The Stacks project

Lemma 86.14.1. Let $S$ be a scheme. Let $X$ be an affine scheme over $S$. Let $T \subset |X|$ be a closed subset. Then the functor

\[ (\mathit{Sch}/S)_{fppf} \longrightarrow \textit{Sets},\quad U \longmapsto \{ f : U \to X \mid f(|U|) \subset T\} \]

is a McQuillan affine formal algebraic space.

Proof. Say $X = \mathop{\mathrm{Spec}}(A)$ and $T$ corresponds to the radical ideal $I \subset A$. Let $U = \mathop{\mathrm{Spec}}(B)$ be an affine scheme over $S$ and let $f : U \to X$ be an element of $F(U)$. Then $f$ corresponds to a ring map $\varphi : A \to B$ such that every prime of $B$ contains $\varphi (I) B$. Thus every element of $\varphi (I)$ is nilpotent in $B$, see Algebra, Lemma 10.17.2. Setting $J = \mathop{\mathrm{Ker}}(\varphi )$ we conclude that $I/J$ is a locally nilpotent ideal in $A/J$. Equivalently, $V(J) = V(I) = T$. In other words, the functor of the lemma equals $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/J)$ where the colimit is over the collection of ideals $J$ with $V(J) = T$. Thus our functor is an affine formal algebraic space. It is McQuillan (Definition 86.9.7) because the maps $A \to A/J$ are surjective and hence $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/J \to A/J$ is surjective, see Lemma 86.9.6. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 86.14: Completion along a closed subset

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Your email address will not be published.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 0AIY. Beware of the difference between the letter 'O' and the digit '0'.