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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{Varieties and rational maps}
\label{section-varieties-rational-maps}
\noindent
Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$.
We will use the phrase {\it rational map of varieties from $X$ to $Y$}
to mean a $\Spec(k)$-rational map from the scheme $X$ to the scheme $Y$
as defined in Morphisms, Definition \ref{morphisms-definition-rational-map}.
As is customary, the phrase ``rational map of varieties''
does not refer to the (common) base field of the varieties,
even though for general schemes we make the distinction between
rational maps and rational maps over a given base.
\medskip\noindent
The title of this section refers to the following fundamental theorem.
\begin{theorem}
\label{theorem-varieties-rational-maps}
Let $k$ be a field. The category of varieties and
dominant rational maps is equivalent to the category of
finitely generated field extensions $K/k$.
\end{theorem}
\begin{proof}
Let $X$ and $Y$ be varieties with generic points $x \in X$ and $y \in Y$.
Recall that dominant rational maps from $X$ to $Y$ are exactly those
rational maps which map $x$ to $y$
(Morphisms, Definition \ref{morphisms-definition-dominant-rational}
and discussion following).
Thus given a dominant rational map $X \supset U \to Y$ we obtain a map of
function fields
$$
k(Y) = \kappa(y) = \mathcal{O}_{Y, y}
\longrightarrow
\mathcal{O}_{X, x} = \kappa(x) = k(X)
$$
Conversely, such a $k$-algebra map (which is automatically local as the
source and target are fields) determines (uniquely) a dominant rational
map by Morphisms, Lemma \ref{morphisms-lemma-rational-map-finite-presentation}.
In this way we obtain a fully faithful functor.
To finish the proof it suffices to show that every finitely generated
field extension $K/k$ is in the essential image.
Since $K/k$ is finitely generated, there exists a finite type
$k$-algebra $A \subset K$ such that $K$ is the fraction field of $A$.
Then $X = \Spec(A)$ is a variety whose function field is $K$.
\end{proof}
\noindent
Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$.
We will use the phrase {\it $X$ and $Y$ are birational varieties}
to mean $X$ and $Y$ are $\Spec(k)$-birational
as defined in Morphisms, Definition \ref{morphisms-definition-birational}.
As is customary, the phrase ``birational varieties''
does not refer to the (common) base field of the varieties,
even though for general irreducible schemes we make the distinction between
being birational and being birational over a given base.
\begin{lemma}
\label{lemma-birational-varieties}
Let $X$ and $Y$ be varieties over a field $k$.
The following are equivalent
\begin{enumerate}
\item $X$ and $Y$ are birational varieties,
\item the function fields $k(X)$ and $k(Y)$ are isomorphic,
\item there exist nonempty opens of $X$ and $Y$ which are isomorphic
as varieties,
\item there exists an open $U \subset X$ and a birational morphism
$U \to Y$ of varieties.
\end{enumerate}
\end{lemma}
\begin{proof}
This is a special case of
Morphisms, Lemma \ref{morphisms-lemma-criterion-birational-finite-presentation}.
\end{proof}
\section{Change of fields and local rings}
\label{section-local-rings}
\noindent
Some preliminary results on what happens to local rings under
an extension of ground fields.
\begin{lemma}
\label{lemma-change-fields-flat}
Let $K/k$ be an extension of fields. Let $X$ be scheme over $k$
and set $Y = X_K$. If $y \in Y$ with image $x \in X$, then
\begin{enumerate}
\item $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is a
faithfully flat local ring homomorphism,
\item with $\mathfrak p_0 = \Ker(\kappa(x) \otimes_k K \to \kappa(y))$
we have $\kappa(y) = \kappa(\mathfrak p_0)$,
\item $\mathcal{O}_{Y, y} = (\mathcal{O}_{X, x} \otimes_k K)_\mathfrak p$
where $\mathfrak p \subset \mathcal{O}_{X, x} \otimes_k K$ is the inverse
image of $\mathfrak p_0$.
\item we have
$\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y} =
(\kappa(x) \otimes_k K)_{\mathfrak p_0}$
\end{enumerate}
\end{lemma}
\begin{proof}
We may assume $X = \Spec(A)$ is affine. Then $Y = \Spec(A \otimes_k K)$.
Since $K$ is flat over $k$, we see that $A \to A \otimes_k K$ is flat.
Hence $Y \to X$ is flat and we get the first statement if we also
use Algebra, Lemma \ref{algebra-lemma-local-flat-ff}.
The second statement follows from
Schemes, Lemma \ref{schemes-lemma-points-fibre-product}.
Now $y$ corresponds to a prime ideal $\mathfrak q \subset A \otimes_k K$
and $x$ to $\mathfrak r = A \cap \mathfrak q$. Then $\mathfrak p_0$
is the kernel of the induced map
$\kappa(\mathfrak r) \otimes_k K \to \kappa(\mathfrak q)$.
The map on local rings is
$$
A_\mathfrak r \longrightarrow (A \otimes_k K)_\mathfrak q
$$
We can factor this map through
$A_\mathfrak r \otimes_k K = (A \otimes_k K)_{\mathfrak r}$
to get
$$
A_\mathfrak r \longrightarrow A_\mathfrak r \otimes_k K
\longrightarrow (A \otimes_k K)_\mathfrak q
$$
and then the second arrow is a localization at some prime. This prime ideal
is the inverse image of $\mathfrak p_0$ (details omitted) and this
proves (3). To see (4) use (3) and that localization and $- \otimes_k K$
are exact functors.
\end{proof}
\begin{lemma}
\label{lemma-change-fields-algebraic-dim}
Notation as in Lemma \ref{lemma-change-fields-flat}.
Assume $X$ is locally of finite type over $k$. Then
$$
\dim(\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y}) =
\text{trdeg}_k(\kappa(x)) - \text{trdeg}_K(\kappa(y)) =
\dim(\mathcal{O}_{Y, y}) - \dim(\mathcal{O}_{X, x})
$$
\end{lemma}
\begin{proof}
This is a restatement of Algebra, Lemma
\ref{algebra-lemma-inequalities-under-field-extension}.
\end{proof}
\begin{lemma}
\label{lemma-change-fields-algebraic-unramified}
Notation as in Lemma \ref{lemma-change-fields-flat}.
Assume $X$ is locally of finite type over $k$,
that $\dim(\mathcal{O}_{X, x}) = \dim(\mathcal{O}_{Y, y})$
and that $\kappa(x) \otimes_k K$ is reduced
(for example if $\kappa(x)/k$ is separable or $K/k$ is separable).
Then $\mathfrak m_x \mathcal{O}_{Y, y} = \mathfrak m_y$.
\end{lemma}
\begin{proof}
(The parenthetical statement follows from
Algebra, Lemma \ref{algebra-lemma-separable-extension-preserves-reducedness}.)
Combining Lemmas \ref{lemma-change-fields-flat} and
\ref{lemma-change-fields-algebraic-dim}
we see that $\mathcal{O}_{Y, y}/\mathfrak m_x \mathcal{O}_{Y, y}$
has dimension $0$ and is reduced. Hence it is a field.
