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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 $p$)

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  • $\begingroup$ 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. $\endgroup$ – Jason Starr Feb 7 at 11:58
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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.

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    $\begingroup$ 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 $v_0=0$. $\endgroup$ – YCor Feb 7 at 17:54

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