and $\mathfrak m_i = \mathfrak m_{x_i}$.
Since $x_i \not \in U$ we see that the open
$U \times_X \Spec(B_i)$ of $\Spec(B_i)$ has only one point, i.e.,
$U \times_X \Spec(B_i) = \Spec(K)$.
Since $X$ is separated, we find that $\Spec(K)$ is a closed subscheme
of $U \times \Spec(B_i)$, i.e., the map $A \otimes B_i \to K$ is a surjection.
By Lemma \ref{lemma-create-globally-generated} we can find a nonzero
$f \in A$ such that $f^{-1} \in \mathfrak m_i$ for $i = 1, \ldots, r$.
Pick opens $x_i \in U_i \subset X$ such that $f^{-1} \in \mathcal{O}(U_i)$.
Then
$$
\mathcal{U} : X = U \cup \bigcup U_i
$$
is an open covering of $X$. Consider the $2$-cocycle with values
in $\mathcal{O}_X^*$ given by $f$ on $U \cap U_i$ and by
$1$ on $U_i \cap U_j$. This defines a line bundle $\mathcal{L}$
with two sections:
\begin{enumerate}
\item a section $s$ defined by $1$ on $U$
and $f^{-1}$ on $U_i$ is as in the statement of the lemma, and
\item a section $t$ defined by $f$ on $U$ and
$1$ on $U_i$.
\end{enumerate}
Note that $X_t \supset U_1 \cup \ldots \cup U_r$.
Hence $s, t$ generate $\mathcal{L}$ and the lemma is proved.
\end{proof}

\begin{lemma}
\label{lemma-enough-globally-generated-ample}
Let $X$ be a quasi-compact scheme. If for every $x \in X$
there exists a pair $(\mathcal{L}, s)$ consisting of a globally generated
invertible sheaf $\mathcal{L}$ and a global section $s$ such that
$x \in X_s$ and $X_s$ is affine, then $X$ has an ample invertible
sheaf.
\end{lemma}

\begin{proof}
Since $X$ is quasi-compact we can find a finite collection
$(\mathcal{L}_i, s_i)$, $i = 1, \ldots, n$
of pairs such that $X_{s_i}$ is affine and $X = \bigcup X_{s_i}$.
Again because $X$ is quasi-compact we can find, for each $i$, a finite
collection of sections $t_{i, j}$, $j = 1, \ldots, m_i$
such that $X = \bigcup X_{t_{i, j}}$. Set $t_{i, 0} = s_i$.
Consider the invertible sheaf
$$
\mathcal{L} = \mathcal{L}_1 
\otimes_{\mathcal{O}_X} \ldots
\otimes_{\mathcal{O}_X} \mathcal{L}_n
$$
and the global sections
$$
\tau_J = t_{1, j_1} \otimes \ldots \otimes t_{n, j_n}
$$
By Properties, Lemma \ref{properties-lemma-affine-cap-s-open}
the open $X_{\tau_J}$ is affine as soon as $j_i = 0$ for some $i$.
It is a simple matter to see that these opens cover $X$.
Hence $\mathcal{L}$ is ample by definition.
\end{proof}

\begin{lemma}
\label{lemma-dim-1-noetherian-integral-separated-has-ample}
Let $X$ be a Noetherian integral separated scheme of dimension $1$.
Then $X$ has an ample invertible sheaf.
\end{lemma}

\begin{proof}
Choose an affine open covering $X = U_1 \cup \ldots \cup U_n$.
Since $X$ is Noetherian, each of the sets $X \setminus U_i$ is finite.
Thus by Lemma \ref{lemma-find-globally-generated}
we can find a pair $(\mathcal{L}_i, s_i)$
consisting of a globally generated invertible sheaf $\mathcal{L}_i$
and a global section $s_i$ such that $U_i = X_{s_i}$.
We conclude that $X$ has an ample invertible sheaf by
Lemma \ref{lemma-enough-globally-generated-ample}.
\end{proof}