\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}
Assume $X$ locally Noetherian and geometrically reduced at $x$.
By Properties, Lemma
\ref{properties-lemma-ring-affine-open-injective-local-ring}
we can find an affine open neighbourhood $U \subset X$ of $x$ such that
$R = \mathcal{O}_X(U) \to \mathcal{O}_{X, x}$
is injective. By
Lemma \ref{lemma-geometrically-reduced-at-point} the assumption
means that $\mathcal{O}_{X, x}$ 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 $\mathcal{O}_{(X_{k'})_{red}, x'_i}$ 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 $X_{\overline{k}}$ 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 $X_{\overline{k}} = \overline{U} \amalg \overline{V}$
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 $W_{k'} = U' \amalg V'$.
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
$W_{k'} = U'' \amalg V''$.
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
$W = U_W \amalg V_W$.
\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 $X = U \amalg V$.
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}
\noindent
The following lemma will be superseded by the stronger
Lemma \ref{lemma-image-connected-component} below.
\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{lemma}[Gabber]
\label{lemma-image-connected-component}
\begin{reference}
Email from Ofer Gabber dated June 4, 2016
\end{reference}
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.
Then $p(\overline{T})$ is a connected component of $X$.
\end{lemma}
\begin{proof}
When $k \subset K$ is finite this is
Lemma \ref{lemma-image-connected-component-finite-extension}.
In general the proof is more difficult.
\medskip\noindent
Let $T \subset X$ be the connected component of $X$ containing
the image of $\overline{T}$. We may replace $X$ by $T$
(with the induced reduced subscheme structure). Thus we
may assume $X$ is connected. Let $A = H^0(X, \mathcal{O}_X)$.
Let $L \subset A$ be the maximal weakly \'etale $k$-subalgebra, see
More on Algebra, Lemma \ref{more-algebra-lemma-max-weakly-etale-subalgebra}.
Since $A$ does not have any nontrivial idempotents we see
that $L$ is a field and a separable algebraic extension of $k$ by
More on Algebra, Lemma \ref{more-algebra-lemma-class-weakly-etale-over-field}.
Observe that $L$ is also the maximal weakly \'etale $L$-subalgebra of $A$
(because any weakly \'etale $L$-algebra is weakly \'etale over $k$
by More on Algebra, Lemma \ref{more-algebra-lemma-composition-weakly-etale}).
By Schemes, Lemma \ref{schemes-lemma-morphism-into-affine}
we obtain a factorization $X \to \Spec(L) \to \Spec(k)$
of the structure morphism.
\medskip\noindent
Let $L'/L$ be a finite separable extension. By
Cohomology of Schemes, Lemma
\ref{coherent-lemma-finite-locally-free-base-change-cohomology}
we have
$$
A \otimes_L L' =
H^0(X \times_{\Spec(L)} \Spec(L'), \mathcal{O}_{X \times_{\Spec(L)} \Spec(L')})
$$
The maximal weakly \'etale $L'$-subalgebra of $A \otimes_L L'$
is $L \otimes_L L' = L'$ by More on Algebra, Lemma
\ref{more-algebra-lemma-change-fields-max-weakly-etale-subalgebra}.
In particular $A \otimes_L L'$ does not have nontrivial idempotents
(such an idempotent would generate a weakly \'etale subalgebra)
and we conclude that $X \times_{\Spec(L)} \Spec(L')$ is connected.
By Lemma \ref{lemma-characterize-geometrically-disconnected}
we conclude that $X$ is geometrically connected over $L$.
\medskip\noindent
Let's give $\overline{T}$ the reduced induced scheme structure
and consider the composition
$$
\overline{T} \xrightarrow{i} X_K = X \times_{\Spec(k)} \Spec(K)
\xrightarrow{\pi}
\Spec(L \otimes_k K)
$$
The image is contained in a connected component of $\Spec(L \otimes_k K)$.
Since $K \to L \otimes_k K$ is integral we see that
the connected components of $\Spec(L \otimes_k K)$
are points and all points are closed, see
Algebra, Lemma \ref{algebra-lemma-integral-over-field}.
Thus we get a quotient field $L \otimes_k K \to E$
such that $\overline{T}$ maps into $\Spec(E) \subset \Spec(L \otimes_k K)$.
Hence $i(\overline{T}) \subset \pi^{-1}(\Spec(E))$. But
$$
\pi^{-1}(\Spec(E)) =
(X \times_{\Spec(k)} \Spec(K)) \times_{\Spec(L \otimes_k K)} \Spec(E) =
X \times_{\Spec(L)} \Spec(E)
$$
which is connected because $X$ is geometrically connected over $L$.
Then we get the equality
$\overline{T} = X \times_{\Spec(L)} \Spec(E)$ (set theoretically)
and we conclude that $\overline{T} \to X$ is surjective as desired.
\end{proof}
\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}).
Thus $X_{\overline{k}}$ has finitely many irreducible components
(Properties, Lemma \ref{properties-lemma-Noetherian-irreducible-components})
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}
\begin{slogan}
Geometric irreductibility is Zariski local modulo connectedness.
\end{slogan}
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 nonempty 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 $X = \bigcup_{i \in I} X_i$
with $I \not = \emptyset$ 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 nonempty
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}
\begin{slogan}
Geometric irreducibility can be tested over a separable algebraic
closure of the base field.
\end{slogan}
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 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}
\begin{lemma}
\label{lemma-proper-geometrically-reduced-global-sections}
Let $k$ be a field. Let $X$ be a proper scheme over $k$.
\begin{enumerate}
\item $A = H^0(X, \mathcal{O}_X)$ is a finite dimensional $k$-algebra,
\item $A = \prod_{i = 1, \ldots, n} A_i$ is a product of Artinian
local $k$-algebras, one factor for each connected component of $X$,
\item if $X$ is reduced, then $A = \prod_{i = 1, \ldots, n} k_i$
is a product of fields, each a finite extension of $k$,
\item if $X$ is geometrically reduced, then $k_i$ is finite separable
over $k$,
\item if $X$ is geometrically irreducible, then $A$ is geometrically
irreducible over $k$,
\item if $X$ is geometrically integral, then $A = k$.
\end{enumerate}
\end{lemma}
\begin{proof}
By Cohomology of Schemes, Lemma
\ref{coherent-lemma-proper-over-affine-cohomology-finite}
we see that $A = H^0(X, \mathcal{O}_X)$ is a finite dimensional
$k$-algebra. This proves (1).
\medskip\noindent
Then $A$ is a product of local rings by
Algebra, Lemma \ref{algebra-lemma-finite-dimensional-algebra} and
Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring}.
If $X = Y \amalg Z$ with $Y$ and $Z$ open in $X$, then we obtain
an idempotent $e \in A$ by taking the section of $\mathcal{O}_X$
which is $1$ on $Y$ and $0$ on $Z$. Conversely, if $e \in A$
is an idempotent, then we get a corresponding decomposition of $X$.
Finally, as $X$ has a Noetherian underlying topological space
its connected components are open. Hence the connected components
of $X$ correspond $1$-to-$1$ with primitive idempotents of $A$.
