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In the following all rings are assumed to be commutative and unitary.

Preliminaries:

For any topological ring $R$ we can form its completion $\widehat{R}$ by taking all Cauchy sequences modulo null sequences. This is again a ring and we can consider completion as a functor $\textbf{RingTop} \to \textbf{Ring}$.

If $R$ is even a linear topological ring - i.e. it admits a fundamental system of neighborhoods of 0 consisting of ideals, say $R \supseteq I_1 \supseteq I_2 \supseteq \dots$ - then $\widehat{R}$ carries a canonical linear topology given by $\widehat{R} \supseteq \widehat{I_1} \supseteq \widehat{I_2} \supseteq \dots$ which turns completion into a functor $\textbf{LRingTop} \to \textbf{LRingTop}$ between linear topological rings.

Question:

Is it possible to spare linearity and turn completion into a functor $\textbf{RingTop} \to \textbf{RingTop}$ in a canonical way which covers the above considerations and additionally has $\widehat{\mathbb{Q}} \cong \mathbb{R}$ as special case?

If not: is there a counterexample which illustrates the difficulties in defining a canonical topology on $\widehat{R}$ which turns completion into a functor in the non-linear case?

Thanks for any help!

Dune
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    You could perhaps add how a Cauchy sequence is meant in this case, as we don't have metric (but only a uniform structure). And, isn't there already a topology on $\widehat R$? Where is the problem? – Berci Feb 23 '13 at 12:01
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    @Berci I'd say that $(a_n)$ is Cauchy if for any open neighbourhood $U$ of $0$ we have $a_n-a_m\in U$ if $n,m>N$ for some $N$, no $|\cdot|$ needed. – Hagen von Eitzen Feb 23 '13 at 12:32
  • Yes, I found it after, but was not immediate. – Berci Feb 23 '13 at 12:54
  • @Berci With Cauchy sequences I meant exactly that what Hagen von Eitzen wrote. Which topology on $\widehat{R}$ do you mean? I can imagine one but does it satisfy the needs? In literature (known to me) no natural topology is mentioned. – Dune Feb 23 '13 at 14:16
  • Probably using Cauchy nets instead of sequences.. – Berci Feb 23 '13 at 15:15
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    It is problematic to define the completion via Cauchy nets since we cannot bound the size of the directed set on which these nets are defined. Thus we won't get a set. @Dune: Interesting question! – Martin Brandenburg Feb 27 '13 at 09:20
  • @Dune could you please cite some reference (textbook or online article preferably) about topological rings and/or this completion matter in general – magma Feb 27 '13 at 12:46
  • @magma [Atiyah, Macdonald. Introduction to commutative algebra (1969)] introduces completions very well for the purposes of commutative algebra but almost ignores the topological aspect. Online references, I found, are this or that. – Dune Feb 27 '13 at 13:54

1 Answers1

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Separated completions are available more generally for uniform spaces, a reference is Bourbaki's Topologie générale, Chapter II, Paragraph 3. Here are some details:

Let $X$ be a uniform space. If $V$ is an entourage of $X$, then a subset $A \subseteq X$ is called small of order $V$ when $A \times A \subseteq V$. A filter on $X$ is called a Cauchy filter if it contains a small set of order $V$ for every entourage $V$. Cauchy filters which are minimal with respect to $\subseteq$ constitute a set $\widehat{X}$, which actually carries a uniform structure. A fundamental system of entourages is given by those sets of minimal Cauchy filters which contain a given small set of order $V$, where $V$ is an entourage on $X$. Then $X \mapsto \widehat{X}$ is a functor which is left adjoint to the functor from separated complete uniform spaces to uniform spaces.

The completion is compatible with products, so that it restricts to the completion of uniform algebraic structures. The case of topological abelian groups (which admit a canonical uniform structure) is treated in loc. cit., Chapter III, Paragraph 3. This can also be applied to topological rings.

Martin Brandenburg
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  • Thank you for this great answer! Just to make sure that I understand correctly: the answer to my question is "no" because Cauchy sequences are too weak, but with "small" filters instead we will get the desired functor? Does it coincide with the above one in the case of linear topological rings (groups) and does $\widehat{\mathbb{Q}} \cong \mathbb{R}$ hold? – Dune Feb 27 '13 at 13:34
  • Adjoint functors are unique up to isomorphism. – Martin Brandenburg Feb 27 '13 at 15:08
  • I will take this as "yes" :) – Dune Feb 27 '13 at 15:42