Showing $\cal{\overline{O}}/\overline{m} \cong O/m$ where $\cal{\overline{O}}$ Let $K$ be a field with a a non-archimedean valuation $|\cdot|$. Given the valuation ring $\cal{O} = \left\{x\in K : |x|\le 1\right\}$ and the unique maximal ideal $\cal{m}=\left\{x\in K : |x| <1\right\}$, I want to show that $\cal{\overline{O}}/\overline{m} \cong  O/m$ where $\cal{\overline{O}}$ and $\overline{\cal{m}}$ both denote completion under the given valuation. My idea is that for each $x\in\cal{\overline{O}}$ I have $(x_n)\subset \cal{O}$ such that $x_n \to x$. But I'm not sure how to work off of this and construct an isomorphism $\cal{\overline{O}} \to  O/m$ with kernel $\overline{m}$. Any hints would be appreciated.
 A: froggie's answer is certainly correct and it is the standard proof of this useful fact.
Let me add the following observation that may help to understand better what is going on. The maximal ideal $\it m$ is open in the topology induced by the valuation. In fact it is an open ball. Thus the quotient topology in ${\cal O}/{\it m}$ is the discrete topology (the inverse image of points are translates of $\it m$, thus open balls themselves). In a discrete space Cauchy sequences, or images of Cauchy sequences under a continuous map, are definitely constant. This says that completing $\cal O$ one should expect the quotient to remain unchanged.
A: Let $x\in \overline{\mathcal{O}}$. As you point out, there is a sequence $x_n$ of elements of $\mathcal{O}$ converging to $x$. You can now consider the residue classes $\overline{x}_n\in \mathcal{O}/m$ for each $n$. 
Note that the sequence $\overline{x}_n$ is eventually constant in $\mathcal{O}/m$, since for $n,k$ large enough you have $|x_n - x_k|<1$. You can then define your map $\phi\colon\overline{\mathcal{O}}\to \mathcal{O}/m$ to be the map $x\mapsto \lim \overline{x}_n$. Of course, you need to check this is well-defined, and a homomorphism, which is easy to do. The map $\phi$ is also clearly surjective.
Now what is the kernel of $\phi$? It consists precisely of those elements $x\in \overline{\mathcal{O}}$ that are limits $x = \lim x_n$ of elements $x_n\in m$, i.e., the kernel of $\phi$ is the completion $\overline{m}$ of $m$.
