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.


  • $\bar{\chi}$ 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.

Could you please help me to establish the remaining connections? Thank you!

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

  • $\begingroup$ $\mathbb C^*$ 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. $\endgroup$ Oct 28, 2018 at 1:08
  • $\begingroup$ @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. $\endgroup$
    – Diglett
    Oct 28, 2018 at 6:52
  • $\begingroup$ @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? $\endgroup$
    – Diglett
    Oct 28, 2018 at 7:29
  • $\begingroup$ 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 ?) $\endgroup$
    – reuns
    Oct 28, 2018 at 8:17
  • $\begingroup$ @reuns: Where does $O_F$ all of the sudden come from? And what does the set have to do with all of this? $\endgroup$
    – Diglett
    Oct 29, 2018 at 17:11

1 Answer 1


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.

  • $\begingroup$ Thanks for your response, I will check my notations/statements. But as I said before, I am rather interested in the equivalences from 1. (or 2.) to 3. (or 4.) and vice versa. I changed my post a bit, hopefully it is clearer now. $\endgroup$
    – Diglett
    Oct 31, 2018 at 10:07
  • $\begingroup$ 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". $\endgroup$ Oct 31, 2018 at 13:05
  • $\begingroup$ ...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. $\endgroup$ Oct 31, 2018 at 13:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .