3
$\begingroup$

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 $\mathbb K$?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $|c_\nu|=1$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.