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 $|x_i|\le 1$.

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

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 $\mathbb K$?

  • $\begingroup$ In one variable, the zeros of a nonzero $f$ have to be isolated, just as in the archimedean case. This answers the question positively, by restricting to discs. $\endgroup$ – Piotr Achinger Jan 6 at 9:46
  • $\begingroup$ Thanks! Could you briefly explain why the zeros are isolated in one variable case? Is it possible to apply your idea for more than variables? $\endgroup$ – Hang Jan 6 at 14:12

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 $|c_\nu|=1$.

  • $\begingroup$ I think what Piotr said about the zeroes in the one variable case being isolated follows from the complete local rings version of the Weiersrtass preparation theorem. $f$ is a scalar from $\mathbb K$ times a monic polynomial times a power series where the leading coefficient is has norm $1$ and all higher coefficients are smaller than $1$, and the number of roots is at most the degree of this monic polynomial $\endgroup$ – Will Sawin Jan 6 at 20:20
  • $\begingroup$ Thank you, Will. Very great answer! It seems that your idea can achieve more. e.g. $\mathbb K\langle X,X^{-1}\rangle$ or even an arbitrary affinoid algebra $A=T_n/ \mathfrak a$. We just lift $f\in A$ to $\tilde f\in T_n$, and apply your above idea to $\tilde f$. Is this right? $\endgroup$ – Hang Jan 6 at 23:15

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