6
$\begingroup$

I am following Scholze's and Weinstein's notes on $p$-adic geometry on http://www.math.uni-bonn.de/people/scholze/Berkeley.pdf, example 2.3.6 (page 18 as of this moment bar any updates).

$\operatorname{Spa}(\mathbb{Z},\mathbb{Z})$ is the space consisting of points

  • $\eta$, sending all nonzero integers to $1$
  • $s_p : \mathbb{Z} \rightarrow \mathbb{F}_p \rightarrow \{0,1\}$, where the second arrow sends all nonzero elements to $1$
  • and $\eta_p : \mathbb{Z} \rightarrow \mathbb{Z}_p \rightarrow p^{\mathbb{Z}_{\leq 0}} \cup \{0\}$, where the second arrow is the $p$-adic absolute value.

One of the closed sets is then $\overline{\{\eta_p\}} =\{ \eta_p , s_p\}$, let me call the complement of this $U$, which is an open set. The notes proceed to define two maps (where I have replaced $R = \mathbb{Z}$ in the notes)

  • $\textrm{Spec}(\mathbb{Z}) \rightarrow \textrm{Spa}(\mathbb{Z},\mathbb{Z})$, sending $\mathfrak{p}$ to the valuation $R \rightarrow \textrm{Frac}(R/\mathfrak{p}) \rightarrow \{0,1\}$, and
  • $\textrm{Spa}(\mathbb{Z}, \mathbb{Z}) \rightarrow \textrm{Spec}(\mathbb{Z})$, sending a valuation to its kernel.

Finally, the notes claim that (again I have replaced $R$ by $\mathbb{Z}$)

if $U \subset \textrm{Spa}(\mathbb{Z},\mathbb{Z})$ is any open subset, the pullback along the composite $\textrm{Spa}(\mathbb{Z},\mathbb{Z}) \rightarrow \textrm{Spec}(\mathbb{Z}) \rightarrow \textrm{Spa}(\mathbb{Z},\mathbb{Z})$ is a subset $V \subset \textrm{Spa}(\mathbb{Z},\mathbb{Z})$ with $V \subset U$.

My question: So I have tried to plug in the example where $U = \textrm{Spa}(\mathbb{Z},\mathbb{Z}) - \{\eta_p, s_p\}$ (as defined above), and I seem to arrive at $V= \textrm{Spa}(\mathbb{Z},\mathbb{Z}) - \{s_p\}$, which contradicts what I should have expected. What went wrong?

I have worked out that (which is possibly incorrect) the first map $\textrm{Spec}(\mathbb{Z}) \rightarrow \textrm{Spa}(\mathbb{Z},\mathbb{Z})$ sends $0$ to $\eta$ and $(p)$ to $s_p$, and the second map $\textrm{Spa}(\mathbb{Z},\mathbb{Z}) \rightarrow \textrm{Spec}(\mathbb{Z})$ sends $\eta \mapsto 0$, and $s_p \mapsto (p)$, and $\eta_p \mapsto 0$.

Remark: the notes finally proceed to claim that

In particular, any open cover of $\textrm{Spa}(R,R)$ is refined by the pullback of an open cover of $\textrm{Spec}(R)$

which to me suggests that it's not simply a misprint on the notes, that's why I would like to clear up the confusion I am having.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.