This post is closely related to the previous one here.

But more generally, we want to study an affinoid algebra $A:=T_n/\mathfrak a$. Let's assume $\mathfrak a= (f_1,\dots,f_r)$ for some $f_i\in T_n$. The question is almost identical but replacing the Tate algebra $T_n$ there by $A$. Denote $x=(x_1,\dots,x_n)\in \mathbb K^n $

Question: Suppose $f(x)=0$ for every $x$ with $|x_i|=1$, then can we conclude $f\equiv 0$ as an element of $A$?

Alternatively, the question can be re-interpreted as follows: Suppose $f(x)=0$ for every common zero of $f_k$'s with $|x_i|=1$, then can we conclude $f\in(f_1,\dots,f_r)$?

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    $\begingroup$ No, this does not work: it is possible that there are no points with $|x_i|=1$ for all $I$, that all such points lie into one irreducible component of your space, etc. $\endgroup$ – Jérôme Poineau Feb 3 at 21:22

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