\begin{lemma}
\label{lemma-glue-invertible-sheaves}
Let $X$ be a scheme. Let $Z_1, \ldots, Z_n \subset X$ be closed
subschemes. Let $\mathcal{L}_i$ be an invertible sheaf on $Z_i$.
Assume that
\begin{enumerate}
\item $X$ is reduced,
\item $X = \bigcup Z_i$ set theoretically, and
\item $Z_i \cap Z_j$ is a discrete topological space for $i \not = j$.
\end{enumerate}
Then there exists an invertible sheaf $\mathcal{L}$ on $X$ whose restriction
to $Z_i$ is $\mathcal{L}_i$. Moreover, if we are given sections
$s_i \in \Gamma(Z_i, \mathcal{L}_i)$ which are nonvanishing at the
points of $Z_i \cap Z_j$, then we can choose $\mathcal{L}$ such
that there exists a $s \in \Gamma(X, \mathcal{L})$ with
$s|_{Z_i} = s_i$ for all $i$.
\end{lemma}

\begin{proof}
Set $T = \bigcup_{i \not = j} Z_i \cap Z_j$. As $X$ is reduced we have
$$
X \setminus T = \bigcup (Z_i \setminus T)
$$
as schemes. Assumption (3) implies $T$ is a discrete subset of $X$.
Thus for each $t \in T$ we can find an open $U_t \subset X$
with $t \in U_t$ but $t' \not \in U_t$ for $t' \in T$, $t' \not = t$.
By shrinking $U_t$ if necessary, we may assume that there exist isomorphisms
$\varphi_{t, i} : \mathcal{L}_i|_{U_t \cap Z_i} \to
\mathcal{O}_{U_t \cap Z_i}$. Furthermore, for each $i$ choose an open covering
$$
Z_i \setminus T = \bigcup\nolimits_j U_{ij}
$$
such that there exist isomorphisms
$\varphi_{i, j} : \mathcal{L}_i|_{U_{ij}} \cong \mathcal{O}_{U_{ij}}$.
Observe that
$$
\mathcal{U} : X = \bigcup U_t \cup \bigcup U_{ij}
$$
is an open covering of $X$. We claim that we can use the isomorphisms
$\varphi_{t, i}$ and $\varphi_{i, j}$ to define a $2$-cocycle with values
in $\mathcal{O}_X^*$ for this covering that defines $\mathcal{L}$ as
in the statement of the lemma.

\medskip\noindent
Namely, if $i \not = i'$, then $U_{i, j} \cap U_{i', j'} = \emptyset$
and there is nothing to do. For $U_{i, j} \cap U_{i, j'}$ we have
$\mathcal{O}_X(U_{i, j} \cap U_{i, j'}) =
\mathcal{O}_{Z_i}(U_{i, j} \cap U_{i, j'})$ by the first remark of the proof.
Thus the transition function for $\mathcal{L}_i$ (more precisely
$\varphi_{i, j} \circ \varphi_{i, j'}^{-1}$) defines the value of our
cocycle on this intersection.
For $U_t \cap U_{i, j}$ we can do the same thing as before.
Finally, for $t \not = t'$ we have
$$
U_t \cap U_{t'} = \coprod (U_t \cap U_{t'}) \cap Z_i
$$
and moreover the intersections $U_t \cap U_{t'} \cap Z_i$ is contained
in $Z_ti \setminus T$. Hence by the same reasoning as before we see that
$$
\mathcal{O}_X(U_t \cap U_{t'}) =
\prod \mathcal{O}_{Z_i}(U_t \cap U_{t'} \cap Z_i)
$$
and we can use the transition functions fpr $\mathcal{L}_i$ (more precisely
$\varphi_{t, i} \circ \varphi_{t', i}^{-1}$) to define the value of
our cocycle on $U_t \cap U_{t'}$. This finishes the proof of existence
of $\mathcal{L}$.

\medskip\noindent
Given sections $s_i$ as in the last assertion of the lemma, in the argument
above, we choose $U_t$ such that $s_i|_{U_t \cap Z_i}$ is nonvanishing and
we choose $\varphi_{t, i}$ such that $\varphi_{t, i}(s_i|_{U_t \cap Z_i}) = 1$.
Then using $1$ over $U_t$ and $\varphi_{i, j}(s_i|_{U_{i, j}})$ over
$U_{i, j}$ will define a section of $\mathcal{L}$ which restricts
to $s_i$ over $Z_i$.
\end{proof}

