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Suppose $\mathbb{Q_p} $ is the fraction field of $\mathbb{Z_p}$ ($p$-adic integers) i.e.

$$\mathbb{Q_p} = \left\lbrace\frac{x}{y} \space \bigg{|} \space x,y \in \mathbb{Z_p} , y\neq 0 \right\rbrace$$

Now with respect to the topology defined by $d(x,y) = e^{-v_p(x-y)}$ ($v_p$ is the $p$-adic valuation) , we need to show that $\mathbb{Q_p} $ is locally compact.

Any suggestions?

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2 Answers 2

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Every point has a fundamental system of neighborhoods given by $\{x+p^n\Bbb Z_p\}_{n\in\Bbb N}$ which are compact.

In essence this is just the fact that $p^n\Bbb Z_p$ is a fundamental system of compact neighborhoods of $0$: since we are in a vector space--really a topological group is enough, but not everyone is familiar with structures of that generality--we can translate these sets anywhere to form a compact (and open) neighborhood of any point.

If you have any trouble seeing this recall that open balls generate the topology and all open balls are of the form $x+p^n\Bbb Z_p$ for some $n\in\Bbb Z$.

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The accepted answer is not complete because it does not argue why $p^n\mathbb{Z}_p$ is compact. I think the following idea deal with this:

To prove that $\mathbb{Q}_p$ is locally compact is enough to prove that $\mathbb{Z}_p$ is compact. For this, thinking in the $p$-adic integers as sequences we have $$\mathbb{Z}_p=\{(a_0,a_1,a_2,...)\mid a_n\in \mathbb{Z}/p\mathbb{Z} \}=(\mathbb{Z}/p\mathbb{Z})^\mathbb{N}$$

Now, by Tychonoff $(\mathbb{Z}/p\mathbb{Z})^\mathbb{N}$ is compact with the product topology and also is easy to see that $x+p^n\mathbb{Z}_p$ is an open set of this topology. So every cover of $\mathbb{Z}_p$ with sets of the form $x+p^n\mathbb{Z}_p$ will have a finite subcover and then $\mathbb{Z}_p$ is compact.

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  • $\begingroup$ you also did not mention why $\mathbb{Z}/p \mathbb{Z}$ is compact ? $\endgroup$
    – MAS
    Jun 15, 2019 at 13:59
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    $\begingroup$ @M.A.SARKAR That is because it is a finite set and finite sets are always compact. $\endgroup$ Jun 16, 2019 at 22:20

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