## Tag `0B82`

## 63.7. Schematic locus and field extension

It can happen that a nonrepresentable algebraic space over a field $k$ becomes representable (i.e., a scheme) after base change to an extension of $k$. See Spaces, Example 56.14.2. In this section we address this issue.

Lemma 63.7.1. Let $k$ be a field. Let $X$ be an algebraic space over $k$. If there exists a purely inseparable field extension $k \subset k'$ such that $X_{k'}$ is a scheme, then $X$ is a scheme.

Proof.The morphism $X_{k'} \to X$ is integral, surjective, and universally injective. Hence this lemma follows from Limits of Spaces, Lemma 61.15.4. $\square$Lemma 63.7.2. Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a quasi-separated algebraic space over $k$.

- If there exists a field extension $k \subset K$ such that $X_K$ is a scheme, then $X_{\overline{k}}$ is a scheme.
- If $X$ is quasi-compact and there exists a field extension $k \subset K$ such that $X_K$ is a scheme, then $X_{k'}$ is a scheme for some finite separable extension $k'$ of $k$.

Proof.Since every algebraic space is the union of its quasi-compact open subspaces, we see that the first part of the lemma follows from the second part (some details omitted). Thus we assume $X$ is quasi-compact and we assume given an extension $k \subset K$ with $K_K$ representable. Write $K = \bigcup A$ as the colimit of finitely generated $k$-subalgebras $A$. By Limits of Spaces, Lemma 61.5.11 we see that $X_A$ is a scheme for some $A$. Choose a maximal ideal $\mathfrak m \subset A$. By the Hilbert Nullstellensatz (Algebra, Theorem 10.33.1) the residue field $k' = A/\mathfrak m$ is a finite extension of $k$. Thus we see that $X_{k'}$ is a scheme. If $k' \supset k$ is not separable, let $k' \supset k'' \supset k$ be the subextension found in Fields, Lemma 9.14.6. Since $k'/k''$ is purely inseparable, by Lemma 63.7.1 the algebraic space $X_{k''}$ is a scheme. Since $k''|k$ is separable the proof is complete. $\square$Lemma 63.7.3. Let $k \subset k'$ be a finite Galois extension with Galois group $G$. Let $X$ be an algebraic space over $k$. Then $G$ acts freely on the algebraic space $X_{k'}$ and $X = X_{k'}/G$ in the sense of Properties of Spaces, Lemma 57.34.1.

Proof.Omitted. Hints: First show that $\mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(k')/G$. Then use compatibility of taking quotients with base change. $\square$Lemma 63.7.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ and let $G$ be a finite group acting freely on $X$. Set $Y = X/G$ as in Properties of Spaces, Lemma 57.34.1. For $y \in |Y|$ the following are equivalent

- $y$ is in the schematic locus of $Y$, and
- there exists an affine open $U \subset X$ containing the preimage of $y$.

Proof.It follows from the construction of $Y = X/G$ in Properties of Spaces, Lemma 57.34.1 that the morphism $X \to Y$ is surjective and étale. Of course we have $X \times_Y X = X \times G$ hence the morphism $X \to Y$ is even finite étale. It is also surjective. Thus the lemma follows from Decent Spaces, Lemma 59.10.3. $\square$Lemma 63.7.5. Let $k$ be a field. Let $X$ be a quasi-separated algebraic space over $k$. If there exists a purely transcendental field extension $k \subset K$ such that $X_K$ is a scheme, then $X$ is a scheme.

Proof.Since every algebraic space is the union of its quasi-compact open subspaces, we may assume $X$ is quasi-compact (some details omitted). Recall (Fields, Definition 9.26.1) that the assumption on the extension $K/k$ signifies that $K$ is the fraction field of a polynomial ring (in possibly infinitely many variables) over $k$. Thus $K = \bigcup A$ is the union of subalgebras each of which is a localization of a finite polynomial algebra over $k$. By Limits of Spaces, Lemma 61.5.11 we see that $X_A$ is a scheme for some $A$. Write $$ A = k[x_1, \ldots, x_n][1/f] $$ for some nonzero $f \in k[x_1, \ldots, x_n]$.If $k$ is infinite then we can finish the proof as follows: choose $a_1, \ldots, a_n \in k$ with $f(a_1, \ldots, a_n) \not = 0$. Then $(a_1, \ldots, a_n)$ define an $k$-algebra map $A \to k$ mapping $x_i$ to $a_i$ and $1/f$ to $1/f(a_1, \ldots, a_n)$. Thus the base change $X_A \times_{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(k) \cong X$ is a scheme as desired.

