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Let $R$ denote the ring of convergent power series over $\mathbb{C}$ in $n$ variables (which is a Noetherian, excellent, local ring). For any finite set of prime ideals one has that the $\mathfrak{m}$-adic closure commutes with taking intersections, i.e \begin{equation}\widehat{\cap_{i=1}^{n} p_{i}} = \cap_{i =1}^{n} \hat{p_{i}}, \end{equation} which can be shown using the fact the the completion of radical ideals of $R$ are again radical ideals (see Is the $\mathfrak m$-adic completion of a radical ideal again a radical ideal?): If $x \in \cap_{i =1}^{n} \hat{p_{i}}$ and $(x_{i,k})_{k \in \mathbb{N}} \in p_{i}$ are sequences converging to $x$, then the product sequences $(x_{1,k} \cdots x_{n,k})_{k\in \mathbb{N}}$ is contained in $\cap_{i=1}^{n} p_{i}$ and converges to $x^{n}$, which is hence contained in $\widehat{\cap_{i=1}^{n} p_{i}}$, and consequently $x$ is then contained in $\widehat{\cap_{i=1}^{n} p_{i}}$, giving $\widehat{\cap_{i=1}^{n} p_{i}} \supset \cap_{i =1}^{n} \hat{p_{i}}$ and thus equality between the two sets.

My question: Is it also true that for a sequence $(p_{i})_{i \in \mathbb{N}}$ of prime ideals \begin{equation}\widehat{\cap_{i=1}^{\infty} p_{i}} = \cap_{i =1}^{\infty} \hat{p_{i}}\end{equation} holds? Even in $\mathbb{R}$ interchanging the closure operator with countable intersections does not give the same set (see for example If the closure operator interchanges with tacking finite intersections, is this then true for a countable intersection?), but maybe in this situation one could work with exact sequences. Set $I_{n} = \cap_{i=1}^{n} p_{i}$ and $I = \cap_{i=1}^{\infty} p_{i}$ . Then $(I_{n})_{n \in \mathbb{N}}$ is a descending chain of radical ideals and $(R/I_{n})_{n \in \mathbb{N}}$ together with the canonical quotient maps $\pi_{n} \colon (R/I_{n}) \to (R/I_{n-1})$ forms an inverse system of rings. $I$ is the kernel of the natural map $\phi \colon R \to \varprojlim (R/I_{n})$ (mapping an element to its sequence of quotient classes , i.e. $\phi(x) = ([x_{n}])_{n\in \mathbb{N}}$), which we can describe as the exact sequence \begin{equation} 0 \xrightarrow{\iota} I \xrightarrow{\iota} R \xrightarrow{\phi} \varprojlim (R/I_{n}), \end{equation}

My hope is that can be shown that this lifts to an exact sequence \begin{equation} 0 \xrightarrow{\iota} \hat{I} \xrightarrow{\iota} \hat{R} \xrightarrow{\tilde{\phi}} ({\varprojlim(R/I_{n}))}^{\wedge}\end{equation} and further that \begin{equation} ({\varprojlim(R/I_{n}))}^{\wedge} = \varprojlim \widehat{(R/I_{n})}, \end{equation} which would show the claim, since $\ker \tilde{\phi} = \cap_{n=1}^{\infty} \hat{I_{n}}$. Maybe someone has an idea wether or not this works - thanks for all replies.

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