Let $A$ be the ring of convergent power series over $\mathbf{Q}_p$ in one variable. That is, elements of $A$ are formal series $$f(t) = \sum_{m=0}^\infty a_m t^m$$ such that the coefficients $a_m$ tend to zero in $\mathbf{Q}_p$ as $m \rightarrow \infty$.

We define the Gauss norm on $A$ by setting $$\Vert f(t) \Vert = \sup_m |a_m|$$ It is obvious that for any point $z \in \mathbf{Z}_p$, we have $|f(z)| \leq \Vert f(t) \Vert$.

Failed attempt (see edit below): In fact, this upper bound should be attained i.e. there should be some $z \in \mathbf{Z}_p$ with $|f(z)| = \Vert f(t) \Vert$.

How do we show the existence of such a $z$?

Edit: So it turns out the above is not true. The supremum is not always necessarily attained. Here is a revised question that is hopefully more true.

Let's replace $\mathbf{Q}_p$ with its completed algebraic closure $\mathbf{C}_p$. The definition of convergent power series and Gauss norms for $\mathbf{C}_p$ are unchanged. Now we have

$$\sup_{z \in \mathcal{O}_{\mathbf{C}_p}} |f(z)| \leq \Vert f(t) \Vert$$

Can we upgrade this inequality to an equality?

I have dropped the requirement that the upper bound on the right side be attained by some $z \in \mathcal{O}_{\mathbf{C}_p}$.

  • $\begingroup$ The max is not attained for $z \in \mathbb{Z}_p$. Counterexample, $f(t)= t^p-t$. $\endgroup$
    – Merosity
    Dec 3, 2020 at 14:53
  • $\begingroup$ @Merosity What if we replace $\mathbf{Q}_p$ with its algebraic closure (or the completion of its algebraic closure) $K$? Will it be attained in $\mathcal{O}_K$ then? $\endgroup$
    – Alex
    Dec 3, 2020 at 15:08

1 Answer 1


$$\|f\| = \sup_n |a_n|_p = \sup_{b\in O_K} |f(b)|_p, \qquad O_K=\bigcup_{n\ge 1}\Bbb{Z}_p[\zeta_{p^n-1}]$$ Proof: $\sup_{b\in O_K} |f(b)|_p\le \|f\|$, and $\|f\| = |p^m|_p$ means that $f=p^m g+p^{m+1}h$ where $g\in \Bbb{Z}_p[x],h\in A\cap \Bbb{Z}_p[[x]]$ and $g\bmod p $ is a non-zero, this polynomial has finitely many roots in the field $O_K/(p)$ thus for some $b\in O_K$, $g(b) \in O_K^\times$ and $|f(b)|_p = |p^m g(b)|_p = \|f\|$.

We can replace $K$ by any extension of $\Bbb{Q}_p$ with an infinite residue field.


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