In this paragraph we finish the proof in case $k$ is finite. In this case we write $X = \mathop{\mathrm{lim}}\nolimits X_i$ with $X_i$ of finite presentation over $k$ and with affine transition morphisms (Limits of Spaces, Lemma 61.10.1). Using Limits of Spaces, Lemma 61.5.11 we see that $X_{i, A}$ is a scheme for some $i$. Thus we may assume $X \to \mathop{\mathrm{Spec}}(k)$ is of finite presentation. Let $x \in |X|$ be a closed point. We may represent $x$ by a closed immersion $\mathop{\mathrm{Spec}}(\kappa) \to X$ (Decent Spaces, Lemma 59.13.6). Then $\mathop{\mathrm{Spec}}(\kappa) \to \mathop{\mathrm{Spec}}(k)$ is of finite type, hence $\kappa$ is a finite extension of $k$ (by the Hilbert Nullstellensatz, see Algebra, Theorem 10.33.1; some details omitted). Say $[\kappa : k] = d$. Choose an integer $n \gg 0$ prime to $d$ and let $k \subset k'$ be the extension of degree $n$. Then $k'/k$ is Galois with $G = \text{Aut}(k'/k)$ cyclic of order $n$. If $n$ is large enough there will be $k$-algebra homomorphism $A \to k'$ by the same reason as above. Then $X_{k'}$ is a scheme and $X = X_{k'}/G$ (Lemma 63.7.3). On the other hand, since $n$ and $d$ are relatively prime we see that $$ \mathop{\mathrm{Spec}}(\kappa) \times_{X} X_{k'} = \mathop{\mathrm{Spec}}(\kappa) \times_{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k') = \mathop{\mathrm{Spec}}(\kappa \otimes_k k') $$ is the spectrum of a field. In other words, the fibre of $X_{k'} \to X$ over $x$ consists of a single point. Thus by Lemma 63.7.4 we see that $x$ is in the schematic locus of $X$ as desired. $\square$

Remark 63.7.6. Let $k$ be finite field. Let $K \supset k$ be a geometrically irreducible field extension. Then $K$ is the limit of geometrically irreducible finite type $k$-algebras $A$. Given $A$ the estimates of Lang and Weil [LW], show that for $n \gg 0$ there exists an $k$-algebra homomorphism $A \to k'$ with $k'/k$ of degree $n$. Analyzing the argument given in the proof of Lemma 63.7.5 we see that if $X$ is a quasi-separated algebraic space over $k$ and $X_K$ is a scheme, then $X$ is a scheme. If we ever need this result we will precisely formulate it and prove it here.

Lemma 63.7.7. Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be an algebraic space over $k$ such that

- $X$ is decent and locally of finite type over $k$,
- $X_{\overline{k}}$ is a scheme, and
- any finite set of $\overline{k}$-rational points of $X_{\overline{k}}$ are contained in an affine.
Then $X$ is a scheme.

