The Stacks project

Proposition 67.3.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

  1. The morphism $f$ is a morphism of algebraic spaces which is locally of finite presentation, see Morphisms of Spaces, Definition 64.28.1.

  2. The morphism $f : X \to Y$ is limit preserving as a transformation of functors, see Definition 67.3.1.

Proof. Assume (1). Let $T$ be a scheme and let $y \in Y(T)$. We have to show that $T \times _ Y X$ is limit preserving over $T$ in the sense of Definition 67.3.1. Hence we are reduced to proving that if $X$ is an algebraic space which is locally of finite presentation over $S$ as an algebraic space, then it is limit preserving as a functor $X : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. To see this choose a presentation $X = U/R$, see Spaces, Definition 62.9.3. It follows from Morphisms of Spaces, Definition 64.28.1 that both $U$ and $R$ are schemes which are locally of finite presentation over $S$. Hence by Limits, Proposition 32.6.1 we have

\[ U(T) = \mathop{\mathrm{colim}}\nolimits U(T_ i), \quad R(T) = \mathop{\mathrm{colim}}\nolimits R(T_ i) \]

whenever $T = \mathop{\mathrm{lim}}\nolimits _ i T_ i$ in $(\mathit{Sch}/S)_{fppf}$. It follows that the presheaf

\[ (\mathit{Sch}/S)_{fppf}^{opp} \longrightarrow \textit{Sets}, \quad W \longmapsto U(W)/R(W) \]

is limit preserving. Hence by Lemma 67.3.5 its sheafification $X = U/R$ is limit preserving too.

Assume (2). Choose a scheme $V$ and a surjective ├ętale morphism $V \to Y$. Next, choose a scheme $U$ and a surjective ├ętale morphism $U \to V \times _ Y X$. By Lemma 67.3.4 the transformation of functors $V \times _ Y X \to V$ is limit preserving. By Morphisms of Spaces, Lemma 64.39.8 the morphism of algebraic spaces $U \to V \times _ Y X$ is locally of finite presentation, hence limit preserving as a transformation of functors by the first part of the proof. By Lemma 67.3.3 the composition $U \to V \times _ Y X \to V$ is limit preserving as a transformation of functors. Hence the morphism of schemes $U \to V$ is locally of finite presentation by Limits, Proposition 32.6.1 (modulo a set theoretic remark, see last paragraph of the proof). This means, by definition, that (1) holds.

Set theoretic remark. Let $U \to V$ be a morphism of $(\mathit{Sch}/S)_{fppf}$. In the statement of Limits, Proposition 32.6.1 we characterize $U \to V$ as being locally of finite presentation if for all directed inverse systems $(T_ i, f_{ii'})$ of affine schemes over $V$ we have $U(T) = \mathop{\mathrm{colim}}\nolimits V(T_ i)$, but in the current setting we may only consider affine schemes $T_ i$ over $V$ which are (isomorphic to) an object of $(\mathit{Sch}/S)_{fppf}$. So we have to make sure that there are enough affines in $(\mathit{Sch}/S)_{fppf}$ to make the proof work. Inspecting the proof of (2) $\Rightarrow $ (1) of Limits, Proposition 32.6.1 we see that the question reduces to the case that $U$ and $V$ are affine. Say $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$. By construction of $(\mathit{Sch}/S)_{fppf}$ the spectrum of any ring of cardinality $\leq |B|$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$. Hence it suffices to observe that in the "only if" part of the proof of Algebra, Lemma 10.126.3 only $A$-algebras of cardinality $\leq |B|$ are used. $\square$


Comments (2)

Comment #2298 by Eric Ahlqvist on

Hi! I think you should switch and in the fiber product on the first line of the proof: 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 04AK. Beware of the difference between the letter 'O' and the digit '0'.