# Characterization of finite cyclic totally ramified extension of local fields with prime power degree

Definition Let $$G_K$$ be the absolute Galois group of a local field $$K$$. We will call a group homomorphism $$\chi: G_K \to \mathbb{C}^*$$ with finite image a character on $$K$$.

Since every finite subgroup of $$\mathbb{C}^*$$ is cyclic, it is generated by a primitive root of unity. So in our case, every character $$\chi$$ corresponds to a unique cyclic Galois extension $$F/K$$ of degree $$n$$, the cardinality of the image of $$\chi$$, and an isomorphism $$\bar{\chi}: \operatorname{Gal}(F/K) \xrightarrow{\sim} \langle \xi_n \rangle \subseteq \mathbb{C}^*$$ where $$\xi_n$$ is a primitive $$n$$-th root of unity.

Let $$\chi$$ be a character which is induced by a cyclic Galois extension $$F/K$$ with prime power degree. Furthermore, let $$\operatorname{Frob}_K$$ be a Frobenius element in $$G_K$$, i.e. an element whose image is $$x \mapsto x^{|\kappa(K)|}$$ under the restriction homomorphism $$G_K \to G_{\kappa(K)}$$ where $$\kappa(K)$$ is the residue field of $$K$$. By $$F^{nr}$$, we denote the maximal unramified subextension of $$F/K$$.

I want to show that the following statements are equivalent:

1. $$F/K$$ is totally ramified,
2. $$F^{nr} = K$$,
3. the image of $$\operatorname{Frob}_K$$ in $$\operatorname{Gal}(F/K)$$ is the identity element,
4. $$\chi(\operatorname{Frob}_K)=1$$.

I think I was able to show the equivalences 1. $$\Leftrightarrow$$ 2. and 3. $$\Leftrightarrow$$ 4., so I am interested in the characterization from 1./2. to 3./4 and vice versa.

Ideas:

• $$\bar{\chi}$$ is injective on $$\operatorname{Gal}(F/K)$$.
• The image of $$\operatorname{Frob}_K$$ under $$G_K \to \operatorname{Gal}(F/K) \simeq G_K/\operatorname{Gal}(\bar{K}/F)$$ is the Frobenius element in $$\operatorname{Gal}(F/K)$$ (Usually, a Frobenius element is unique up to conjugacy. But since $$F/K$$ is cyclic, it is unique indeed.). It should be the generator of $$\operatorname{Gal}(F/K)$$.
• The inertia subgroup $$I_{F/K}$$ of $$\operatorname{Gal}(F/K)$$ is the unique cyclic subgroup of order $$e$$, the ramification index of $$F/K$$.
• We can identify $$\operatorname{Gal}(F/K)/I_{F/K}$$ with $$\operatorname{Gal}(F^{nr}/K)$$. A generator of this group is the image of $$\operatorname{Frob}_K$$, I think.

Edit: I found a crucial mistake! $$F/K$$ must have prime power degree, otherwise the equivalences won't work!

• $\mathbb C^*$ isn't cyclic, it is true that its finite subgroups are (as it's a field), but there are a huge number of completely independent elements. – Alex J Best Oct 28 '18 at 1:08
• @AlexJBest: Thanks, I fixed that! I'm still not able to find a good way to show that $\mathbb{C}^\times$ is not cyclic though. – Diglett Oct 28 '18 at 6:52
• @reuns: Let $\varphi: G_K \to G_{\kappa(K)}$ be the restriction homomorphism where $\kappa(K)$ is the residue field of $K$. A Frobenius element is any element $\operatorname{Frob}_K$ in $G_K$ such that $\varphi(\operatorname{Frob}_K): x \mapsto x^{|\kappa(K)|}$. I think I mentioned it in the post, maybe I have not emphasized enough that it is arbitrarily chosen. Let me see how I can improve my post. Could you let me know which part was not exact enough? – Diglett Oct 28 '18 at 7:29
• The unramified extensions are obtained by adjoining the primitive roots of $x^{q^m}-x$. With that unique Frobenius it seems you are thinking that $O_F = \{ \sum_{j \ge 0} \zeta^{n_j}\, \pi^j\}$ with $\zeta$ a primitive root of $x^{q^m}-x$ and $Frob_K$ restricted to $F$ being $\phi(\sum_{j \ge 0} \zeta^{n_j} \pi^j) = \sum_{j \ge 0} \zeta^{qn_j} \pi^j$. If it is the case, you should try constructing such a $\pi$ and see what you get (what is $\phi$'s fixed field ?) – reuns Oct 28 '18 at 8:17
• @reuns: Where does $O_F$ all of the sudden come from? And what does the set have to do with all of this? – Diglett Oct 29 '18 at 17:11

Your notation $$F^{nr}$$ = maximal unramified subextension of $$F/K$$ is awful. I even suspect it's at the origin of your trouble (see below). I'll not use it and introduce instead $$K_{nr}$$ = maximal unramified extension of $$K$$, so that your $$F^{nr}$$ is $$K_{nr} \cap F$$. Then :
Equivalence 1./2. For any finite extension $$F/K$$, practically by definition, $$F/(K_{nr} \cap F)$$ is totally ramified, so $$K=(K_{nr} \cap F)$$ iff $$F/K$$ is totally ramified.
Equivalence 3./4. The absolute Galois group $$G_K$$ is a profinite group, and its quotient $$Gal(K_{nr}/K)$$ is procyclic, topologically generated by $$Frob_K$$ (hence isomorphic to the profinite completion $$\hat {\mathbf Z}$$ of $$\mathbf Z$$, see Serre's "Local Fields", chap. XII, but we'll not use this fact). Whereas, in your notations, the natural projection from $$G_K$$ onto $$Gal(F/K)$$ has kernel $$Ker \chi$$, which is a closed subgroup of $$G_K$$. Consequently, your statement 4., as written, makes no sense. Perhaps you were thinking of the projection of $$Gal(K_{nr}/K)$$ onto $$Gal(F/(K_{nr} \cap F))$$, which sends $$Frob_K$$ to the relative Frobenius automorphism $$\phi$$ of $$Gal((K_{nr} \cap F)/K)$$. But even in this case, 3. doesn't make sense, because $$\phi$$ doesn't live in $$Gal(F/K)$$. The only natural way out is to replace the projection of $$G_K$$ onto $$Gal(F/K)$$ by that of $$Gal(K_{nr}/K)$$ onto $$Gal((K_{nr} \cap F)/K)$$, but then the desired equivalence becomes trivial.
• Sorry, I can't see any real change. By definition, the Frobenius automorphism (absolute or relative) is defined only for an unramified extension, hence lives in a relevant quotient Galois group. Consequently, as stated, the assertions 3. and 4., where Frobenius lives in *subgroups", don't make sense. The point lies in your definition of $Frob_K$, which you view as "an element whose image is $x↦x^{|κ(K)|}$ under the restriction homomorphism $G_K→G_{κ(K)}$ where κ(K) is the residue field of K". – nguyen quang do Oct 31 '18 at 13:05
• ...This "restriction", necessarily from $G_K$ to a subgroup, doesn't exist, because it requires a "lift" $G_{κ(K)} \to G_K$ wich doesn't exist. More precisely, given a finite Galois extension of local fields $F/K$, there is a canonical isomorphism between $Gal((K_{nr} \cap F)/K)$ and the Galois group of the residual field extension (which is generated by a Frob in the sense of finite fields), so your "lift" exists iff $F/K$ is unramified. – nguyen quang do Oct 31 '18 at 13:11