# 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!

• $\mathbb{Z}_2[\zeta_3]/\mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $\mathbb{Z}_2[\zeta_3]$ so $[\mathbb{Z}_2[\zeta_3]/(2) : \mathbb{Z}_2/(2)]=[\mathbb{Z}_2[\zeta_3]:\mathbb{Z}_2]$). $\ \ \mathbb{Z}_2[\zeta_3,2^{1/3}]/\mathbb{Z}_2[\zeta_3]$ is totally ramified of degree $3$ ($\pi=2^{1/3}$ then $(\pi)$ is the unique prime ideal of $\mathbb{Z}_2[\zeta_3,2^{1/3}]$ and $\pi^3 / 2 \in \mathbb{Z}_2[\zeta_3,2^{1/3}]^\times$). $\mathbb{Z}_2[\zeta_32^{1/3}]$ is isomorphic to $\mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},\zeta_32^{1/3}$ have the same minimal polynomial – reuns Dec 23 '18 at 1:43
• @reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation. – Diglett Dec 23 '18 at 11:52

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 $$[L:F]=e(L/F)f(L/F)=3$$.