Lemma 86.19.13. Let $S$ be a scheme. Let $Y$ be a McQuillan affine formal algebraic space over $S$, i.e., $Y = \text{Spf}(B)$ for some weakly admissible topological $S$-algebra $B$. Then there is an equivalence of categories between

the category of morphisms $f : X \to Y$ of affine formal algebraic spaces which are representable by algebraic spaces and étale, and

the category of topological $B$-algebras of the form $A^\wedge $ where $A$ is an étale $B$-algebra and $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/JA$ with $J \subset B$ running over the weak ideals of definition of $B$.

The equivalence is given by sending $A^\wedge $ to $X = \text{Spf}(A^\wedge )$. In particular, any $X$ as in (1) is McQuillan.

**Proof.**
Let $A$ be an étale $B$-algebra. Then $B/J \to A/JA$ is étale for every open ideal $J \subset B$. Hence the morphism $\text{Spf}(A^\wedge ) \to Y$ is representable and étale. The functor $\text{Spf}$ is fully faithful by Lemma 86.9.10. To finish the proof we will show in the next paragraph that any $X \to Y$ as in (1) is in the essential image.

Choose a weak ideal of definition $J_0 \subset B$. Set $Y_0 = \mathop{\mathrm{Spec}}(B/J_0)$ and $X_0 = Y_0 \times _ Y X$. Then $X_0 \to Y_0$ is an étale morphism of affine schemes (see Lemma 86.19.7). Say $X_0 = \mathop{\mathrm{Spec}}(A_0)$. By Algebra, Lemma 10.143.10 we can find an étale algebra map $B \to A$ such that $A_0 \cong A/J_0A$. Consider an ideal of definition $J \subset J_0$. As above we may write $\mathop{\mathrm{Spec}}(B/J) \times _ Y X = \mathop{\mathrm{Spec}}(\bar A)$ for some étale ring map $B/J \to \bar A$. Then both $B/J \to \bar A$ and $B/J \to A/JA$ are étale ring maps lifting the étale ring map $B/J_0 \to A_0$. By More on Algebra, Lemma 15.11.2 there is a unique $B/J$-algebra isomorphism $\varphi _ J : A/JA \to \bar A$ lifting the identification modulo $J_0$. Since the maps $\varphi _ J$ are unique they are compatible for varying $J$. Thus

\[ X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J) \times _ Y X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/JA) = \text{Spf}(A) \]

and we see that the lemma holds.
$\square$

## Comments (2)

Comment #1952 by Brian Conrad on

Comment #2007 by Johan on