spaces-over-fields.tex

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\title{Algebraic Spaces over Fields}


\maketitle

\phantomsection
\label{section-phantom}

\tableofcontents

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

\noindent
This chapter is the analogue of the chapter on varieties in the setting
of algebraic spaces. A reference for algebraic spaces is
\cite{K}.


\section{Conventions}
\label{section-conventions}

\noindent
The standing assumption is that all schemes are contained in
a big fppf site $\Sch_{fppf}$. And all rings $A$ considered
have the property that $\Spec(A)$ is (isomorphic) to an
object of this big site.

\medskip\noindent
Let $S$ be a scheme and let $X$ be an algebraic space over $S$.
In this chapter and the following we will write $X \times_S X$
for the product of $X$ with itself (in the category of algebraic
spaces over $S$), instead of $X \times X$.




\section{Geometric components}
\label{section-geometric-components}


\begin{lemma}
\label{lemma-minimal-primes-tensor-strictly-henselian}
Let $k$ be an algebraically closed field. Let $A$, $B$ be strictly
henselian local $k$-algebras with residue field equal to $k$.
Let $C$ be the strict henselization of $A \otimes_k B$ at the maximal
ideal $\mathfrak m_A \otimes_k B + A \otimes_k \mathfrak m_B$.
Then the minimal primes of $C$ correspond $1$-to-$1$ to pairs of
minimal primes of $A$ and $B$.
\end{lemma}

\begin{proof}
First note that a minimal prime $\mathfrak r$ of $C$ maps to a minimal
prime $\mathfrak p$ in $A$ and to a minimal prime $\mathfrak q$ of $B$
because the ring maps $A \to C$ and $B \to C$ are flat (by going down for
flat ring map
Algebra, Lemma \ref{algebra-lemma-flat-going-down}).
Hence it suffices to show that the strict henselization of
$(A/\mathfrak p \otimes_k B/\mathfrak q)_{
\mathfrak m_A \otimes_k B + A \otimes_k \mathfrak m_B}$
has a unique minimal prime ideal. By
Algebra, Lemma \ref{algebra-lemma-quotient-strict-henselization}
the rings $A/\mathfrak p$, $B/\mathfrak q$ are strictly henselian.
Hence we may assume that $A$ and $B$ are strictly henselian
local domains and our goal is to show that $C$ has a unique minimal prime. By
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-geometrically-unibranch}.
we see that the integral closure $A'$ of $A$ in its fraction field
is a normal local domain with residue field $k$ and similarly for the
integral closure $B'$ of $B$ into its fraction field. By
Algebra, Lemma \ref{algebra-lemma-geometrically-normal-tensor-normal}
we see that $A' \otimes_k B'$ is a normal ring. Hence its localization
$$
R = (A' \otimes_k B')_{
\mathfrak m_{A'} \otimes_k B' + A' \otimes_k \mathfrak m_{B'}}
$$
is a normal local domain. Note that $A \otimes_k B \to A' \otimes_k B'$
is integral (hence gong up holds --
Algebra, Lemma \ref{algebra-lemma-integral-going-up})
and that $\mathfrak m_{A'} \otimes_k B' + A' \otimes_k \mathfrak m_{B'}$
is the unique maximal ideal of $A' \otimes_k B'$
lying over $\mathfrak m_A \otimes_k B + A \otimes_k \mathfrak m_B$.
Hence we see that
$$
R = (A' \otimes_k B')_{
\mathfrak m_A \otimes_k B + A \otimes_k \mathfrak m_B}
$$
by
Algebra, Lemma \ref{algebra-lemma-unique-prime-over-localize-below}.
It follows that
$$
(A \otimes_k B)_{
\mathfrak m_A \otimes_k B + A \otimes_k \mathfrak m_B}
\longrightarrow
R
$$
is integral. We conclude that $R$ is the integral closure of
$(A \otimes_k B)_{
\mathfrak m_A \otimes_k B + A \otimes_k \mathfrak m_B}$
in its fraction field, and by
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-geometrically-unibranch}
once again we conclude that $C$ has a unique prime ideal.
\end{proof}








\section{Schematic locus}
\label{section-schematic}


\begin{lemma}
\label{lemma-locally-finite-type-dim-zero}
Let $k$ be a field. Let $X$ be an algebraic space over $\Spec(k)$.
If $X$ is locally of finite type over $k$ and has dimension $0$, then $X$
is a scheme.
\end{lemma}

\begin{proof}
Let $U$ be an affine scheme and let $U \to X$ be an \'etale morphism.
Set $R = U \times_X U$. Note that the two projection morphisms
$s, t : R \to U$ are \'etale morphisms of schemes. By
Properties of Spaces, Definition \ref{spaces-properties-definition-dimension}
we see that $\dim(U) = 0$ and similarly $\dim(R) = 0$. On the other hand,
the morphism $U \to \Spec(k)$ is locally of finite type as the
composition of the \'etale morphism $U \to X$ and $X \to \Spec(k)$, see
Morphisms of Spaces,
Lemmas \ref{spaces-morphisms-lemma-composition-finite-type} and
\ref{spaces-morphisms-lemma-etale-locally-finite-type}.
Similarly, $R \to \Spec(k)$ is locally of finite type.
Hence by
Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0}
we see that $U$ and $R$ are disjoint unions of spectra of
local Artinian $k$-algebras $A$ finite over $k$. In particular, as
$$
R = U \times_X U \longrightarrow U \times_{\Spec(k)} U
$$
is a monomorphism, we see that $R$ is a finite union of spectra of
finite $k$-algebras. It follows that $R$ is affine, see
Schemes, Lemma \ref{schemes-lemma-disjoint-union-affines}.
Applying
Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0}
once more we see that $R$ is finite over $k$. Hence $s, t$
are finite, see
Morphisms, Lemma \ref{morphisms-lemma-finite-permanence}.
Thus
Groupoids, Proposition \ref{groupoids-proposition-finite-flat-equivalence}
shows that the open subspace $U/R$ of $X$ is an affine scheme. Since the
schematic locus of $X$ is an open subspace (see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}),
and since $U \to X$ was an arbitrary \'etale morphism from an affine scheme
we conclude that $X$ is a scheme.
\end{proof}

