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This is a follow-up to this question.

A finitely generated subgroup of $(\mathbf Q, +)$ is isomorphic to the direct limit of the system $$\mathbf Z\xrightarrow{1}\mathbf Z\xrightarrow{1}\mathbf Z\xrightarrow{1}\mathbf Z\to\cdots$$

The rationals with squarefree denominator are isomorphic to the direct limit of the system $$\mathbf Z\xrightarrow{2}\mathbf Z\xrightarrow{3}\mathbf Z\xrightarrow{5}\mathbf Z\to\cdots$$

Would the set $\mathbf Z[\frac{1}{2}]$ (i.e. the set of all rationals with denominator $2^n$) then be isomorphic to the direct limit of the system

$$\mathbf Z\xrightarrow{2}\mathbf Z\xrightarrow{2}\mathbf Z\xrightarrow{2}\mathbf Z\to\cdots$$

What about the localization of $\mathbf Z$ at 2, how would that be expressed as a direct limit?

Also, where can I read more about this?

node196884
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1 Answers1

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Yes, that's right. The general pattern is that the directed colimit of

$$\mathbb{Z} \xrightarrow{n_1} \mathbb{Z} \xrightarrow{n_2} \mathbb{Z} \xrightarrow{n_3} \dots $$

computes an increasing union of subgroups of $\mathbb{Q}$, namely the increasing union

$$\mathbb{Z} \hookrightarrow \frac{1}{n_1} \mathbb{Z} \hookrightarrow \frac{1}{n_1 n_2} \mathbb{Z} \hookrightarrow \dots $$

which means it is isomorphic to the rational numbers with denominator dividing some term of the sequence $n_1, n_1 n_2, \dots $. In particular, you can get $\mathbb{Z} \left[ \frac{1}{p} \right]$ by taking $n_i = p$ for every $i$, and you can get $\mathbb{Z}_{(p)}$ by taking $n_i$ to be an enumeration of the powers of every prime other than $p$. (You can be a little cleverer than this also; it suffices to take $n_i$ to consist of primes other than $p$ but in such a way that every prime appears infinitely often, using dovetailing. But there's no particular need to be this clever.)

Qiaochu Yuan
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