\begin{remark}
\label{remark-conductor}
Let $A$ be a reduced ring. Let $I, J$ be ideals of $A$
such that $V(I) \cup V(J) = \Spec(A)$. Set $B = A/J$.
Then $I \to IB$ is an isomorphism of $A$-modules. Namely, we
have $IB = I + J/J = I/(I \cap J)$ and $I \cap J$ is zero because
$A$ is reduced and $\Spec(A) = V(I) \cup V(J) = V(I \cap J)$.
Thus for any projective $A$-module $P$ we also have $IP = I(P/JP)$.
\end{remark}

\begin{lemma}
\label{lemma-dim-1-noetherian-reduced-separated-has-ample}
Let $X$ be a Noetherian reduced separated scheme of dimension $1$.
Then $X$ has an ample invertible sheaf.
\end{lemma}

\begin{proof}
Let $Z_i$, $i = 1, \ldots, n$ be the irreducible components of $X$.
We view these as reduced closed subschemes of $X$.
By Lemma \ref{lemma-dim-1-noetherian-integral-separated-has-ample}
there exist ample invertible sheaves $\mathcal{L}_i$ on $Z_i$.
Set $T = \bigcup_{i \not = j} Z_i \cap Z_j$. As $X$ is Noetherian
of dimension $1$, the set $T$ is finite and consists of closed
points of $X$. For each $i$ we may, possibly after replacing
$\mathcal{L}_i$ by a power, choose $s_i \in \Gamma(Z_i, \mathcal{L}_i)$
such that $(Z_i)_{s_i}$ is affine and contains $T \cap Z_i$, see
Properties, Lemma \ref{properties-lemma-ample-finite-set-in-principal-affine}.

\medskip\noindent
By Lemma \ref{lemma-glue-invertible-sheaves} we can find an invertible sheaf
$\mathcal{L}$ on $X$ and $s \in \Gamma(X, \mathcal{L})$
such that $(\mathcal{L}, s)|_{Z_i} = (\mathcal{L}_i, s_i)$.
Observe that $X_s$ contains $T$ and is set theoretically equal
to the affine closed subschemes $(Z_i)_{s_i}$. Thus it is affine by
Limits, Lemma \ref{limits-lemma-affines-glued-in-closed-affine}.
To finish the proof, it suffices to find for every $x \in X$, $x \not \in T$
an integer $m > 0$ and a section $t \in \Gamma(X, \mathcal{L}^{\otimes m})$
such that $X_t$ is affine and $x \in X_t$. Since $x \not \in T$
we see that $x \in Z_i$ for some unique $i$, say $i = 1$.
Let $Z \subset X$ be the reduced closed subscheme whose underlying
topological space is $Z_2 \cup \ldots \cup Z_n$.
Let $\mathcal{I} \subset \mathcal{O}_X$ be the ideal
sheaf of $Z$. Denote that $\mathcal{I}_1 \subset \mathcal{O}_{Z_1}$
the inverse image of this ideal sheaf under the inclusion
morphism $Z_1 \to X$. Observe that
$$
\Gamma(X, \mathcal{I}\mathcal{L}^{\otimes m}) =
\Gamma(Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m})
$$
see Remark \ref{remark-conductor}. Thus it suffices to find $m > 0$
and $t \in \Gamma(Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m})$
with $x \in (Z_1)_t$ affine. Since $\mathcal{L}_1$ is ample
and since $x$ is not in $Z_1 \cap T = V(\mathcal{I}_1)$
we can find a section
$t_1 \in \Gamma(Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m_1})$
with $x \in (Z_1)_{t_1}$, see Properties, Proposition
\ref{properties-proposition-characterize-ample}.
Since $\mathcal{L}_1$ is ample we can find a section
$t_2 \in \Gamma(Z_1, \mathcal{L}_1^{\otimes m_2})$
with $x \in (Z_1)_{t_2}$ and $(Z_1)_{t_2}$ affine, see
Properties, Definition \ref{properties-definition-ample}.
Set $m = m_1 + m_2$ and $t = t_1 t_2$. Then
$t \in \Gamma(Z_1, \mathcal{I}_1 \mathcal{L}_1^{\otimes m})$
with $x \in (Z_1)_t$ by construction and
$(Z_1)_t$ is affine by Properties, Lemma
\ref{properties-lemma-affine-cap-s-open}.
\end{proof}

