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$\DeclareMathOperator\Spa{Spa}$What are the points of $\Spa\mathbb{Z}_p$? I read in Scholze-Weinstein that this adic spectrum consists of 2 points, a special point, which corresponds to the pullback of the trivial valuation on $\mathbb{F}_p$ and a generic point (corresponding to the $p$-adic absolute value?), but what about the trivial valuation?

Let's denote these 3 valuations by $v_s, v_p, v_t$ for specialized, $p$-adic, and trivial. Then I claim that these 3 valuations are pairwise unequivalent.

$(v_s,v_p)$: consider $0,p\in \mathbb{Z}_p$, then $|0|_s\ge |p|_s$ as both are $0$, but $|p|_p = 1/p > 0 = |0|_p$, implying that the two absolute values are inequivalent.

$(v_s,v_t)$: consider $0,p\in \mathbb{Z}_p$, then $|0|_s\ge |p|_s$ as both are $0$, but $|p|_t = 1 > 0 = |0|_t$, implying that the two absolute values are inequivalent.

$(v_p,v_t)$: consider $p,p^2\in \mathbb{Z}_p$, then $|p^2|_t \ge |p|_t$ as both are $1$, but $|p^2|_p = 1/p^2 < 1/p = |p|_p$.

Then my question is, doesn't this mean that $\Spa\mathbb{Z}_p$ has at least 3 points?

EDIT: just realized that the trivial valuation is not continuous.

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    $\begingroup$ I think you mean $\mathop{Spa} (\mathbb{Z}_p,\mathbb{Z}_p)$ here. The trivial valuation needs to be excluded because of the boundedness condition with respect to the second $\mathbb{Z}_p$. $\endgroup$
    – Jérôme Poineau
    Commented Mar 17, 2023 at 21:11
  • $\begingroup$ Thanks for the reply, the notation $\text{Spa}\mathbb{Z}_p$ is short for the affinoid adic space you wrote. This abbreviation is very commonly used by Scholze and Weinstein. $\endgroup$
    – kindasorta
    Commented Mar 18, 2023 at 12:37

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