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Given $R$ an absolute valued ring (with unit), sometimes $\overline{U(0,1)}=B(0,1)$ (for example, $\mathbb{Q},\mathbb{R},\mathbb{C},\mathbb{H}$) and sometimes $\overline{U(0,1)}\neq B(0,1)$ (for example, $\mathbb{Q}_p, \mathbb{Z}$).

Which properties must we impose to $R$ to ensure the identity holds or not?

Smells like the archimedean property has something to say here (p-adic example). But also $B(0,1)$ must be "full" in some sense (compare $\mathbb{Z}$ with $\mathbb{Z}[1/2]$, the latter satisfying the identity). Not sure if inverses really matter.

Also, all my counterexamples satisfy that the open ball is clopen. I wonder if there's a counterexample with a nonclosed open ball or if it is a necessary condition.

Edit: I forgot to say that $U(0,1)=\{x\in R\,:\,|x|<1\}$ and $B(0,1)=\{x\in R\,:\,|x|\leq 1\}$.

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    $\begingroup$ It would be useful if you add the definiton of $U(0,1)$ and $B(0,1)$. Are they the "$< 1$" and "$\leq 1$" discs centered at the origin, respectively, I guess? $\endgroup$
    – Qfwfq
    Mar 17, 2015 at 13:55
  • $\begingroup$ Yes, you are right, I forgot it. $\endgroup$
    – Pablo P.
    Mar 18, 2015 at 14:18

1 Answer 1

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I believe a complete algebraic closure of the p-adic numbers has the property that the closed unit ball is the closure of the open.

If so, no sort of archimedean property is involved.

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  • $\begingroup$ The open unit ball in a non-Archimedean field (such as $\mathbf{C}_p$ or $\mathbf{Q}_p$) is closed in the metric topology. So to say that the closure of the open unit ball equals the closed unit ball (which is the ring of integers $\mathscr{O}_{\mathbf{C}_p}$) is to say that the open ball coincides with the closed ball. But the open ball is the unique maximal ideal of the closed unit ball, so in particular, a proper ideal, and the two can never coincide. $\endgroup$ Mar 17, 2015 at 20:18
  • $\begingroup$ If the value group is dense, the open ball is not closed $\endgroup$
    – user35486
    Mar 18, 2015 at 14:25
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    $\begingroup$ In any metric space whose metric satisfies the strong triangle inequality, open and closed balls are clopen sets. $\endgroup$ Mar 18, 2015 at 15:17
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    $\begingroup$ and "value group dense" does not change the fact that all open balls are closed sets. $\endgroup$ Apr 17, 2015 at 16:07

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