Let $K$ be a local field, that is, complete discrete valuation field, with finite residue field.
Then, integer ring of $K$ and valuation ring of $K$ corresponds?
And to what extent can I extend to another field?
In trivial case, integer ring of $\Bbb Q_p$ is $\Bbb Z_p$, and valuation ring is also$\Bbb Z_p$ .
Thank you in advance.
Best Answer
For a discretely valued field "the ring of integers" and "the valuation ring" means the same thing: $O_K=\{ a\in K,v(a)\ge 0\}$.
This is different to the ring of integers of a number field understood as the integral closure of $\Bbb{Z}$.
Complete discretely valued field with finite residue field (a local field) gives that $K$ is a finite extension of $\Bbb{Q}_p$ or $\Bbb{F}_p((t))$.
For any finite extension $K/F$ then $F$ is discretely valued and complete and $O_K$ is the integral closure of $O_F$ in $K$.
(take the normal closure $L$ of $K/F$, there is a standard proof that the discrete valuation on $L$ is unique because it is recovered from algebraic properties such as "the elements having a $n$-th roots in $L$ for all $p\nmid n$", whence $Aut(L/F)$ acts continuously from which its fixed subfield $E$ is closed (complete), if $L/F$ is not separable then $F=E^{p^r}$ is complete too, so we are in the standard case of $K/F$ a finite extension of local field...)
The same should hold for any complete discretely valued field. It doesn't hold for non-complete ones: try with $O_K=\Bbb{Z}[i]_{(2+i)}$, the integral closure of $O_F=\Bbb{Z}_{(5)}$ is $\Bbb{Z}[i]_{(2+i)}\cap \Bbb{Z}[i]_{(2-i)}$.