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Suppose that $R$ is a complete DVR with field of fractions $K$, uniformiser $\pi$ and residue field $k$.

Let $B$ be a subring of the ring $K(t)$ of rational functions over $K$. Moreover assume that $B$ is a discrete valuation ring such that $B[\pi^{-1}]=K(t)$.

Can the residue field of $B$ be an algebraic extension of $k$? If yes then can it be an infinite algebraic extension of $k$?

I am only really interested in the case where $K$ has characteristic $0$ and $k$ has characteristic $p>0$.

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I suppose that you want $B$ to dominate $R$ ? If so, the answer is yes and yes. I will write the details later. – Qing Liu Dec 14 '10 at 17:04
Yes. Thank you. I look forward to it. – Simon Wadsley Dec 14 '10 at 17:49
up vote 8 down vote accepted

I will suppose $k$ is perfect for simplicity.

Statement: There exists $B$ with algebraic residue extension $k_B/k$ if and only if $k$ has infinite index in its algebraic closure. And this implies that $[k_B : k]=+\infty$.

First we prove the following facts:

(1) Let $F/K$ be an extension of discrete valuation fields. If the residue extension $k_F/k$ is finite, then $F/K$ is finite.

(2) Let $K^{sh}$ be a strict henselization of $K$. If $k$ has infinite index in its algebraic closure, then there exists $\theta\in\widehat{K^{sh}}$ (completion) which is transcendental over $K$.

Proof of (1). Let $k_F$ be the residue field of $F$. Lift a basis of $k_F/k$ to $a_1,...,a_r\in O_F$ (valuation ring of $F$). Let $e$ be the ramification index of $O_F$ over $R$ and let $\pi_F$ be a uniformizing element of $O_F$. Then the finite set $${\lbrace a_i \pi_F^{j} \rbrace}_{1\le i\le r, \ 0 \le j \le e-1}$$ generates a submodule $M$ of $O_F$ such that $O_F=M+\pi^n O_F$ for all $n\ge 1$ and one can check that $M\cap \pi^n O_F = \pi^n M$. As $M$ is automatically complete for the $\pi$-adic topology (it is finitely generated), this implies that $O_F=M$, thus $F/K$ is finite.

Proof of (2). For any $n\ge 1$, there exists a subextension of $\bar{k}$ of degree at least $n$. By primitive element theorem, there exists $x_n\in \bar{k}$ of degree at least $n$ over $k$. Lift $x_n$ to a $\theta_n\in K^{sh}$ with $[K(\theta_n):K]=[k(x_n):k]$ (we use the fact that $K$ is henselian here). Set $$\theta= \sum_{n\ge 1} \theta_n \pi^{n-1} \in \widehat{K^{sh}}.$$ We are going to show that $\theta$ is transcendental over $K$. Suppose the contrary. Then $\theta\in K^{sh}$. Let $L=K[\theta] \subseteq K^{sh}$. As $\theta = \theta_1 \mod \pi$ and $L$ is henselian, we have $\theta_1 \in L$. Similarly, $(\theta-\theta_1)/\pi \in L$ is equal to $\theta_2 \mod \pi$, hence $\theta_2 \in L$ and so on. Therefore $\theta_n\in L$ for all $n$. Hence $[L : K] \ge n$ for all $n$. Contradiction.

Now we prove the statement. If $k_B$ is algebraic over $k$, then $k_B/k$ is infinite by (1). In particular, $k$ has infinite index in its algebraic closure. Conversely, suppose $k$ satisfies this property. Let $\theta\in \widehat{K^{sh}}$ given by (2). Consider the field $K(\theta)\subseteq \widehat{K^{sh}}$. It is isomorphic to $K(t)$. We endow it with the discrete valuation induced by that of $\widehat{K^{sh}}$. This valuation extends that of $K$. So its valuation ring $B$ dominates $R$ and $k_B$ is contained in $\bar{k}$, hence algebraic over $k$.

Remark 1 By Artin-Schreier Theorem (that I learnd from D. Harbater), the condition $k$ has finite index in $\bar{k}$ is equivalent to $k$ is algebraically closed or real closed.

Remark 2 If $k$ is not necessarily perfect, (1) is still true (same proof). Suppose $k$ has infinite index in its algebraic closure, then the existence of $B$ with $k_B$ algebraic over $k$ is still true, but we have to work with (finite extension of) $W(\bar{k})$ or $\bar{k}((t))$ instead of $\widehat{K^{sh}}$.

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Great answer. Thank you very much. – Simon Wadsley Dec 15 '10 at 8:57

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