I want to show that the algebraic closure $L:= \overline{K((T))}$ of Puiseaux field $K((T))$ for $K$ alg. closed of char $K=0$ equals the union $\bigcup_{n \ge 1} K((T^{1/n}))$.
The closure clearly contains the union. For the other direction we have to chow that every finite alg extension of $K((T))$ is contained in $K((T^{1/n}))$ for certain $n$.
We asking which different types of finite extensions for local fields can occure? There are tamely ramified, wildly ramified and unramified extensions.
Since $K$ is already algebraically closed the "fundamental formula" (what is it's true name?) $[M:K((T))]=e f$ tells that no non trivial unramified extension could exist. Also, wildly extensions no exist as the characteritic of residue field $K$ is $0$. That is we only have have to deal with finite tamely ramified extensions.
Questions: What do we know about such finite tamely ramified extensions? The residue field not changes. How can I conclude that each such extension is contained in a $K((T^{1/n}))$ for appropriate $n$? Any result (I actually not know) from theory of local field?
Addendum: I noticed that this question Algebraic closure of $k((t))$ dealed with same problem. The answer is beautiful but attacked the problem from another viewpoint. Can the claim be proved only with methods from class field theory?
an outlook: although I haven't finished the proof yet, it is known to to true for $K$ alg closed of char $0$. Can the claim be generalized under weaker assumptions? Ie if we assume char $K>0$ or $K$ is not more closed?
My guess is not since we obtain in any or these two cases a new class of non traivial finite extensions: the unramified or the wildly ramified which we before excluded. But that's just my conjecture. Are some concrete conterexamples known?
Best Answer
$F=K((t))$ with $K$ algebraically closed of characteristic $0$.
For a finite extension $E/F$, with corresponding complete DVR $O_F=K[[t]]\subset O_E$ and uniformizers $\pi_F=t,\pi_E$ and residue field $O_F/(\pi_F)=O_E/(\pi_E)=K$.
The completeness says that $$O_E=\sum_{n\ge 0} \pi_E^n K=\sum_{n= 0}^{[E:F]-1} \pi_E^n O_F$$
ie. $\pi_F = u \pi_E^{[E:F]}$ where $u\in O_F^\times$.
Let $a\in K^\times$ such that $au \in 1+\pi_F O_F$.
Then $(au)^{1/[E:F]}=\sum_{k\ge 0} {[E:F]\choose k} (au-1)^k \in O_F$ and $a^{1/[E:F]}\in O_F$ which means that $u^{1/[E:F]}\in O_F$ ie. $\pi_F^{1/[E:F]} \in O_F$.
But $\pi_F^{1/[E:F]} $ is also an uniformizer so that $$O_E=\sum_{n\ge 0} (\pi_F^{1/[E:F]})^n K=\sum_{n= 0}^{[E:F]-1} (\pi_F^{1/[E:F]})^n O_F$$ and
Whence $$\overline{F}=\bigcup_{[E:F]<\infty} E=\bigcup_{[E:F]<\infty}F(\pi_F^{1/[E:F]})=\bigcup_m K((t))(t^{1/m})$$