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

  • 1
    $\begingroup$ You need to assume that $k$ is infinite ($k$ finite with all nonzero element of norm one would yield counterexamples with $n=1$). $\endgroup$
    – YCor
    Commented Oct 4, 2023 at 23:59
  • $\begingroup$ 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. $\endgroup$ Commented Oct 18, 2023 at 14:21

1 Answer 1


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 $f$.


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