\begin{lemma}
\label{lemma-lift-line-bundle-from-reduction-dimension-1}
Let $i : Z \to X$ be a closed immersion of schemes
inducing a homeomorphism on underlying topological spaces.
If the underlying topological space of $X$ is Noetherian and
$\dim(X) \leq 1$, then $\text{Pic}(X) \to \text{Pic}(Z)$ is surjective.
\end{lemma}

\begin{proof}
Consider the short exact sequence
$$
0 \to (1 + \mathcal{I})^* \to \mathcal{O}^*_X \to \mathcal{O}^*_Z \to 0
$$
of sheaves of abelian groups on $X$. Since $\dim(X) \leq 1$ we see that
$H^2(X, \mathcal{F}) = 0$ for any abelian sheaf $\mathcal{F}$, see
Cohomology, Lemma \ref{cohomology-lemma-vanishing-Noetherian}.
Hence the map $H^1(X, \mathcal{O}^*_X) \to H^1(Z, \mathcal{O}_Z^*)$
is surjective. This proves the lemma by
Cohomology, Lemma \ref{cohomology-lemma-h1-invertible}.
\end{proof}

\begin{proposition}
\label{proposition-dim-1-noetherian-separated-has-ample}
Let $X$ be a Noetherian separated scheme of dimension $1$.
Then $X$ has an ample invertible sheaf.
\end{proposition}

\begin{proof}
Let $Z \subset X$ be the reduction of $X$. By
Lemma \ref{lemma-dim-1-noetherian-reduced-separated-has-ample}
the scheme $Z$ has an ample invertible sheaf.
Thus by Lemma \ref{lemma-lift-line-bundle-from-reduction-dimension-1}
there exists an invertible $\mathcal{O}_X$-module $\mathcal{L}$
on $X$ whose restriction to $Z$ is ample.
Then $\mathcal{L}$ is ample by an application of
Cohomology of Schemes, Lemma \ref{coherent-lemma-ample-on-reduction}.
\end{proof}

\begin{remark}
\label{remark-useless-generalization}
In fact, if $X$ is a scheme whose reduction is a Noetherian
separated scheme of dimension $1$, then $X$ has an ample invertible
sheaf. The argument to prove this is the same as the proof of
Proposition \ref{proposition-dim-1-noetherian-separated-has-ample}
except one uses
Limits, Lemma \ref{limits-lemma-ample-on-reduction}
instead of
Cohomology of Schemes, Lemma \ref{coherent-lemma-ample-on-reduction}.
\end{remark}















\section{Finding affine opens}
\label{section-finding-affine-opens}

\noindent
We continue the discussion started in
Properties, Section \ref{properties-section-finding-affine-opens}.
It turns out that we can find affines containing a finite given set
of codimension $1$ points on a separated scheme. See
Proposition \ref{proposition-finite-set-of-points-of-codim-1-in-affine}.

\medskip\noindent
We will improve on the following lemma in
Descent, Lemma \ref{descent-lemma-characterize-open-immersion}.

\begin{lemma}
\label{lemma-characterize-open-immersion}
Let $f : X \to Y$ be a morphism of schemes. Let $X^0$ denote the set
of generic points of irreducible components of $X$. If
\begin{enumerate}
\item $f$ is separated,
\item there is an open covering $X = \bigcup U_i$ such that
$f|_{U_i} : U_i \to X$ is an open immersion, and
\item if $\xi, \xi' \in X^0$, $\xi \not = \xi'$, then $f(\xi) \not = f(\xi')$,
\end{enumerate}
then $f$ is an open immersion.
\end{lemma}

\begin{proof}
Suppose that $y = f(x) = f(x')$. Pick a specialization $y_0 \leadsto y$
where $y_0$ is a generic point of an irreducible component of $Y$.
Since $f$ is locally on the source an isomorphism we can pick specializations
$x_0 \leadsto x$ and $x'_0 \leadsto x'$ mapping to $y_0 \leadsto y$.
Note that $x_0, x'_0 \in X^0$. Hence $x_0 = x'_0$ by assumption (3).
As $f$ is separated we conclude that $x = x'$. Thus $f$ is an open immersion.
\end{proof}

\begin{lemma}
\label{lemma-local-isomorphism}
Let $X \to S$ be a morphism of schemes. Let $x \in X$ be a point with
image $s \in S$. If
\begin{enumerate}
\item $\mathcal{O}_{X, x} = \mathcal{O}_{S, s}$,
\item $X$ is reduced,
\item $X \to S$ is of finite type, and
\item $S$ has finitely many irreducible components,
\end{enumerate}
then there exists an open neighbourhood $U$
of $x$ such that $f|_U$ is an open immersion.
\end{lemma}

