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.


1 Answer 1

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

  • $\begingroup$ 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 $$O_K \subseteq \text{ integral closure of $O_F$ in $K$.}$$ $\endgroup$
    – zxx
    Commented Feb 24, 2023 at 14:27
  • 1
    $\begingroup$ @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 $K$. $\endgroup$
    – reuns
    Commented Feb 24, 2023 at 14:36
  • $\begingroup$ Thanks a lot! I see. $\endgroup$
    – zxx
    Commented Feb 24, 2023 at 17:19

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