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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?

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$F=K((t))$ 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

$E = F(\pi_F^{1/[E:F]})$

Whence $$\overline{F}=\bigcup_{[E:F]<\infty} E=\bigcup_{[E:F]<\infty}F(\pi_F^{1/[E:F]})=\bigcup_m K((t))(t^{1/m})$$

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  • $\begingroup$ Hi, thank you for the alternative proof. Intend next days to study also your approach from the linked question. Interesting new viepoint I wasn't familar with. two questions on this proof: how to you get $(au)^{1/[E:F]}=\sum_{k\ge 0} {[E:F]\choose k} (au-1)^k$? and why is a $F(\pi_F^{1/[E:F]})=K((T))(\pi_F^{1/[E:F]})$ contained in a $K((T))(T^{1/[E:F]})?$ Or how $\pi_F$ and $T$ are related? $\endgroup$
    – user780527
    May 12, 2020 at 20:18
  • $\begingroup$ ...ohhh yes, $\pi_F$ IS $T$ :) That was foolish $\endgroup$
    – user780527
    May 12, 2020 at 20:19
  • $\begingroup$ It is the binomial series $(1+x)^{1/n}=..$ $\endgroup$
    – reuns
    May 12, 2020 at 20:24
  • $\begingroup$ I understand. And $E = F(\pi_F^{1/[E:F]})$ follows agian by general degree formula for finite extensions of local fields $[L:K]= ef$, right? This would force the extension $ F(\pi_F^{1/[E:F]}) \subset E$ to have degree $1$ by construction and considerations about their uniformizers and coinciding residue fields, I think. $\endgroup$
    – user780527
    May 13, 2020 at 22:03
  • $\begingroup$ a nitpick: do you know if this result also holds if we weaken one of the the two assumtions. ie if we eg consider alg closed $K$ with char $K > 0$ or a $K$ of char $K$=0 but not alg closed. I think I both cases the result above is not more true. I'm still haven't found conterexamples for these two cases. Do you know one? $\endgroup$
    – user780527
    May 13, 2020 at 22:03
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Because $K$ is algebraically closed of characteristic $0$, every finite extension of $K((t))$ is a tamely totally ramified extension: tamely ramified because the residue field $K$ of $K((t))$ has characteristic is $0$, and totally ramified because the residue field $K$ of $K((t))$ has no proper finite extensions.

A totally ramified extension of a complete discretely valued field $F$ is generated by a root of an Eisenstein polynomial, and a tamely totally ramified extension is generated by a root of an Eisenstein polynomial of binomial type $x^n - c$ where $c$ is a uniformizer of $F$, but typically we can't force $c$ to be a particular uniformizer. A standard example of this issue is the cyclotomic extension $\mathbf Q_p(\zeta_p)$ of $\mathbf Q_p$: this is tamely totally ramified of degree $p-1$, so $\mathbf Q_p(\zeta_p)$ equals $\mathbf Q_p(\sqrt[p-1]{up})$ for some $u \in \mathbf Z_p^\times$. It turns out we can use $u = -1$: $\mathbf Q_p(\zeta_p)$ is $\mathbf Q_p(\sqrt[p-1]{-p})$ but it is not $\mathbf Q_p(\sqrt[p-1]{p})$ when $p > 2$.

When the residue field of $F$ is algebraically closed of characteristic $0$ then we can control the constant term of that binomial Eisenstein polynomial: it can be any uniformizer of $F$.

Theorem. Let $F$ be a complete discretely valued field with uniformizer $\pi$ and $n$ be a positive integer.

(1) Every tamely totally ramified extension of $F$ with degree $n$ has the form $F(\sqrt[n]{u\pi})$ for some $u \in \mathcal O_F^\times$ and some $n$th root of $u\pi$.

(2) If the residue field of $F$ is algebraically closed of characteristic $0$ then the only extension of $F$ with degree $n$ is $F(\sqrt[n]{\pi})$, where this field is independent of the choice of $n$th root of $\pi$.

Proof.

(1) Let $E/F$ be totally ramified of degree $n$ and $\Pi$ be a uniformizer of $E$. Then $\pi$ and $\Pi^n$ have the same valuation in $E$, so $\Pi^n = v\pi$ for some unit $v \in \mathcal O_E^\times$. A totally ramified extension of $F$ is generated by any of its uniformizers, so $E = F(\Pi)$. (Warning. Since $\Pi^n = v\pi$, we could write $E = F(\sqrt[n]{v\pi})$, but this is not what the theorem is about because $v$ here is a unit up in $E$, not down in $F$, so $v\pi$ is not a uniformizer in $F$. The fact that we didn't even use the tameness property also shows the poor quality of such an argument.)

