# Formal spectrum and completing the localization

Let's suppose I have a Noetherian ring $$R$$ with an ideal $$I$$ such that $$R=\varprojlim R/I^{n+1}$$ ($$\varprojlim$$ denotes inverse limit). I can form the formal spectrum $$\operatorname{Spf}R$$ which has the same topological space as $$\operatorname{Spec}(R/I^{n+1})$$, for any $$n$$, and it has a sheaf of topological rings given by the inverse limits of the structure sheaves of $$\operatorname{Spec}(R/I^{n+1})$$ (induce the inverse limit topology). The topological space is also the set of open prime ideals. (An prime ideal which is open contains $$I$$ and vice versa.)

I read that, if $$D_f$$ denotes the distinguished open set of open prime ideals of $$R$$ which do not contain $$f$$, then $$\mathcal{O}_{\operatorname{SpfR}}(D_f)=\widehat{A_f}$$, where the hat denotes completion and the subscript of $$f$$ denotes localization.

My question is: what does it mean to complete $$A_f$$? First, what is the topology on $$A_f$$? $$I$$ does not necessarily correspond to an ideal of $$A_f$$ because $$I$$ is not apparently required to be prime. Is it possible to find a prime ideal $$p$$ which generates the same topology? If so, then we can take the corresponding prime ideal in $$A_f$$, and then we can complete $$A_f$$ with respect to this ideal.

Alternatively, we can figure out a way to induce a topology on $$A_f$$. For instance, we can take the so called final topology, i.e. the finest topology on $$A_f$$ such that the canonical map $$A\to A_f$$ is continuous. Now we have a topological ring $$A_f$$, but this is not necessarily induced by an ideal. For instance, it seems to me that the inclusion map $$\mathbb{Z}_p \to \mathbb{Q}_p$$ (which is induced by localization at $$p$$) induces the standard metric topology on $$\mathbb{Q}_p$$, but of course the only ideal of $$\mathbb{Q}_p$$ is $$(0)$$ and the metric topology is not discrete.

So, I have a few questions. What does it mean to complete a topological ring whose topology is not induced by an ideal? In the case that $$I\subset A$$ is prime, does the corresponding ideal in $$A_f$$ induce the same topology as the topology induced by $$A\to A_f$$?

• $\varprojlim$ produces $\varprojlim$ - I've added this to your post. Aug 6 '21 at 0:40

Given any multiplicatively closed subset $$S$$ of a ring $$A$$ and an ideal $$I\subset A$$, $$I$$ induces an ideal $$S^{-1}I$$ of $$S^{-1}A$$ which is a proper ideal as long as $$I\cap S=\emptyset$$, and this is the ideal that we complete along to form $$\widehat{A_f}$$.
• This bijection only holds for prime ideals. What if $I$ is not prime?
• This seems reasonable. Certainly if $I$ contains some power of $f$ then $D(f)$ is empty and $S^{-1}I$ is not proper so you should get the zero ring for the sections over the empty set. It passes the sanity check, so I think this scans. Thanks.