I have some doubts regarding this statement, which I don't know if it's true:
Statement:
Let $K_{\mathfrak p}$ be the completion of a number field w.r.t. the $\mathfrak p$-adic valuation, and let $K_0$ be the maximal unramified extension of $K_{\mathfrak p}$. Let $L/K_0$ be a finite extension of degree $n$. Then, $L/K_0$ is tamely ramified if and only if $L=K_0(\sqrt[n]{\pi})$, where $\pi$ is a uniformizer in $K_0$, and $n$ is coprime to the characteristic of the residue field of $K_0$ (which I assume it's $p>0$).
Remarks: This statement is true in the finite case (see theorem 11.8 in https://math.mit.edu/classes/18.785/2017fa/LectureNotes11.pdf). More specifically: assume $L/K_{\mathfrak p}$ is a finite extension of degree $n$, with $n$ coprime to $\mathrm{char}(k_{\mathfrak p})=p>0$. (I denote $k_{\mathfrak p}$ the residue field of $K_{\mathfrak p}$, and $k_L$ the residue field of $L$). Then, $L/K_{\mathfrak p}$ is totally tamely ramified if and only if $L=K(\sqrt[n]{\pi})$, where $\pi$ is a uniformizer in $K_{\mathfrak p}$.
The problem is that I am not sure if one can "extend" it to $K_0$, which is an infinite extension of $K_{\mathfrak p}$. I tried to prove it directly, trying to follow the proof of proposition 12 in page 52 of Serge Lang's book "Algebraic Number Theory".
My attempt: Assume $L/K_0$ is tamely ramified. Since $k_{\mathfrak p}^{\mathrm{unr}}=\overline{\mathbb{F}}_p=k_L$, then $[L:K_0]=n=e$, by the fundamental identity. Since $L/K_0$ is tamely ramified, then $e$ is coprime to $p$. Let $\pi_0$ and $\pi_L$ be uniformizers in $K_0$ and $L$, respectively. Then we can express $\pi_0$ uniquely as
$$ \pi_0 \mathcal{O}_L = u^{-1}\pi_L^e\text{ with } u\in\mathcal{O}_L^\times.$$
(I take the inverse of $u$ by convenience for what comes next). Since $k_{\mathfrak p}=k_L$, one can find a unit $u_0\in (\mathcal{O}_{\mathfrak p}^{\mathrm{unr}})^\times$ such that $u_0\equiv u\bmod{\mathfrak{M}_L}$, where $\mathfrak{M}_L$ denotes the maximal ideal in the valuation ring $\mathcal{O}_L$. Then, taking $z:=uu_0^{-1}-1$, we have that
$$\pi_L^e=u\pi_0 = \pi+\pi z, $$
and $z\equiv\bmod{\mathfrak{M}_L}$, so $|x|<1$. Therefore,
$$ |\pi_L^e-\pi|=|\pi z|<|\pi| \quad(*). $$
Let $g(x)=x^e-\pi$, and $\alpha_1,\dots,\alpha_e$ its roots. Then,
$$ |g(\pi_L)| = |\pi_L^e-\pi| = \prod_{i=1}^e |\pi_L-\alpha_i|. $$
We have that $|\pi_L|=|\alpha_i|$ for all $i=1,\dots,e$. This is because $|\alpha_i|^e=|\pi| = |\pi_L|^e$ for all $i=1,\dots,e$. This implies that there exists some $i$ (put $i=1$) such that
$$ |\pi_L-\alpha_1| <|\alpha_1|,$$
otherwise you get a contradiction using $(*)$.
On the other hand,
$$ |g'(\alpha_1)|=|\alpha_1|^{e-1} = \prod_{i=2}^e |\alpha_1-\alpha_i|, $$
and $|\alpha_1-\alpha_i|\leq |\alpha_1|$ for all $i\neq 1$, by the triangle inequality. This implies (I'm not sure why) that $|\alpha_1-\alpha_i|=|\alpha_1|$ for all $i\neq 1$. Therefore, one can use Krasner's lemma to get that $K_{0}(\alpha_1)\subset K_0(\pi_L)$.
Now I'm not sure how to conclude, because I would have to use a result like theorem 11.5 in https://math.mit.edu/classes/18.785/2017fa/LectureNotes11.pdf, I think. But every finite extension of $K_0$ is totally ramified, so I'm not really sure what's the point of this proof…
Any help will be appreciated, thank you very much.
Best Answer
The proof in your attempt can be completed to a proof of this fact. I will elaborate on the two parts that you said were unclear.
Suppose we for contradiction did not have equality (say at $i=2$.) Then we would have by the equation directly above the quote
$$|g'(\alpha_1)|=|\alpha_1|^{e-1}=\prod_{i=2}^e|\alpha_1-\alpha_i|=|\alpha_1-\alpha_2|\prod_{i=3}^{e}|\alpha_1-\alpha_i|<|\alpha_1||\alpha_1|^{e-2}=|\alpha_1|^{e-1}$$
which is a contradiction.
What you want here is a bit weaker than theorem 11.5. All you really need is that $x^e-\pi$ is irreducible. This is true, as any root of it has $\pi_0$-adic valuation $1/e$, and therefore generates an extension of degree at least $e$. It follows that $[K_0(\alpha_1):K_0]=n$, which along with $[L:K_0]=n$ and $K_0(\alpha_1)\subseteq K_0(\pi_L)\subseteq L$ implies $K_0(\alpha_1)=L$.
Two remarks:
First, there is a more general statement of theorem 11.8 which applies in this situation. See http://alpha.math.uga.edu/~pete/8410FULL.pdf theorem 2.60. In particular, you need only assume that the base field is henselian.
Second, I came up with an alternate proof of this fact in this specific case that may be interesting. It uses more high-powered results but (IMO) is slightly more intuitive, so I will briefly sketch it here. By considering the ramification filtration on the maximal tamely ramified extension of $K_0$, $K_0^{tr}/K_0$, one can deduce that the Galois group of $K_0^{tr}/K_0$ is procyclic. In particular, this implies that all tamely ramified extensions of $K_0$ are cyclic. We can now use Kummer theory to deduce that $L=K_0(\sqrt[n]{\alpha})$ for some $\alpha\in K_0$, and by explicit discriminant calculations we can deduce that we may choose $\alpha$ to be a uniformizer.