# number of valuation ring of $\Bbb Q_p((T))$

I would like to know the number of valuation rings of $$\Bbb Q_p((T))$$. I know $$\Bbb Q_p$$ has $$2$$ valuation rings, that is,$$\Bbb Q_p$$ and $$\Bbb Z_p$$. Every algebraic extension of $$\Bbb Q_p$$ has more than $$2$$ valuation rings because of extension theorem on valuation.But $$\Bbb Q_p((T))$$ is not algebraic over $$\Bbb Q_p$$,I am at a loss.

How many valuation rings of $$\Bbb Q_p((T))$$ are there?

Thank you in advance.

• What definition of "valuation ring" are you working with here? I would have expected something like $\mathbb Q_p[[T]]$. Commented Mar 25, 2021 at 15:00
• Commented Mar 25, 2021 at 15:07
• Ofcourse I know that. Zp and Qp are example of valuation ring of Qp, as I wrote. Commented Mar 25, 2021 at 15:45
• Ah sorry, I misread your example there. Commented Mar 25, 2021 at 16:33
• Every algebraic extension of $\Bbb{Q}_p$ has only one discrete valuation. Commented Mar 25, 2021 at 18:21

There is only one (non-trivial) discrete valuation on $$\Bbb{Q}_p((T))$$.

For all $$f\in 1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]]$$ the binomial series gives that $$f^{1/n}\in \Bbb{Q}_p((T))$$ whenever $$p\nmid n$$. Therefore, that $$v$$ is discrete implies $$v(f)=0$$.

Decompose $$\Bbb{Q}_p((T))^\times = T^\Bbb{Z} p^\Bbb{Z} \langle \zeta_{p-1}\rangle \ (1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]])$$ Thus, it remains to find $$v(\zeta_{p-1}),v(p)$$ and $$v(T)$$.

Since $$(\zeta_{p-1})^{p-1}=1$$ we must have $$v(\zeta_{p-1})=0$$.

• If $$v(p)\ne 0$$, then $$p$$ is a multiple of $$1$$ so we must have $$v(p)>0$$. For $$r$$ large enough we have $$v(p^{-r} T)<0$$ so that $$v(1+p^{-r} T)<0$$, a contradiction.

• Whence $$v(p)=0$$. We must have $$v(T)\ne 0$$ for $$v$$ being non-trivial.

$$v(T)<0$$ would contradict $$v(1+T)=0$$ so we must have $$v(T)>0$$ and hence ​$$\{ x\in \Bbb{Q}_p((T))^\times,v(x)\ge 0\} = \Bbb{Q}_p[[T]]-0$$

• Would not the following hold for $F$ any field of characteristic $0$: $F((T))^\times \simeq T^{\mathbb Z} \times F^\times \times (1+TF[[T]])$, the last factor is $n$-divisible for infinitely many $n$ so has valuation $0$. Then for any $a \in F^\times$ if $v(a) \neq 0$ analogous to your argument one would get $v(1+a^rT) < 0$ for $r$ big or small enough, contradiction, so $v(F^\times)=0$. Then for $v$ to be non-trivial $v(T) > 0$ and proceeding as you do. Or where's my mistake? Commented Nov 7, 2022 at 3:50
• @TorstenSchoeneberg Seems right. As a sanity check $k[[x]][[y]]=k[[y]][[x]]$ but $k((x))((y))\ne k((y))((x))$ and $\inf v(a_n)$ isn't a valuation on $\Bbb{Q}_p((T))$ Commented Nov 7, 2022 at 8:33

A valuation ring as you define it in a comment can also be described via a valuation, i.e. a surjective homomorphism $$v: K^\times \twoheadrightarrow \Gamma$$ onto a totally ordered abelian group $$\Gamma$$.

You should first of all note (also w.r.t. your previous question) that if we stay in this generality, almost all fields you can reasonably think of have infinitely many valuation rings, although you might not be able to see them:

Namely, let $$K$$ be any (!) field that embeds into $$\mathbb C$$ (and unless you have a good point against the axiom of choice, this includes all $$\mathbb Q_p$$ and also the example $$\mathbb Q_p((T))$$ in this question). Now there are isomorphisms (unless you have a good point against the axiom of choice) $$\mathbb C \simeq \mathbb C_\ell$$ where $$\mathbb C_\ell$$ is the completion of the algebraic closure of the $$\ell$$-adic numbers $$\mathbb Q_\ell$$ for your second favourite prime $$\ell$$. Said field $$\mathbb C_\ell$$ comes naturally equipped with its $$\ell$$-adic valuation $$v_\ell$$ with value group (written multiplicatively) $$\Gamma = \ell^\mathbb Q$$. The restriction of this valuation along the isomorphisms and to the field $$K$$ is a valuation on $$K$$, which when restricted to $$\mathbb Q$$ gives the $$\ell$$-adic valuation. In particular, all these valuations are distinct and thus we have (modulo choice) infinitely many distinct valuation rings in $$K$$.

