Seminorms on tensor products of affinoid algebras

Let $A \leftarrow C \rightarrow B$ be affinoid $K$-algebras, where $K$ is a non-archimedean field with non-trivial absolute value. Equipping $A$, $B$, $C$ with the supremum seminorms, there is a canonical seminorm $\nu_1$ on $A \otimes_C B$: $$\nu_1(f) = \inf \max_i |a_i|_{\sup} |b_i|_{\sup},$$ the infimum taken over all representations $f = \sum a_i \otimes b_i$. Then $\nu_1$ extends by continuity to a seminorm on the completed tensor product $A \widehat{\otimes}_C B$ (where the latter is endowed with any residue norm). On the other hand, $A \widehat{\otimes}_C B$ is known to be an affinoid algebra in its own right, so has its own supremum seminorm $\nu_2$.

Does $\nu_1 = \nu_2$?

I'm hopeful this is true, but doubt it (although I have no specific counter-example). It's clear that $\nu_2 \le \nu_1$, and the equality holds if and only if $\nu_1$ is power-multiplicative (i.e. $\nu_1(f^n) = \nu_1(f)^n$).

If it helps, we may assume $C$ is a Tate algebra and $\text{Sp} \, A$, $\text{Sp} \, B$ are affinoid subdomains of $\text{Sp} \, C$.

• This fails for $C=K$ and reduced $A$ and $B$ with $K$-finite $A$ such that $A \otimes_K B$ is non-reduced (nonzero nilpotents have $\nu_2=0$, $\nu_1\ne 0$). Sup-norm $\le 1$ is power-boundedness, so it seems relevant to note that $\widetilde{A} \otimes_{\widetilde{K}} \widetilde{B} \rightarrow (A \widehat{\otimes}_K B)^{\sim}$ is an isomorphism for algebraically closed $K$ by Satz 5 in section 6 of Bosch's Orthonormalbasen in der nichtarchimedischen Funktionentheorie in Manuscripta Math 1. Maybe that paper yields $\nu_1=\nu_2$ for $K$ algebraically closed? – nfdc23 Sep 3 '17 at 21:08
• Also see 7.2.6/3 in the BGR book for how the "reduction" functor $\widetilde{(\cdot)}$ interacts with the formation of certain Laurent domains, since you mention being interested in the special case that $A$ and $B$ correspond to affinoid subdomains of ${\rm{Sp}}(C)$. – nfdc23 Sep 4 '17 at 2:20