Since the residue field degree is $1$, $v \equiv u \bmod \Pi$ for some $u \in \mathcal O_F^\times$, so $v = uw$ where $w \equiv 1 \bmod \Pi$. The polynomial $g(x) = x^n - w$ in $\mathcal O_E[x]$ has a root $\varepsilon$ in $\mathcal O_E^\times$ by Hensel's lemma ($|g(1)|_E < 1$ and $|g'(1)|_E = |n|_E = 1$; that $|n|_E = 1$ is where we use the tameness condition). Thus $v = uw = u\varepsilon^n$, so $\Pi^n = v\pi = u\varepsilon^n\pi$, so $(\Pi/\varepsilon)^n = u\pi$. Since $\Pi/\varepsilon$ is a uniformizer of $E$, $E = F(\Pi/\varepsilon) = F(\sqrt[n]{u\pi})$.

(2) First we develop some background about extensions of $F$ by $n$th roots. Since the residue field of $F$ is algebraically closed of characteristic $0$, $F$ has characteristic $0$. The polynomial $x^n - 1$ has $n$ distinct roots in the residue field of $F$, so $x^n - 1$ has $n$ distinct roots in $F$ by Hensel's lemma. Therefore for $a \in F^\times$, the field $F(\sqrt[n]{a})$ is a Galois extension of $F$ (a Kummer extension) and it is independent of the choice of $n$th root of $a$: there is only one of these fields in an algebraic closure of $F$, unlike the three extensions of $\mathbf Q$ by a cube root of $2$ in an algebraic closure of $\mathbf Q$.

Now let's look at an extension field $E$ of $F$ with degree $n$. Since $F$ is complete with respect to a discrete valuation and its residue field is perfect, we have $n = e(E/F)f(E/F)$, and $f(E/F) = 1$ since the residue field is algebraically closed: every finite extension of $F$ is totally ramified. And since $n \not= 0$ in the residue field of $F$, every finite extension $E$ of $F$ is tamely totally ramified. Therefore if we pick a uniformizer $\pi$ of $F$, (1) tells us $E = F(\sqrt[n]{u\pi})$ for some unit $u \in \mathcal O_F^\times$ and some $n$th root of $u\pi$, although the previous paragraph tells us $E$ is the same for all choices of $n$th root of $u\pi$.

In the residue field of $F$, $u$ is congruent to an $n$th power (since the residue field is algebraically closed of characteristic $0$), so $u$ is an $n$th power in $\mathcal O_F^\times$ by Hensel's lemma. Therefore $E = F(\sqrt[n]{u\pi}) = F(\sqrt[n]{\pi})$. QED

Corollary. If $F$ is a complete discretely valued field whose residue field is algebraically closed of characteristic $0$, and $\pi$ is a uniformizer of $F$, then the algebraic closure of $F$ is $\bigcup_{n \geq 1} F(\sqrt[n]{\pi})$.

Proof. The algebraic closure of $F$ is a union of finite extensions of $F$, and the only extension of $F$ of a degree $n$ is $F(\sqrt[n]{\pi})$ (which is independent of the choice of $n$th root of $\pi$. QED

When $K$ is an algebraically closed field of characteristic $0$, the field $K((T))$ has residue field $K$ and it is complete with respect to the $T$-adic valuation, with uniformizer $T$. So if you apply the corollary to the field $K((T))$ with $\pi = T$, then the corollary says the algebraic closure of $K((T))$ is $\bigcup_{n \geq 1} K((T))(\sqrt[n]{T})$. We have $K((\sqrt[n]{T})) = K((T))(\sqrt[n]{T})$: for a power $\sqrt[n]{T}^j$, write $j = kn + r$ where $0 \leq r \leq n$, so $\sqrt[n]{T}^j = T^k\sqrt[n]{T}^r$ and that lets you rewrite a Laurent series in $\sqrt[n]{T}$ as a $K((T))$-linear combination of $1$, $\sqrt[n]{T}, \ldots, \sqrt[n]{T}^{n-1}$. Therefore the algebraic closure of $K((T))$ is $\bigcup_{n \geq 1} K((\sqrt[n]{T}))$.

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