**Proof.**
Let $I_\lambda \subset A$, $\lambda \in \Lambda $ and $J_\mu \subset C$, $\mu \in M$ be fundamental systems of open ideals, then by definition

\[ A \widehat{\otimes }_ B C = \mathop{\mathrm{lim}}\nolimits _{\lambda , \mu } A/I_\lambda \otimes _ B C/J_\mu \]

with the limit topology. Thus a fundamental system of open ideals is given by the kernels $K_{\lambda , \mu }$ of the maps $A \widehat{\otimes }_ B C \to A/I_\lambda \otimes _ B C/J_\mu $. Note that $K_{\lambda , \mu }$ is the closure of the ideal $I_\lambda (A \widehat{\otimes }_ B C) + J_\mu (A \widehat{\otimes }_ B C)$. Finally, we have a ring homomorphism $\tau : A \otimes _ B C \to A \widehat{\otimes }_ B C$ with dense image.

Proof of (1). If $I_\lambda $ and $J_\mu $ consist of topologically nilpotent elements, then so does $K_{\lambda , \mu }$ by Lemma 86.4.10. Hence $A \widehat{\otimes }_ B C$ is weakly admissible by definition.

Proof of (2). Assume for some $\lambda _0$ and $\mu _0$ the ideals $I = I_{\lambda _0} \subset A$ and $J_{\mu _0} \subset C$ are ideals of definition. Thus for every $\lambda $ there exists an $n$ such that $I^ n \subset I_\lambda $. For every $\mu $ there exists an $m$ such that $J^ m \subset J_\mu $. Then

\[ \left(I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C)\right)^{n + m} \subset I_\lambda (A \widehat{\otimes }_ B C) + J_\mu (A \widehat{\otimes }_ B C) \]

It follows that the open ideal $K = K_{\lambda _0, \mu _0}$ satisfies $K^{n + m} \subset K_{\lambda , \mu }$. Hence $K$ is an ideal of definition of $A \widehat{\otimes }_ B C$ and $A \widehat{\otimes }_ B C$ is admissible by definition.

Proof of (3). If $\Lambda $ and $M$ are countable, so is $\Lambda \times M$.

Proof of (4). Assume $\Lambda = \mathbf{N}$ and $M = \mathbf{N}$ and we have finitely generated ideals $I \subset A$ and $J \subset C$ such that $I_ n = I^ n$ and $J_ n = J^ n$. Then

\[ I(A \widehat{\otimes }_ B C) + J(A \widehat{\otimes }_ B C) \]

is a finitely generated ideal and it is easily seen that $A \widehat{\otimes }_ B C$ is the completion of $A \otimes _ B C$ with respect to this ideal. Hence (4) follows from Algebra, Lemma 10.96.3.

Proof of (5). Let $\mathfrak c \subset C$ be the ideal of topologically nilpotent elements. Since $A$ and $C$ are adic Noetherian, we see that $\mathfrak a$ and $\mathfrak c$ are ideals of definition (details omitted). From part (4) we already know that $A \widehat{\otimes }_ B C$ is adic and that $\mathfrak a(A \widehat{\otimes }_ B C) + \mathfrak c(A \widehat{\otimes }_ B C)$ is a finitely generated ideal of definition. Since

\[ A \widehat{\otimes }_ B C / \left(\mathfrak a(A \widehat{\otimes }_ B C) + \mathfrak c(A \widehat{\otimes }_ B C)\right) = A/\mathfrak a \otimes _{B/\mathfrak b} C/\mathfrak c \]

is Noetherian as a finite type algebra over the Noetherian ring $C/\mathfrak c$ we conclude by Algebra, Lemma 10.97.5.
$\square$

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