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

is a McQuillan affine formal algebraic space.

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.
$\square$

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