Proof.If $k \subset K$ is an extension, then the base change $X_K$ is decent (Decent Spaces, Lemma 59.6.5) and locally of finite type over $K$ (Morphisms of Spaces, Lemma 58.23.3). By Lemma 63.7.1 it suffices to prove that $X$ becomes a scheme after base change to the perfection of $k$, hence we may assume $k$ is a perfect field (this step isn't strictly necessary, but makes the other arguments easier to think about). By covering $X$ by quasi-compact opens we see that it suffices to prove the lemma in case $X$ is quasi-compact (small detail omitted). In this case $|X|$ is a sober topological space (Decent Spaces, Proposition 59.11.9). Hence it suffices to show that every closed point in $|X|$ is contained in the schematic locus of $X$ (use Properties of Spaces, Lemma 57.13.1 and Topology, Lemma 5.12.8).Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma 59.13.6 we can find a closed immersion $\mathop{\mathrm{Spec}}(l) \to X$ representing $x$. Then $\mathop{\mathrm{Spec}}(l) \to \mathop{\mathrm{Spec}}(k)$ is of finite type (Morphisms of Spaces, Lemma 58.23.2) and we conclude that $l$ is a finite extension of $k$ by the Hilbert Nullstellensatz (Algebra, Theorem 10.33.1). It is separable because $k$ is perfect. Thus the scheme $$ \mathop{\mathrm{Spec}}(l) \times_X X_{\overline{k}} = \mathop{\mathrm{Spec}}(l) \times_{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(\overline{k}) = \mathop{\mathrm{Spec}}(l \otimes_k \overline{k}) $$ is the disjoint union of a finite number of $\overline{k}$-rational points. By assumption (3) we can find an affine open $W \subset X_{\overline{k}}$ containing these points.

By Lemma 63.7.2 we see that $X_{k'}$ is a scheme for some finite extension $k'/k$. After enlarging $k'$ we may assume that there exists an affine open $U' \subset X_{k'}$ whose base change to $\overline{k}$ recovers $W$ (use that $X_{\overline{k}}$ is the limit of the schemes $X_{k''}$ for $k' \subset k'' \subset \overline{k}$ finite and use Limits, Lemmas 31.4.11 and 31.4.13). We may assume that $k'/k$ is a Galois extension (take the normal closure Fields, Lemma 9.16.3 and use that $k$ is perfect). Set $G = \text{Gal}(k'/k)$. By construction the $G$-invariant closed subscheme $\mathop{\mathrm{Spec}}(l) \times_X X_{k'}$ is contained in $U'$. Thus $x$ is in the schematic locus by Lemmas 63.7.3 and 63.7.4. $\square$

The following two lemmas should go somewhere else. Please compare the next lemma to Decent Spaces, Lemma 59.17.8.

Lemma 63.7.8. Let $k$ be a field. Let $X$ be an algebraic space over $k$. The following are equivalent

- $X$ is locally quasi-finite over $k$,
- $X$ is locally of finite type over $k$ and has dimension $0$,
- $X$ is a scheme and is locally quasi-finite over $k$,
- $X$ is a scheme and is locally of finite type over $k$ and has dimension $0$, and
- $X$ is a disjoint union of spectra of Artinian local $k$-algebras $A$ over $k$ with $\dim_k(A) < \infty$.

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 58.34.6 we see that (1) and (2) are equivalent. Hence it follows from Lemma 63.6.1 (and trivial implications) that (1) – (4) are equivalent. Finally, Varieties, Lemma 32.20.2 shows that (1) – (4) are equivalent with (5). $\square$Lemma 63.7.9. 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 $X$.

Proof.Assume $Y$ is locally quasi-finite over $k$. By Lemma 63.7.8 we see that $Y = \coprod \mathop{\mathrm{Spec}}(A_i)$ where each $A_i$ is an Artinian local ring finite over $k$. By Decent Spaces, Lemma 59.18.1 we see that $X$ is a scheme. Consider $X_i = f^{-1}(\mathop{\mathrm{Spec}}(A_i))$. 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 $X_i = \mathop{\mathrm{Spec}}(B_i)$ 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 10.19.1 (because $\mathfrak m_{A_i}$ is a nilpotent ideal!). Thus $B_i$ is a finite local $k$-algebra, and we conclude by Lemma 63.7.8 that $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. $\square$