This proves (2).
\medskip\noindent
If $X$ is reduced, then $A$ is reduced. Hence the local rings $A_i = k_i$
are reduced and therefore fields (for example by
Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}).
This proves (3).
\medskip\noindent
If $X$ is geometrically reduced, then same thing is true for
$A \otimes_k \overline{k} =
H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$
(see Cohomology of Schemes, Lemma
\ref{coherent-lemma-flat-base-change-cohomology} for equality).
This implies that $k_i \otimes_k \overline{k}$ is a product
of fields and hence $k_i/k$ is separable for example by
Algebra,
Lemmas \ref{algebra-lemma-characterize-separable-field-extensions} and
\ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}.
This proves (4).
\medskip\noindent
If $X$ is geometrically irreducible, then $A \otimes_k \overline{k} =
H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$
is a zero dimensional local ring by part (2) and hence its
spectrum has one point, in particular it is irreducible.
Thus $A$ is geometrically irreducible. This proves (5).
\medskip\noindent
If $X$ is geometrically irreducible and geometrically reduced, then
$A = k_1$ is a field and the extension $k_1/k$ is finite separable and
geometrically irreducible. However, then $k_1 \otimes_k \overline{k}$
is a product of $[k_1 : k]$ copies of $\overline{k}$ and we conclude
that $k_1 = k$.
\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 $k'/k$,
\item $X_{k'}$ is a normal scheme for every finitely generated field
extension $k'/k$,
\item $X_{k'}$ is a normal scheme for every finite purely inseparable
field extension $k'/k$,
\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}), and
\item $X_{k^{perf}}$ is a normal scheme.
\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.
Hence (4) implies (5).
\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.
\medskip\noindent
The equivalence of (5) and (6) follows from the definition
of geometrically normal algebras and the equivalence (just proved)
of (3) and (4).
\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}
\begin{lemma}
\label{lemma-base-change-normal-by-separable}
Let $k$ be a field. Let $X$ be a normal scheme over $k$. Let $K/k$
be a separable field extension. Then $X_K$ is a normal scheme.
\end{lemma}
\begin{proof}
Follows from Lemma \ref{lemma-fibre-product-normal} and
Algebra, Lemma
\ref{algebra-lemma-separable-field-extension-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{Change of fields and ample invertible sheaves}
\label{section-change-fields-ample}
\noindent
The following result is typical for the results in this section.
\begin{lemma}
\label{lemma-ample-after-field-extension}
Let $k$ be a field. Let $X$ be a scheme over $k$.
If there exists an ample invertible sheaf on $X_K$ for some
field extension $k \subset K$, then $X$ has an ample invertible
sheaf.
\end{lemma}
\begin{proof}
Let $k \subset K$ be a field extension such that $X_K$ has
an ample invertible sheaf $\mathcal{L}$.
The morphism $X_K \to X$ is surjective. Hence $X$ is quasi-compact
as the image of a quasi-compact scheme (Properties, Definition
\ref{properties-definition-ample}). Since $X_K$ is quasi-separated
(by Properties, Lemma
\ref{properties-lemma-affine-s-opens-cover-quasi-separated})
we see that $X$ is quasi-separated: If $U, V \subset X$ are
affine open, then $(U \cap V)_K = U_K \cap V_K$ is quasi-compact
and $(U \cap V)_K \to U \cap V$ is surjective. Thus
Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} applies.
\medskip\noindent
Write $K = \colim A_i$ as the colimit of the subalgebras of $K$
which are of finite type over $k$. Denote
$X_i = X \times_{\Spec(k)} \Spec(A_i)$.
Since $X_K = \lim X_i$ we find an $i$ and an invertible sheaf'
$\mathcal{L}_i$ on $X_i$ whose pullback to $X_K$ is $\mathcal{L}$
(Limits, Lemma \ref{limits-lemma-descend-invertible-modules};
here and below we use that $X$ is quasi-compact and quasi-separated as
just shown). By Limits, Lemma \ref{limits-lemma-limit-ample}
we may assume $\mathcal{L}_i$ is ample after possibly increasing $i$.
Fix such an $i$ and let $\mathfrak m \subset A_i$ be a maximal
ideal. By the Hilbert Nullstellensatz
(Algebra, Theorem \ref{algebra-theorem-nullstellensatz})
the residue field $k' = A_i/\mathfrak m$ is a finite
extension of $k$. Hence $X_{k'} \subset X_i$ is a closed subscheme
hence has an ample invertible sheaf
(Properties, Lemma \ref{properties-lemma-ample-on-closed}).
Since $X_{k'} \to X$ is finite locally free we conclude
that $X$ has an ample invertible sheaf by
Divisors, Proposition \ref{divisors-proposition-push-down-ample}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-affine-after-field-extension}
Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is quasi-affine
for some field extension $k \subset K$, then $X$ is quasi-affine.
\end{lemma}
\begin{proof}
Let $k \subset K$ be a field extension such that $X_K$ is quasi-affine.
The morphism $X_K \to X$ is surjective. Hence $X$ is quasi-compact
as the image of a quasi-compact scheme (Properties, Definition
\ref{properties-definition-quasi-affine}). Since $X_K$ is quasi-separated
(as an open subscheme of an affine scheme)
we see that $X$ is quasi-separated: If $U, V \subset X$ are
affine open, then $(U \cap V)_K = U_K \cap V_K$ is quasi-compact
and $(U \cap V)_K \to U \cap V$ is surjective. Thus
Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} applies.
\medskip\noindent
Write $K = \colim A_i$ as the colimit of the subalgebras of $K$
which are of finite type over $k$. Denote
$X_i = X \times_{\Spec(k)} \Spec(A_i)$.
Since $X_K = \lim X_i$ we find an $i$ such that $X_i$ is quasi-affine
(Limits, Lemma \ref{limits-lemma-limit-quasi-affine};
here we use that $X$ is quasi-compact and quasi-separated as
just shown). By the Hilbert Nullstellensatz
(Algebra, Theorem \ref{algebra-theorem-nullstellensatz})
the residue field $k' = A_i/\mathfrak m$ is a finite
extension of $k$. Hence $X_{k'} \subset X_i$ is a closed
subscheme hence is quasi-affine (Properties, Lemma
\ref{properties-lemma-quasi-affine-locally-closed}).
Since $X_{k'} \to X$ is finite locally free we conclude by
Divisors, Lemma \ref{divisors-lemma-push-down-quasi-affine}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-projective-after-field-extension}
Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is quasi-projective
over $K$ for some field extension $k \subset K$, then $X$ is quasi-projective
over $k$.
\end{lemma}
\begin{proof}
By definition a morphism of schemes $g : Y \to T$ is quasi-projective
if it is locally of finite type, quasi-compact, and there exists
a $g$-ample invertible sheaf on $Y$.
Let $k \subset K$ be a field extension such that $X_K$ is quasi-projective
over $K$. Let $\Spec(A) \subset X$ be an affine open. Then $U_K$ is an
affine open subscheme of $X_K$, hence $A_K$ is a $K$-algebra of finite type.
