Lemma 85.17.1. Let $A \in \mathop{\mathrm{Ob}}\nolimits (\textit{WAdm})$. Let $A \to A'$ be a ring map (no topology). Let $(A')^\wedge = \mathop{\mathrm{lim}}\nolimits _{I \subset A\text{ w.i.d}} A'/IA'$ be the object of $\textit{WAdm}$ constructed in Example 85.15.11.

1. If $A$ is in $\textit{WAdm}^{count}$, so is $(A')^\wedge$.

2. If $A$ is in $\textit{WAdm}^{adic*}$, so is $(A')^\wedge$.

3. If $A$ is in $\textit{WAdm}^{Noeth}$ and $A'$ is Noetherian, then $(A')^\wedge$ is in $\textit{WAdm}^{Noeth}$.

Proof. Part (1) is clear from the construction. Assume $A$ has a finitely generated ideal of definition $I \subset A$. Then $I^ n(A')^\wedge = \mathop{\mathrm{Ker}}((A')^\wedge \to A'/I^ nA')$ by Algebra, Lemma 10.96.3. Thus $I(A')^\wedge$ is a finitely generated ideal of definition and we see that (2) holds. Finally, assume that $A$ is Noetherian and adic. By (2) we know that $(A')^\wedge$ is adic. By Algebra, Lemma 10.97.6 we see that $(A')^\wedge$ is Noetherian. Hence (3) holds. $\square$

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