Lemma 85.13.1. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated formal algebraic space over $S$. Then $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ for a system of algebraic spaces $(X_\lambda , f_{\lambda \mu })$ over a directed set $\Lambda $ where each $f_{\lambda \mu } : X_\lambda \to X_\mu $ is a thickening.

## 85.13 Quasi-compact and quasi-separated formal algebraic spaces

The following result is due to Yasuda, see [Proposition 3.32, Yasuda].

**Proof.**
By Lemma 85.12.1 we may choose an affine formal algebraic space $Y$ and a representable surjective étale morphism $Y \to X$. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda $ as in Definition 85.5.1.

Pick $\lambda \in \Lambda $. Then $Y_\lambda \times _ X Y$ is a scheme by Lemma 85.5.11. The reduction (Lemma 85.7.2) of $Y_\lambda \times _ X Y$ is equal to the reduction of $Y_{red} \times _{X_{red}} Y_{red}$ which is quasi-compact as $X$ is quasi-separated and $Y_{red}$ is affine. Therefore $Y_\lambda \times _ X Y$ is a quasi-compact scheme. Hence there exists a $\mu \geq \lambda $ such that $\text{pr}_2 : Y_\lambda \times _ X Y \to Y$ factors through $Y_\mu $, see Lemma 85.5.4. Let $Z_\lambda $ be the scheme theoretic image of the morphism $\text{pr}_2 : Y_\lambda \times _ X Y \to Y_\mu $. This is independent of the choice of $\mu $ and we can and will think of $Z_\lambda \subset Y$ as the scheme theoretic image of the morphism $\text{pr}_2 : Y_\lambda \times _ X Y \to Y$. Observe that $Z_\lambda $ is also equal to the scheme theoretic image of the morphism $\text{pr}_1 : Y \times _ X Y_\lambda \to Y$ since this is isomorphic to the morphism used to define $Z_\lambda $. We claim that $Z_\lambda \times _ X Y = Y \times _ X Z_\lambda $ as subfunctors of $Y \times _ X Y$. Namely, since $Y \to X$ is étale we see that $Z_\lambda \times _ X Y$ is the scheme theoretic image of the morphism

by Morphisms of Spaces, Lemma 65.16.3. By the same token, $Y \times _ X Z_\lambda $ is the scheme theoretic image of the morphism

The claim follows. Then $R_\lambda = Z_\lambda \times _ X Y = Y \times _ X Z_\lambda $ together with the morphism $R_\lambda \to Z_\lambda \times _ S Z_\lambda $ defines an étale equivalence relation. In this way we obtain an algebraic space $X_\lambda = Z_\lambda /R_\lambda $. By construction the diagram

is cartesian (because $X$ is the coequalizer of the two projections $R = Y \times _ X Y \to Y$, because $Z_\lambda \subset Y$ is $R$-invariant, and because $R_\lambda $ is the restriction of $R$ to $Z_\lambda $). Hence $X_\lambda \to X$ is representable and a closed immersion, see Spaces, Lemma 63.11.5. On the other hand, since $Y_\lambda \subset Z_\lambda $ we see that $(X_\lambda )_{red} = X_{red}$, in other words, $X_\lambda \to X$ is a thickening. Finally, we claim that

We have $Y \times _ X X_\lambda = Z_\lambda \supset Y_\lambda $. Every morphism $T \to X$ where $T$ is a scheme over $S$ lifts étale locally to a morphism into $Y$ which lifts étale locally into a morphism into some $Y_\lambda $. Hence $T \to X$ lifts étale locally on $T$ to a morphism into $X_\lambda $. This finishes the proof. $\square$

Remark 85.13.2. In this remark we translate the statement and proof of Lemma 85.13.1 into the language of formal schemes à la EGA. Looking at Remark 85.8.4 we see that the lemma can be translated as follows

Every quasi-compact and quasi-separated formal scheme has a fundamental system of ideals of definition.

To prove this we first use the induction principle (reformulated for quasi-compact and quasi-separated formal schemes) of Cohomology of Schemes, Lemma 30.4.1 to reduce to the following situation: $\mathfrak X = \mathfrak U \cup \mathfrak V$ with $\mathfrak U$, $\mathfrak V$ open formal subschemes, with $\mathfrak V$ affine, and the result is true for $\mathfrak U$, $\mathfrak V$, and $\mathfrak U \cap \mathfrak V$. Pick any ideals of definition $\mathcal{I} \subset \mathcal{O}_\mathfrak U$ and $\mathcal{J} \subset \mathcal{O}_\mathfrak V$. By our assumption that we have a fundamental system of ideals of definition on $\mathfrak U$ and $\mathfrak V$ and because $\mathfrak U \cap \mathfrak V$ is quasi-compact, we can find ideals of definition $\mathcal{I}' \subset \mathcal{I}$ and $\mathcal{J}' \subset \mathcal{J}$ such that

Let $U \to U' \to \mathfrak U$ and $V \to V' \to \mathfrak V$ be the closed immersions determined by the ideals of definition $\mathcal{I}' \subset \mathcal{I} \subset \mathcal{O}_\mathfrak U$ and $\mathcal{J}' \subset \mathcal{J} \subset \mathcal{O}_\mathfrak V$. Let $\mathfrak U \cap V$ denote the open subscheme of $V$ whose underlying topological space is that of $\mathfrak U \cap \mathfrak V$. By our choice of $\mathcal{I}'$ there is a factorization $\mathfrak U \cap V \to U'$. We define similarly $U \cap \mathfrak V$ which factors through $V'$. Then we consider

and

Since taking scheme theoretic images of quasi-compact morphisms commutes with restriction to opens (Morphisms, Lemma 29.6.3) we see that $Z_ U \cap \mathfrak V = \mathfrak U \cap Z_ V$. Thus $Z_ U$ and $Z_ V$ glue to a scheme $Z$ which comes equipped with a morphism $Z \to \mathfrak X$. Analogous to the discussion in Remark 85.8.3 we see that $Z$ corresponds to a weak ideal of definition $\mathcal{I}_ Z \subset \mathcal{O}_\mathfrak X$. Note that $Z_ U \subset U'$ and that $Z_ V \subset V'$. Thus the collection of all $\mathcal{I}_ Z$ constructed in this manner forms a fundamental system of weak ideals of definition. Hence a subfamily gives a fundamental system of ideals of definition, see Remark 85.8.4.

Lemma 85.13.3. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Then $X$ is an affine formal algebraic space if and only if its reduction $X_{red}$ (Lemma 85.7.2) is affine.

**Proof.**
By Lemmas 85.11.1 and 85.12.1 and Definitions 85.11.3 and 85.12.2 we see that $X$ is quasi-compact and quasi-separated. By Yasuda's lemma (Lemma 85.13.1) we can write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as a filtered colimit of thickenings of algebraic spaces. However, each $X_\lambda $ is affine by Limits of Spaces, Lemma 68.15.3 because $(X_\lambda )_{red} = X_{red}$. Hence $X$ is an affine formal algebraic space by definition.
$\square$

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