The code snippet corresponding to this tag is a part of the file `spaces-over-fields.tex` and is located in lines 808–1120 (see updates for more information).

```
\section{Schematic locus and field extension}
\label{section-schematic-and-field-extension}
\noindent
It can happen that a nonrepresentable algebraic space over a field $k$
becomes representable (i.e., a scheme) after base change to an extension
of $k$. See Spaces, Example \ref{spaces-example-non-representable-descent}.
In this section we address this issue.
\begin{lemma}
\label{lemma-scheme-after-purely-inseparable-base-change}
Let $k$ be a field. Let $X$ be an algebraic space over $k$.
If there exists a purely inseparable field extension $k \subset k'$
such that $X_{k'}$ is a scheme, then $X$ is a scheme.
\end{lemma}
\begin{proof}
The morphism $X_{k'} \to X$ is integral, surjective, and
universally injective. Hence this lemma follows from
Limits of Spaces, Lemma
\ref{spaces-limits-lemma-integral-universally-bijective-scheme}.
\end{proof}
\begin{lemma}
\label{lemma-when-scheme-after-base-change}
Let $k$ be a field with algebraic closure $\overline{k}$.
Let $X$ be a quasi-separated algebraic space over $k$.
\begin{enumerate}
\item If there exists a field extension $k \subset K$ such that
$X_K$ is a scheme, then $X_{\overline{k}}$ is a scheme.
\item If $X$ is quasi-compact and there exists a field extension
$k \subset K$ such that $X_K$ is a scheme, then $X_{k'}$
is a scheme for some finite separable extension $k'$ of $k$.
\end{enumerate}
\end{lemma}
\begin{proof}
Since every algebraic space is the union of its quasi-compact open
subspaces, we see that the first part of the lemma follows from
the second part (some details omitted). Thus we assume $X$ is quasi-compact
and we assume given an extension $k \subset K$ with $K_K$ representable.
Write $K = \bigcup A$ as the colimit of finitely generated $k$-subalgebras
$A$. By Limits of Spaces, Lemma \ref{spaces-limits-lemma-limit-is-scheme}
we see that $X_A$ is a scheme for some $A$. Choose a maximal ideal
$\mathfrak m \subset A$. By the Hilbert Nullstellensatz
(Algebra, Theorem \ref{algebra-theorem-nullstellensatz})
the residue field $k' = A/\mathfrak m$ is a finite extension of $k$.
Thus we see that $X_{k'}$ is a scheme. If $k' \supset k$ is not
separable, let $k' \supset k'' \supset k$ be the subextension
found in Fields, Lemma \ref{fields-lemma-separable-first}.
Since $k'/k''$ is purely inseparable, by
Lemma \ref{lemma-scheme-after-purely-inseparable-base-change}
the algebraic space $X_{k''}$ is a scheme. Since $k''|k$ is separable
the proof is complete.
\end{proof}
\begin{lemma}
\label{lemma-base-change-by-Galois}
Let $k \subset k'$ be a finite Galois extension with Galois group $G$.
Let $X$ be an algebraic space over $k$. Then $G$ acts freely on the
algebraic space $X_{k'}$ and $X = X_{k'}/G$ in the sense of
Properties of Spaces, Lemma \ref{spaces-properties-lemma-quotient}.
\end{lemma}
\begin{proof}
Omitted. Hints: First show that $\Spec(k) = \Spec(k')/G$.
Then use compatibility of taking quotients with base change.
\end{proof}
\begin{lemma}
\label{lemma-when-quotient-scheme-at-point}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$ and
let $G$ be a finite group acting freely on $X$. Set $Y = X/G$ as
in Properties of Spaces, Lemma \ref{spaces-properties-lemma-quotient}.
For $y \in |Y|$ the following are equivalent
\begin{enumerate}
\item $y$ is in the schematic locus of $Y$, and
\item there exists an affine open $U \subset X$
containing the preimage of $y$.
\end{enumerate}
\end{lemma}
\begin{proof}
It follows from the construction of $Y = X/G$ in
Properties of Spaces, Lemma \ref{spaces-properties-lemma-quotient}
that the morphism $X \to Y$ is surjective and \'etale.
