Algebraic closure of $k((t))$

Question

Let $k$ be an algebraic closed field of characteristic $0$.

I want to understand why the algebraic closure of the field $k((t))$ is $\underset{n\geq 1}{\bigcup k((t^{1/n}))}$.

Obviously, $\underset{n\geq 1}{\bigcup k((t^{1/n}))} \subset \overline{k((t))}$ as $x^n-t$ are irreducible over $k((t))$.

Question 1: Why for every element $\alpha \in \overline{k((t))}$ there exist a $n = n(\alpha) \in \mathbb{N}$ (<- "the common denominator") such that $\alpha = \sum_{r \in \mathbb{Z}} f_{r} t^{r/n}$ ?

Question 2: How are the algebraic closures of $k(t)$ and $k((t))$ related?

Answer 1

Your last question strongly suggests to look at the case $k=\Bbb{C}$.

• For $L/\Bbb{C}(z)$ a finite Galois extension then its elements are locally meromorphic on $\Bbb{C}$ minus a few branch points (the zeros of the discriminant of the minimal polynomials) and $Gal(L/\Bbb{C}(z))$ consists of the analytic continuations along closed loops enclosing some of those branch points.

Proof: with $\gamma_1(a),\ldots,\gamma_m(a)$ the analytic continuations of $a$ then the coefficients of $h(X)=\prod_{l=1}^m(X-\gamma_m(a))$ stay the same under analytic continuation, thus they are meromorphic on the Riemann sphere, ie. they are in $\Bbb{C}(z)$ so $h(X)$ is the minimal polynomial of $a$.

• Let $\gamma$ be a closed-loop enclosing once only one of those branch points $z_0$. Wlog assume $z_0=0$. Then the subfield $E\subset L$ fixed by $\gamma$ is a field of meromorphic functions at $0$, thus it is a field of Laurent series in $z$ convergent for $0<|z|<\epsilon$. And $L/E$ is Galois with cyclic Galois group generated by $\gamma$. The primitive element theorem gives $L=E(f)$, $n=[L:E]$, letting $w=\sum_{l=1}^n e^{-2i\pi l/n}\gamma^l(f)$ then $\gamma(w)=e^{2i\pi /n}w$ thus $w^n\in E$, $w^n = z^k h(z)$ with $h$ analytic at $0$ and $h(0)\ne 0$ so that $w=z^{k/n} h(z)^{1/n}$ where $h(z)^{1/n}$ is analytic ie. $h(z)^{1/n}\in E$. Since $\gamma^l(w)=e^{2i\pi kl/n}z^{k/n}h(z)^{1/n}\ne w$ then $\gcd(k,n)=1$ and $L=E(z^{1/n})$.

With $H(z^{1/n})$ the field of algebraic functions that are Laurent series in $z^{1/n}$ then $\bigcup_n H(z^{1/n})$ is algebraically closed and $\bigcup_n H(z^{1/n})=\overline{\Bbb{C}(z)}$.

$\bigcup_n \Bbb{C}((z^{1/n}))$ is its completion for the non-archimedian absolute value $|z^{k/n} h(z^{1/n})|= 2^{-k/n},h(0)\ne 0$, it is algebraically closed too, the Galois group $Gal(\bigcup_n \Bbb{C}((z^{1/n})) / \Bbb{C}((z)))$ consists of formal closed-loops around $z=0$, where the winding number is a profinite integer $\in\varprojlim \Bbb{Z/(n)}$.