## 66.15 Locally separated spaces

It turns out that a locally separated algebraic space is decent.

Lemma 66.15.1. Let $A$ be a ring. Let $k$ be a field. Let $\mathfrak p_ n$, $n \geq 1$ be a sequence of pairwise distinct primes of $A$. Moreover, for each $n$ let $k \to \kappa (\mathfrak p_ n)$ be an embedding. Then the closure of the image of

\[ \coprod \nolimits _{n \not= m} \mathop{\mathrm{Spec}}(\kappa (\mathfrak p_ n) \otimes _ k \kappa (\mathfrak p_ m)) \longrightarrow \mathop{\mathrm{Spec}}(A \otimes A) \]

meets the diagonal.

**Proof.**
Set $k_ n = \kappa (\mathfrak p_ n)$. We may assume that $A = \prod k_ n$. Denote $x_ n = \mathop{\mathrm{Spec}}(k_ n)$ the open and closed point corresponding to $A \to k_ n$. Then $\mathop{\mathrm{Spec}}(A) = Z \amalg \{ x_ n\} $ where $Z$ is a nonempty closed subset. Namely, $Z = V(e_ n; n \geq 1)$ where $e_ n$ is the idempotent of $A$ corresponding to the factor $k_ n$ and $Z$ is nonempty as the ideal generated by the $e_ n$ is not equal to $A$. We will show that the closure of the image contains $\Delta (Z)$. The kernel of the map

\[ (\prod k_ n) \otimes _ k (\prod k_ m) \longrightarrow \prod \nolimits _{n \not= m} k_ n \otimes _ k k_ m \]

is the ideal generated by $e_ n \otimes e_ n$, $n \geq 1$. Hence the closure of the image of the map on spectra is $V(e_ n \otimes e_ n; n \geq 1)$ whose intersection with $\Delta (\mathop{\mathrm{Spec}}(A))$ is $\Delta (Z)$. Thus it suffices to show that

\[ \coprod \nolimits _{n \not= m} \mathop{\mathrm{Spec}}(k_ n \otimes _ k k_ m) \longrightarrow \mathop{\mathrm{Spec}}(\prod \nolimits _{n \not= m} k_ n \otimes _ k k_ m) \]

has dense image. This follows as the family of ring maps $\prod _{n \not= m} k_ n \otimes _ k k_ m \to k_ n \otimes _ k k_ m$ is jointly injective.
$\square$

Lemma 66.15.2 (David Rydh). A locally separated algebraic space is decent.

**Proof.**
Let $S$ be a scheme and let $X$ be a locally separated algebraic space over $S$. We may assume $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$, see Properties of Spaces, Definition 64.3.1. Unadorned fibre products will be over $\mathbf{Z}$. Let $x \in |X|$. Choose a scheme $U$, an étale morphism $U \to X$, and a point $u \in U$ mapping to $x$ in $|X|$. As usual we identify $u = \mathop{\mathrm{Spec}}(\kappa (u))$. As $X$ is locally separated the morphism

\[ u \times _ X u \to u \times u \]

is an immersion (Morphisms of Spaces, Lemma 65.4.5). Hence More on Groupoids, Lemma 40.11.5 tells us that it is a closed immersion (use Schemes, Lemma 26.10.4). As $u \times _ X u \to u \times _ X U$ is a monomorphism (base change of $u \to U$) and as $u \times _ X U \to u$ is étale we conclude that $u \times _ X u$ is a disjoint union of spectra of fields (see Remark 66.4.1 and Schemes, Lemma 26.23.11). Since it is also closed in the affine scheme $u \times u$ we conclude $u \times _ X u$ is a finite disjoint union of spectra of fields. Thus $x$ can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field, see Lemma 66.4.3.

Next, let $U = \mathop{\mathrm{Spec}}(A)$ be an affine scheme and let $U \to X$ be an étale morphism. To finish the proof it suffices to show that $F = U \times _ X \mathop{\mathrm{Spec}}(k)$ is finite. Write $F = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ as the disjoint union of finite separable extensions of $k$. We have to show that $I$ is finite. Set $R = U \times _ X U$. As $X$ is locally separated, the morphism $j : R \to U \times U$ is an immersion. Let $U' \subset U \times U$ be an open such that $j$ factors through a closed immersion $j' : R \to U'$. Let $e : U \to R$ be the diagonal map. Using that $e$ is a morphism between schemes étale over $U$ such that $\Delta = j \circ e$ is a closed immersion, we conclude that $R = e(U) \amalg W$ for some open and closed subscheme $W \subset R$. Since $j'$ is a closed immersion we conclude that $j'(W) \subset U'$ is closed and disjoint from $j'(e(U))$. Therefore $\overline{j(W)} \cap \Delta (U) = \emptyset $ in $U \times U$. Note that $W$ contains $\mathop{\mathrm{Spec}}(k_ i \otimes _ k k_{i'})$ for all $i \not= i'$, $i, i' \in I$. By Lemma 66.15.1 we conclude that $I$ is finite as desired.
$\square$

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