# Number of rank-one valuation rings

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 $$V$$?

• 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. Jun 25, 2021 at 14:49
• Thank you for you comment. In math.stackexchange.com/questions/4076006/… what is the value group $\ell^\mathbb{Q}$? Jun 26, 2021 at 7:03
• Do you think there are fields with the above property? Jun 26, 2021 at 7:05
• $\ell^\mathbb Q$ 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. Jun 26, 2021 at 7:47
• 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. Jun 26, 2021 at 7:50

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.