The Stacks project

Lemma 85.15.13. 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

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

  2. 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 85.5.10. 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 85.15.7). 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

In the statement of this lemma, in (2) the map should be (not ), in the displayed expression should be , and likewise in the final sentence should be .


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AN8. Beware of the difference between the letter 'O' and the digit '0'.