\begin{lemma}
\label{lemma-locally-quasi-finite-over-field}
Let $k$ be a field. Let $X$ be an algebraic space over $k$.
The following are equivalent
\begin{enumerate}
\item $X$ is locally quasi-finite over $k$,
\item $X$ is locally of finite type over $k$ and has dimension $0$,
\item $X$ is a scheme and is locally quasi-finite over $k$,
\item $X$ is a scheme and is locally of finite type over $k$ and has
dimension $0$, and
\item $X$ is a disjoint union of spectra of Artinian local $k$-algebras
$A$ over $k$ with $\dim_k(A) < \infty$.
\end{enumerate}
\end{lemma}

\begin{proof}
Because we are over a field relative dimension of $X/k$ is the same as
the dimension of $X$. Hence by
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-locally-quasi-finite-rel-dimension-0}
we see that (1) and (2) are equivalent. Hence it follows from
Lemma \ref{lemma-locally-finite-type-dim-zero}
(and trivial implications) that (1) -- (4) are equivalent.
Finally,
Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0}
shows that (1) -- (4) are equivalent with (5).
\end{proof}

\begin{lemma}
\label{lemma-mono-towards-locally-quasi-finite-over-field}
Let $k$ be a field. Let $f : X \to Y$ be a monomorphism of algebraic spaces
over $k$. If $Y$ is locally quasi-finite over $k$ so is $X$.
\end{lemma}

\begin{proof}
Assume $Y$ is locally quasi-finite over $k$. By
Lemma \ref{lemma-locally-quasi-finite-over-field}
we see that $Y = \coprod \Spec(A_i)$ where each $A_i$ is an
Artinian local ring finite over $k$. By
Decent Spaces, Lemma
\ref{decent-spaces-lemma-monomorphism-toward-disjoint-union-dim-0-rings}
we see that $X$ is a scheme. Consider $X_i = f^{-1}(\Spec(A_i))$.
Then $X_i$ has either one or zero points. If $X_i$ has zero points there
is nothing to prove. If $X_i$ has one point, then
$X_i = \Spec(B_i)$ with $B_i$ a zero dimensional local ring
and $A_i \to B_i$ is an epimorphism of rings. In particular
$A_i/\mathfrak m_{A_i} = B_i/\mathfrak m_{A_i}B_i$ and we see that
$A_i \to B_i$ is surjective by Nakayama's lemma,
Algebra, Lemma \ref{algebra-lemma-NAK}
(because $\mathfrak m_{A_i}$ is a nilpotent ideal!).
Thus $B_i$ is a finite local $k$-algebra, and we conclude by
Lemma \ref{lemma-locally-quasi-finite-over-field}
that $X \to \Spec(k)$ is locally quasi-finite.
\end{proof}





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



\begin{lemma}
\label{lemma-smooth-regular}
Let $k$ be a field.
Let $X$ be an algebraic space smooth over $k$.
Then $X$ is a regular algebraic space.
\end{lemma}

\begin{proof}
Choose a scheme $U$ and a surjective \'etale morphism $U \to X$.
The morphism $U \to \Spec(k)$ is smooth as a composition of
an \'etale (hence smooth) morphism and a smooth morphism (see
Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-smooth}
and \ref{spaces-morphisms-lemma-composition-smooth}).
Hence $U$ is regular by
Varieties, Lemma \ref{varieties-lemma-smooth-regular}.
By
Properties of Spaces, Definition
\ref{spaces-properties-definition-type-property}
this means that $X$ is regular.
\end{proof}

\begin{lemma}
\label{lemma-smooth-separable-closed-points-dense}
Let $k$ be a field. Let $X$ be an algebraic space smooth over $\Spec(k)$.
The set of $x \in |X|$ which are image of morphisms $\Spec(k') \to X$
with $k' \supset k$ finite separable is dense in $|X|$.
\end{lemma}

\begin{proof}
Choose a scheme $U$ and a surjective \'etale morphism $U \to X$.
The morphism $U \to \Spec(k)$ is smooth as a composition of
an \'etale (hence smooth) morphism and a smooth morphism (see
Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-smooth}
and \ref{spaces-morphisms-lemma-composition-smooth}).
Hence we can apply Varieties, Lemma
\ref{varieties-lemma-smooth-separable-closed-points-dense} to see that
the closed points of $U$ whose residue fields are finite separable over
$k$ are dense. This implies the lemma by our definition of the
topology on $|X|$.
\end{proof}








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