# integer ring and valuation ring of local fields

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$$\Bbb Z_p$$ .

• 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 $$\Bbb{Z}[i]_{(2+i)}\cap \Bbb{Z}[i]_{(2-i)}$$.

• Could you explain why $O_K$ is the integral closure of $O_F$ in $K$ (in the case that $K/F$ is a finite Galois extension of NA local field)? I got difficulty in proving $$O_K \subseteq \text{ integral closure of O_F in K.}$$
– zxx
Commented Feb 24, 2023 at 14:27
• @zxx It follows from that the valuation of $F$ extends uniquely to $L$, it gives that all the $F$-conjugates $a_j$ of an element $a\in L$ have the same valuation, so $v(a)\ge 0$ impiles that all the coefficients of $\prod_{j=1}^{[L:K]} (x-a_j)$ have $\ge 0$ valuation, ie. $a$ is integral over $K$. Commented Feb 24, 2023 at 14:36
• Thanks a lot! I see.
– zxx
Commented Feb 24, 2023 at 17:19