Lemma 87.14.6. Let $S$ be a scheme. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme over $S$. Let $T \subset X$ be a closed subset. Let $X_{/T}$ be the formal completion of $X$ along $T$.
If $X \setminus T$ is quasi-compact, i.e., $T$ is constructible, then $X_{/T}$ is adic*.
If $T = V(I)$ for some finitely generated ideal $I \subset A$, then $X_{/T} = \text{Spf}(A^\wedge )$ where $A^\wedge $ is the $I$-adic completion of $A$.
If $X$ is Noetherian, then $X_{/T}$ is Noetherian.
Proof. By Algebra, Lemma 10.29.1 if (1) holds, then we can find an ideal $I \subset A$ as in (2). If (3) holds then we can find an ideal $I \subset A$ as in (2). Moreover, completions of Noetherian rings are Noetherian by Algebra, Lemma 10.97.6. All in all we see that it suffices to prove (2).
Proof of (2). Let $I = (f_1, \ldots , f_ r) \subset A$ cut out $T$. If $Z = \mathop{\mathrm{Spec}}(B)$ is an affine scheme and $g : Z \to X$ is a morphism with $g(Z) \subset T$ (set theoretically), then $g^\sharp (f_ i)$ is nilpotent in $B$ for each $i$. Thus $I^ n$ maps to zero in $B$ for some $n$. Hence we see that $X_{/T} = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I^ n) = \text{Spf}(A^\wedge )$. $\square$
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