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!

  • $\begingroup$ $\mathbb{Z}_2[\zeta_3]/\mathbb{Z}_2$ 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 $\endgroup$ – reuns Dec 23 '18 at 1:43
  • $\begingroup$ @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. $\endgroup$ – 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 $[L:F]=e(L/F)f(L/F)=3$.


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