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So we have that $\overline{\mathbb{Z}[\frac{1}{p}]} = \mathbb{R}$ that $\mathbb{Z}[\frac{1}{p}]$ is dense in $\mathbb{R}$. How can it be that $\mathbb{Z}[\frac{1}{p}]$ under the diagonal map $t \mapsto (t,t)$ is now a lattice in $\mathbb{R} \times \mathbb{Q}_p$ ?

First we have the diagonal embedding and then we have the inclusion map: $$ \mathbb{Z}[\tfrac{1}{p}] \to \mathbb{Z}[\tfrac{1}{p}] \times \mathbb{Z}[\tfrac{1}{p}] \to \mathbb{R} \times \mathbb{Q}_p $$ So we have that $\mathbb{Z}[\frac{1}{p}] \subseteq \mathbb{R}$ in one metric and $\mathbb{Z}[\frac{1}{p}] \subseteq \mathbb{Q}_p$ in the $p$-adic metric.

The product of two metric spaces is could be metric space, there is certainly a product topology. Here one metric Euclidean and the other is non-Archimedian.

cactus314
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  • What is a lattice ? – reuns Feb 18 '21 at 23:53
  • You ask how a set $A$ embedding densely into sets $B$ and $C$ (like $\mathbf Z[1/p]$ in $\mathbf R$ and $\mathbf Q_p$) can be discrete in $B \times C$ with the product topology. A simple example of this is $\mathbf Z[\sqrt{2}]$ embedded two ways into $\mathbf R$: its natural embedding $a+b\sqrt{2} \mapsto a+b\sqrt{2}$ and the conjugation $a+b\sqrt{2} \mapsto a-b\sqrt{2}$. Both images are dense in $\mathbf R$, but the map to $\mathbf R^2$ by both embeddings has image ${(a+b\sqrt{2},a-b\sqrt{2}) : a,b\in \mathbf Z}$, which is $\mathbf Z(1,1)+\mathbf Z(\sqrt{2},-\sqrt{2})$ and that's discrete. – KCd Apr 15 '21 at 01:27

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$\Bbb{R\times Q_p}$ with the norm $\|(a,b)\|=|a|_\infty+|b|_p$,

$\iota(c)=(c,c)$,

$\iota(\Bbb{Z}[p^{-1}])$ is a discrete subgroup and $[0,1)\times\Bbb{Z}_p$ is a fundamental domain of $\Bbb{R\times Q_p}/\iota(\Bbb{Z}[p^{-1}])$ which is compact.

reuns
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