Lemma 86.15.8. Let $S$ be a scheme. Let $Y$ be an affine formal algebraic space. Let $f : X \to Y$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ which is representable and affine. Then

1. $X$ is an affine formal algebraic space.

2. if $Y$ is countably indexed, then $X$ is countably indexed.

3. if $Y$ is adic*, then $X$ is adic*,

4. if $Y$ is Noetherian and $f$ is (locally) of finite type, then $X$ is Noetherian.

Proof. Proof of (1). Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda$ as in Definition 86.5.1. Since $f$ is representable and affine, the fibre products $X_\lambda = Y_\lambda \times _ Y X$ are affine. And $X = \mathop{\mathrm{colim}}\nolimits Y_\lambda \times _ Y X$. Thus $X$ is an affine formal algebraic space.

Proof of (2). If $Y$ is countably indexed, then in the argument above we may assume $\Lambda$ is countable. Then we immediately see that $X$ is countably indexed too.

Proof of (3). Assume $Y$ is adic*. Then $Y = \text{Spf}(B)$ for some adic topological ring $B$ which has a finitely generated ideal $J$ such that $\{ J^ n\}$ is a fundamental system of open ideals. Of course, then $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J^ n)$. The schemes $X \times _ Y \mathop{\mathrm{Spec}}(B/J^ n)$ are affine and we can write $X \times _ Y \mathop{\mathrm{Spec}}(B/J^ n) = \mathop{\mathrm{Spec}}(A_ n)$. Then $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A_ n)$. The $B$-algebra maps $A_{n + 1} \to A_ n$ are surjective and induce isomorphisms $A_{n + 1}/J^ nA_{n + 1} \to A_ n$. By Algebra, Lemma 10.98.2 the ring $A = \mathop{\mathrm{lim}}\nolimits A_ n$ is $J$-adically complete and $A/J^ n A = A_ n$. Hence $X = \text{Spf}(A^\wedge )$ is adic*.

Proof of (4). Combining (3) with Lemma 86.6.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 86.6.5. First, it tells us the affine schemes $Y_\lambda$ are Noetherian. Then $X_\lambda \to Y_\lambda$ is of finite type, hence $X_\lambda$ is Noetherian too (Morphisms, Lemma 29.15.6). Then the criterion tells us $X$ is Noetherian and the proof is complete. $\square$

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