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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 $A\to A_f$?

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    $\begingroup$ $\varprojlim$ produces $\varprojlim$ - I've added this to your post. $\endgroup$
    – KReiser
    Aug 6 '21 at 0:40
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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 $\widehat{A_f}$.

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  • $\begingroup$ This bijection only holds for prime ideals. What if $I$ is not prime? $\endgroup$
    – user948537
    Aug 6 '21 at 14:46
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    $\begingroup$ @Peter I've updated the answer. $\endgroup$
    – KReiser
    Aug 6 '21 at 19:08
  • $\begingroup$ 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. $\endgroup$
    – user948537
    Aug 7 '21 at 20:40

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