# Tate algebras and fundamental theorem of algebra

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 $$\mathbb K$$?

• 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. – Piotr Achinger Jan 6 at 9:46
• 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? – 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 $$|c_\nu|=1$$.
• 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 – Will Sawin Jan 6 at 20:20
• 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? – Hang Jan 6 at 23:15