$\DeclareMathOperator\Spa{Spa}$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.
