# Puiseux decomposition over a field with positive characteristic

Let $K$ be an algebraically closed field with characteristic $p>0$, and let $f(t,x)\in K[t,x]$ be a polynomial separable in $x$. Denote: $$\Lambda = \bigcup_{i\in \mathbb{N}} K((t^\frac{1}{i}))$$ I want to prove that $f(t,x)$ as a polynomial in $K[x]$ has a root in $\Lambda$.

If the characteristic of $K$ is zero or if $p$ does not divide $\deg_x(f)$ then it is true (ref. Lectures on Expansion Techniques In Algebraic Geometry by S.S.Abhyankar page 22 theorem 5.5).

I am looking for a proof assuming only that $f$ is separable in $x$.

• It is not true. Commented Dec 29, 2016 at 21:19
• Your extension of $K((t))$ seems to catch none of its wild separable extensions. So not big enough. Commented Dec 29, 2016 at 21:46

The comment of Felipe Voloch does not seem very useful without an example or reference. In fact, if $K$ is the algebraic closure of $\mathbb{F}_p$ then the equation $tx^p-tx-1$ has no solution in $\Lambda$, an observation of Chevalley. For a "generalized Puiseux field" that does contain an algebraic closure of $K((t))$, see S. Vaida, Illinois J. Math 41 (1997), 129-141. These facts are mentioned in the solution to Exercise 6.4 of my book Enumerative Combinatorics, vol. 2.
• You may want to change Chevalley's example to $tx^p - tx - 1$ to match the parameters of the question. Commented Dec 30, 2016 at 8:35
• @RichardStanley Thanks for the answer. So what polynomials would have a root in $\Lambda$? Only polynomials that their splitting field is tamely ramified? Commented Dec 31, 2016 at 11:45