Then $A$ is a $k$-algebra of finite type by
Algebra, Lemma \ref{algebra-lemma-finite-type-descends}.
Hence $X \to \Spec(k)$ is locally of finite type.
Since $X_K \to \Spec(K)$ is quasi-compact, we see that $X_K$ is
quasi-compact, hence $X$ is quasi-compact, hence $X \to \Spec(k)$
is of finite type. By Morphisms, Lemma
\ref{morphisms-lemma-finite-type-over-affine-ample-very-ample}
we see that $X_K$ has an ample invertible sheaf.
Then $X$ has an ample invertible sheaf by
Lemma \ref{lemma-ample-after-field-extension}.
Hence $X \to \Spec(k)$ is quasi-projective by Morphisms, Lemma
\ref{morphisms-lemma-finite-type-over-affine-ample-very-ample}.
\end{proof}
\noindent
The following lemma is a special case of
Descent, Lemma \ref{descent-lemma-descending-property-proper}.
\begin{lemma}
\label{lemma-proper-after-field-extension}
Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is proper
over $K$ for some field extension $k \subset K$, then $X$ is proper
over $k$.
\end{lemma}
\begin{proof}
Let $k \subset K$ be a field extension such that $X_K$ is proper over $K$.
Recall that this implies $X_K$ is separated and quasi-compact
(Morphisms, Definition \ref{morphisms-definition-proper}).
The morphism $X_K \to X$ is surjective. Hence $X$ is quasi-compact
as the image of a quasi-compact scheme (Properties, Definition
\ref{properties-definition-ample}). Since $X_K$ is separated
we see that $X$ is quasi-separated: If $U, V \subset X$ are
affine open, then $(U \cap V)_K = U_K \cap V_K$ is quasi-compact
and $(U \cap V)_K \to U \cap V$ is surjective. Thus
Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} applies.
\medskip\noindent
Write $K = \colim A_i$ as the colimit of the subalgebras of $K$
which are of finite type over $k$. Denote
$X_i = X \times_{\Spec(k)} \Spec(A_i)$.
By Limits, Lemma \ref{limits-lemma-eventually-proper}
there exists an $i$ such that $X_i \to \Spec(A_i)$ is proper.
Here we use that $X$ is quasi-compact and quasi-separated as
just shown. Choose a maximal ideal $\mathfrak m \subset A_i$.
By the Hilbert Nullstellensatz
(Algebra, Theorem \ref{algebra-theorem-nullstellensatz})
the residue field $k' = A_i/\mathfrak m$ is a finite
extension of $k$. The base change $X_{k'} \to \Spec(k')$
is proper (Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}).
Since $k \subset k'$ is finite both $X_{k'} \to X$ and the composition
$X_{k'} \to \Spec(k)$
are proper as well (Morphisms, Lemmas \ref{morphisms-lemma-finite-proper},
\ref{morphisms-lemma-base-change-proper}, and
\ref{morphisms-lemma-composition-proper}).
The first implies that $X$ is separated over $k$ as $X_{k'}$
is separated
(Morphisms, Lemma \ref{morphisms-lemma-image-universally-closed-separated}).
The second implies that $X \to \Spec(k)$ is proper
by Morphisms, Lemma \ref{morphisms-lemma-image-proper-is-proper}.
\end{proof}
\begin{lemma}
\label{lemma-projective-after-field-extension}
Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is projective
over $K$ for some field extension $k \subset K$, then $X$ is projective
over $k$.
\end{lemma}
\begin{proof}
A scheme over $k$ is projective over $k$ if and only if it is
quasi-projective and proper over $k$. See
Morphisms, Lemma \ref{morphisms-lemma-projective-is-quasi-projective-proper}.
Thus the lemma follows from
Lemmas \ref{lemma-quasi-projective-after-field-extension} and
\ref{lemma-proper-after-field-extension}.
\end{proof}
\section{Tangent spaces}
\label{section-tangent-spaces}
\noindent
In this section we define the tangent space of a morphism of schemes
at a point of the source using points with values in dual numbers.
\begin{definition}
\label{definition-dual-numbers}
For any ring $R$ the {\it dual numbers} over $R$ is the
$R$-algebra denoted $R[\epsilon]$. As an $R$-module it is free with
basis $1$, $\epsilon$ and the $R$-algebra structure comes from setting
$\epsilon^2 = 0$.
\end{definition}
\noindent
Let $f : X \to S$ be a morphism of schemes.
Let $x \in X$ be a point with image $s = f(x)$ in $S$.
Consider the solid commutative diagram
\begin{equation}
\label{equation-tangent-space}
\vcenter{
\xymatrix{
\Spec(\kappa(x)) \ar[r] \ar[dr] \ar@/^1pc/[rr] &
\Spec(\kappa(x)[\epsilon]) \ar@{.>}[r] \ar[d]&
X \ar[d] \\
&
\Spec(\kappa(s)) \ar[r] &
S
}
}
\end{equation}
with the curved arrow being the canonical morphism of
$\Spec(\kappa(x))$ into $X$.
\begin{lemma}
\label{lemma-tangent-space}
The set of dotted arrows making (\ref{equation-tangent-space}) commute
has a canonical $\kappa(x)$-vector space structure.
\end{lemma}
\begin{proof}
Set $\kappa = \kappa(x)$. Observe that we have a pushout in the
category of schemes
$$
\Spec(\kappa[\epsilon]) \amalg_{\Spec(\kappa)} \Spec(\kappa[\epsilon])
= \Spec(\kappa[\epsilon_1, \epsilon_2])
$$
where $\kappa[\epsilon_1, \epsilon_2]$ is the $\kappa$-algebra with
basis $1, \epsilon_1, \epsilon_2$ and
$\epsilon_1^2 = \epsilon_1\epsilon_2 = \epsilon_2^2 = 0$.
This follows immediately from the corresponding result for
rings and the description of morphisms from spectra of local rings
to schemes in
Schemes, Lemma \ref{schemes-lemma-morphism-from-spec-local-ring}.
Given two arrows
$\theta_1, \theta_2 : \Spec(\kappa[\epsilon]) \to X$
we can consider the morphism
$$
\theta_1 + \theta_2 :
\Spec(\kappa[\epsilon]) \to
\Spec(\kappa[\epsilon_1, \epsilon_2])
\xrightarrow{\theta_1, \theta_2} X
$$
where the first arrow is given by $\epsilon_i \mapsto \epsilon$.
On the other hand, given $\lambda \in \kappa$ there is a self map
of $\Spec(\kappa[\epsilon])$ corresponding to the $\kappa$-algebra
endomorphism of $\kappa[\epsilon]$ which sends $\epsilon$ to $\lambda \epsilon$.
Precomposing $\theta : \Spec(\kappa[\epsilon]) \to X$
by this selfmap gives $\lambda \theta$. The reader can verify
the axioms of a vector space by verifying the existence
of suitable commutative diagrams of schemes. We omit the details.
(An alternative proof would be to express everything in terms of local
rings and then verify the vector space axioms on the level of ring maps.)
