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 $p>0$. 



How exotic can DVRs be in the ring of rational functions over a local field?
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](http://www.math.uconn.edu/~kconrad/math316/316artinschreier.pdf) (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 $\widehat{K^{sh}}$. 


