Lemma 85.16.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $X_{/T}$ be the formal completion of $X$ along $T$.

1. If $X \setminus T \to X$ is quasi-compact, then $X_{/T}$ is locally adic*.

2. If $X$ is locally Noetherian, then $X_{/T}$ is locally Noetherian.

Proof. Choose a surjective étale morphism $U \to X$ with $U = \coprod U_ i$ a disjoint union of affine schemes, see Properties of Spaces, Lemma 64.6.1. Let $T_ i \subset U_ i$ be the inverse image of $T$. We have $X_{/T} \times _ X U_ i = (U_ i)_{/T_ i}$ (Lemma 85.10.4). Hence $\{ (U_ i)_{/T_ i} \to X_{/T}\}$ is a covering as in Definition 85.7.1. Moreover, if $X \setminus T \to X$ is quasi-compact, so is $U_ i \setminus T_ i \to U_ i$ and if $X$ is locally Noetherian, so is $U_ i$. Thus the lemma follows from the affine case which is Lemma 85.10.6. $\square$

Comment #2029 by Brian Conrad on

Maybe it is worthwhile to include a Remark warning that if $X = {\rm{Spec}}(A)$ is affine and $T$ is the zero locus of a finitely generated ideal $I$ (so $X-T$ is quasi-compact) with radical $J$ then from the definitions $X_{/T} = {\rm{Spf}}(A^{\wedge})$ for the $I$-adic completion $A^{\wedge}$ of $A$ whereas the natural map from $A^{\wedge}$ to the $J$-adic completion of $A$ can fail to be a ring isomorphism.

As an example, to illustrate the issues, one can mention the old standby $A = O_K$ for an algebraically closed field $K$ equipped with a rank-1 valuation, $I = (\pi)$ for a nonzero nonunit $\pi \in A$, and $J = \mathfrak{m}$ the maximal ideal. Since $J^2=J$ the $J$-adic completion of $A$ is the residue field, whereas the $I$-adic completion of $A$ is the valuation ring of the completion of $K$ for the given rank-1 valuation).

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