Of course we have $X \times_Y X = X \times G$ hence the morphism
$X \to Y$ is even finite \'etale. It is also surjective.
Thus the lemma follows from
Decent Spaces, Lemma \ref{decent-spaces-lemma-when-quotient-scheme-at-point}.
\end{proof}
\begin{lemma}
\label{lemma-scheme-after-purely-transcendental-base-change}
Let $k$ be a field. Let $X$ be a quasi-separated
algebraic space over $k$. If there exists a purely transcendental
field extension $k \subset K$ such that $X_K$ is a scheme, then
$X$ is a scheme.
\end{lemma}
\begin{proof}
Since every algebraic space is the union of its quasi-compact open
subspaces, we may assume $X$ is quasi-compact (some details omitted).
Recall (Fields, Definition \ref{fields-definition-transcendence})
that the assumption on the extension $K/k$ signifies that
$K$ is the fraction field of a polynomial ring (in possibly infinitely
many variables) over $k$. Thus $K = \bigcup A$ is the union of subalgebras
each of which is a localization of a finite polynomial algebra over $k$.
By Limits of Spaces, Lemma \ref{spaces-limits-lemma-limit-is-scheme}
we see that $X_A$ is a scheme for some $A$. Write
$$
A = k[x_1, \ldots, x_n][1/f]
$$
for some nonzero $f \in k[x_1, \ldots, x_n]$.
\medskip\noindent
If $k$ is infinite then we can finish the proof as follows: choose
$a_1, \ldots, a_n \in k$ with $f(a_1, \ldots, a_n) \not = 0$.
Then $(a_1, \ldots, a_n)$ define an $k$-algebra map $A \to k$
mapping $x_i$ to $a_i$ and $1/f$ to $1/f(a_1, \ldots, a_n)$.
Thus the base change $X_A \times_{\Spec(A)} \Spec(k) \cong X$ is a
scheme as desired.
\medskip\noindent
In this paragraph we finish the proof in case $k$ is finite. In this
case we write $X = \lim X_i$ with $X_i$ of finite presentation over $k$
and with affine transition morphisms
(Limits of Spaces, Lemma \ref{spaces-limits-lemma-relative-approximation}).
Using Limits of Spaces, Lemma \ref{spaces-limits-lemma-limit-is-scheme}
we see that $X_{i, A}$ is a scheme for some $i$. Thus we may assume
$X \to \Spec(k)$ is of finite presentation. Let $x \in |X|$ be a closed
point. We may represent $x$ by a closed immersion
$\Spec(\kappa) \to X$
(Decent Spaces, Lemma \ref{decent-spaces-lemma-decent-space-closed-point}).
Then $\Spec(\kappa) \to \Spec(k)$ is of finite type, hence $\kappa$
is a finite extension of $k$ (by the Hilbert Nullstellensatz, see
Algebra, Theorem \ref{algebra-theorem-nullstellensatz};
some details omitted). Say $[\kappa : k] = d$. Choose an integer
$n \gg 0$ prime to $d$ and let $k \subset k'$ be the extension
of degree $n$. Then $k'/k$ is Galois with $G = \text{Aut}(k'/k)$
cyclic of order $n$. If $n$ is large enough there will be $k$-algebra
homomorphism $A \to k'$ by the same reason as above.
Then $X_{k'}$ is a scheme and $X = X_{k'}/G$
(Lemma \ref{lemma-base-change-by-Galois}).
On the other hand, since $n$ and $d$ are relatively prime we see that
$$
\Spec(\kappa) \times_{X} X_{k'} =
\Spec(\kappa) \times_{\Spec(k)} \Spec(k') =
\Spec(\kappa \otimes_k k')
$$
is the spectrum of a field. In other words, the fibre of $X_{k'} \to X$
over $x$ consists of a single point. Thus by
Lemma \ref{lemma-when-quotient-scheme-at-point}
we see that $x$ is in the schematic locus of $X$ as desired.
\end{proof}
\begin{remark}
\label{remark-when-does-the-argument-work}
Let $k$ be finite field. Let $K \supset k$ be a geometrically
irreducible field extension. Then $K$ is the limit of geometrically
irreducible finite type $k$-algebras $A$. Given $A$ the estimates
of Lang and Weil \cite{LW}, show that for $n \gg 0$ there exists
an $k$-algebra homomorphism $A \to k'$ with $k'/k$ of degree $n$.
Analyzing the argument given in the proof of
Lemma \ref{lemma-scheme-after-purely-transcendental-base-change}
