# Formal series which are always zero

Let $$(k, |\cdot|)$$ be a complete field with a non-Archimedean norm, not necessarily algebraically closed. Define the Tate algebra as follows:

\begin{align*} k \langle T_1, \dots, T_n \rangle = \{ \sum_{m \in \mathbb{N}^n} a_m T_1^{m_1} \cdots T_n^{m_n} \in k[[T_1, \dots, T_n]] \mid \lim_{m \to \infty} a_m = 0 \} \end{align*}

Suppose that $$f \in k \langle T_1, \dots, T_n \rangle$$ has the property that for every $$x \in k^n$$ in the closed unitary ball, we have $$f(x) = 0$$. Does this imply that $$f = 0$$? I know that this is true if $$k$$ is algebraically closed because of the maximum principle

• You need to assume that $k$ is infinite ($k$ finite with all nonzero element of norm one would yield counterexamples with $n=1$).
– YCor
Commented Oct 4, 2023 at 23:59
• How about something like this: the claim is true if $n=1$. If $n>1$ you can write $f(T_1,...T_n)$ as $\sum_{i \geq 0} f_i(T_1,...,T_{n-1}) \cdot T_n^i$. Then the $n=1$ case will tell you that $f_i(x_1,...,x_{n-1})=0$ for all $x \in k^{n-1}$ and you can finish by induction. Commented Oct 18, 2023 at 14:21

Yes (assuming $$k$$ infinite: $$k$$ finite yields easy counterexamples), by a simple argument of elementary analysis.
• Case $$k$$ infinite, discrete: then $$f$$ is a polynomial vanishing on $$k^n$$, hence is zero.
• Case $$k$$ infinite, nondiscrete. Assuming by contradiction $$f$$ nonzero, write $$f(x)=P(x)+R(x)$$ where $$P$$ is a nonzero homogeneous polynomial of degree $$m$$, and $$R$$ has only terms of degree $$>m$$.
Then $$t^{-m}f(tx)=P(x)+t^{-m}R(tx)$$. Write $$R(x)=\sum_{i>m}R_i(x)$$ with $$R_i$$ homogeneous of degree $$i$$. Then $$t^{-m}R(tx)=\sum_{i>m}t^{i-m}R_i(x)$$. If $$x$$ is fixed in the closed 1-ball and $$t$$ tends to zero, this tends to zero (since the norm is ultrametric and each term in the sum tends to zero). Since $$f$$ vanishes in the closed 1-ball, we deduce that $$P$$ vanishes on the closed 1-ball as well. Homogeneity implies that $$P$$ vanishes on $$k^n$$, and hence $$P=0$$, contradiction.
Actually, the proof shows more precisely that $$t^{-m}f(tx)$$ converges, when $$t\to 0$$ to a nonzero homogeneous polynomial on the closed 1-ball, where $$m$$ the least degree appearing in $$f$$.