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Thanks for your reading. Given an inverse system of rings $\left(A_n, \phi_n: A_n \rightarrow A_{n-1}\right)_{n \in \mathbb{N}}$ where every homomorphism $\phi_n$ is surjective, we study the inverse limit $A:=\varprojlim A_n$. For every $n$ there is an induced surjective projection map $\psi_n:A \rightarrow A_n$, which satisfies $\phi_n\circ\psi_n=\phi_{n-1}$. Let $I_n:=\ker(\psi_n)$.

Do we always have the filtration $I_\bullet$ of $A$ separated? Are there any counterexamples? If it is not right, I think we need to assume $A$ Hausdorff.(On some context I don't see the assumption. Thus I guess some authors use the definition of topological groups/rings as they are always $T_2$? In some extents, maybe this is just a terminology question.)

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    $\begingroup$ The filtration $I_{\bullet}$ is separated by definition: if $x\in \bigcap I_n$, the image of $x$ in each $A_n$ is zero, hence $x=0$ by definition of $\varprojlim A_n$. $\endgroup$ Commented Mar 19 at 14:38
  • $\begingroup$ @abx Thanks. Do you mean $\cap$ not $\cup$? $\endgroup$ Commented Mar 19 at 14:39
  • $\begingroup$ @abx I found this point. Why do we must have if the images is zero then it is zero in $A$? $\endgroup$ Commented Mar 19 at 14:41
  • $\begingroup$ How do you define $A$? $\endgroup$ Commented Mar 19 at 14:45
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    $\begingroup$ The standard definition of $\varprojlim A_n$ is: $\{(x_n)\in\prod A_n\,|\,\phi_n(x_n)=x_{n-1}\ \forall\,n\} $. $\endgroup$ Commented Mar 19 at 15:04

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