Puiseux series over a non algebraically closed field. Suppose that $K$ is a field of characteristic $0$ and $L/K$ is an extension so that every degree $\leq n$ polynomial over $K$ splits in $L$. We do not assume that $L$ is algebraically closed.
It is a classical fact that the algebraic closure of $K[[t]]$ is $K[[t^{1/n} : n \geq 1]]$ when $K$ is algebraically closed. In our set up, is it true that every degree $n$ polynomial over $K[[t]]$ splits completely over $L[[t^{1/n!}]]$?
I think the classical proof of Newton-Puiseux expansions can be modified to prove this but I would be very happy if there were a reference.
 A: Thanks to this question, I found a proof that can be pretty easily modified to show what I want.
We will use a form of Hensel's lemma which says that if we have a polynomial $f$ over $K[[t]]$ that splits into $f = gh$ over $t = 0$ with $\gcd(g,h) = 1$, then we can find a lifting of the splitting to $K[[t]]$.
Now suppose we have a polynomial $f(x)$ of degree $d \leq n$ over $K[[t]]$. We show that it has a non trivial factorization over $L[[t]]$. First, we make a linear change of variables so that:
$$f(x) = x^d + a_2(t)x^{d-1} + \dots + a_0$$
and then define:
$$g(x) = t^{-d\delta}f(xt^{\delta}) = x^d + b_2(t)x^{d-1} + \dots + b_0$$
where we choose $\delta = \mathrm{val}_t(a_i(t)/i)$ so that the $b_i(t)$ have a positive $t$-valuation and are not all zero modulo $t^{1/n!}$. (This step is why we need to adjoin $t^{1/n!}$).
Then, we see that $g(x) \pmod{t^{1/n!}}$ is of the form $x^d + c_2x^{d-2} + \dots + c_0$
over $L$ with the $c_i \in K$ and not all of them $0$. Since the $x^{d-2}$ term is $0$ and one of the other terms is not, we see that $g(x) \pmod{t^{1/n!}}$ is not of the form $(x-\gamma)^d$ so it factors into two non trivial, co prime factors over $L$. (This is where we use the fact that every degree $\leq n$ polynomial over $K$ splits over $L$).
By Hensel's lemma, we can now lift our factorization to $L[[t]]$.
