7
$\begingroup$

The specialization map $sp:\mathfrak{X}_\eta\to \mathfrak{X}_{red}$ has an important role in rigid analytic geometry. I tried looking in Huber's papers ("Continuous Valuations", "A generalization of formal schemes and rigid analytic varieties", and "Etale cohomology of Rigid Analytic Varieties and Adic Spaces") browsing for a treatment of this map and I didn't find much.

Ideally, I would like to find a statement like the following:

Proposition: Given a Tate Huber-pair $(A,A^+)$ consider the topological space $Spa(A,A^+)$. Then there is a map of topological spaces $sp: Spa(A,A^+)\to Spf(A^+)$, this map is a spectral map of spectral spaces and it is a closed map.

I know that in a traditional Noetherian formal scheme context (or probably even topologically finite type over a valuation ring context) this proposition is a consequence of interpreting the $Spa(A,A^+)$ as the limit of admissible blow-ups, but I was wondering if this was known and written in a non-Noetherian context like perfectoid spaces or even in a non-sheafy context.

Any reference to the specialization map in a Huber-like approach will be much appreciated.

$\endgroup$
1
  • 1
    $\begingroup$ What do you mean by $\operatorname{Spf}(A^+)$? If $A^{+}$ isn't bounded (for example, $A^+=A^0$ and $A$ is not uniform), then the topological ring $A^+$ is not an adic ring (and, moreover, I don't see any reason for this to be a pre-admissible ring). So there is no much sense to speak about $\operatorname{Spf}(A^+)$. If you assume that $A^+$ is adic, then there is always a map $\operatorname{Spa}(A^+,A^+) \to \operatorname{Spf}(A^+)$. And the composition map $\operatorname{Spa}(A, A^+) \to \operatorname{Spa}(A^+, A^+) \to \operatorname{Spf}(A^+)$ is the desired specialization map. $\endgroup$
    – gdb
    Jul 26, 2018 at 23:27

1 Answer 1

5
$\begingroup$

In case you're still interested: Bhatt recently proved that for any Tate-Huber pair $(A,A^+)$, the topological space $\mathrm{Spa}(A,A^+)$ is homeomorphic to an inverse limit of admissible blowups in the expected manner (Theorem 8.1.2 here). For fun I'll summarize the construction (which probably won't seem too surprising). Fix a topologically nilpotent unit $\varpi \in A^+$, and let $I$ denote the category of proper birational maps of schemes $f_i:X_i \to \mathrm{Spec}(A^+)$ which restrict to isomorphisms over the open subset $\mathrm{Spec}(A) \subset \mathrm{Spec}(A^+)$. This is cofiltered, because any $X_i \times_{\mathrm{Spec}(A^+)}X_{i'}\to \mathrm{Spec}(A^+)$ is still in $I$. Let $\overline{X_i} \subset X_i$ denote the vanishing loci of $\varpi$, so these are also cofiltered, and they all map compatibly to $\mathrm{Spec}(A^+/\varpi)$.

Now, there is a canonical map $\Phi: \mathrm{Spa}(A,A^+) \to \lim_{\leftarrow} \overline{X_i}$ given by the following recipe: for any point $x \in \mathrm{Spa}(A,A^+)$, let $K_x$ and $K_x^+$ be the associated residue field and valuation subring, respectively, so there are canonical maps $A \to K_x$ and $A^+ \to K_x^+$. We then get canonical compatible maps $\mathrm{Spec}(K_x) \to X_i$ for all $i$, simply because the loci in the $X_i$'s where $\varpi \neq 0$ all identify with $\mathrm{Spec}(A)$ by definition, so we can lift the given map $\mathrm{Spec}(K_x)\to \mathrm{Spec}(A)$ uniquely. But we also have a map $\mathrm{Spec}(K_x^+) \to \mathrm{Spec}(A^+)$, so by the valuative criterion of properness this lifts uniquely along each $f_i$, giving a compatible system of maps $r_i : \mathrm{Spec}(K_x^+) \to X_i$. Applying the $r_i$'s to the unique closed point of $\mathrm{Spec}(K_x^+)$ then gives the desired point in $\lim \overline{X_i}$.

Anyway, Bhatt proves that $\Phi$ is always a homeomorphism. The specialization map $\mathrm{sp}$ you want is just the composition of $\Phi$ with the natural map $\lim \overline{X_i} \to \mathrm{Spec}(A^+/\varpi)$. Of course, you don't need to go through all of this to define the specialization map, but it does seem to be the easiest way to prove the following:

Fact: The map $\mathrm{sp}$ is continuous, spectral, and closed.

Proof sketch: Continuity and spectrality follow from Bhatt's result together with some basic nonsense about limits of spectral spaces. Closedness can be deduced from the following useful criterion: any quasicompact, spectral, specializing map of locally spectral spaces is closed.

$\endgroup$

Your Answer

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