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Answer 4
I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms Z/nZ→Z/mZ,m|n given by sending 1 to 1. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion Z＾ of Z. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.

This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms Z/nZ→Z/mZ,n|m given by sending 1 to mn. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, Q/Z. And this group, in turn, compactifies to the circle group in whichever way you prefer.

(These two diagrams are "dual," though, something which I learned recently when I was asked to prove on an exam that Hom(Q/Z,Q/Z)?Z＾. Just observe that Hom(Z/nZ,Q/Z)?Z/nZ and that contravariant Hom functors send colimits to limits!)

Edit: Let me also say something about the precise meaning of "compactification" here. A compactification of a space T is an embedding T→X into a compact Hausdorff space X with dense image. The embedding being considered here is the obvious one from Q/Z to R/Z, and the fact that it has dense image is essentially what the word "completion" also means. Compactifications are not unique, but it's possible that there is a sense in which as a topological group R/Z is the "most natural" compactification of Q/Z. But I don't know too much about topological groups.
answered Feb 7, 2010 Qiaochu Yuan

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