$\newcommand{\Sp}{\mathrm{Sp}}\newcommand{\abs}[1]{\lvert #1\rvert}\newcommand{\comptensor}{\mathbin{\hat{\otimes}}}$ 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 admissible 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 admissible exact.

(A map $f \colon M \to N$ of Banach $B$-modules is called admissible 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 admissible.)

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 admissible 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 admissible 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 admissible exact). In particular, $M \comptensor B' \to \prod_i M \comptensor B_i'$ is admissible 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 success.

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    $\begingroup$ admissable should be admissible $\endgroup$ Sep 9, 2019 at 15:26
  • $\begingroup$ Note that this is the type of problem which could be addressed by passing to "condensed math" $\endgroup$ Feb 22, 2023 at 6:06

1 Answer 1


This is not true. Assume that $X$ is the closed unit disc (given by $|T| \le 1$, with algebra $B$) and $V$ is a smaller disc (given by $|T| \le r$ for some $r \in (0,1)$, with algebra $B_V$). Consider the annulus $W$ defined by $|T|=1$ with algebra $B_W$. Then the restriction map $B \to B_W$ is injective and admissible (because function on the disc reach their maximum on the boundary).

If we do the completed tensor product with $B_V$, we get the map $B_V \to B_W \hat{\otimes}_B B_V$. But $B_W \hat{\otimes}_B B_V$ is the algebra of $V\cap W = \emptyset$, that is to say 0, so the map is not injective.

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    $\begingroup$ Thank you, Jérôme! It is somewhat unfortunate that this does not hold, but that's alright, I guess. $\endgroup$ Sep 9, 2019 at 21:36
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    $\begingroup$ Yes, I guess this means that flatness in the Banach sense is not such an interesting notion. Let me state a positive result however: Ben-Bassat and Kremnizer (arxiv.org/abs/1312.0338) prove that morphisms of affinoid algebras $A \to B$ corresponding to embedding of affinoid domains may be characterized by the fact that $B \hat{\otimes}^\mathbf{L}_A B \simeq B$. $\endgroup$ Sep 10, 2019 at 9:15
  • $\begingroup$ Hi Jacob, Jerome, thank you for this question, answer and the citation! I want to add that I think that in your example (which shows B_V is not flat with respect to the completed tensor product) , B_W might actually be flat with respect to the completed tensor product but I am not sure how to prove it. I was planning on including this example in my next paper with Kremnizer. Another example is that Z_p is not flat with respect to the completed tensor product over Z (as a Banach ring with the absolute value). On the other hand, I think R is. $\endgroup$ Nov 24, 2019 at 11:19
  • $\begingroup$ I meant Z_p with the usual p-norm. $\endgroup$ Nov 24, 2019 at 11:57
  • $\begingroup$ Tag @JérômePoineau $\endgroup$ Nov 24, 2019 at 13:55

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