\begin{proof}
We may remove the (finitely many) irreducible components of $S$
which do not contain $s$. We may replace $S$ by an affine open
neighbourhood of $s$. We may replace $X$ by an affine open neighbourhood
of $x$. Say $S = \Spec(A)$ and $X = \Spec(B)$. Let $\mathfrak q \subset B$,
resp.\ $\mathfrak p \subset A$ be the prime ideal corresponding to $x$,
resp.\ $s$. As $A$ is a reduced and all of the minimal primes of
$A$ are contained in $\mathfrak p$ we see that $A \subset A_\mathfrak p$.
As $X \to S$ is of finite type, $B$ is of finite type over $A$.
Let $b_1, \ldots, b_n \in B$ be elements which generate $B$ over $A$
Since $A_\mathfrak p = B_\mathfrak q$ we can find
$f \in A$, $f \not \in \mathfrak p$
and $a_i \in A$ such that $b_i$ and $a_i/f$ have the same image
in $B_\mathfrak q$. Thus we can find $g \in B$, $g \not \in \mathfrak q$
such that $g(fb_i - a_i) = 0$ in $B$. It follows that the image of
$A_f \to B_{fg}$ contains the images of $b_1, \ldots, b_n$, in
particular also the image of $g$.
Choose $n \geq 0$ and $f' \in A$ such that $f'/f^n$ maps
to the image of $g$ in $B_{fg}$. Since $A_\mathfrak p = B_\mathfrak q$
we see that $f' \not \in \mathfrak p$.
We conclude that $A_{ff'} \to B_{fg}$ is surjective.
Finally, as $A_{ff'} \subset A_\mathfrak p = B_\mathfrak q$ (see above)
the map $A_{ff'} \to B_{fg}$ is injective, hence an isomorphism.
\end{proof}

\begin{lemma}
\label{lemma-points-in-affine}
Let $f : T \to X$ be a morphism of schemes. Let $X^0$, resp.\ $T^0$
denote the sets of generic points of irreducible components.
Let $t_1, \ldots, t_m \in T$ be a finite set of points
with images $x_j = f(t_j)$. If
\begin{enumerate}
\item $T$ is affine,
\item $X$ is quasi-separated,
\item $X^0$ is finite
\item $f(T^0) \subset X^0$ and $f : T^0 \to X^0$ is injective, and
\item $\mathcal{O}_{X, x_j} = \mathcal{O}_{T, t_j}$,
\end{enumerate}
then there exists an affine open of $X$ containing $x_1, \ldots, x_r$.
\end{lemma}

\begin{proof}
Using Limits, Proposition \ref{limits-proposition-affine}
there is an immediate reduction to the case where $X$ and $T$ are reduced.
Details omitted.

\medskip\noindent
Assume $X$ and $T$ are reduced. We may write $T = \lim_{i \in I} T_i$
as a directed limit of schemes of finite presentation over $X$
with affine transition morphisms, see
Limits, Lemma \ref{limits-lemma-relative-approximation}.
Pick $i \in I$ such that $T_i$ is affine, see
Limits, Lemma \ref{limits-lemma-limit-affine}.
Say $T_i = \Spec(R_i)$ and $T = \Spec(R)$.
Let $R' \subset R$ be the image of $R_i \to R$.
Then $T' = \Spec(R')$ is affine, reduced, of finite type over $X$,
and $T \to T'$ dominant. For $j = 1, \ldots, r$ let $t'_j \in T'$
be the image of $t_j$. Consider the local ring maps
$$
\mathcal{O}_{X, x_j} \to
\mathcal{O}_{T', t'_j} \to
\mathcal{O}_{T, t_j}
$$
Denote $(T')^0$ the set of generic points of irreducible
components of $T'$. Let $\xi \leadsto t'_j$ be a specialization with
$\xi \in (T')^0$. As $T \to T'$ is dominant we can choose $\eta \in T^0$ 
mapping to $\xi$ (warning: a priori we do not know that $\eta$ specializes
to $t_j$). Assumption (3) applied to $\eta$ tells us that the image $\theta$
of $\xi$ in $X$ corresponds to a minimal prime of $\mathcal{O}_{X, x_j}$.
Lifting $\xi$ via the isomorphism of (5) we obtain a specialization
$\eta' \leadsto t_j$ with $\eta' \in X^0$ mapping to $\theta \leadsto x_j$.
The injectivity of (4) shows that $\eta = \eta'$. Thus
every minimal prime of $\mathcal{O}_{T', t'_j}$ lies below
a minimal prime of $\mathcal{O}_{T, t_j}$. We conclude that
$\mathcal{O}_{T', t'_j} \to \mathcal{O}_{T, t_j}$ is injective,
hence both maps above are isomorphisms.

