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I apologize if my question is a bit naive, but I don't have much experience in number theory and sometimes get very confused.

Suppose $K$ is a non-archimedean local field (essentially a completion of a number field at some prime ideal of its ring of integers) and $\mathcal{O}_K$ is its ring of integers, i.e., its valuation ring ($\mathcal{O}_K = \{ x \in K \mid |x| \leq 1 \}$).

It seems to follow immediately from the definition of $\mathcal{O}_K$ that it is a closed subgroup of the additive group of $K$. But then when one defines the adele ring, one takes the restricted product of completions of a number field $F$ with respect to open compact subgroups (definition of restricted product) of those completions, and these subgroups are the rings of integers.

Also, in his proof that a local field $K$ is locally compact and its valuation ring $\mathcal{O}_K$ is compact, Neukirch states that for every $a \in K$, the set $a + \mathcal{O}_K$ is open.

So my question is: what is the topological status of $\mathcal{O}_K$ in $K$? Is it open, closed or both? Why?

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    $\begingroup$ Students should first be familiar with the topology on local fields before they delve into the topology on the adele ring or idele group :) $\endgroup$
    – anon
    Apr 11, 2014 at 18:02
  • $\begingroup$ Since the metric is discrete, closed balls are same as open balls. $\endgroup$
    – user10676
    Apr 11, 2014 at 19:44
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    $\begingroup$ @user10676 No, that's very wrong. Neither the metric nor the underlying topology of a nontrivially-valued field are discrete. $\endgroup$
    – anon
    Apr 12, 2014 at 7:09
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    $\begingroup$ @anon, sorry there is a misunderstanding. The valuation is discrete, implying that balls of radius $\leq R$ are the same than balls of radius $<R+\varepsilon$ (accept for $R=0$) and vice versa. $\endgroup$
    – user10676
    Apr 12, 2014 at 11:46

2 Answers 2

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In any metric space a nonempty closed ball is closed and a open ball is open. The latter is more or less by definition (since the open balls generate the topology of open sets). See if you can prove the first.

Since ${\cal O}_K$ is a closed ball of radius $1$ it is closed. It is also the union of all of the open balls of radius $1$ that it contains. Try to prove this fact. Hence ${\cal O}_K$ is also open. (Obviously it is an additive subgroup.)

I highly recommend becoming familiar with visualizing the topology of a local number field as a tree; see Pictures of Ultrametric Spaces (pdf) and How to picture $\Bbb C_p$ (mathoverflow question).

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  • $\begingroup$ I'm not sure I understand what you mean by the union of all open balls of radius $1$ it contains. Isn't there just one open ball of radius $1$? $\endgroup$
    – Balerion_the_black
    Apr 11, 2014 at 21:42
  • $\begingroup$ But I suppose I understand now why the integers are both open and closed. For example, if we took $K = \mathbb{Q}_p$ then we can define $\mathcal{O}_K$ as $\{ x \in K \mid |x| \leq 1 \}$ or just as well as $\{ x \in K \mid |x| < p \}$, which makes it into an open ball. As I said, I don't have that much experience with number theory so I still tend to get confused over simple things. $\endgroup$
    – Balerion_the_black
    Apr 11, 2014 at 22:02
  • $\begingroup$ @Balerion_the_black No, the number of open balls of radius $1$ is the size of the residue field (which is always greater than one). The open ball of radius one around the origin is precisely the prime ideal of the ring of integers, so in particular this open ball doesn't contain the multiplicative identity. Of course ${\cal O}_K$ is generally the open ball of radius $p^{\large\epsilon}$ for small enough $\epsilon$ if $K$ has finite absolute ramification degree (which is equal to $v_K(p)$ assuming the valuation group is normalized so $v_K(K^\times)=\Bbb Z$). $\endgroup$
    – anon
    Apr 12, 2014 at 7:05
  • $\begingroup$ OK so you basically meant open balls of radius $1$ not necessarily around zero. $\endgroup$
    – Balerion_the_black
    Apr 12, 2014 at 10:27
  • $\begingroup$ I meant exactly what I said: ${\cal O}_K$ is the union of all of the open balls of radius $1$ that it contains. Further commentary: one can pick the radius arbitrarily small and this remains true - this means ${\cal O}_K$ is a totally bounded set. A generalization of Heine-Borel replaces bounded with totally bounded and closed with complete and tells us anything totally bounded and complete is compact, and this tells us ${\cal O}_K$ is compact hence $K$ is locally compact. $\endgroup$
    – anon
    Apr 12, 2014 at 18:45
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$\mathfrak m\subset \mathscr O_K$ coincides with the open unit ball of radius 1. $$\mathscr O_K=\bigcup_{a\in \mathscr O_K}a+\mathfrak m.$$ $a+\mathfrak m$ coincides with the open ball of radius 1 about $a$. Hence $\mathscr O_K$ is open, as it is covered by such balls. As you point out, as a profinite set it is also compact. The topology on a local field is Hausdorff, hence $\mathscr O_K$ is indeed closed.

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