The Stacks project

Proof. It is clear that (5) implies (4).

Assume (4) for $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. Set $A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$. Let $T \to X$ be a closed immersion with $T$ a scheme (note that $T \to X$ is representable by Lemma 87.9.2). Since $X_\lambda \to X$ is a thickening, so is $X_\lambda \times _ X T \to T$. On the other hand, $X_\lambda \times _ X T \to X_\lambda $ is a closed immersion, hence $X_\lambda \times _ X T$ is affine. Hence $T$ is affine by Limits, Proposition 32.11.2. Then $T \to X$ factors through $X_\lambda $ for some $\lambda $ by Lemma 87.9.4. Thus $A \to \Gamma (X_\lambda , \mathcal{O}) \to \Gamma (T, \mathcal{O})$ is surjective. In this way we see that (3) holds.

It is clear that (3) implies (2).

Assume (2) for $A$ and $X \to \mathop{\mathrm{Spec}}(A)$. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. Then $A_\lambda = \Gamma (X_\lambda , \mathcal{O})$ is a quotient of $A$ by assumption (2). Hence $A^\wedge = \mathop{\mathrm{lim}}\nolimits A_\lambda $ is a complete topological ring, see discussion in More on Algebra, Section 15.36. The maps $A^\wedge \to A_\lambda $ are surjective as $A \to A_\lambda $ is. We claim that for any $\lambda $ the kernel $I_\lambda \subset A^\wedge $ of $A^\wedge \to A_\lambda $ is a weak ideal of definition. Namely, it is open by definition of the limit topology. If $f \in I_\lambda $, then for any $\mu \in \Lambda $ the image of $f$ in $A_\mu $ is zero in all the residue fields of the points of $X_\mu $. Hence it is a nilpotent element of $A_\mu $. Hence some power $f^ n \in I_\mu $. Thus $f^ n \to 0$ as $n \to 0$. Thus $A^\wedge $ is weakly admissible. Finally, suppose that $I \subset A^\wedge $ is a weak ideal of definition. Then $I \subset A^\wedge $ is open and hence there exists some $\lambda $ such that $I \supset I_\lambda $. Thus we obtain a morphism $\mathop{\mathrm{Spec}}(A^\wedge /I) \to \mathop{\mathrm{Spec}}(A_\lambda ) \to X$. Then it follows that $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A^\wedge /I)$ where now the colimit is over all weak ideals of definition. Thus (1) holds.

Assume (1). In this case it is clear that $X$ is an affine formal algebraic space. Let $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be any presentation as in Definition 87.9.1. For each $\lambda $ we can find a weak ideal of definition $I \subset A$ such that $X_\lambda \to X$ factors through $\mathop{\mathrm{Spec}}(A/I) \to X$, see Lemma 87.9.4. Then $X_\lambda = \mathop{\mathrm{Spec}}(A/I_\lambda )$ with $I \subset I_\lambda $. Conversely, for any weak ideal of definition $I \subset A$ the morphism $\mathop{\mathrm{Spec}}(A/I) \to X$ factors through $X_\lambda $ for some $\lambda $, i.e., $I_\lambda \subset I$. It follows that each $I_\lambda $ is a weak ideal of definition and that they form a cofinal subset of the set of weak ideals of definition. Hence $A = \mathop{\mathrm{lim}}\nolimits A/I = \mathop{\mathrm{lim}}\nolimits A/I_\lambda $ and we see that (5) is true and moreover that $A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$. $\square$