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Let $A$ be a (commutative with unit) noetherian ring. If $I$ is an ideal of $A$, the $I$-adic completion of $A$ is by definition $$ \widehat{A} := \underset{\leftarrow}\lim A/I^n. $$ This operation is well known and has a lot of good properties, for example it is $I$-adically complete.

In an article I'm reading I found the sentence "completion at all prime ideals such that...". My question is: what is the completion with respect to a family of ideals?

My guess is the following: suppose we have finetely many ideals, say $I_1,\ldots,I_n$. Then the final result is obtained taking the completion wrt $I_1$, then the completion wrt (the ideal generated by) $I_2$ and so on. If we have infinitely many ideals we have to take a direct limit.

Is this construction discussed somewhere? Has it good properties? For example it's not totally clear to me that the order of the ideals doesn't affect the final result.

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Can you please fill in the rest of the sentence you cite so it is clear that the author doesn't simply mean "completion at each prime ideal such that..."? –  BCnrd Aug 5 '10 at 14:23
    
Let $R$ be a $p$-adically complete $\mathbb Z_p$-algebra. The sentence is "taking the completion of $R$ at the prime ideals of the special fiber, we can assume that it is a discrete valutation ring". –  Ricky Aug 5 '10 at 14:30
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Let me hypothesise that he does not mean to complete R at these prime ideals all at once, but rather one at a time (i.e. reduce to applying the case of a discrete valuating ring but possibly infinitely many times). –  Kevin Buzzard Aug 5 '10 at 14:40
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@Ricky: Kevin's interpretation is definitely the correct one. The construction in your question seems not interesting for its own sake. (Of course, sometimes in proofs one may pass to a completion, and then later to another, and so on, such as mixtures of formal and restricted power series, but I've never seen this general concept arise in a form for which the iterated process is intrinsically a topic of interest. The order of completions makes a huge difference; just try $x$-adic and $y$-adic completion of $k[x,y]$ for a field $k$.) –  BCnrd Aug 5 '10 at 14:58
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Just for reference: In other contexts, completion with respect to a family of ideals does occur, where one takes the directed system of finite (possibly repeated) products of ideals in the family, and then takes the limit of the resulting diagram. This nicely generalizes the completion wrt a single ideal. –  Ulrik Buchholtz Aug 5 '10 at 18:12

2 Answers 2

It is highly presumptuous of me to call this an answer rather than a comment. But there is a straightforward construction suggested by the language.

  1. Take the intersection of the complements of the prime ideals.

  2. Localize the ring away from that monoid, getting a topological ring.

  3. Form the Hausdorff completion of the corresponding uniform space.

Noetherian is not required, nor is minimality of any of the primes.

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Here is an idea (which is probably wrong). If I have an ideal $I\subset J$ in a ring $R$ then this induces a map $R/I\rightarrow R/J.$ Thus if we have a collection of ideals and various containments between them (aka some poset) we can take the direct limit of this system.

In the case of the $I$-adics, this system is just a linear ordering.

Edit: This seems to be exactly a comment made already...

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