Talk:Characterisation of Non-Archimedean Division Ring Norms

In the proof of the sufficient condition, the computation introduces the term:
 * $\size{B \paren{n + 1}}$

but this term is, in part, a summation over the real numbers and so the premise:
 * $\forall n \in \N_{>1}: \left\vert{n \cdot 1_k}\right\vert \le 1$

does not hold and can't be used in the next line, and so you are left with the $\paren{n + 1}$ term. You have to take the nth-root of the terms and then let $n \to \infty$ to get the desired result.

Also the bound $B$ is a bound in the real numbers, but not in the field $k$ --Leigh.Samphier (talk) 16:40, 26 January 2019 (EST)


 * Can it be put right? This dates back to work done by User:Linus44 (sadly no longer active) and went a bit over my head, so unfortunately I can't help. --prime mover (talk) 16:59, 26 January 2019 (EST)