we see that if $X$ is a quasi-separated algebraic space over $k$
and $X_K$ is a scheme, then $X$ is a scheme. If we ever need this
result we will precisely formulate it and prove it here.
\end{remark}
\begin{lemma}
\label{lemma-scheme-over-algebraic-closure-enough-affines}
Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$
be an algebraic space over $k$ such that
\begin{enumerate}
\item $X$ is decent and locally of finite type over $k$,
\item $X_{\overline{k}}$ is a scheme, and
\item any finite set of $\overline{k}$-rational points of $X_{\overline{k}}$
are contained in an affine.
\end{enumerate}
Then $X$ is a scheme.
\end{lemma}
\begin{proof}
If $k \subset K$ is an extension, then the base change $X_K$ is
decent (Decent Spaces, Lemma
\ref{decent-spaces-lemma-representable-named-properties})
and locally of finite type
over $K$ (Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-base-change-finite-type}).
By Lemma \ref{lemma-scheme-after-purely-inseparable-base-change}
it suffices to prove that $X$ becomes a scheme after base change to
the perfection of $k$, hence we may assume $k$ is a perfect field
(this step isn't strictly necessary, but makes the other arguments
easier to think about).
By covering $X$ by quasi-compact opens we see that it suffices to prove
the lemma in case $X$ is quasi-compact (small detail omitted).
In this case $|X|$ is a sober topological space
(Decent Spaces, Proposition
\ref{decent-spaces-proposition-reasonable-sober}).
Hence it suffices to show that every closed point in $|X|$
is contained in the schematic locus of $X$
(use Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme} and
Topology, Lemma \ref{topology-lemma-quasi-compact-closed-point}).
\medskip\noindent
Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma
\ref{decent-spaces-lemma-decent-space-closed-point}
we can find a closed immersion $\Spec(l) \to X$ representing $x$.
Then $\Spec(l) \to \Spec(k)$ is of finite type (Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-composition-finite-type}) and we
conclude that $l$ is a finite extension of $k$
by the Hilbert Nullstellensatz (Algebra, Theorem
\ref{algebra-theorem-nullstellensatz}). It is separable because
$k$ is perfect. Thus the scheme
$$
\Spec(l) \times_X X_{\overline{k}} =
\Spec(l) \times_{\Spec(k)} \Spec(\overline{k}) =
\Spec(l \otimes_k \overline{k})
$$
is the disjoint union of a finite number of $\overline{k}$-rational points.
By assumption (3) we can find an affine open $W \subset X_{\overline{k}}$
containing these points.
\medskip\noindent
By Lemma \ref{lemma-when-scheme-after-base-change} we see that $X_{k'}$
is a scheme for some finite extension $k'/k$. After enlarging
$k'$ we may assume that there exists an affine open $U' \subset X_{k'}$
whose base change to $\overline{k}$ recovers $W$
(use that $X_{\overline{k}}$ is the limit of the schemes $X_{k''}$
for $k' \subset k'' \subset \overline{k}$ finite and use
Limits, Lemmas \ref{limits-lemma-descend-opens} and
\ref{limits-lemma-limit-affine}). We may assume
that $k'/k$ is a Galois extension (take the normal closure
Fields, Lemma \ref{fields-lemma-normal-closure} and use
that $k$ is perfect). Set $G = \text{Gal}(k'/k)$.
By construction the $G$-invariant closed subscheme
$\Spec(l) \times_X X_{k'}$ is contained in $U'$.
Thus $x$ is in the schematic locus by
Lemmas \ref{lemma-base-change-by-Galois} and
\ref{lemma-when-quotient-scheme-at-point}.
\end{proof}
\noindent
The following two lemmas should go somewhere else.
Please compare the next lemma to
Decent Spaces, Lemma \ref{decent-spaces-lemma-conditions-on-space-over-field}.
\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}
```

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