ag.algebraic geometry – Open immersion of affinoid adic spaces

If $R$ and $S$ are complete Huber rings with $varphi: R to S$ a continuous map, then is it true in general that if $mathrm{Spa}(S, S^circ) to mathrm{Spa}(R, R^circ)$ is an open immersion of adic spaces (here $S^circ$ and $R^circ$ are the power-bounded subrings) then $mathrm{Spec}(S) to mathrm{Spec}(R)$ is injective?

For example, this is true if $R$ and $S$ both have the discrete topology, because if $frak p$ and $frak q$ are two prime ideals in $S$ which are equal after restricting to $R$ then $(frak p, |cdot|_{rm triv})$ and $(frak q, |cdot|_{rm triv})$ (trivial valuations), which are both points in $mathrm{Spa}(S,S)$, restrict to the trivial valuation on $R/varphi^{-1}(frak p)$.

But I’m not sure how generally to expect that this is true.

ag.algebraic geometry – A (not necessarily finite) etale map of reduced, irreducible affinoid spaces that is a bijection on classical points

Let me mention that I’m working with ‘classical’ rigid analytic spaces.

Let $f : Arightarrow B$ be an etale map of affinoid algebras over $mathbb{C}_p$.
Since there are possibly various notions of `etale’ maps of affinoids out there let me clarify what I mean. Let $Y = operatorname{Sp}(B)$ and $X = operatorname{Sp}(A)$. For every point $y in Y$ corresponding to the maximal ideal $mathfrak{n} subset B$, if its image in $X$ is denoted $x$ and corresponds to the maximal ideal $mathfrak{m} subset A$, then the induced map on local rings $mathcal{O}_{X,x} rightarrow mathcal{O}_{Y,y}$ is a flat map and $mathfrak{m}mathcal{O}_{Y,y} = mathfrak{n}mathcal{O}_{Y,y}$ (and the residue field extension is separable – but this is automatic over $mathbb{C}_p$). Note that the definition does not include “finite” hypothesis for the map.

Back to the question now:

Question: So $f : A rightarrow B$ is an etale map of affinoid algebras over $mathbb{C}_p$ such that the induced map $Y:=operatorname{Sp}(B)rightarrow operatorname{Sp}(A)=:X$, is a set-theoretic bijection. Suppose also that $A$ and $B$ are integral domains. Can I conclude that $Arightarrow B$ is an isomorphism. If in addition, $f :A rightarrow B$ were known to be a finite map (i.e. making $B$ a finite $A$-module) then indeed it is fairly straightforward to prove that such a map must be an isomorphism. However, apriori not knowing whether or not the map is finite leads to some difficulty. I wonder if someone could help me arrive at a proof.

Some random observations of mine that might or might not be useful:

(1) There is a cool theorem which says basically that etale maps of affinoids can be written locally as a composition of an open immersion and then a finite etale map. In our situation, this basically means that we can reduce to the case that $f : A rightarrow B$ has a factorization as $A xrightarrow{f’} B’ rightarrow B$ where $A rightarrow B’$ is a finite etale map of affinoid algebras, and $operatorname{Sp}(B) hookrightarrow operatorname{Sp}(B’)$ is an embedding as an affinoid subdomain. We can even assume that $B’$ is an integral domain.

(2) I can prove that $f :A hookrightarrow B$ is a faithfully flat ring map, and hence it must be an inclusion. Also, it follows fairly easily that for every maximal ideal $mathfrak{m}subset A$, that $mathfrak{m}B$ is maximal in $B$. Similarly, $A hookrightarrow B’$ is a finite-etale map (in the usual ring-theoretic sense now) and is injective (being faithfully flat). If I am able to show that $mathfrak{m}B’$ is a maximal ideal in $B’$, I would also be done.

(3) The map $operatorname{Spec}(B) rightarrow operatorname{Spec}(A)$ being faithfully flat, is submersive. Affinoids are Noetherian, Jacobson and so closed points are dense in every closed subset of Specs. Somehow my intuition now seems to be telling me that if on closed points $operatorname{Max}(B)rightarrow operatorname{Max}(A)$ is a bijection then maybe, just maybe, $operatorname{Spec}(B) rightarrow operatorname{Spec}(A)$ is also a bijection. If it were a bijection then it would also be a homeomorphism (being a submersion), which I feel is a step in the right direction, (but again that doesn’t still finish the proof since we want the ring map $Arightarrow B$ to be an isomorphism).

I am fairly convinced that the answer should be that $f$ is an isomorphism under the given hypotheses, but feel free to assume that $A$ is regular at every closed point if that helps.

Thanks in advance.

ag.algebraic geometry – Are maps corresponding to affinoid subdomains flat in the Banach sense?

$newcommand{Sp}{mathrm{Sp}}newcommand{abs}(1){lvert #1rvert}newcommand{comptensor}{mathbin{hat{otimes}}}$
Let $k$ be a complete non-archimedian field and let $X = Sp(B)$ be a $k$-affinoid space. Let $V = Sp(B’) subseteq X$ be an affinoid subdomain. It is well-known that the corresponding map $B to B’$ is a flat ring homomorphism; see e.g. Cor. 7.3.2/6 in Bosch-Güntzer-Remmert.

Let us say that $B’$ is Banach-flat over $B$ if whenever $M to N to P$ is an admissable exact sequence of Banach $B$-modules then the completed tensor product sequence $M comptensor_B B’ to N comptensor_B B’ to P comptensor_B B’$ is also admissable exact.

(A map $f colon M to N$ of Banach $B$-modules is called admissable if there is a constant $C>0$ such that any $n in f(M)$ there is a preimage $m in M$ such that $f(m) = n$ and $abs{m} le C abs{n}$. By Banach’s open mapping theorem, this condition is equivalent to $f(M)$ being closed in $N$. Generally exact sequences of Banach modules can only be expected to behave well, if all mappings are admissable.)

Is it true that if $V = Sp(B’) subseteq X = Sp(B)$ is an affinoid subdomain, then $B’$ is a flat Banach algebra over $B$?

Note that for a functor being exact in the Banach sense, it suffices to consider short exact sequences $0 to M to N to P to 0$. Taking the completed tensor products is always Banach-right exact (because being admissable right exact is equivalent to being a cokernel diagram and because completed tensor product is left adjoint to Banach-Hom), so it suffices here to show that tensoring with $B’$ preserves admissable injective maps of Banach modules.

One can show that fixing $M$, the association $V = Sp(B’) mapsto M comptensor_B B’$ is a sheaf on $X$ (more precisely, its Čech
complex is admissable exact). In particular, $M comptensor B’ to prod_i M comptensor B_i’$ is admissable injective, if $Sp(B’) = bigcup_i Sp(B_i’)$. By the theorem of Gerritzen and Grauert we can therefore assume that $V = X(f_1/f, dots, f_r/f)$ is a rational subdomain of $X$, for which we have a rather explicit description of the algebra $B’$. But so far I had no succes.