0
$\begingroup$

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*}

$\endgroup$
3
$\begingroup$

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 $\sqrt I=\bar I$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.