6
$\begingroup$

In Cassels' article "Global Fields", he uses the term "ring of integers" and the notation $\mathcal{O}_K$, where $K$ is a field with a non-archimedean valuation, to denote the ring of elements $x \in K$ with $|x| \leq 1$. How justified is this notation?

In the case that $K = \mathbb{Q}$ and $| \cdot |$ is a $p$-adic valuation, $\mathcal{O}_K$ is the localization of $\mathbb{Z}$ at $p$, i.e. $(\mathbb{Z})_p$. In the case that $K$ is a number field with a $\mathfrak{p}$-adic valuation, this is similarly the localization at $\mathfrak{p}$ of its ring of integers, so the definition is reasonable here.

What about complete fields? If $K = \mathbb{Q}_p$, $\mathcal{O}_K = \mathbb{Z}_p$. Is it true that for $K$ a finite extension of $\mathbb{Q}_p$, $\mathcal{O}_K$ is the integral closure of $\mathbb{Z}_p$ in $K$? Is this integral closure - which I'll denote $\mathbb{Z}_K$ - already a local ring, or is localization necessary?

I think I can prove it if $\mathbb{Z}_K$ is a DVR - as $(\mathcal{O}_K)_\mathfrak{p}$ is in the number field case - but I don't know why (or if) this should be true.

Here's my attempt to extend the proof from the number field case: let $\mathbb{Z}_K$ be the integral closure of $\mathbb{Z}_p$ in $K$. Then, it's easy to see that it is contained in $\mathcal{O}_K$ by applying the ultrametric inequality to an integral equation for an element. Also, $K$ is the fraction field of $\mathbb{Z}_K$: for any integral domain $A$ with field of fractions $F$, the field of fractions of the integral closure of $A$ in some finite extension $K$ of $F$ is just $K$ (as shown here). If we knew that $\mathbb{Z}_K$ were a DVR with uniformizer $\pi$, we could write an element $x$ of $K$ as $\frac{\pi^m u}{\pi^n v}$ with $u, v$ units of $\mathbb{Z}_K \subseteq \mathcal{O}_K$, and thus $|u| = |v| = 1$. Since the valuation is non-trivial on $\mathbb{Z}_K$, we'd have $|\pi|< 1$, so if $x \in \mathcal{O}_K$, we'd have $m \geq n$ and thus $x = \pi^k w \in \mathbb{Z}_K$ where $k \geq 0$ and $w$ is a unit of $\mathbb{Z}_K$.

EDIT Is the integral closure $\mathbb{Z}_K$ of $\mathbb{Z}_p$ in a finite extension $K$ of $\mathbb{Q}_p$ a DVR? The answer below seems to show that the $\mathfrak{p}$-adic completion of the ring of integers of a number field is a DVR. But are these the same? I know that the finite extension $K$ of $\mathbb{Q}_p$ is the $\mathfrak{p}$-adic completion of some number field $K_0$. Is the same true for the rings of integers?

$\endgroup$
2
+50
$\begingroup$

I think there is a flaw in your attempt : you are assuming that the topology (induced by the DVR structure) on $\mathbb{Z}_K$ is the same as the norm topology, which has to be proved.

I have asked the same question here. The proof can be found in : Neukirch, 'Algebraic number theory' (first step of the proof of Theorem 4.8 of Chapter II). I can give some detail if you don't have access to the book.

$\endgroup$
3
$\begingroup$

The answer to your questions is basically "yes, this is fine." Let's take them in turn. First of all, all elements of a local field are of the form $x=u\pi^k$ for some $k\in\Bbb Z$ and $u$ a unit of the integer ring. Your use of "field" makes it hard to tell if you mean local or global, but if you mean "global," then the answer is that no, $\mathcal{O}_K$ in this context is not the usual ring of integers, even when $K=\Bbb Q$, since there, for example if I take $p=2$ then ${1\over 3}$ is a unit relative to the valuation, so is in the set $\{x : |x|\le 1\}$ and is clearly not an integer.

For extensions of $\Bbb Q_p$, this is fine, one can easily argue that $\{x\in K : |x|\le 1\}$ is a PID in a DVR, hence is integrally closed (because it is a UFD).

To show the integer ring completed is a DVR is straightforward, you know that you have the maximal ideal is generated by a single element which is not nilpotent and that it is a Noetherian local ring. Serre's Local Fields has a proof that this is equivalent to being a DVR in the early pages as I recall.

$\endgroup$
  • 1
    $\begingroup$ Why does the fact that the valuation ring is integrally closed imply that it is the integral closure of $\mathbb{Z}_p$ in $K$? And why is the integral closure of $\mathbb{Z}_p$ in $K$ a DVR? (I'm not sure what you're referring to by "the integer ring completed" - what integer ring?) $\endgroup$ – Dorebell Oct 23 '15 at 6:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.