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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.

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    $\begingroup$ The valuation ring is $O=\{ a\in k, |a|\le 1\}$ not the closure of $\mathfrak{m}=\{ a\in k, |a|< 1\}$. Absolute value means that the valuation is real valued. $\mathfrak{m}$ is principal iff $\{ |a|,a\in k^*\}$ has a largest element $<1$ ie. iff it is a discrete subgroup of $\Bbb{R}^*$ iff it is of the form $r^\Bbb{Z}$ for some $r>0$ iff $O$ is a DVR. We often not assume that the valuation is real-valued, if can map to any ordered abelian group. $\endgroup$
    – reuns
    Dec 15, 2021 at 18:38
  • $\begingroup$ Did not mean it as closure, meant it as "closed" ball notation. If you feel its necessary I can fix it. I don't get why $\{ \mid a \mid: a \in \mathcal{m}\}$ having a maximum means it is a principal ideal. Because there are many elements that can have that maximum as absolute value. $\endgroup$
    – Bajo Fondo
    Dec 15, 2021 at 18:46
  • $\begingroup$ They will be a unit time the other: elements of absolute value $1$ are units of $O$. As you see sometimes the closure of $B(0,1)$ is not the same as $B[0,1]$ so clearly your notation is not ok. $\endgroup$
    – reuns
    Dec 15, 2021 at 18:49
  • $\begingroup$ @reuns I have seen this notation in books, and it is actually the notation that is used in the book from which this problem arose, and given they made an emphasis as to explain that it was not the closure, they kept that notation, I'll give you that without explanation the notation is misleading, but aside from that, if you explain, it is ok. Not everyone has to use the notation you use, that's the thing about notation, it is just that. $\endgroup$
    – Bajo Fondo
    Dec 15, 2021 at 18:52
  • $\begingroup$ There is no reason to be upset, it is just that don't use an ambiguous notation when you can write $O=\{ a\in k, |a|\le 1\}$ $\endgroup$
    – reuns
    Dec 15, 2021 at 18:53

1 Answer 1

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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.

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    $\begingroup$ For any $a \in \mathcal B$ with value $\lvert a \rvert = \alpha$, the principal ideal $a \mathcal O$ is exactly $\{x \in \mathcal O: \lvert x \rvert \le \alpha\}$. So if $\sup_{x \in \mathfrak B} \lvert x \rvert =1$, the ideal $\mathfrak B$ cannot be principal, because if it were, its generator would have some value $\alpha < 1$, and it could not contain any element of value between $\alpha$ and $1$ (which exist because that supremum is $1$). -- A classic example for a nonarchimedean field with non-discrete value group is $\mathbb C_p$, the completion of an algebraic closure of $\mathbb Q_p$. $\endgroup$
    – Torsten Schoeneberg
    Dec 20, 2021 at 18:33
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    $\begingroup$ Torsten Schoeneberg. Oh, thanks. Seems so simple now that you mention it. Going to check your example out, I am guessing you are taking the p-adic absolute value and checking that it extends nicely. $\endgroup$
    – Bajo Fondo
    Dec 20, 2021 at 19:34

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