So that kind of dooms your hope to find a field with three (or any finite number of) valuation rings, at least in characteristic $$0$$. (Update: As reuns comments and I spelled out in an answer to your previous question, actually every field that is not algebraic over some $$\mathbb F_p$$ has infinitely many valuation rings -- and those that are algebraic over some $$\mathbb F_p$$ of course have only one, namely themselves.)

Good news is, if you restrict your search to valuation rings whose value group $$\Gamma$$ (from now on written additively) is discrete in the sense that it has a smallest element $$>0_\Gamma$$ (which unfortunately is not the same as what we call a DVR; the difference is that here I allow "discrete" value groups of higher rank, i.e. basically $$\mathbb Z^n$$), then I think you've found a match here.

To see that, adapt reuns' beautiful answer: The $$n$$-divisibility of that big subgroup $$(1+p\Bbb{Z}_p+T \Bbb{Q}_p[[T]], \cdot) \subset K^\times$$ still shows (because of our "dsicreteness" assumption) that all its elements must have value $$0$$, and of course so do all roots of unity; if $$v(p)=0$$ continue as in his proof and either get the trivial valuation $$w_0(K^\times)=0$$ i.e.

$$R_0=K$$

or the discrete (rank 1) valuation $$w_1(\sum a_i T^i)=\min(i: a_i \neq 0)$$ i.e.

$$R_1= \mathbb Q_p[[T]].$$

But now we could also have the option that $$v(p) > 0$$ as long as we make sure (cf. reuns' first bullet) that $$v(p^{-r}T)$$ is always $$\ge 0$$, meaning that the value of $$T$$ has to be "infinitely bigger" than that of $$p$$. That forces us basically to have $$\mathbb Z \oplus \mathbb Z$$ as value group, with the generator $$v(T) := (1,0)$$ "infinitely bigger" than the other generator $$v(p)=(0,1)$$, i.e. the lexicographic order. Call this rank 2 valuation $$w_2 : K^\times \rightarrow \mathbb Z \oplus \mathbb Z$$, it is explicitly given on $$x= \sum a_i T^i$$ as

$$w_2(x) := \left(w_1(x), v_p(a_{w_1(x)})\right).$$

Its valuation ring is

$$R_2 = \{x \in K: w_2(x) \ge (0,0)\} = \{\sum_{i \ge 0} a_i X^i \in \mathbb Q_p[[X]]: a_0 \in \mathbb Z_p\}$$

and there you got your three possible valuation rings with "discrete" value groups (of ranks $$0,1,2$$) $$R_0, R_1, R_2$$.

Compare the other answers to Concrete examples of valuation rings of rank two. for another example.

• Sorry, but why you restrict value group to discrete ones? As you say, I agree that almost all fields has infinitely many values except finite field（with 1 valuation ring）, local fields（with 2 valuation ring）. So I'm searching for a field with finite valuation（with more than ３ valuation ring）. Commented Mar 26, 2021 at 5:17
• Have you read the third paragraph? If we don't make that restriction, e.g. all algebraic extensions of $\mathbb Q_p$ do not (as you claim) have 2, but infinitely many valuation rings (and even of rank 1: but with dense value group), and the same is true for the field in this question. I understand user reuns has told you the same thing before. Commented Mar 26, 2021 at 5:33
• I understand your claim but my problem awareness is to find a field which has 'finitely' many more than 3 valuation ring（not restricted to discrete valuation）,as I said again and again. So, the titled field of this question and algebraic extension of Qp is out of the field I'm looking for. This is a fruit you and reuns gave me, thank you so much.But, although I'm asking for a field which has 'finitely' many（more than 3）valuation ring field, you and reuns always shows me an example of a field which has infinitely many valuation ring. Commented Mar 26, 2021 at 6:52
• @bellow $K$ has finitely many valuations (discrete and non-discrete, any kind of value group) iff it is contained in $\overline{\Bbb{F}}_p$ for some $p$ (having only the trivial valuation). Otherwise it contains $\Bbb{F}_p(x)$ or $\Bbb{Q}$ whose infinitely many valuations extend to $K$. Commented Mar 26, 2021 at 21:36
• Thank you, but the field K contained in a algebraic closure of Fp only has '1' valuation ring. As I said again and again,I' m looking for a field with 'finitely more than 3' valuation ring. Commented Mar 27, 2021 at 1:46