First off, some notation/definitions. (EDITED)
Given $k$ a field with an non-archimedean absolute value:
Valuation ring:= $\mathcal{O}$:=$\overline{B}(0,1)$ (As in closed ball, not closure)
Valuation ideal:= $\mathcal{B}$:=$B(0,1)$
Ok, so I am reading this book and it tells us to conjecture if given any field the valuation ideal is a principal ideal of $\mathcal{O}$.
If $k$ is finite, the absolute value must be the trivial one, and $\mathcal{B} = \{0\}$.
given $k$ infinite, it also works for the p-adic absolute value on $\mathbb{Q}$ and on $\mathbb{C}(t)$.
But I cannot seem to prove or refute this with an example. Examples are hard because its hard to make a non-archemedean abs value (at least for me), and I am guessing that if the idea is to stop "messing around" with divisibility. Because if you take away the correlation between your absolute value (like p-adic absolute values) with divisibility then I see no reason for there to be a single generating element.
Any help would be appreciated.