Rational power series and extensions

Let $F$ be a field, let $F(x)$ the field of rational functions, and let $F((x))$ the field of Laurent series (which contains $F(x)$). One may ask: which series $\sum_i a_i x^i$ lie in $F(x)$? The answer is well-known: it's those whose coefficients satisfy a recurrence relation.

I'm wondering about the following variant of this question (I actually care about the case where $F$ is finite, but I keep the question general and still try to not say wrong things). Let $k$ be a finite extension of $F(x)$, let $\nu$ be a valuation that extends the trivial valuation on $F$ and let $K$ be the completion of $k$ at $\nu$. Question: how are the elements of $k$ characterized as series in $K$?

The question roughly decomposes into two cases:

• if $\nu$ is unramified (over its restriction to $F(x)$), then $K = E((x))$ is a field of Laurent-series over a finite extension $E$ of $F$. A plausible answer in this case might be that the elements of $k$ are the series satisfying a recurrence relation with some restriction on the coefficients.
• if $\nu$ is totally ramified then $K = F((y))$ where $y^\ell = x$. A plausible answer in this case might be that the coefficients whose index is congruent modulo $\ell$ satisfy a recurrence relation.

Is there a good reference for this question?

• 1. Another variant of the question would be: what is a characterization of power series in $F((x))$ that are algebraic over $F(x)$? 2. I am implicitly assuming that $\nu$ has degree $1$ over $F(x)$. I am also interested in the answer for valuations of higher degree even over $F(x)$. – Stefan Witzel Nov 21 '17 at 7:47
• Here are two ways to connect the question to more "classical" maths: 1. Replace $F(x)$ by a number field (e.g. $\mathbb{Q}$) and $K$ by a non-Archimedean completion (e.g. $\mathbb{Q}_p$). In that case I don't quite know how to make sense of coefficients'' if the degree of $K$ is not $1$. 2. Take $F = \mathbb{C}$. Then the question is about Taylor series of meromorphic functions on Riemann surfaces. – Stefan Witzel Dec 1 '17 at 9:48