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Let $K$ be a local field with non-trivial absolute value $| \cdot |:K \rightarrow \mathbb{R}$. There is an induced metric $d$ on $K$ defined by $d(a,b):=|b-a|$ for $a,b \in K$.

So for any natural number $n\geq 1$, the cartesian product $K^n$ is a metric space together with the sup metric $d_{\infty}$ defined by

$$ d_{\infty}((a_1,\cdots,a_n),(b_1,\cdots,b_n)):=max\{|b_1-a_1|,\cdots,|b_n-a_n|\}. $$

The norm topology on $K^n$ is the metric topology on $K^n$ induced by $d_{\infty}$.

On the other hand, $K^n$ also admits its Zariski topology. My question is the following:

Suppose that $|\cdot|$ is non-Archimedean. Is the Zariski topology on $K^n$ coarser than its norm topology?

Remark. In working towards a positive answer for the above question, it is enough to check that elements in some basis $\mathcal{B}$ for the Zariski topology are open in the norm topology. Choose the basis $\mathcal{B}$ of distinguished open sets.

Let $f \in K[x_1,\cdots,x_n]$. In order to show $D_f:=\{(a_i) \in K^n \; | \; f(a_i) \neq 0\}$ is open in the norm topology, it is enough to find an open ball around each $(x_i) \in D_f $ which is contained in $D_f$. More precisely, for each $(x_i) \in D_f$ there should exist $\delta > 0$ such that for any $(y_i) \in K^n$:

$$ d_{\infty}((y_i),(x_i)) < \delta \implies f(y_i) \neq 0 \; \; \; (1)$$

Condition (1) is reminiscing of "continuity of f:$K^n \rightarrow K$ at $(x_i) \in K^n$". More precisely, points in $K$ are closed in its norm topology, so the existence of such $\delta$ would follow if polynomial functions $g:K^n \rightarrow K$ were continuous w.r.t. the norm topologies in $K^n$ and $K$.

I bet that polynomial functions $g:K^n \rightarrow K$ are continuous w.r.t. the norm topologies in the case where the absolute value $|\cdot|$ is non-Archimedean; in this case, $d_{\infty}$ satisfies the ultra-metric inequality

$$ d(a,c) \leq max\{d(a,b),d(b,c)\}. $$

Unfortunately, I am not familiar with "ultra-metric analysis", so any help would be appreciated in the case that it is possible to conclude this reasoning.

Thank you in advance.

Alex Ravsky
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Pedro A. Matos
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    Thought about it a little more, and you are right, at least for the case $n=1$. Since a Zariski closed set is a finite number of points (or empty set or all) and in an ultrametric space singletons are closed, we have the result. – take008 Nov 11 '19 at 06:31

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Since the polynomials $f_j$ are continuous for $\|.\|_\infty$ then $Z(f_1,\ldots,f_m)$ (the common zeros of the $f_j$, a closed set in Zariski topology) is closed in the $\|.\|_\infty$ topology. Thus $X$ closed in Zariski topology implies $X$ closed in $\|.\|_\infty$ topology, and $U=K^n-X$ open in Zariski topology implies $U$ open in $\|.\|_\infty$ topology. The converse is not true as $\Bbb{Z}_p^n$ is closed for $\|.\|_\infty$ but its Zariski closure is $\Bbb{Q}_p^n$.

reuns
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