1
$\begingroup$

I don't understand how to prove a conclusion in the Theorem.

When k is $p$-adic, the subgroups 1+$p^{v}$, $v>0$, of $u$ $(|u|=1)$ form a fundamental system of neighborhoods of $1$ in $u$, We must have therefore $\tilde{c}(1+p^{v})$ = 1 for sufficiently large $v$, $\tilde{c}$ is a character of $u$.

Why do the characters of $u$ equal to 1 in a neighborhood of $1$ in $u$?

Is it related to topology of $u$ ?

Thank you in advance.

$\endgroup$
3
  • $\begingroup$ yeah you can see at the bottom of page (2.05) [proquest version] aka the start of the second paragraph of section 2.3 that he means complex-valued characters, and then yeah exactly it follows from the topology of $u$ (“$\mathbb{C}^\times$ has no small subgroups”) $\endgroup$
    – alpoge
    Sep 22, 2021 at 1:20
  • $\begingroup$ I got it! Thank you for your comment! $\endgroup$
    – Fuutorider
    Sep 22, 2021 at 2:26
  • $\begingroup$ Yay!! good luck w the rest of it!! $\endgroup$
    – alpoge
    Sep 22, 2021 at 2:26

0

Browse other questions tagged or ask your own question.