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Are we «allowed» to think of an embedding $\mathbb{Q}_p$ into $\mathbb{C}$ as something geometric like in the picture enter image description here

(Picture description: The $5$-adic integers $\mathbb{Z}_5$ are the pentagon dots inside and on the unit circle).

The $p$-adic valuation on elements in $\mathbb{Q}_p$ is at odds with this.

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    $\begingroup$ Do you mean an isometric embedding? This is not doable. If you meant a topological embedding then you can send $\sum_{k\ge -K} a_k p^k\in \Bbb{Q}_p$ ($a_k\in 0\ldots p-1$) to $\sum_{k\ge -K} a_k p^{-2k}\in \Bbb{R}$. $\endgroup$
    – reuns
    Commented Jul 6, 2022 at 9:47
  • $\begingroup$ You "natrally" embed $\mathbb Q_p$ into $\mathbb C_p$. But $\mathbb{C}_p\simeq\mathbb{C}$ is non-construcive and non-unique. Also (unilke $\mathbb C / \mathbb R$) the extension $\mathbb C_p/ \mathbb Q_p$ is infinite-dimensional. $\endgroup$
    – Gerald Edgar
    Commented Sep 22, 2022 at 0:29

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Short answer: no, it's a big mess. The field $\mathbb C$ and the field $\mathbb{Q}_p$ have nice geometric structure, but the map from one to the other is only obtained by completely forgetting geometry, and just thinking about abstract field theory.

In particular, you can cannot tell from any finite truncation of a $p$-adic number any information about where it is going on the complex numbers. The kind of thing you should think about is this: $\mathbb{Q}_p$ contains lots of interesting number fields. For example, lots of negative integers have square roots in $\mathbb{Q}_p$ (if $m$ is a square mod $p$, it has a square root in $\mathbb{Q}_p$), and of course, these have to go to complex numbers on the imaginary axis. But, you can't tell from any finite truncation of the $p$-adic expansion that this number goes on the imaginary axis; its truncation is the same as lots of rational numbers.

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