70.10 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 63.14.2. In this section we address this issue.

Lemma 70.10.1. 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.

Proof. The morphism $X_{k'} \to X$ is integral, surjective, and universally injective. Hence this lemma follows from Limits of Spaces, Lemma 68.15.4. $\square$

Lemma 70.10.2. Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a quasi-separated algebraic space over $k$.

1. If there exists a field extension $k \subset K$ such that $X_ K$ is a scheme, then $X_{\overline{k}}$ is a scheme.

2. 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 68.5.11 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 70.10.1 the algebraic space $X_{k''}$ is a scheme. Since $k''|k$ is separable the proof is complete. $\square$

Lemma 70.10.3. 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 64.34.1.

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 70.10.4. 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 64.34.1. For $y \in |Y|$ the following are equivalent

1. $y$ is in the schematic locus of $Y$, and

2. 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 64.34.1 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 66.10.3. $\square$

Lemma 70.10.5. 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 68.5.11 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 68.10.1). Using Limits of Spaces, Lemma 68.5.11 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 66.14.6). 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 70.10.3). 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 70.10.4 we see that $x$ is in the schematic locus of $X$ as desired. $\square$

Remark 70.10.6. 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 70.10.5 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 70.10.7. Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be an algebraic space over $k$ such that

1. $X$ is decent and locally of finite type over $k$,

2. $X_{\overline{k}}$ is a scheme, and

3. 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 66.6.5) and locally of finite type over $K$ (Morphisms of Spaces, Lemma 65.23.3). By Lemma 70.10.1 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 66.12.4). Hence it suffices to show that every closed point in $|X|$ is contained in the schematic locus of $X$ (use Properties of Spaces, Lemma 64.13.1 and Topology, Lemma 5.12.8).

Let $x \in |X|$ be a closed point. By Decent Spaces, Lemma 66.14.6 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 65.23.2) 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 70.10.2 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 32.4.11 and 32.4.13). 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 70.10.3 and 70.10.4. $\square$

The following two lemmas should go somewhere else. Please compare the next lemma to Decent Spaces, Lemma 66.18.8.

Lemma 70.10.8. Let $k$ be a field. Let $X$ be an algebraic space over $k$. The following are equivalent

1. $X$ is locally quasi-finite over $k$,

2. $X$ is locally of finite type over $k$ and has dimension $0$,

3. $X$ is a scheme and is locally quasi-finite over $k$,

4. $X$ is a scheme and is locally of finite type over $k$ and has dimension $0$, and

5. $X$ 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 65.34.6 we see that (1) and (2) are equivalent. Hence it follows from Lemma 70.9.1 (and trivial implications) that (1) – (4) are equivalent. Finally, Varieties, Lemma 33.20.2 shows that (1) – (4) are equivalent with (5). $\square$

Lemma 70.10.9. 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 70.10.8 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 66.19.1 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 70.10.8 that $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. $\square$

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