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.
Ok, I am going to partially answer my question just in case there ever will be future enquirers asking the same question. If you ask that there exist a maximum absolute value less than $1$ then $\mathcal{B}$ is a principal ideal.
I.E if there exist $ M= max\{\mid x \mid : x \in \mathcal{B }\}$ then just take $a$ such that $\mid \, a \mid = M$. (Lets just suppose $M \neq 0$ since if it was you are just in the trivial case.)
If $b \in \mathcal{O}$ then $b=a\cdot \frac{b}{a}$ (since $\mathbb{k}$ is a field) and there are two things to check out:
First, if $\mid \, b \mid = M$ then since $\mid \, \frac{b}{a}\mid = 1$, $\frac{b}{a} \in \mathcal{O}$
Secondly if $b \in \mathcal{B}$ then $\mid b \mid < M < 1$ and $ \mid \frac{b}{a} \mid = \frac{\mid b \mid}{M} < 1$ and $\frac{b}{a} \in \mathcal{B}$
So $<a > = \mathcal{B}$.
Now something seems off about the next part, but cannot seem to see where I am wrong: Now if there is no maximum, so $sup\{\mid x \mid : x \in \mathcal{B }\}=1$, i.e. $1$ is an accumulation point of $\{\mid x \mid : x \in \mathcal{B }\}$.
All I need now is a non-archimedean absolute value that fulfills the above criteria so I can get (if its possible) a counter-example for the general case.