Computing degrees and ramification indices of some extensions of $\mathbb{Q}_2$ Let $K=\mathbb{Q}_2$ and $F = K(\zeta_3,\alpha)$ where $\zeta$ is a primitive third root of unity and $\alpha$ is a cubic root of $2$, i.e. $\alpha^3 = 2$.Let $K_1 = K(\zeta_3)$, $K_2 = K(\alpha)$ and $L=K_1(\beta)$ where $\beta$ is defined by $\beta^3 = \zeta_3 \alpha$. 
Now I would like to compute the degrees and ramification indices of $L/K_1$.
Discoveries and attempts:


*

*I noticed that $L$ contains $F$ since $\beta^3/\zeta_3 = \alpha$.

*I know that $F/K_1$ is a totally ramified extension of degree $3$. This can be seen by taking a look at the extensions $F/K_1/K$ and $F/K_2/K$ and noticing that $K_1/K$ is unramified of degree $2$ and $K_2/K$ is totally ramified of degree $3$.


Could you please explain what can I do now to get information about $L/K_1$ resp. $L/F$? Thank you!
 A: Let $L_{1}=\mathbb{Q}_{2}(\beta)$ and note that $\beta$ is a root of $x^{9}-2$ over $K$, which is an Eisenstein polynomial. Therefore $L_{1}/K$ is totally ramified of degree 9 (see e.g. Proposition 3.6 of Local Fields and Their Extensions by Fesenko & Vostokov for a proof if you don't know this result). Now, $K_{1}/K$ is Galois of degree 2, therefore we have by Galois theory
$$[L:L_{1}]=[K_{1}:K],$$
because $L=L_{1}K_{1}$. Therefore we conclude that $[L:L_{1}]=2$. So $[L:K]=18$ by the tower formula, whence $[L:K_{1}]=9$.
Now, note that
$$f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}),$$
because $K_{1}/K$ is unramified (where $f(L/K)$ means the inertia degree of the extension $L/K$). Also
$$e(L/K)=e(L/L_{1})e(L_{1}/K)=9e(L/L_{1}),$$
because $L_{1}/K$ is totally ramified (where $e(L/K)$ means the ramification index of the extension $L/K$). Thus we see that $2\mid f(L/K)$ and $9\mid e(L/K)$. By the fundamental identity we have $$18=[L:K]=e(L/K)f(L/K),$$
whence $f(L/K)=2$ and $e(L/K)=9$. This implies that $f(L/K_{1})=1$ because
$$2=f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}).$$
It also implies that $e(L/K_{1})=9$ because
$$9=e(L/K)=e(L/K_{1})e(K_{1}/K)=e(L/K_{1}).$$
Now for the extension $L/F$ we have
$$2=f(L/K)=f(L/F)f(F/K)=f(L/F)f(F/K_{1})f(K_{1}/K)=2f(L/F)f(F/K_{1}),$$
whence $f(L/F)=1$. Furthermore
$$9=e(L/K)=e(L/F)e(F/K)=e(L/F)e(F/K_{1})e(K_{1}/K)=3e(L/F),$$
whence $e(L/F)=3$ and we conclude that $[L:F]=e(L/F)f(L/F)=3$.
