Lemma 86.21.1. 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 86.19.11.

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

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

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

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

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

Proof. Recall that $A \to (A')^\wedge$ is taut, see discussion in Example 86.19.11. Hence statements (1), (2), (3), and (4) follow from Lemmas 86.5.7, 86.5.9, 86.7.5, and 86.6.5. Finally, assume that $A$ is Noetherian and adic. By (4) we know that $(A')^\wedge$ is adic. By Algebra, Lemma 10.97.6 we see that $(A')^\wedge$ is Noetherian. Hence (5) holds. $\square$

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