Remark 85.16.7 (Warning). Suppose $X = \mathop{\mathrm{Spec}}(A)$ and $T \subset X$ is the zero locus of a finitely generated ideal $I \subset A$. Let $J = \sqrt{I}$ be the radical of $I$. Then from the definitions we see that $X_{/T} = \text{Spf}(A^\wedge )$ where $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/I^ n$ is the $I$-adic completion of $A$. On the other hand, the map $A^\wedge \to \mathop{\mathrm{lim}}\nolimits A/J^ n$ from the $I$-adic completion to the $J$-adic completion can fail to be a ring isomorphisms. As an example let

$A = \bigcup \nolimits _{n \geq 1} \mathbf{C}[t^{1/n}]$

and $I = (t)$. Then $J = \mathfrak m$ is the maximal ideal of the valuation ring $A$ and $J^2 = J$. Thus the $J$-adic completion of $A$ is $\mathbf{C}$ whereas the $I$-adic completion is the valuation ring described in Example 85.9.2 (but in particular it is easy to see that $A \subset A^\wedge$).

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