Given a commutative ring $R$ with proper ideal $I$, we can let $N_r=\{r+I^n\}_{n\in\Bbb N}$ and $\mathcal{B}=\bigcup_{r\in R} N_r$, and take the smallest topology containing these sets. I believe this is the $I$-adic topology?
- Am I correctly interpreting the definition of the $I$-adic topology?
- This definition seems highly dependent on the choice of ideal $I$, in particular, one could take the $0$ ideal? I guess that would give the discrete topology? Are there any classes of ideals where the topology is equivalent?
- Is it true that for any proper ideal $I\subseteq R$ that $\hat{R_I}=\displaystyle\projlim_{\longleftarrow} (R/I^n)$ is isomorphic as a topological space to the completion of $R$ with the topology of 1?
What may be confusing me with 1 is that all over the place they say 'take a system of neighbourhoods of 0' (and then use the translations of these), where we haven't yet even picked a topology, so the word neighbourhood means nothing yet (since otherwise you presuppose the notion of openness in your non-topological ring)... But then it further seems like you don't want to take the smallest topology containing $\mathcal{B}$ as I did above, since then you're pretending these sets in $\mathcal{B}$ are open, when maybe you only want them to be neighbourhoods?
I will update the question after reflecting on this reference.
– AlgebraQuestion May 29 '22 at 13:37As I understand what you wrote, yes, you interpret definition of $I$-adic topology correctly. Yes, of course, it is highly dependent on the ideal $I$, it even has it in the name: $\mathbf{I}$-adic topology! :)
I didn't quite get the notation in the third question you asked, but to me it seems that the answer is yes, the construction you mention, topologically, is precisely the completion of $R$
– Aleksei Kubanov May 31 '22 at 13:42