\end{proof}
\begin{definition}
\label{definition-tangent-space}
Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. The set of
dotted arrows making (\ref{equation-tangent-space}) commute with
its canonical $\kappa(x)$-vector space structure is called
the {\it tangent space of $X$ over $S$ at $x$} and we denote it $T_{X/S, x}$.
An element of this space is called a {\it tangent vector} of $X/S$ at $x$.
\end{definition}
\noindent
Since tangent vectors at $x \in X$ live in the scheme theoretic fibre
$X_s$ of $f : X \to S$ over $s = f(x)$, we get a canonical identification
\begin{equation}
\label{equation-tangent-space-fibre}
T_{X/S, x} = T_{X_s/s, x}
\end{equation}
This pleasing definition involving the functor of points has the following
algebraic description, which suggests defining the
{\it cotangent space of $X$ over $S$ at $x$} as the $\kappa(x)$-vector
space
$$
T^*_{X/S, x} = \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)
$$
simply because it is canonically $\kappa(x)$-dual to the tangent space of
$X$ over $S$ at $x$.
\begin{lemma}
\label{lemma-tangent-space-cotangent-space}
Let $f : X \to S$ be a morphism of schemes. Let $x \in X$.
There is a canonical isomorphism
$$
T_{X/S, x} = \Hom_{\mathcal{O}_{X, x}}(\Omega_{X/S, x}, \kappa(x))
$$
of vector spaces over $\kappa(x)$.
\end{lemma}
\begin{proof}
Set $\kappa = \kappa(x)$.
Given $\theta \in T_{X/S, x}$ we obtain a map
$$
\theta^*\Omega_{X/S} \to
\Omega_{\Spec(\kappa[\epsilon])/\Spec(\kappa(s))} \to
\Omega_{\Spec(\kappa[\epsilon])/\Spec(\kappa)}
$$
Taking sections we obtain an $\mathcal{O}_{X, x}$-linear map
$\xi_\theta: \Omega_{X/S, x} \to \kappa \text{d}\epsilon$, i.e.,
an element of the right hand side of the
formula of the lemma. To show that $\theta \mapsto \xi_\theta$ is
an isomorphism we can replace $S$ by $s$ and $X$ by the
scheme theoretic fibre $X_s$. Indeed, both sides of the
formula only depend on the scheme theoretic fibre;
this is clear for $T_{X/S, x}$ and for the RHS see
Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}.
We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$
as this does not change $T_{X/S, x}$
(Schemes, Lemma \ref{schemes-lemma-morphism-from-spec-local-ring})
nor $\Omega_{X/S, x}$
(Modules, Lemma \ref{modules-lemma-stalk-module-differentials}).
\medskip\noindent
Let $(A, \mathfrak m, \kappa)$ be a local ring over a field $k$.
To finish the proof we have to show that any $A$-linear map
$\xi : \Omega_{A/k} \to \kappa$ comes from a unique $k$-algebra
map $\varphi : A \to \kappa[\epsilon]$ agreeing with the canonical
map $c : A \to \kappa$ modulo $\epsilon$. Write
$\varphi(a) = c(a) + D(a) \epsilon$
the reader sees that $a \mapsto D(a)$ is a $k$-derivation.
Using the universal property of $\Omega_{A/k}$ we see that each
$D$ corresponds to a unique $\xi$ and vice versa. This finishes the proof.
\end{proof}
\begin{lemma}
\label{lemma-tangent-space-rational-point}
Let $f : X \to S$ be a morphism of schemes.
Let $x \in X$ be a point and let $s = f(x) \in S$.
Assume that $\kappa(x) = \kappa(s)$. Then there are canonical isomorphisms
$$
\mathfrak m_x/(\mathfrak m_x^2 + \mathfrak m_s\mathcal{O}_{X, x})
=
\Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)
$$
and
$$
T_{X/S, x} =
\Hom_{\kappa(x)}(
\mathfrak m_x/(\mathfrak m_x^2 + \mathfrak m_s\mathcal{O}_{X, x}),
\kappa(x))
$$
This works more generally if $\kappa(x)/\kappa(s)$ is a separable
algebraic extension.
\end{lemma}
\begin{proof}
The second isomorphism follows from the first by
Lemma \ref{lemma-tangent-space-cotangent-space}.
For the first, we can replace $S$ by $s$ and $X$ by $X_s$, see
Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}.
We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$, see
Modules, Lemma \ref{modules-lemma-stalk-module-differentials}.
Thus we have to show the following algebra fact: let
$(A, \mathfrak m, \kappa)$ be a local ring over a field $k$
such that $\kappa/k$ is separable algebraic. Then the canonical map
$$
\mathfrak m/\mathfrak m^2
\longrightarrow
\Omega_{A/k} \otimes \kappa
$$
is an isomorphism. Observe that
$\mathfrak m/\mathfrak m^2 = H_1(\NL_{\kappa/A})$. By
Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}
it suffices to show that $\Omega_{\kappa/k} = 0$ and
$H_1(\NL_{\kappa/k}) = 0$. Since $\kappa$ is the union of
its finite separable extensions in $k$ it suffices to prove
this when $\kappa$ is a finite separable extension of $k$
(Algebra, Lemma \ref{algebra-lemma-colimits-NL}).
In this case the ring map $k \to \kappa$ is \'etale
and hence $\NL_{\kappa/k} = 0$ (more or less by definition, see
Algebra, Section \ref{algebra-section-etale}).
\end{proof}
\begin{lemma}
\label{lemma-map-tangent-spaces}
Let $f : X \to Y$ be a morphism of schemes over a base scheme $S$.
Let $x \in X$ be a point. Set $y = f(x)$. If $\kappa(y) = \kappa(x)$,
then $f$ induces a natural linear map
$$
\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}.
$$
which is dual to the linear map
$\Omega_{Y/S, y} \otimes \kappa(y) \to \Omega_{X/S, \kappa(x)}$
via the identifications of Lemma \ref{lemma-tangent-space-cotangent-space}.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-tangent-space-product}
Let $X$, $Y$ be schemes over a base $S$. Let $x \in X$ and $y \in Y$ with
the same image point $s \in S$ such that $\kappa(s) = \kappa(x)$ and
$\kappa(s) = \kappa(y)$. There is a canonical isomorphism
$$
T_{X \times_S Y/S, (x, y)} = T_{X/S, x} \oplus T_{Y/S, y}
$$
The map from left to right is induced by the maps on tangent spaces coming
from the projections $X \times_S Y \to X$ and $X \times_S Y \to Y$.
The map from right to left is induced by the maps
$1 \times y : X_s \to X_s \times_s Y_s$ and
$x \times 1 : Y_s \to X_s \times_s Y_s$ via the identification
(\ref{equation-tangent-space-fibre}) of
tangent spaces with tangent spaces of fibres.
\end{lemma}
\begin{proof}
The direct sum decomposition follows from
Morphisms, Lemma \ref{morphisms-lemma-differential-product}
via Lemma \ref{lemma-tangent-space-rational-point}. Compatibility
with the maps comes from Lemma \ref{lemma-map-tangent-spaces}.
