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How do I find the direct limit:

$lim \frac{1}{n}\mathbb{Z}$?


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So e.g. the map from $\mathbb{Z} \rightarrow \frac{1}{2} \mathbb{Z}$ is $f(n) = n$? If so, what on earth is the map $\frac{1}{2}\mathbb{Z} \rightarrow \frac{1}{3} \mathbb{Z}$? – Daniel McLaury Sep 21 '11 at 3:41
what is the map from $\frac{1}{2}\mathbb{Z}$ to $\frac{1}{3}\mathbb{Z}?$ – jspecter Sep 21 '11 at 3:43
The only way this makes sense as written is to take the maps as isomorphisms, in which case the direct limit is just $\mathbb{Z}$. What maybe seems a little more reasonable would be to have all the maps be $f_{ij}(x) = x$ and take the direct system where this makes sense, in which case the direct limit should just be $\mathbb{Q}$. – Daniel McLaury Sep 21 '11 at 3:48
OP, have a look at this question and see if that's what you want: – Daniel McLaury Sep 21 '11 at 3:52
up vote 7 down vote accepted

Your question, as stated, doesn't make sense. What are the maps $n^{-1}\mathbb{Z}\to (n+1)^{-1}\mathbb{Z}$? We do get natural maps $n^{-1}\mathbb{Z}\to m^{-1}\mathbb{Z}$ if $n|m$, though (what are they?), and so we can take the colimit of this diagram. You should be able to check that it is isomorphic to $\mathbb{Q}$ because the colimit will identify different ways of writing the same fraction. That is, $\frac{a}{b}\in b^{-1}\mathbb{Z}$ and $\frac{ka}{kb}\in (kb)^{-1}\mathbb{Z}$ will get identiified because the map $b^{-1}\mathbb{Z}\to(kb)^{-1}\mathbb{Z}$ will take $\frac{a}{b}$ to $\frac{ka}{kb}$, meaning they will represent the same element in $\operatorname{colim} n^{-1}\mathbb{Z}$.

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I am still confused: if "lim" were "colim" (implicitly filtered), then, sure, indexed by the directed set of integers ordered by multiplication, we'd get $\mathbb Q$. However, if the question really is about a (projective) limit, which is a bit dubious, then with maps being inclusions (still with the poset of integers by divisibility... no other sane choice), probably the limit is (isomorphic to) the nested intersection of $N\mathbb Z$'s, which is $\{0\}$, it seems to me at this moment. So, really, is it "lim", or "colim"? – paul garrett Sep 21 '11 at 23:38
@paul: The notation $\lim$ is sometimes used for colimits as well, though usually in the form $\displaystyle \underset{\longrightarrow}{\lim}$. – Zhen Lin Sep 22 '11 at 1:20

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