# Reference Request: Specialization map in Huber's Context

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.

• 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.
– gdb
Jul 26 '18 at 23:27

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.