Algebraic closure of Puiseaux field $K((T))$ equals $\bigcup_{n \ge 1} K((T^{1/n}))$

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?

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

• 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?
– user780527
May 12, 2020 at 20:18
• ...ohhh yes, $\pi_F$ IS $T$ :) That was foolish
– user780527
May 12, 2020 at 20:19
• It is the binomial series $(1+x)^{1/n}=..$ May 12, 2020 at 20:24
• 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.
– user780527
May 13, 2020 at 22:03
• 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?
– user780527
May 13, 2020 at 22:03

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 $$\bigcup_{n \geq 1} K((\sqrt[n]{T}))$$.