Lemma 64.10.1. 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.

## 64.10 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 57.14.2. In this section we address this issue.

**Proof.**
The morphism $X_{k'} \to X$ is integral, surjective, and universally injective. Hence this lemma follows from Limits of Spaces, Lemma 62.15.4.
$\square$

Lemma 64.10.2. 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 62.5.11 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 64.10.1 the algebraic space $X_{k''}$ is a scheme. Since $k''|k$ is separable the proof is complete.
$\square$

Lemma 64.10.3. 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 58.34.1.

**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 64.10.4. 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 58.34.1. 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 58.34.1 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 60.10.3.
$\square$

Lemma 64.10.5. 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 62.5.11 we see that $X_ A$ is a scheme for some $A$. Write

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 62.10.1). Using Limits of Spaces, Lemma 62.5.11 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 60.14.6). 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 64.10.3). On the other hand, since $n$ and $d$ are relatively prime we see that

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 64.10.4 we see that $x$ is in the schematic locus of $X$ as desired. $\square$

Remark 64.10.6. 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 64.10.5 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 64.10.7. 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 60.6.5) and locally of finite type over $K$ (Morphisms of Spaces, Lemma 59.23.3). By Lemma 64.10.1 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 60.12.4). Hence it suffices to show that every closed point in $|X|$ is contained in the schematic locus of $X$ (use Properties of Spaces, Lemma 58.13.1 and Topology, Lemma 5.12.8).

Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma 60.14.6 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 59.23.2) 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

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 64.10.2 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 64.10.3 and 64.10.4. $\square$

The following two lemmas should go somewhere else. Please compare the next lemma to Decent Spaces, Lemma 60.18.8.

Lemma 64.10.8. 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 59.34.6 we see that (1) and (2) are equivalent. Hence it follows from Lemma 64.9.1 (and trivial implications) that (1) – (4) are equivalent. Finally, Varieties, Lemma 32.20.2 shows that (1) – (4) are equivalent with (5).
$\square$

Lemma 64.10.9. 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 64.10.8 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 60.19.1 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 64.10.8 that $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)