Let $\mathbb{K}$ be an algebraically closed field with a complete absolute value and denote by $R$ its valuation ring.

Consider a power series $$f(X)=\sum_J a_JX^J\in \mathbb{k}[[X_1,\dots,X_n]]$$ such that $|a_J|\rightarrow 0$ as $\| J\|\rightarrow \infty$.

As a series $\sum_{i=0}^nr_n$ converges in $\mathbb{K}$ iff $|r_n|\rightarrow 0$ we have that the series above converges over $R^n$ and hence it defines a function $f:R^n\rightarrow R$. The problem is the following

Prove that $|f(x)|$ attains its maximum when $x$ varies in $R^n$ and that this maximum is equal to $\max_J |a_J|$

The sourse of this problem is Exercise 1.1.3. in Brian Conrad notes Several approaches to non-archimedean geometry.


2 Answers 2


In the case $R = \overline{\mathbb{Z}_p}$ (for other non-archimedian fields the idea should be the same, relying on $R$ containing a local field whose residue field is infinite)

Pick one $J$ such that $|a_J| =1$ (if $\max|a_j| < 1$ divide by some maximal $a_J$)

Discard all the $a_j X^j$ such that $|a_j| < 1$.

If $J = 0$ set $X= 0$ and you are done, otherwise let $g(t)=F(t,t,\ldots)= \sum_{i=0}^d b_i t^i \in R[t]$.

Let $F=\mathbb{Q}_p(b_0,\ldots,b_d)$ and $L = F(\zeta_{p^\infty-1})$.

The polynomial $g(t) \bmod (\varpi_L)$ has at most $d$ roots in $O_L/(\varpi_L) = \{0\} \cup \{ \zeta_{p^k-1}^l,k \ge 0, l \ge 0\}$ which is infinite

Thus for some $c\in O_L$, $g(c) \in \zeta_{p^k-1}^r+\varpi_L O_L$ and $|f(c,c,\ldots)|=|g(c)| = 1$

  • $\begingroup$ That was a good idea. Thank you! $\endgroup$ Feb 7, 2019 at 20:25

If $a_J=0$ for all $J$, then there is nothing to prove. Suppose now that not all $a_J=0$. Since $|a_J|\to 0$, the set $S=\{K:|a_K|=\displaystyle\max_J|a_J|\}$ is finite. If $|\cdot|$ is non-archimedean, then $|a+b|\leq\max\{|a|,|b|\}$. Let $ X=(x_1,\dots,x_n)$ be in the unit ball of $R^n$. On the one hand, $$|f(X)|=\vert\sum_{J}a_JX^J\vert\leq\max_{J}|a_JX^J|\leq\max_{J\in S}|a_J|$$ On the other hand, $\vert\sum_{J\not\in S}a_JX^J\vert\leq\max_{J\not\in S}|a_JX^J|<\max_{J\in S}|a_JX^J|$. It is enough to choose $X\in\mathbb{Z}^n$ such that $$\bigg\vert\sum_{J\in S}a_JX^J\bigg\vert=\max_{J\in S}|a_JX^J|=\max_{J\in S}|a_J|.$$ By the implication $|a|<|b|\Rightarrow|a+b|=|b|$, it follows that $$|f(X)|=\bigg\vert\sum_{J\not\in S}a_JX^J+\sum_{J\in S}a_JX^J\bigg\vert=\bigg\vert\sum_{J\in S}a_JX^J\bigg\vert=\max_{J\in S}|a_JX^J|=\max_{J\in S}|a_J|$$

  • $\begingroup$ Why $\max_{J\in S}|a_JX^J|=\vert\sum_{J\in S}a_JX^J\vert$? I think that at some point you need to use that the field is algebraically closed. $\endgroup$ Feb 7, 2019 at 16:17
  • $\begingroup$ What is the definition of $|\cdot|$ in $\mathbb{k}[[X_1,\dots,X_n]]$? $\endgroup$
    – Chilote
    Feb 7, 2019 at 16:21
  • $\begingroup$ There is no norm in that ring. What you have is that each time you replace $X_1,\dots,X_n$ by elements $x=(x_1,\dots,x_n)\in R^n$ the series converges, and the norm of $|f(x)| =| f(x_1,\dots,x_n)|$ is the absolute value of the limit of the series, this is an element in $\mathbb{K}$. $\endgroup$ Feb 7, 2019 at 16:34

You must log in to answer this question.