The number of valuation rings of a given field $K$ is either $1$ (in case when $K$ is an algebraic extension of a finite field) or $\infty$ (else) as it is discussed in this article.

In famous examples like algebraic number fields or in a function field $k(X)$ over an arbitrary field $k$, there are even always infinitely many discrete rank-one valuation rings (that is, their value group is isomorphic to $\mathbb Z$ or, equivalently, the valuation ring is Noetherian).

A construction using ultraproducts of fields also should give a field having infinitely many valuation rings, non of them of rank-one.

Now I am interested in a special case that lies in between.

Question 1: Is there a field $K$ containing a unique rank-one valuation ring $V$? What about the field $\mathbb Q_p$ of $p$-adic numbers with valuation ring $\mathbb Z_p$?

Here rank-one does not necessarily mean that $V$ be Noetherian. It only says that the value group is isomorphic to a subgroup of $(\mathbb R, + , \leq)$ or, equivalently, that the value group has no proper non-zero isolated subgroup or, equivalently, that $V$ has Krull dimension $1$.

I do in addition want to pick an element in $V$ that lies in no other proper valuation ring of $K$. So what about the following

Question 2: Can we choose $K$ and $V$ in such a way that $V$ is not the union of all valuation rings that are strictly contained in $V$?

Thank you in advance for your help!

  • 1
    $\begingroup$ Cf. answers to math.stackexchange.com/q/4076006/96384 and math.stackexchange.com/q/4074221/96384 for related discussion. Every field that embeds into $\mathbb C$ (this includes $\mathbb Q_p$ via axiom of choice) has infinitely many rank-one valuations. $\endgroup$ Jun 25, 2021 at 14:49
  • $\begingroup$ Thank you for you comment. In math.stackexchange.com/questions/4076006/… what is the value group $\ell^\mathbb{Q}$? $\endgroup$
    – Daniel W.
    Jun 26, 2021 at 7:03
  • $\begingroup$ Do you think there are fields with the above property? $\endgroup$
    – Daniel W.
    Jun 26, 2021 at 7:05
  • 1
    $\begingroup$ $\ell^\mathbb Q$ denotes all rational powers of the prime number $\ell$, as a subgroup of the multiplicative group $\mathbb R^\times$: The image of the absolute value $\lvert \cdot \rvert_\ell$ on $\mathbb C_\ell$. If you prefer to write the valuation additively, then the value group here is just the additive group $\mathbb Q$, of rank $1$ (contained in $\mathbb R$) but not discrete. $\endgroup$ Jun 26, 2021 at 7:47
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    $\begingroup$ I do not know an answer to your question, but I do think if such fields exist they need to be very exotic. I admit I am also not sure if I follow your supposed example with ultraproducts of fields without any rank-$1$ valuations. $\endgroup$ Jun 26, 2021 at 7:50

1 Answer 1


The answer to question 1 is "no". Let $K$ be any field that is not algebraic over a finite field. Such a field either contains $\mathbb{Q}$ or a rational function field $k(t)$ over a subfield $k$ of $K$. Both fields possess infinitely many valuations of rank $1$. Hence it suffices to show that a rank-$1$-valuation $v$ on a subfield $K_0$ of $K$ can always be extended to a rank-1-valuation of $K$ itself. Let $T$ be a transcendence basis of $K|K_0$. Then one can extend $v$ to the rational function field $K(T)$ (in possibly infinitely many variables) through

$ w(\sum\limits_{i_1,\ldots ,i_r\in S}a_{i_1,\ldots ,i_r}t_1^{i_1}\cdot\ldots\cdot t_r^{i_r}):=\min(v(a_{i_1,\ldots ,i_r}) :i_1,\ldots ,i_r\in S) $

for arbitrary pairwise distinct $t_1,\ldots ,t_r\in T$, finite sets $S\subset\mathbb{N}$ and coefficents $a_{i_1,\ldots ,i_r}\in K_0$ (so-called Gauss extension). The extension $w$ has rank $1$.

The field extension $K|K(T)$ is algebraic, hence any extension of $w$ to $K$ has rank 1 as well.


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