Question about $Q = \varinjlim \frac{1}{n} \mathbb{Z}$ I am a little confused by one of Vakil's exercises, which reads 

Interpret the statement $Q = \varinjlim \frac{1}{n} \mathbb{Z}$ 

Does it mean $Q = \bigcup \frac{1}{n} \mathbb{Z}$? But what is the morphism $\frac{1}{p} \mathbb{Z} \to \frac{1}{q} \mathbb{Z}$? inclusion?
Or $\ Q = \bigcup\limits_{p\text{ is prime}} \varinjlim\frac{1}{p} \mathbb{Z}$.
I appreciate all your help. 
 A: This statement can be interpreted in many ways. 
The most naive one is a statement about subgroups of $\mathbb{Q}$ : in this case $\frac{1}{n}\mathbb{Z}$ is simply a subset of $\mathbb{Q}$, and we have inclusions $\frac{1}{n}\mathbb{Z} \subset \frac{1}{m}\mathbb{Z}$ if and only if $n\mid m$; so that you may see $\varinjlim \frac{1}{n}\mathbb{Z} = \bigcup \frac{1}{n}\mathbb{Z}$ and this union is quite clearly equal to $\mathbb{Q}$. 
But this is a bit unsatisfying because we're relying too much on some set-theoretic notions : that the maps be literal inclusions is a very rare phenomenon and so it can't quite be generalized or duplicated. So driven by our naive analysis we may give a more abstract analysis of the situation : $\frac{1}{n}\mathbb{Z}$ is clearly isomorphic to $\mathbb{Z}$, so we should be able to just write $\varinjlim \mathbb{Z}$ : the problem with this is that now we don't know what the maps are anymore ! 
That's where writing $\frac{1}{n}\mathbb{Z}$ instead of $\mathbb{Z}$ warns us. We're actually looking at a directed system, indexed by $\mathbb{N}$ ordered by division, and the map $\mathbb{Z}\to \mathbb{Z}$ from position $n$ to position $m$ ($n\mid m$) should correspond to the inclusion from our naive analysis. It's not hard to convince yourself that this map should be mulitplication by $\frac{m}{n}$. 
We now have an abstract directed system of groups (all $\mathbb{Z}$) indexed by $(\mathbb{N}, \mid)$ with maps $f_{mn} : \mathbb{Z}\to \mathbb{Z}, 1\mapsto \frac{m}{n}$ when $n\mid m$; and now we can't just say that the directed limit is the union in a set theoretic sense (because it wouldn't make sense). Once we're here we now have to actually compute this abstractly and show that the result is (isomorphic to ) $\mathbb{Q}$. 
Recall that the directed limit of a directed system satisfies a certain universal property, defined here for instance, and it so happens that in nice situations such as groups, we actually have a very concrete construction of something that satisfies this universal property, as described here. Now we can safely say that our direct limit is simply $\displaystyle\coprod_{n\in \mathbb{N}}\mathbb{Z} / \sim$ where $(x,n) \sim (y,m) \iff \exists k, n\mid k \land m \mid k \land f_{kn}(x) = f_{km}(y)$
But we know explicitly what $f_{kn}, f_{km}$ are. That is, in the disjoint union, $x$ at rank $n$ and $y$ at rank $m$ are identified if and only if $\frac{k}{n}x = \frac{k}{m}y$ for some $k$, but this just means $mx= ny$. 
It's either a bit of guess work, or you're driven by the naive analysis, or you remember precisely how $\mathbb{Q}$ was defined to begin with, but you can now see that this looks a lot like $\mathbb{Q}$, where you figure that $x$ at rank $n$ is interpreted as $\frac{x}{n}$; and you see that $x$ at rank $n$ is identified with $y$ at rank $m$ precisely if $\frac{x}{n} = \frac{y}{m}$.
It's now just a matter of verifying that addition is the correct one, but that's just a routine check: $(x,n) + (y,m)$ is defined in the direct limit as follows : take $k$ such that $n\mid k, m\mid k$, then push $x$ and $y$ to make them land at rank $k$, that is, apply $f_{kn}$ and $f_{km}$ to them, and then add them there : the result doesn't depend on the chosen $k$ up to $\sim$. 
So taking $k=nm$ for instance, we get that $f_{kn}$ is multiplication by $m$ and $f_{km}$ is multiplication by $n$, so that $(x,n)+(y,m)$ is defined as $(mx, nm)+ (ny, nm)$, that is $(mx+ny, nm)$, which corresponds to $\frac{mx+ny}{nm}$, which is exactly $\frac{x}{n}+\frac{y}{m}$. 
You are now convinced that the result is $\mathbb{Q}$ and can finally set up your isomorphism : define $\iota_n : \mathbb{Z}\to \mathbb{Q}$ by $\iota_n(x) =\frac{x}{n}$, get a (set) map $\iota: \displaystyle\coprod_n \mathbb{Z} \to \mathbb{Q}$; check that $\iota(\alpha) = \iota(\beta) \iff \alpha \sim \beta$ and that $\iota$ is surjective, so that it factors through a bijection $s: \displaystyle\coprod_{n\in \mathbb{N}}\mathbb{Z} / \sim\to \mathbb{Q}$. The check we did about $+$ proves that this bijection is also a morphism, hence an isomorphism. 
This gives a general method for "computing" direct limits : if you know explicitly the maps or have at leats a very good idea of what they do, and you guess correctly what the direct limit is, you can then just set up a morphism between the actual limit and the guessed limit, and if your guess was correct and justified enough you can usually easily prove that it's an isomorphism. 
