Basic question on p-adic numbers I have a very basic question about $p$-adic numbers that arose while reading about $p$-divisible group-schemes. On the very first page of this paper for instance, as well as in other sources, it is suggested that 
$$\lim_{\rightarrow}\mathbb{Z}/p^n\mathbb{Z}\simeq\mathbb{Q}_p/\mathbb{Z}_p$$
which surprizes me because as far as I know, this inductive limit actually is the definition of $\mathbb{Z}_p$ (as I see it on Wikipedia).  
Could someone please confirm whether the above isomorphism is true (ie. exists), and if yes, explain why we have this identity? 
I thank you very much.
 A: I don't know about the direct (inductive) limit, but the definition of $\mathbf Z_p$ is that it is the inverse or projective limit:
$$\mathbf Z_p=\varprojlim \mathbf Z/p^n \mathbf Z=\Bigl\{(\bar x_n)\in\prod_n\mathbf Z/p^n \mathbf Z\mid x_{n+1}\equiv x_n\bmod p^n \Bigr\}$$
A: The inductive limit is indeed what you said. First you need to say how do you send $\mathbb{Z}/p^n\mathbb{Z}$ to $\mathbb{Z}/p^{n+1}\mathbb{Z}$. This is with a map as abelian groups (not as a rings) given by sending $1$ to $p$. So we identify $\mathbb{Z}/p^n\mathbb{Z}$ with the subgroup $p\mathbb{Z}/p^{n+1}\mathbb{Z}$. 
Now the inductive limit becomes the union when you put everything inside the correct place: $\mathbb{Q}_p/\mathbb{Z}_p$. This is done by sending $\mathbb{Z}/p^n\mathbb{Z}$ to the subgroup $\frac 1{p^n}\mathbb{Z}_p/\mathbb{Z}_p$ form by the elements killed for multiplication by ${p^n}$, sending $1$ to the class of this $p$-adic number $\frac 1{p^n}$. Clearly this subgroup has order $p^n$, so the map is an isomorphism. Finally, any element of $\mathbb{Q}_p/\mathbb{Z}_p$ has order $p^n$ for some $n\ge 0$, so 
$$\mathbb{Q_p}/\mathbb{Z_p}=\bigcup_{n\ge 1} \frac 1{p^n}\mathbb{Z}_p/\mathbb{Z}_p\cong\lim_{\to} \mathbb{Z}/p^n\mathbb{Z}$$
