I am trying to compute the extension degree and intertial degree (or ramification index) of the local field extension $\mathbb{Q}_3(\sqrt[4]{-3}, i, \xi_3, \sqrt[3]{2})/\mathbb{Q}_3(\xi_3, \sqrt[3]{2})$ where $\xi_3$ is a third root of unity.
Progress
- I was able to show that $\mathbb{Q}_3(\xi_3, \sqrt[3]{2})/\mathbb{Q}_3$ is totally ramified of degree $6$.
- The extension $\mathbb{Q}_3(i)$ is unramified of degree $2$.
- Since $\xi_3 = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$, we can write $\mathbb{Q}_3(\xi_3, \sqrt[3]{2}) = \mathbb{Q}_3(\sqrt{-3}, \sqrt[3]{2})$ which is more similar to our given extension.
- For examining how to work with $\sqrt[4]{-3}$, we observe that $$ x^4 + 3 = (x^2+\sqrt{-3})(x^2-\sqrt{-3}), $$ so the last factor is likely to be the minimal polynomial of $\sqrt[4]{-3}$ over $\mathbb{Q}_3(\sqrt{-3}, \sqrt[3]{2})$. I cannot find a good argument why this is true though.
Could you please help me with this calculation? Thank you in advance!
Use the following trick:
let $L/F$ be a fied extension of degree $m$, and let $\alpha\in F_{alg}$ of degree $n$. If $m$ and $n$ are coprime, then $\alpha$ is algebraic of degree $n$ over $L$.
Proof. We have $[L(\alpha):F]=[L(\alpha):L]m=n [L(\alpha):F(\alpha)]$.
Hence $n\mid [L(\alpha):L]m$, and by Gauss lemma, $n\mid [L(\alpha):L]$. In particular, $[L(\alpha):L]\geq n$. But he other inequality is obvious, since the minimal polynomial $P$ of $\alpha$ over $F$ has degree $n$.
Now take $\alpha=\sqrt[4]{-3}, F=\mathbb{Q}_3(\sqrt{-3}), L=\mathbb{Q}_3(\sqrt{-3},\sqrt[3]{2})$