Let $K=\mathbb{Q}_p$ and $L,F$ be extensions of $K$ such that

  • $[L:K] = [F:K]$ is a prime power,
  • $L/K$ is not totally ramified,
  • $F/K$ is unramified,
  • $L/K$ is cyclic (and $F/K$ too which is implied by the fact that it's unramified).

Question: Can one limit the possibilites for $\operatorname{Gal}(LF/K)$?

Thoughts and Remarks

  • Since $L/K$ and $F/K$ are both cyclic of the same prime power degree and since both are not totally ramified, the image of a Frobenius element (i.e. a lift of $x \mapsto x^p$ over the residue field of $K$) is a generator in $\operatorname{Gal}(L/K)$ resp. $\operatorname{Gal}(F/K)$.
  • The degree of $LF/K$ lies between $e(L/K) \cdot n$ (where $n$ is the mentioned prime power degree of $L/K$ and $F/K$, and $e(L/K)$ is the ramification index of $L/K$) and $n^2$.
  • The possible choices for $\operatorname{Gal}(LF/K)$ should "lie between" $C_e \times C_n$ and $C_n \times C_n$.

Could you help me advancing with my line of thought? Thank you!

  • $\begingroup$ What do you mean when you say "the image of a Frobenius element is a generator in ${\rm Gal}(L/K)$"? The mapping from ${\rm Gal}(L/K)$ to the Galois group of the residue field extension is onto, but not $1$-$1$ when $L/K$ is ramified (whether or not it is totally ramified), and that is true for all finite Galois extensions of $K$ no matter what the degree is over $K$ and thus doesn't say anything special at all. $\endgroup$
    – KCd
    Dec 3, 2020 at 4:05
  • $\begingroup$ @KCd: I mean a Frobenius element in the absolute Galois group $G_K$ which is defined as a lift of $x \mapsto x^p \in G_k$ to $G_K$ (where $k$ is the residue field of $K$). Since the last map is also called a Frobenius element, I understand that my explanation was unclear. I hope I could resolve that issue with that explanation. $\endgroup$
    – Diglett
    Dec 3, 2020 at 4:13
  • $\begingroup$ So I considered a Frobenius element $\operatorname{Frob}_K$ in $G_K$ and considered its image in $\operatorname{Gal}(L/K)$ resp. $\operatorname{Gal}(F/K)$. In both cases, the image in the respective groups is a generator. $\endgroup$
    – Diglett
    Dec 3, 2020 at 4:17
  • $\begingroup$ I think my answer is not too wrong now, at least it fits with KCd's example $L=\Bbb{Q}_3(\zeta_{16} 3^{1/2}), F=\Bbb{Q}_3(\zeta_{16} )$ $\endgroup$
    – reuns
    Dec 3, 2020 at 4:31
  • $\begingroup$ Lifting from $G_k$ to $G_K$ has an enormous kernel: you can say for sure about a lift to $G_K$ what its effect is on the maximal unramified extension of $K$, but that is very far from all of $\overline{K}$. So you have not really pinned down a particular element of ${\rm Gal}(L/K)$ by speaking of a Frobenius (lift of $x \mapsto x^p$ on $k$). $\endgroup$
    – KCd
    Dec 3, 2020 at 4:40

1 Answer 1


$q$ is prime, $L/K$ is cyclic of degree $q^r$, $L'/K$ is unramified of degree $q^{r-s}>1$ and $L/L'$ is totally ramified of degree $q^s>1$, then any automorphism $\sigma\in Gal(L/K)$ such that $\sigma|_{L'}$ is the Frobenius will be a generator of $Gal(L/K)$.

Since $F/K$ is unramified of degree $q^r$ then $[LF:L]=[F:L'] = q^s$ and $[LF:K] = q^r q^s$

$$Gal(L/L')\times Gal(F/L') \cong Gal(LF/L') = \langle a\rangle \times \langle b\rangle = C_{q^s}\times C_{q^s}$$ where $b$ is the Frobenius of $LF/L$ and $a$ is a generator of $Gal(LF/F)$.

Let $\phi$ be an extension of the Frobenius of $F/K$ to $LF$, then $$\phi^{q^{r-s}}|_F=b|_F$$ so $\phi^{q^{r-s}} = a^n b$, the order of $\phi$ is $q^r$ so that $\langle a\rangle \cap \langle \phi \rangle = \{1\}$ and $$ Gal(LF/K) = \langle a\rangle \times \langle \phi \rangle = C_{q^s} \times C_{q^ r}$$

  • $\begingroup$ I have trouble understanding why $\phi^{q^{r-s}}|_F = b_F$ and why that does imply that $\phi^{q^{r-s}} = a^n b$ (what is $n$, by the way) and $\langle a \rangle \cap \langle \phi \rangle = \{ 1 \}$. Could you please elaborate on these points? $\endgroup$
    – Diglett
    Dec 7, 2020 at 22:13
  • $\begingroup$ $|O_K/(\pi_K)|=t, |O_L/(\pi_L)| = t^{q^{r-s}}=T $, $\phi$ is the Frobenius $x\to x^t$ of $O_{FL}/(\pi_{FL})/O_K/(\pi_K)$ so $\phi^{q^{r-s}} : x \to x^T$ is the Frobenius of $O_{FL}/(\pi_{FL})/O_L/(\pi_L)$. Thus it is in $Aut(FL/L')$ and for some $n$, $\phi^{q^{r-s}}= a^n b$. The residue field implies that the order of $\phi$ is at least $q^r$ and since $a,b$ have order $q^s$ then $\phi^{q^r}=1$. Finally $\phi^m = a^l$ implies that $\phi^m$ acts trivially on $O_{FL}/(\pi_{FL})$ so $q^r | m$ ie. $\phi^m=1$ and $\langle a\rangle \cap \langle \phi \rangle = \{1\}$. $\endgroup$
    – reuns
    Dec 7, 2020 at 22:41
  • $\begingroup$ Thank you! I have another question: Shouldn't $b$ be a Frobenius element of $FL/L$? $\endgroup$
    – Diglett
    Dec 10, 2020 at 0:36
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$
    – reuns
    Dec 10, 2020 at 1:31
  • 1
    $\begingroup$ For an abelian group $G$ with two subgroups $A,B$ such that $A\times B\to G,(a,b)\to ab$ is surjective then $G=A\times B$ iff the kernel of this map is trivial iff $A\cap B=\{1\}$ (for the case $G$ non-abelian: we are not using that $A,B$ are abelian, only that $ab=ba$) $\endgroup$
    – reuns
    Feb 23, 2021 at 3:52

You must log in to answer this question.

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