\medskip\noindent
By Lemma \ref{lemma-local-isomorphism} there exists an open
$U \subset T'$ containing all the points $t'_j$ such that
$U \to X$ is a local isomorphism as in
Lemma \ref{lemma-characterize-open-immersion}.
By that lemma we see that $U \to X$ is an open immersion.
Finally, by
Properties, Lemma \ref{properties-lemma-ample-finite-set-in-affine}
we can find an open $W \subset U \subset T'$
containing all the $t'_j$. The image of $W$ in $X$ is the
desired affine open.
\end{proof}

\begin{lemma}
\label{lemma-finite-set-codim-1-points-in-affine}
Let $X$ be an integral separated scheme. Let $x_1, \ldots, x_r \in X$
be a finite set of points such that $\mathcal{O}_{X, x_i}$
is Noetherian of dimension $\leq 1$. Then there exists an affine
open subscheme of $X$ containing all of $x_1, \ldots, x_r$.
\end{lemma}

\begin{proof}
Let $K$ be the field of rational functions of $X$.
Set $A_i = \mathcal{O}_{X, x_i}$. Then $A_i \subset K$ and $K$
is the fraction field of $A_i$. Since $X$ is separated, and
$x_i \not = x_j$ there cannot be a valuation ring $\mathcal{O} \subset K$
dominating both $A_i$ and $A_j$. Namely, considering the
diagram
$$
\xymatrix{
\Spec(\mathcal{O}) \ar[r] \ar[d] & \Spec(A_1) \ar[d] \\
\Spec(A_2) \ar[r] & X
}
$$
and applying the valuative criterion of separatedness
(Schemes, Lemma \ref{schemes-lemma-separated-implies-valuative})
we would get $x_i = x_j$. Thus we see by
Lemma \ref{lemma-glue-separated}
that $A_i \otimes A_j \to K$ is surjective for all $i \not = j$.
By Lemma \ref{lemma-glue-a-bunch-of-local-rings}
we see that $A = A_1 \cap \ldots \cap A_r$ is a Noetherian
semi-local rings with exactly $r$ maximal ideals
$\mathfrak m_1, \ldots, \mathfrak m_r$ such that $A_i = A_{\mathfrak m_i}$.
Moreover,
$$
\Spec(A) = \Spec(A_1) \cup \ldots \cup \Spec(A_r)
$$
is an open covering and the intersection of any two pieces of this
covering is $\Spec(K)$. Thus the given morphisms $\Spec(A_i) \to X$
glue to a morphism of schemes
$$
\Spec(A) \longrightarrow X
$$
mapping $\mathfrak m_i$ to $x_i$ and inducing isomorphisms of local rings.
Thus the result follows from Lemma \ref{lemma-points-in-affine}.
\end{proof}

\begin{lemma}
\label{lemma-extra-silly}
Let $A$ be a ring, $I \subset A$ an ideal,
$\mathfrak p_1, \ldots, \mathfrak p_r$ primes of $A$, and
$\overline{f} \in A/I$ an element. If $I \not \subset \mathfrak p_i$
for all $i$, then there exists an $f \in A$, $f \not \in \mathfrak p_i$
which maps to $\overline{f}$ in $A/I$.
\end{lemma}

\begin{proof}
We may assume there are no inclusion relations among the $\mathfrak p_i$
(by removing the smaller primes). First pick any $f \in A$ lifting
$\overline{f}$. Let $S$ be the set $s \in \{1, \ldots, r\}$ such
that $f \in \mathfrak p_s$. If $S$ is empty we are done. If not,
consider the ideal $J = I \prod_{i \not \in S} \mathfrak p_i$.
Note that $J$ is not contained in $\mathfrak p_s$ for $s \in S$
because there are no inclusions among the $\mathfrak p_i$ and because
$I$ is not contained in any $\mathfrak p_i$.
Hence we can choose $g \in J$, $g \not \in \mathfrak p_s$ for
$s \in S$ by Algebra, Lemma \ref{algebra-lemma-silly}.
Then $f + g$ is a solution to the problem posed by the lemma.
\end{proof}

