Number of valuation ring of a given field

Question

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.

Thank you in advance.

Answer 1

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.

Answer 2

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.