\end{proof}
\begin{lemma}
\label{lemma-injective-tangent-spaces-unramified}
Let $f : X \to Y$ be a morphism of schemes locally of finite type over a
base scheme $S$. Let $x \in X$ be a point. Set $y = f(x)$ and assume
that $\kappa(y) = \kappa(x)$. Then the following are equivalent
\begin{enumerate}
\item $\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}$ is injective, and
\item $f$ is unramified at $x$.
\end{enumerate}
\end{lemma}
\begin{proof}
The morphism $f$ is locally of finite type by
Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}.
The map $\text{d}f$ is injective, if and only if
$\Omega_{Y/S, y} \otimes \kappa(y) \to \Omega_{X/S, x} \otimes \kappa(x)$
is surjective (Lemma \ref{lemma-map-tangent-spaces}).
The exact sequence $f^*\Omega_{Y/S} \to \Omega_{X/S} \to \Omega_{X/Y} \to 0$
(Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials})
then shows that this happens if and only if
$\Omega_{X/Y, x} \otimes \kappa(x) = 0$.
Hence the result follows from
Morphisms, Lemma \ref{morphisms-lemma-unramified-at-point}.
\end{proof}
\section{Generically finite morphisms}
\label{section-generically-finite}
\noindent
In this section we revisit the notion of a generically finite
morphism of schemes as studied in
Morphisms, Section \ref{morphisms-section-generically-finite}.
\begin{lemma}
\label{lemma-quasi-finite-in-codim-1}
Let $f : X \to Y$ be locally of finite type. Let $y \in Y$ be a point
such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$.
Assume in addition one of the following conditions is satisfied
\begin{enumerate}
\item for every generic point $\eta$ of an irreducible component
of $X$ the field extension $\kappa(\eta) \supset \kappa(f(\eta))$
is finite (or algebraic),
\item for every generic point $\eta$ of an irreducible component
of $X$ such that $f(\eta) \leadsto y$ the field extension
$\kappa(\eta) \supset \kappa(f(\eta))$ is finite (or algebraic),
\item $f$ is quasi-finite at every generic point of an
irreducible component of $X$,
\item $Y$ is locally Noetherian and $f$
is quasi-finite at a dense set of points of $X$,
\item add more here.
\end{enumerate}
Then $f$ is quasi-finite at every point of $X$ lying over $y$.
\end{lemma}
\begin{proof}
Condition (4) implies $X$ is locally Noetherian
(Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}).
The set of points at which morphism is quasi-finite is open
(Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-points-open}).
A dense open of a locally Noetherian scheme contains all generic
point of irreducible components, hence (4) implies (3).
Condition (3) implies condition
(1) by Morphisms, Lemma \ref{morphisms-lemma-residue-field-quasi-finite}.
Condition (1) implies condition (2).
Thus it suffices to prove the lemma in case (2) holds.
\medskip\noindent
Assume (2) holds. Recall that $\Spec(\mathcal{O}_{Y, y})$
is the set of points of $Y$ specializing to $y$, see
Schemes, Lemma \ref{schemes-lemma-specialize-points}.
Combined with
Morphisms, Lemma \ref{morphisms-lemma-base-change-quasi-finite}
this shows we may replace $Y$ by $\Spec(\mathcal{O}_{Y, y})$.
Thus we may assume $Y = \Spec(B)$ where $B$ is a Noetherian
local ring of dimension $\leq 1$ and $y$ is the closed point.
\medskip\noindent
Let $X = \bigcup X_i$ be the irreducible components of $X$ viewed
as reduced closed subschemes. If we can show each fibre $X_{i, y}$
is a discrete space, then $X_y = \bigcup X_{i, y}$ is discrete as
well and we conclude that $X \to Y$ is quasi-finite at all points
of $X_y$ by Morphisms, Lemma
\ref{morphisms-lemma-quasi-finite-at-point-characterize}.
Thus we may assume $X$ is an integral scheme.
\medskip\noindent
If $X \to Y$ maps the generic point $\eta$ of $X$ to $y$, then $X$
is the spectrum of a finite extension of $\kappa(y)$ and the
result is true. Assume that $X$ maps $\eta$ to a point corresponding
to a minimal prime $\mathfrak q$ of $B$ different from $\mathfrak m_B$.
We obtain a factorization $X \to \Spec(B/\mathfrak q) \to \Spec(B)$.
Let $x \in X$ be a point lying over $y$.
By the dimension formula
(Morphisms, Lemma \ref{morphisms-lemma-dimension-formula})
we have
$$
\dim(\mathcal{O}_{X, x}) \leq \dim(B/\mathfrak q) +
\text{trdeg}_{\kappa(\mathfrak q)}(R(X)) - \text{trdeg}_{\kappa(y)} \kappa(x)
$$
We know that $\dim(B/\mathfrak q) = 1$, that the generic point of $X$
is not equal to $x$ and specializes to $x$ and that $R(X)$ is algebraic
over $\kappa(\mathfrak q)$. Thus we get
$$
1 \leq 1 - \text{trdeg}_{\kappa(y)} \kappa(x)
$$
Hence every point $x$ of $X_y$ is closed in $X_y$ by
Morphisms, Lemma
\ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}
and hence $X \to Y$ is quasi-finite at every point $x$ of $X_y$ by
Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}
(which also implies that $X_y$ is a discrete topological space).
\end{proof}
\begin{lemma}
\label{lemma-finite-in-codim-1}
Let $f : X \to Y$ be a proper morphism. Let $y \in Y$ be a point
such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$.
Assume in addition one of the following conditions is satisfied
\begin{enumerate}
\item for every generic point $\eta$ of an irreducible component
of $X$ the field extension $\kappa(\eta) \supset \kappa(f(\eta))$
is finite (or algebraic),
\item for every generic point $\eta$ of an irreducible component
of $X$ such that $f(\eta) \leadsto y$ the field extension
$\kappa(\eta) \supset \kappa(f(\eta))$ is finite (or algebraic),
\item $f$ is quasi-finite at every generic point of $X$,
\item $Y$ is locally Noetherian and $f$
is quasi-finite at a dense set of points of $X$,
\item add more here.
\end{enumerate}
Then there exists an open neighbourhood $V \subset Y$ of $y$ such that
$f^{-1}(V) \to V$ is finite.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-quasi-finite-in-codim-1} the morphism $f$ is
quasi-finite at every point of the fibre $X_y$. Hence
$X_y$ is a discrete topological space
(Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}).
As $f$ is proper the fibre $X_y$ is quasi-compact, i.e., finite.
Thus we can apply Cohomology of Schemes, Lemma
\ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood}
to conclude.
\end{proof}
\begin{lemma}
\label{lemma-modification-normal-iso-over-codimension-1}
Let $X$ be a Noetherian scheme. Let $f : Y \to X$ be a birational proper
morphism of schemes with $Y$ reduced. Let $U \subset X$ be the
maximal open over which $f$ is an isomorphism. Then $U$ contains
\begin{enumerate}
\item every point of codimension $0$ in $X$,
\item every $x \in X$ of codimension $1$ on $X$ such that
$\mathcal{O}_{X, x}$ is a discrete valuation ring,
\item every $x \in X$ such that the fibre of $Y \to X$ over $x$ is
finite and such that $\mathcal{O}_{X, x}$ is normal, and
\item every $x \in X$ such that $f$ is quasi-finite at some
$y \in Y$ lying over $x$ and $\mathcal{O}_{X, x}$ is normal.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) follows from Morphisms, Lemma
\ref{morphisms-lemma-birational-isomorphism-over-dense-open}.
