We say that the ring $R$ is topological, when we are given neighbourhoods $\{O_{\lambda};\lambda \in \Lambda\}$ of $0_R$.

We say ${\cal I}$ is a closed ideal of $R$ if and only if for any sub-index set $\Delta \subset \Lambda$ such that

\begin{equation*} O_{\lambda} \subset {\cal I}, \phantom{A} \underset{\lambda \in \Delta}\cap O_{\lambda} = 0_R, \end{equation*} any Cauchy sequence converges in ${\cal I}$ as follows$\colon$ \begin{equation*} \underset{c_{\lambda} \in R,\,a_{\lambda} \in O_{\lambda} \\ \phantom{AAA}\lambda \in \Delta}{\Sigma} c_{\lambda}a_{\lambda} \in {\cal I}. \end{equation*}

Let ${\cal I}$ be a non-closed ideal of the compact ring $R$, and $\overline{{\cal I}}$ be the closure of ${\cal I}$. We have ${\cal I} \subsetneq \overline{\cal I}$. We define the radical ${\sqrt{\cal I}} \colon= \{r \in R ; r^n \in {\cal I}~{\mathrm{for ~some}}~n > 0\}$.

Q. Is it possible for a non-closed ideal ${\cal I}$ of the compact ring $R$ to hold the following equality?$\colon$

\begin{equation*} \sqrt{\cal I} = \overline{\cal I}. \end{equation*}


Let $k$ be a finite field and $X_1$, $X_2$,...,$X_n$,... infinitely many independent variables. Let $A_n:=k[[X_1,...,X_n]]$. Let $M_n$ denote the maximal ideal of $A_n$. Let $\pi_n:A_{n+1}\rightarrow A_n$ be the homomorphism which sends $X_{n+1}$ to 0 and $X_i$ to $X_i$ for $i\le n$. In this way, the rings $A_i$ form a projective system. Let $A:=\lim\limits_\leftarrow A_i$. Elements of the ring $A$ can be thought of as arbitrary (possibly infinite) sums of the form $f_0+f_1+f_2+...+f_n+...$, where $f_n\in(X_n)A_n$.

Endow $A$ with the topology of the projective limit. A typical open neighbourhood of 0, which we denote by $O_{m,n}$, $m,n\in\mathbb N$, consists of those series above for which $f_i\in M_i^m$ for all $i\le n$.

Let $M$ denote the maximal ideal of $A$ (it consists of all the series above for which $f_0=0$). Let $I$ denote the non-closed ideal of $A$ generated by all the variables $X_i$ and all the elements of the form

$f^2$, $f\in M$. Then $\sqrt I=\bar I$.


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