The Stacks project

Lemma 87.29.3. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Assume $X$ is McQuillan and let $A$ be the weakly admissible topological ring associated to $X$. Then there is an anti-equivalence of categories between

  1. the category $\mathcal{C}$ introduced above, and

  2. the category of maps $Y \to X$ of finite type of affine formal algebraic spaces.

Proof. Let $(I_\lambda )$ be a fundamental system of weakly admissible ideals of definition in $A$. Consider $Y$ as in (2). Then $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ is affine (Definition 87.24.1 and Lemma 87.19.7). Say $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$. The ring map $A/I_\lambda \to B_\lambda $ is of finite type because $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A/I_\lambda )$ is of finite type (by Definition 87.24.1). Then $(B_\lambda )$ is an object of $\mathcal{C}$.

Conversely, given an object $(B_\lambda )$ of $\mathcal{C}$ we can set $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$. This is an affine formal algebraic space. We claim that

\[ Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda ) \]

To show this it suffices we get the same values if we evaluate on a quasi-compact scheme $U$. A morphism $U \to \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ comes from a morphism $U \to \mathop{\mathrm{Spec}}(B_\mu ) \times _{\mathop{\mathrm{Spec}}(A/I_\mu )} \mathop{\mathrm{Spec}}(A/I_\lambda )$ for some $\mu \geq \lambda $ (use Lemma 87.9.4 two times). Since $\mathop{\mathrm{Spec}}(B_\mu ) \times _{\mathop{\mathrm{Spec}}(A/I_\mu )} \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$ by our second assumption on objects of $\mathcal{C}$ this proves what we want. Using this we can show the morphism $Y \to X$ is of finite type. Namely, we note that for any morphism $U \to X$ with $U$ a quasi-compact scheme, we get a factorization $U \to \mathop{\mathrm{Spec}}(A/I_\lambda ) \to X$ for some $\lambda $ (see lemma cited above). Hence

\[ Y \times _ X U = Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )) \times _{\mathop{\mathrm{Spec}}(A/I_\lambda )} U = \mathop{\mathrm{Spec}}(B_\lambda ) \times _{\mathop{\mathrm{Spec}}(A/I_\lambda )} U \]

is a scheme of finite type over $U$ as desired. Thus the construction $(B_\lambda ) \mapsto \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$ does give a functor from category (1) to category (2).

To finish the proof we show that the above constructions define quasi-inverse functors between the categories (1) and (2). In one direction you have to show that

\[ \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda ) \]

for any object $(B_\lambda )$ in the category $\mathcal{C}$. This we proved above. For the other direction you have to show that

\[ Y = \mathop{\mathrm{colim}}\nolimits (Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )) \]

given $Y$ in the category (2). Again this is true by evaluating on quasi-compact test objects and because $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I_\lambda )$. $\square$

Comments (4)

Comment #1970 by Brian Conrad on

In the text preceding this lemma, "" should be "", "systems " should be "inverse systems ", and "systems of homomorphisms" should be "compatible systems of homomorphisms".

On the first line of the proof, "" should be "", and just after the first sentence insert "Consider as in (2)".

A bit more seriously, since the details are omitted at the end I wonder how you justify that the two constructions are really inverse to each other when isn't McQuillan. That is, if there is a counterexample to part 3 of the subsequent Remark (Tag 0AJK) then wouldn't we run into a problem when starting with (1), making the construction in (2), and then coming back to a construction as in (1) (i.e., seems we wouldn't get what we start with).

Comment #1979 by on

OK, I wrote out the details to make sure the construction works. See here or wait till I update the site later today.

Comment #1980 by jojo on

At the end there is a type quasi-comnpact should be quasi-compact.

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