# Example of computing a direct limit

Let $$(\mathbb{N},\leq$$) be a directed set where $$m\leq n$$ if an only if $$m$$ divides $$n$$.

We define a directed system of groups where $$G_{n}=\mathbb{Z}$$ for all $$n\in \mathbb{N}$$ and $$f_{mn}\colon \mathbb{Z} \to \mathbb{Z}$$ is multiplication by $$\frac{n}{m}$$.

I am trying to compute the directed limit of that system. The definition that I know is $$\dot{\bigcup}_{n\in \mathbb{N}} G_{n} / \sim,$$ where $$z \sim y$$ if $$f_{mn}(z)=f_{kn}(y)$$ where $$n$$ divides $$m$$ and $$n$$ divides $$k$$.

What I have tried is the following: in $$G_{2}=\mathbb{Z}$$, $$2n=n$$ where $$n\in G_{1}$$. In $$G_{3}=\mathbb{Z}$$, $$3n=n$$ where $$n\in G_{1}$$. In $$G_{4}$$, $$4n=n=2n$$ where $$n\in G_{1}$$ and $$2n\in G_{2}$$. In general, in $$G_{k}$$, $$kn=n=k_{2}n=k_{3}n=\cdots=k_{m}n$$ where $$k_{2},\cdots,k_{m}$$ are the divisors of $$k$$, $$n\in G_{1}$$ and $$k_{i}n\in G_{i}$$. Nevertheless, I do not know how to continue.

• Everything divides $0$. – Derek Elkins Mar 14 at 21:20
• @DerekElkins I am assuming that $0$ is not a natural number – Karen Mar 14 at 21:20
• $\le$ is the worst possible symbol to denote divisibility in $\mathbf{N}$... – YCor Mar 14 at 22:29

$$\underrightarrow{\lim}G_n\simeq \mathbf{Q},$$ the additive group of rational numbers.
Indeed, denote by $$|$$ divisibility in $$\mathbf{N}_{>0}$$. Since $$\{n!:n\ge 0\}$$ is cofinal in $$(\mathbf{N}_{>0},|)$$, the direct limit is the same as the direct limit restricted to this cofinal sequence $$H_n=G_{n!}=\mathbf{Z}$$ with given map $$H_{n-1}\to H_{n}$$ given by multiplication by $$n$$.
It is equivalent to write $$H'_n=\frac{1}{n!}$$ and consider the direct limit using identity maps. This makes clear that the direct limit is the union of this increasing sequence of subgroups of $$\mathbf{Q}$$, namely $$\mathbf{Q}$$ itself.
• Why is the direct limit equivalent to just taking the direct limit of cofinal terms? You have the same ordering for the construction of $\hat{\mathbb{Z}}$ but certainly this argument shouldn’t work there. – Santana Afton Mar 15 at 0:55
• @SantanaAfton Think twice, one indeed has $\widehat{\mathbf{Z}}=\underleftarrow{\lim}\mathbf{Z}/n!\mathbf{Z}$. – YCor Mar 15 at 16:58
• Hm. Is the isomorphism $(n_i)\mapsto ( n_1n_2\dots n_i )$? – Santana Afton Mar 15 at 21:06