Part (2) follows from part (3) and Lemma \ref{lemma-finite-in-codim-1}
(and the fact that finite morphisms have finite fibres).
\medskip\noindent
Part (3) follows from part (4) and
Morphisms, Lemma \ref{morphisms-lemma-finite-fibre}
but we will also give a direct proof.
Let $x \in X$ be as in (3). By
Cohomology of Schemes, Lemma
\ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood}
we may assume $f$ is finite. We may assume $X$ affine.
This reduces us to
the case of a finite birational morphism of Noetherian affine schemes
$Y \to X$ and $x \in X$ such that $\mathcal{O}_{X, x}$ is a
normal domain. Since $\mathcal{O}_{X, x}$ is a domain and $X$
is Noetherian, we may replace $X$ by an affine open of $x$ which
is integral. Then, since $Y \to X$ is birational and $Y$ is reduced
we see that $Y$ is integral. Writing $X = \Spec(A)$ and $Y = \Spec(B)$
we see that $A \subset B$ is a finite inclusion of domains having the same
field of fractions. If $\mathfrak p \subset A$ is the prime corresponding
to $x$, then $A_\mathfrak p$ being normal implies that
$A_\mathfrak p \subset B_\mathfrak p$ is an equality.
Since $B$ is a finite $A$-module, we see there exists an
$a \in A$, $a \not \in \mathfrak p$ such that $A_a \to B_a$
is an isomorphism.
\medskip\noindent
Let $x \in X$ and $y \in Y$ be as in (4). After replacing $X$
by an affine open neighbourhood we may assume $X = \Spec(A)$
and $A \subset \mathcal{O}_{X, x}$, see
Properties, Lemma \ref{properties-lemma-ring-affine-open-injective-local-ring}.
Then $A$ is a domain and hence $X$ is integral.
Since $f$ is birational and $Y$ is reduced
it follows that $Y$ is integral too. Consider the ring map
$\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$. This is a ring map
which is essentially of finite type, the residue field extension
is finite, and $\dim(\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y}) = 0$
(to see this trace through the definitions of quasi-finite
maps in
Morphisms, Definition \ref{morphisms-definition-quasi-finite} and
Algebra, Definition \ref{algebra-definition-quasi-finite}). By
Algebra, Lemma \ref{algebra-lemma-essentially-finite-type-fibre-dim-zero}
$\mathcal{O}_{Y, y}$ is the localization of a finite
$\mathcal{O}_{X, x}$-algebra $B$. Of course we may replace
$B$ by the image of $B$ in $\mathcal{O}_{Y, y}$ and assume
that $B$ is a domain with the same fraction field as $\mathcal{O}_{Y, y}$.
Then $\mathcal{O}_{X, x} \subset B$
have the same fraction field as $f$ is birational. Since
$\mathcal{O}_{X, x}$ is normal, we conclude that
$\mathcal{O}_{X, x} = B$ (because finite implies integral),
in particular, we see that $\mathcal{O}_{X, x} = \mathcal{O}_{Y, y}$. By
Morphisms, Lemma \ref{morphisms-lemma-morphism-defined-local-ring}
after shrinking $X$ we may assume there is a section
$X \to Y$ of $f$ mapping $x$ to $y$ and inducing the given
isomorphism on local rings. Since $X \to Y$ is closed
(by Schemes, Lemma \ref{schemes-lemma-section-immersion})
necessarily maps the generic point of $X$ to the generic point of $Y$
it follows that the image of $X \to Y$ is $Y$.
Then $Y = X$ and we've proved what we wanted to show.
\end{proof}
\section{Variants of Noether normalization}
\label{section-noether-normalization}
\noindent
Noether normalization is the statement that if $k$ is a field
and $A$ is a finite type $k$ algebra of dimension $d$, then
there exists a finite injective $k$-algebra homomorphism
$k[x_1, \ldots, x_d] \to A$. See
Algebra, Lemma \ref{algebra-lemma-Noether-normalization}.
Geometrically this means there
is a finite surjective morphism $\Spec(A) \to \mathbf{A}^d_k$
over $\Spec(k)$.
\begin{lemma}
\label{lemma-noether-normalization}
Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ with
image $s \in S$. Let $V \subset S$ be an affine open neighbourhood
of $s$. If $f$ is locally of finite type and $\dim_x(X_s) = d$,
then there exists an affine open $U \subset X$ with
$x \in U$ and $f(U) \subset V$ and a factorization
$$
U \xrightarrow{\pi} \mathbf{A}^d_V \to V
$$
of $f|_U : U \to V$ such that $\pi$ is quasi-finite.
\end{lemma}
\begin{proof}
This follows from
Algebra, Lemma \ref{algebra-lemma-quasi-finite-over-polynomial-algebra}.
\end{proof}
\begin{lemma}
\label{lemma-noether-normalization-affine}
Let $f : X \to S$ be a finite type morphism of affine schemes.
Let $s \in S$. If $\dim(X_s) = d$, then there exists a factorization
$$
X \xrightarrow{\pi} \mathbf{A}^d_S \to S
$$
of $f$ such that the morphism $\pi_s : X_s \to \mathbf{A}^d_{\kappa(s)}$
of fibres over $s$ is finite.
\end{lemma}
\begin{proof}
Write $S = \Spec(A)$ and $X = \Spec(B)$ and let $A \to B$ be the ring
map corresponding to $f$. Let $\mathfrak p \subset A$ be the prime ideal
corresponding to $s$. We can choose a surjection
$A[x_1, \ldots, x_r] \to B$. By
Algebra, Lemma \ref{algebra-lemma-Noether-normalization}
there exist elements $y_1, \ldots, y_d \in A$ in the $\mathbf{Z}$-subalgebra
of $A$ generated by $x_1, \ldots, x_r$ such that the $A$-algebra homomorphism
$A[t_1, \ldots, t_d] \to B$ sending $t_i$ to $y_i$ induces a finite
$\kappa(\mathfrak p)$-algebra homomorphism
$\kappa(\mathfrak p)[t_1, \ldots, t_d] \to B \otimes_A \kappa(\mathfrak p)$.
This proves the lemma.
\end{proof}
\begin{lemma}
\label{lemma-geometric-structure-unramified}
Let $f : X \to S$ be a morphism of schemes. Let $x \in X$.
Let $V = \Spec(A)$ be an affine open neighbourhood of $f(x)$ in $S$.
If $f$ is unramified at $x$, then there exist exists an affine open
$U \subset X$ with $x \in U$ and $f(U) \subset V$
such that we have a commutative diagram
$$
\xymatrix{
X \ar[d] & U \ar[l] \ar[rd] \ar[r]^-j &
\Spec(A[t]_{g'}/(g)) \ar[d] \ar[r] &
\Spec(A[t]) = \mathbf{A}^1_V \ar[ld] \\
Y & & V \ar[ll]
}
$$
where $j$ is an immersion, $g \in A[t]$ is a monic polynomial, and
$g'$ is the derivative of $g$ with respect to $t$. If $f$ is \'etale
at $x$, then we may choose the diagram such that $j$ is an open immersion.
