0
$\begingroup$

Let $k$ be an algebraically closed complete non-archimedean field. Let $\mathcal{O}_k$ be its ring of integers. Suppose that $A$ is a $k$-Banach algebra, and $B$ is its closed unitary ball. Note that $B$ is a Banach $\mathcal{O}_k$-algebra.

$\textbf{Question 1:}$ Is it true that $\mathrm{Hom}_{\mathcal{O}_k}(B, \mathcal{O}_k)$ is nonempty? I think this is similar to the result for complex Banach algebras where the space of characters is nonempty. Here, the Hom is taken on continuous $\mathcal{O}_k$-algebra morphisms.

$\textbf{Question 2:}$ Suppose that for all $\sigma \in \mathrm{Hom}_{\mathcal{O}_k}(B, \mathcal{O}_k)$ we have that $\sigma(a) = 0$. Does this imply that $a = 0$?

$\endgroup$
2
  • 1
    $\begingroup$ Do you mean Hom(B, OK)? The Hom-space you describe is always zero. $\endgroup$ Commented Sep 30, 2023 at 13:19
  • $\begingroup$ yes, sorry. It is Hom(B, Ok) $\endgroup$ Commented Sep 30, 2023 at 13:50

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.