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