Let $k$ be a field and $|\cdot|$ is a absolute value of $k$. Let $\bar k$ be a completion of $k$. We define the ring of integers
$$ O=\{a \in k : |a|\leq 1\} $$
and define an ideal
$$ P=\{a \in k : |a|<1\}. $$
$P$ is a maximal ideal of $O$, so $O/P$ is a field. Now define $\bar O$ and $\bar P$ in a similar way in $\bar k$.
My question is : why $\bar O/\bar P$ is isomorphic to $O/P$? I have no ideal how to prove this.
(I'm studying Cassels and Frohlich's book, and they say that it is clear.)
We have the map $O \to \bar{O}/ \bar {P}$, the kernel is clearly $\bar{P}$, so we wish to show it is onto.
The point is $O$ is dense in $\bar{O}$, and so the image is dense in $O \to \bar{O}/ \bar {P}$, but the latter has the discrete topology.
You can see this more explicitly if you want: given $o \in \bar O$, find $o'\in O$ that is very close to it, then $|o-o'| < 1$, so $o$ maps to $[o']$ like we wanted.