# Galois group of the compositum of a non-totally-ramified and a the unramified extension of $\mathbb{Q}_p$ of prime power degree

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!

• 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.
– KCd
Dec 3, 2020 at 4:05
• @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. Dec 3, 2020 at 4:13
• 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. Dec 3, 2020 at 4:17
• I think my answer is not too wrong now, at least it fits with KCd's example $L=\Bbb{Q}_3(\zeta_{16} 3^{1/2}), F=\Bbb{Q}_3(\zeta_{16} )$ Dec 3, 2020 at 4:31
• 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 $k$).
– KCd
Dec 3, 2020 at 4:40

$$q$$ 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 $$Gal(LF/K) = \langle a\rangle \times \langle \phi \rangle = C_{q^s} \times C_{q^ r}$$

• 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? Dec 7, 2020 at 22:13
• $|O_K/(\pi_K)|=t, |O_L/(\pi_L)| = t^{q^{r-s}}=T$, $\phi$ 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 $\langle a\rangle \cap \langle \phi \rangle = \{1\}$. Dec 7, 2020 at 22:41
• Thank you! I have another question: Shouldn't $b$ be a Frobenius element of $FL/L$? Dec 10, 2020 at 0:36
• Yes. ${}{}{}{}$ Dec 10, 2020 at 1:31
• 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 $ab=ba$) Feb 23, 2021 at 3:52