Valuation ideal on non-archimedean field is a principal ideal?

Solution 1:

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.