Let $\mathbb K$ be an algebraically-closed complete non-archimedean field whose absolute value is non-trivial. Consider the Tate algebra $T\_n=\mathbb K\langle X\_1,\dots, X\_n \rangle$ and fix $f\in T\_n$. We use $|\cdot|$ to denote the non-archimedean norm.


Then $f(x\_1,\dots,x\_n)$ converges for every $x=(x\_1,\dots,x\_n)\in B(\mathbb K)$ where $B(\mathbb K)$ denotes the unit ball consisting of $x$ with $|x\_i|\le 1$.



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> **Question**: Suppose $f(x)=0$ for any $x$ with $|x\_i|=1$, then can we conclude $f\equiv 0$ as an element of $T\_n$?
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For a polynomial this is obviously true due to the fundamental theorem of algebra. I am just curious about whether there is something similar in the non-archimedean world. More generally, let's assume $f=0$ on some subset $E$ of $T\_n$. What conditions for $E$ can ensure $f\equiv 0$ in $T\_n$?


**EDIT:** When $\mathbb K = \mathbb C((q))$ with $|q^n|=e^{-n}$, it seems that we can argue as follows: First note that $T\_n \subset \mathbb C((q))[[X\_1,\dots, X\_n]]$. Suppose $f=\sum\_{\nu\in \mathbb N^n} c\_\nu X^\nu$ and also write $f=\sum\_{n\ge -n\_0} q^n f\_n$ with $f\_n\in \mathbb C[[X\_1,\dots, X\_n]]$. Actually we must have $f\_n \in \mathbb C[X\_1,\dots, X\_n]$ as $|c\_\nu|\to 0$ and there is at most finite terms $c\_\nu X^\nu$ contributing to $q^nf\_n$. Finally, $|x\_i|=1$ is equivalent to $x\_i\in \mathbb C$ and hence the assumption of the question means each $f\_n\equiv 0$; so $f\equiv 0$ (maybe I had some mistakes) If my argument here was right, can we apply it for more general $\mathbb K$?



Tate algebras and fundamental theorem of algebra
Yes, the argument matches exactly your edit. We can reason as follows. Let $f$ be a nonzero element of the Tate algebra. By definition, $|c\_\nu| \to 0$ as $|\nu|\to \infty$. Thus $|c\_\nu|$ attains a maximal value for some $\nu$. By dividing by $c\_\nu$, we may assume that the maximum value is $1$. Then the coefficients lie in the ring of elements of $\mathbb K$ of norm $\leq 1$, and we can mod out by the ideal of elements of norm $<1$, obtaining a power series over the residue field $k$, which must be a polynomial. If the original power series is identically zero on elements of norm $1$ then this is identically zero on nonzero elements of $k$, hence must be the zero polynomial, which contradicts the claim that $|c\_\nu|=1$.


