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Obviously, $\mathbf{Z}_p \not\cong \mathbf{Z}_q$ ($p$-adic integers: $\mathbf{Z}_p = \varprojlim_n\mathbf{Z}/p^n$) for $p \neq q$ as (topological or abstract) groups, but are they homeomorphic as topological (profinite) spaces?

user3267
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  • are you talking about modulo p groups or p-adic? – N. S. Sep 16 '18 at 02:40
  • $p$-adics. I've edited my question. – user3267 Sep 16 '18 at 02:41
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    Any $\mathbb{Z}_p$ is homeomorphic to the Cantor set. (This follows from general properties of Cantor spaces; alternatively, convert $\sum a_n p^n\in \mathbb{Z}_p$ to a base-$p$ decimal expansion in $[0, 1]$ in a way that is continuous with respect to the $p$-adic metric. – anomaly Sep 16 '18 at 03:03
  • Thanks. Since $C = \prod_{\mathbf{N}}{0,1}$, this also shows all $\mathcal{O}_K$ for $K$ a $p$-adic local field are homeomorphic. – user3267 Sep 16 '18 at 03:07

1 Answers1

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Every $\mathbb{Z}_p \cong C$, where $C$ is the (classical, deleting middle thirds) Cantor set. Therefore, $\mathbb{Z}_p \cong \mathbb{Z}_q$ for primes $p,q$.

This is easy to see in $\mathbb{Z}_2$ by taking $$\sum_{i\geq 0} a_i 2^i \mapsto \sum_{i\geq 0}(2 a_i)3^{-(n+1)} \text{.}$$ For the others, it is a little easier to construct slightly different Cantor sets for each $p$, then show those are all homeomorphic to $C$. You can find details in section 2 of Trautwein, Roder, and Barozzi, Topological properties of $\mathbb{Z}_p$ and $\mathbb{Q}_p$ and Euclidean models.

Eric Towers
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