# Does there exist a discrete valuation subring $R$ of $K((t))$ ($K$ a number field) of residue characteristic $p$ with $\mathrm{Frac}(R) = K((t))$?

Let $$K$$ be a number field, and let $$K((t))$$ be the field of formal Laurent series. Let $$p > 0$$ be a prime.

I have two questions:

1. Does there exist a discrete valuation subring $$R$$ of $$K((t))$$ of residue characteristic $$p$$ satisfying $$\mathrm{Frac}(R) = K((t))$$?
2. Given a discrete valuation subring $$A\subset K((t))$$, is it always possible to find a discrete valuation subring $$R$$ dominating $$A$$ and satisfying $$\mathrm{Frac}(R) = K((t))$$?

(Of course (2) implies (1) since we may pick $$A$$ to be the localization of $$\mathbb{Z}$$ at $$p$$)

• Let $L=K(\{s_i:i\in I\})$ be a maximal purely transcendental subextension of $K((t))/K$. Denote the integer ring of $K$ by $\mathfrak{o}_K$, and let $\mathfrak{p}$ be an ideal over $p\mathbb{Z}$. For the infinite polynomial ring $S=\mathfrak{o}_K[\{s_i:i\in I\}]$, the ideal $\mathfrak{p}S$ is a height one prime. The localization $S_{\mathfrak{p}S}$ is a DVR. For the algebraic field extension $K((t))/L$, there is a valuation ring $R$ dominating $S_{\mathfrak{p}S}$ whose fraction field equals $K((t))$. By Krull-Akizuki, you can choose $R$ to be a colimit of DVRs, probably not a DVR itself. – Jason Starr Feb 7 '19 at 11:58

No. Consider the associated discrete valuation $$v$$ as a homomorphism $$K((t))^\times \to \mathbb Z$$. We have $$K((t))^\times = K^\times \times t^{\mathbb Z} \times (1+ t K[[t]])$$. Elements of $$1+t K[[t]]$$ have arbitrarily high $$n$$th power roots in $$K((t))$$, hence they must be sent to $$0$$ by $$v$$. When restricted to $$K$$, $$v$$ must be a discrete valuation $$v_0$$ of $$K$$. Then we must have
$$v( a_d t^d + a_{d+1} t^{d+1} +\dots ) = v_0(a_d) + c d$$ for some $$c \in \mathbb Z$$.
But valuations of this form clearly do not satisfy the inequality for the valuation of a sum unless $$v_0$$ is trivial, because we can make $$v_0(a_d)$$ very large and $$v_0(a_{d+1})$$ very small. So the only discrete valuation is the standard one $$v( a_d t^d + a_{d+1} t^{d+1} +\dots ) =d$$. But this has residue characteristic zero.
• Just for exposition, the middle argument simplifies as follows: $v(1+at)=0$ for all $a\in K$; assuming by contradiction $v_0\neq 0$ and choosing $a$ with $v_0(a)<-v(t)$, we have $v(at)<0$ while $v(1+at)=v(1)=0$, contradicting the ultrametric axiom on $v$. So $v_0=0$. – YCor Feb 7 '19 at 17:54