Tamely ramified extensions of $K_{\mathfrak p}^{\mathrm{unr}}$

Question

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.

Answer 1

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.

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 $i\neq 1$.

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.

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 any finite extension of $K_0$ is totally ramified, so I'm not really sure what's the point of this proof...

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 $K_0(\alpha_1)=L$.

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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.