# $p$-adic power series and its maximum in the unit ball

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.

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 $$|f(c,c,\ldots)|=|g(c)| = 1$$

• That was a good idea. Thank you! 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 $$|f(X)|=\bigg\vert\sum_{J\not\in S}a_JX^J+\sum_{J\in S}a_JX^J\bigg\vert=\bigg\vert\sum_{J\in S}a_JX^J\bigg\vert=\max_{J\in S}|a_JX^J|=\max_{J\in S}|a_J|$$

• 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. Feb 7, 2019 at 16:17
• What is the definition of $|\cdot|$ in $\mathbb{k}[[X_1,\dots,X_n]]$? Feb 7, 2019 at 16:21
• 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 $\mathbb{K}$. Feb 7, 2019 at 16:34