# Number of valuation ring of a given field

Let $$K$$ be a field. I'd like to know the number of valuation ring of $$K$$.

My conjecture; The number of valuation ring of $$K$$ is $$1$$ or $$2$$ or $$∞$$.

Let $$K$$ be $$\mathbb{Q}$$, then $$\mathbb{Z}_{(p)}$$ is a valuation ring of $$\mathbb{Q}$$, so there are infinitely many valuation ring according to the number of prime $$p$$.

Let $$K$$ be a local field, then valuation ring is just $$K$$ and its integer ring, so in this case the number is $$2$$.

Let $$K$$ be a finite field,I believe the number is $$1$$ because $$K$$ itself is the only valuation ring of finite field.

I couldn't find a field $$K$$ whose number of valuation ring is natural number more than $$3$$.

I know there is a $$1$$ to $$1$$ correspondence between the places and valuation rings of a field. So I can confirm above claims from the view point of $$places$$, but I still cannot find field $$K$$ which contains more than $$3$$ valuation rings.

• You meant discrete valuation ring as otherwise there are many more for $\Bbb{Q}_p$. Commented Mar 24, 2021 at 6:42
• Ok my answer was flawed, I'm trying to repair. Commented Mar 24, 2021 at 7:47

Just to expand a little on comments by reuns:

The number of valuation rings of a field $$K$$ is either $$1$$ or infinity. The first is the case if and only if $$K$$ is contained in some $$\overline{\mathbb F_p}$$. In particular, your claim that local fields have $$2$$ valuation rings is false, they have infinitely many; and there is no field which has a finite number (except $$1$$) of valuation rings.

Namely, a valuation ring can equivalently be described via a valuation, i.e. a surjective homomorphism $$v: K^\times \twoheadrightarrow \Gamma$$ onto a totally ordered abelian group $$\Gamma$$.

First case: $$K$$ embeds into some $$\overline{\mathbb F_p}$$. Then every element of $$K^\times$$ has finite order, so the valuation has to be trivial, so the only valuation ring in $$K$$ is $$K$$ itself.

Second case: $$K$$ does not embed into any $$\overline{\mathbb F_p}$$. Then it either contains $$\mathbb Q$$, or, if $$\mathrm{char}(K)=p$$, it contains an element which is transcendental over $$\mathbb F_p$$, so it contains a subfield isomorphic to $$\mathbb F_p(T)$$. You write in your question you know that $$\mathbb Q$$ has infinitely many distinct valuations. So does $$\mathbb F_p(T)$$ (because there are infinitely many irreducible polynomials over $$\mathbb F_p$$). But it is a general theorem (credited to Chevalley) that if $$L$$ is a field with a valuation, and $$K$$ is any field extension of it, then one can extend the valuation to one on $$K$$ (for this theorem in full generality, the axiom of choice is needed). But that means that our $$K$$ has infinitely many distinct valuations (because already their restrictions to valuations on either $$L=\mathbb Q$$ or $$L=\mathbb F_p(T)$$ are distinct; of course in general, the extensions of one of the infinitely many valuations on $$L$$ might not be unique, but that only could give us "even more" valuations). So $$K$$ has infinitely many mutually different valuation rings.

• For a discretely valued field "the ring of integers" and "the valuation ring" means the same thing: $O_K=\{ a\in K,v(a)\ge 0\}$.SO, integer ring of Qp is just Zp、where am I missing? Commented Mar 27, 2021 at 4:24
• That not every valuation is a discrete valuation. Compare reuns' first comment to this question. Commented Mar 27, 2021 at 4:43
• Oh, thank you, that is the first my misunderstanding. Sorry for bother, but what is an example of valuation ring of Qp but Zp and Qp? Commented Mar 27, 2021 at 5:28
• @bellow: Well there you put your finger on something, as far as I know examples come from the axiom of choice, thus cannot be given in total explicitness. I could just say that e.g. for each prime $\ell$ there is one (actually, infinitely many) whose intersection with $\mathbb Q$ is exactly $\mathbb Z_{(\ell)}$. Actually, if you give any number field $K$ which embeds into $\mathbb Q_p$, you can extend any valuation of $K$ to $\mathbb Q_p$. You also have great freedom on transcendental elements. That's the issue: Too much freedom to be explicit here. Commented Mar 28, 2021 at 22:12
• One thing one can be sure about is that no such extension can remain discrete, cf. math.stackexchange.com/a/3390398/96384. For a related discussion about extending values to $\mathbb R$, cf. math.stackexchange.com/a/2563837/96384 Commented Mar 28, 2021 at 22:17

For a non-negative integer $$a$$ let

$$K=\Bbb{Q}(B), \qquad B=\{ (1+ma)^{1/n}, n\ge 1,\gcd(n,a)=1,m\ge 0\}$$

If $$K=Frac(R)$$ with $$R$$ a DVR then the valuation $$v$$ on $$R$$ extends the $$p$$-adic valuation for some $$p$$.

• If $$p\nmid a$$ then $$p$$ divides some $$1+ma$$ and $$v((1+ma)^{1/n})=v(1+ma)/n$$ gives that $$v(K)=\Bbb{Q}$$ ie. $$v$$ is not discrete.

• Whence $$p\ |\ a$$.

With $$O_K$$ the algebraic integers $$\subset K$$ then $$O_K$$ has a lot of prime ideals above $$p$$, and since $$K$$ is unramified at $$p$$ we get that any valuation above $$p$$ is discrete, ie. infinitely many choices for $$v$$.

To get only one choice for $$v$$ we need to add more elements to $$B$$:

For each $$b\in B$$, $$f\in \Bbb{Z}[x]$$,$$n\ge 1,\gcd(n,a)=1$$: add $$(1+(b-1)f(b))^{1/n}$$ to $$B$$, and then repeat iteratively, choosing $$b$$ also in the newly added elements.

This time only one valuation above $$p$$ is discrete: the one such that $$v(b-1)>0$$ for all $$b\in B$$, which gives a natural embedding $$K\to \Bbb{Q}_p$$.

The other valuations are not discrete because if $$v(b-1)=0$$ then for some $$f\in \Bbb{Z}[x]$$ we'll have $$v(1+(b-1)f(b))>0$$ and when adding the $$n$$-th roots of $$1+(b-1)f(b)$$ we'll get a non-discrete valuation.

• Thank you so much, but how do you confirm the number of $K$'s valuation ring is just $n$? Commented Mar 24, 2021 at 7:06
• Sorry to bother, but It seems not obvious to me that $K$ has just $n$ valuation ring. You just say $K$ has more than $3$ valuation ring? Commented Mar 24, 2021 at 7:27