Does it hold that the $p$-adic completion of the integers equals the completion of the localization in $p$? maybe this is a stupid question, but I could not solve it even for the ordinary integers $\mathbb{Z}$. Furthermore, I don't have to much knowledge on algebraic number theory and ramifications.
Let $K$ be a number field and $v$ a discrete valuation on it. Now let $\mathcal{O}_{K,v}$ be the localization at $v$, with maximal ideal $\mathfrak{p}_v = \mathfrak{p}\mathcal{O}_{K,v}$. Is it true that $\widehat{\mathcal{O}_{K,v}} \cong \varprojlim_n \; \mathcal{O}_K/\mathfrak{p}^n$ ? 

In general, if $\mathfrak m$ is a maximal ideal in the commutative ring $A$,
then the natural map from the $\mathfrak m$-adic completion of $A$ to 
the $\mathfrak m$-adic completion of the localization $A_{\mathfrak m}$ is an isomorphism, the reason being that at each finite stage, the morphism
$$A/\mathfrak m^n \to A_{\mathfrak m}/\mathfrak m^n A_{\mathfrak m}$$
is an isomorphism, because any element of $A\setminus \mathfrak m$ has invertible
image in $A/\mathfrak m$, and hence in $A/\mathfrak m^n$.
This last assertion is where we use that $\mathfrak m$ is maximal, and the statement
is not true if $\mathfrak m$ is not maximal.