\end{lemma}
\begin{proof}
The unramified case is a translation of
Algebra, Proposition \ref{algebra-proposition-unramified-locally-standard}.
In the \'etale case this is a translation of
Algebra, Proposition \ref{algebra-proposition-etale-locally-standard}
or equivalently it follows from
Morphisms, Lemma \ref{morphisms-lemma-etale-locally-standard-etale}
although the statements differ slightly.
\end{proof}
\begin{lemma}
\label{lemma-unramfied-over-affine}
Let $f : X \to S$ be a finite type morphism of affine schemes.
Let $x \in X$ with image $s \in S$. Let
$$
r =
\dim_{\kappa(x)} \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x) =
\dim_{\kappa(x)} \Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) =
\dim_{\kappa(x)} T_{X/S, x}
$$
Then there exists a factorization
$$
X \xrightarrow{\pi} \mathbf{A}^r_S \to S
$$
of $f$ such that $\pi$ is unramified at $x$.
\end{lemma}
\begin{proof}
By Morphisms, Lemma \ref{morphisms-lemma-finite-type-differentials}
the first dimension is finite.
The first equality follows as the restriction of
$\Omega_{X/S}$ to the fibre is the module of differentials
from Morphisms, Lemma
\ref{morphisms-lemma-base-change-differentials}.
The last equality follows from Lemma \ref{lemma-tangent-space-cotangent-space}.
Thus we see that the statement makes sense.
\medskip\noindent
To prove the lemma write $S = \Spec(A)$ and $X = \Spec(B)$ and let
$A \to B$ be the ring map corresponding to $f$. Let $\mathfrak q \subset B$
be the prime ideal corresponding to $x$. Choose a surjection of $A$-algebras
$A[x_1, \ldots, x_t] \to B$. Since $\Omega_{B/A}$ is generated by
$\text{d}x_1, \ldots, \text{d}x_t$ we see that their images in
$\Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$ generate
this as a $\kappa(x)$-vector space. After renumbering we may assume
that $\text{d}x_1, \ldots, \text{d}x_r$ map to a basis of
$\Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$.
We claim that $P = A[x_1, \ldots, x_r] \to B$ is unramified at $\mathfrak q$.
To see this it suffices to show that $\Omega_{B/P, \mathfrak q} = 0$
(Algebra, Lemma \ref{algebra-lemma-unramified}).
Note that $\Omega_{B/P}$ is the quotient of $\Omega_{B/A}$ by the
submodule generated by $\text{d}x_1, \ldots, \text{d}x_r$.
Hence
$\Omega_{B/P, \mathfrak q} \otimes_{B_\mathfrak q} \kappa(\mathfrak q) = 0$
by our choice of $x_1, \ldots, x_r$.
By Nakayama's lemma, more precisely Algebra, Lemma \ref{algebra-lemma-NAK}
part (2) which applies as $\Omega_{B/P}$ is finite (see reference above),
we conclude that $\Omega_{B/P, \mathfrak q} = 0$.
\end{proof}
\begin{lemma}
\label{lemma-immersion-into-affine}
Let $f : X \to S$ be a morphism of schemes.
Let $x \in X$ with image $s \in S$. Let $V \subset S$ be
an affine open neighbourhood of $s$. If $f$ is locally of
finite type and
$$
r =
\dim_{\kappa(x)} \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x) =
\dim_{\kappa(x)} \Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) =
\dim_{\kappa(x)} T_{X/S, x}
$$
then there exist
\begin{enumerate}
\item an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a
factorization
$$
U \xrightarrow{j} \mathbf{A}^{r + 1}_V \to V
$$
of $f|_U$ such that $j$ is an immersion, or
\item an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a
factorization
$$
U \xrightarrow{j} D \to V
$$
of $f|_U$ such that $j$ is a closed immersion and $D \to V$
is smooth of relative dimension $r$.
\end{enumerate}
\end{lemma}
\begin{proof}
Pick any affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$.
Apply Lemma \ref{lemma-unramfied-over-affine} to $U \to V$ to
get $U \to \mathbf{A}^r_V \to V$ as in the statement of that lemma.
By Lemma \ref{lemma-geometric-structure-unramified}
we get a factorization
$$
U \xrightarrow{j} D \xrightarrow{j'} \mathbf{A}^{r + 1}_V
\xrightarrow{p} \mathbf{A}^r_V \to V
$$
where $j$ and $j'$ are immersions, $p$ is the projection, and
$p \circ j'$ is standard \'etale. Thus we see in particular that
(1) and (2) hold.
\end{proof}
\section{Dimension of fibres}
\label{section-dimension-fibres}
\noindent
We have already seen that dimension of fibres of finite type morphisms
typically jump up. In this section we discuss the phenomenon that in
codimension $1$ this does not happen. More generally, we discuss how
much the dimension of a fibre can jump. Here is a list of related results:
\begin{enumerate}
\item For a finite type morphism $X \to S$ the set of
$x \in X$ with $\dim_x(X_{f(x)}) \leq d$ is open, see
Algebra, Lemma \ref{algebra-lemma-dimension-fibres-bounded-open-upstairs}
and
Morphisms, Lemma \ref{morphisms-lemma-openness-bounded-dimension-fibres}.
\item We have the dimension formula, see
Algebra, Lemma \ref{algebra-lemma-dimension-formula} and
Morphisms, Lemma \ref{morphisms-lemma-dimension-formula}.
\item Constant fibre dimension for an integral finite type scheme
dominating a valuation ring, see Algebra, Lemma
\ref{algebra-lemma-finite-type-domain-over-valuation-ring-dim-fibres}.
\item If $X \to S$ is of finite type and is quasi-finite at every
generic point of $X$, then $X \to S$ is quasi-finite in codimension $1$, see
Algebra, Lemma \ref{algebra-lemma-finite-in-codim-1} and
Lemma \ref{lemma-quasi-finite-in-codim-1}.
\end{enumerate}
The last result mentioned above generalizes as follows.
\begin{lemma}
\label{lemma-dimension-fibre-in-codim-1}
Let $f : X \to Y$ be locally of finite type. Let $x \in X$ be a point
with image $y \in Y$ such that $\mathcal{O}_{Y, y}$ is Noetherian of
dimension $\leq 1$. Let $d \geq 0$ be an integer such that for every
generic point $\eta$ of an irreducible component of $X$ which contains
$x$, we have $\dim_\eta(X_{f(\eta)}) = d$. Then $\dim_x(X_y) = d$.
\end{lemma}
\begin{proof}
Recall that $\Spec(\mathcal{O}_{Y, y})$
is the set of points of $Y$ specializing to $y$, see
Schemes, Lemma \ref{schemes-lemma-specialize-points}.
Thus we may replace $Y$ by $\Spec(\mathcal{O}_{Y, y})$