\begin{lemma}
\label{lemma-finite-set-codim-1-points-in-affine-per-component}
Let $X$ be a scheme. Let $T \subset X$ be finite set of points. Assume
\begin{enumerate}
\item $X$ has finitely many irreducible components $Z_1, \ldots, Z_t$, and
\item $Z_i \cap T$ is contained in an affine open of the reduced
induced subscheme corresponding to $Z_i$.
\end{enumerate}
Then there exists an affine open subscheme of $X$ containing $T$.
\end{lemma}

\begin{proof}
Using Limits, Proposition \ref{limits-proposition-affine}
there is an immediate reduction to the case where $X$
is reduced. Details omitted. In the rest of the proof we
endow every closed subset of $X$ with the induced reduced closed
subscheme structure.

\medskip\noindent
We argue by induction that we can find an affine open
$U \subset Z_1 \cup \ldots \cup Z_r$ containing
$T \cap (Z_1 \cup \ldots \cup Z_r)$. For $r = 1$ this holds by assumption.
Say $r > 1$ and let $U \subset Z_1 \cup \ldots \cup Z_{r -  1}$
be an affine open containing $T \cap (Z_1 \cup \ldots \cup Z_{r - 1})$.
Let $V \subset X_r$ be an affine open containing $T \cap Z_r$ (exists by
assumption). Then $U \cap V$ contains
$T \cap ( Z_1 \cup \ldots \cup Z_{r - 1} ) \cap Z_r$.
Hence
$$
\Delta = (U \cap Z_r) \setminus (U \cap V)
$$
does not contain any element of $T$. Note that $\Delta$ is a closed
subset of $U$. By prime avoidance (Algebra, Lemma \ref{algebra-lemma-silly}),
we can find a standard open $U'$ of $U$ containing $T \cap U$ and avoiding
$\Delta$, i.e., $U' \cap Z_r \subset U \cap V$.
After replacing $U$ by $U'$ we may assume that $U \cap V$ is
closed in $U$.

\medskip\noindent
Using that by the same arguments as above also the set
$\Delta' = (U \cap (Z_1 \cup \ldots \cup Z_{r - 1})) \setminus (U \cap V)$
does not contain any element of $T$ we find a $h \in \mathcal{O}(V)$
such that $D(h) \subset V$ contains $T \cap V$ and such that
$U \cap D(h) \subset U \cap V$. Using that $U \cap V$ is closed in $U$
we can use Lemma \ref{lemma-extra-silly}
to find an element $g \in \mathcal{O}(U)$ whose restriction to $U \cap V$
equals the restriction of $h$ to $U \cap V$ and such that
$T \cap U \subset D(g)$. Then we can replace $U$ by $D(g)$
and $V$ by $D(h)$ to reach the situation where $U \cap V$ is
closed in both $U$ and $V$. In this case the scheme $U \cup V$ is
affine by Limits, Lemma \ref{limits-lemma-affines-glued-in-closed-affine}.
This proves the induction step and thereby the lemma.
\end{proof}

\noindent
Here is a conclusion we can draw from the material above.

\begin{proposition}
\label{proposition-finite-set-of-points-of-codim-1-in-affine}
Let $X$ be a separated scheme such that every quasi-compact open
has a finite number of irreducible components. Let $x_1, \ldots, x_r \in X$
be points such that $\mathcal{O}_{X, x_i}$ is Noetherian
of dimension $\leq 1$. Then there exists an affine open subscheme
of $X$ containing all of $x_1, \ldots, x_r$.
\end{proposition}

\begin{proof}
We can replace $X$ by a quasi-compact open containing $x_1, \ldots, x_r$
hence we may assume that $X$ has finitely many irreducible components.
By Lemma \ref{lemma-finite-set-codim-1-points-in-affine-per-component}
we reduce to the case where $X$ is integral. This case is
Lemma \ref{lemma-finite-set-codim-1-points-in-affine}.
\end{proof}








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