My question 1 is how to calculate the ramification index and inertia degree of $K_1 := \mathbb{Q}_2(\sqrt{3}, \sqrt{7})$ and $K_2 := \mathbb{Q}_2(\sqrt{3}, \sqrt{2})$ over $\mathbb{Q}_2$.
My attempts 1: I have proved that $\mathbb{Q}_2(\sqrt{2})$, $\mathbb{Q}_2(\sqrt{3})$ and $\mathbb{Q}_2(\sqrt{7})$ are totally ramified over $\mathbb{Q}_2$ by noting that
- In $\mathbb{Q}_2(\sqrt{2})$, the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}_2$ is $T^2 – 2$, which is Eisensteinian over $\mathbb{Q}_2$;
- In $\mathbb{Q}_2(\sqrt{3}) = \mathbb{Q}_2(\sqrt{3}+1)$, the minimal polynomial of $\sqrt{3}+1$ over $\mathbb{Q}_2$ is $(T-1)^2 – 3 = T^2-2T-2$, which is Eisensteinian over $\mathbb{Q}_2$;
- In $\mathbb{Q}_2(\sqrt{7}) = \mathbb{Q}_2(\sqrt{7}+1)$, the minimal polynomial of $\sqrt{7}+1$ over $\mathbb{Q}_2$ is $(T-1)^2 – 7 = T^2-2T-6$, which is Eisensteinian over $\mathbb{Q}_2$.
But I got stuck on considering extensions as $K_1 \mid \mathbb{Q}_2(\sqrt{3})$. There seems to be only two choices: unramified and totally ramified. And I only know two facts on distinguishing them:
Fact 1: If $E=F(\alpha)$ for $\alpha \in \mathcal{O}_E$, and $g(T) \in \mathcal{O}_F[T]$ is a monic polynomial such that $\overline{g(T)}$ has no multiple roots in the algebraic closure of $\kappa_F$, where $\kappa_F$ is the residue field of $F$. Then $E \mid F$ is unramified (i.e. $[E:F] = f(E:F)$).
Fact 2: If $E=F(\alpha)$ for $\alpha \in \mathcal{O}_E$, and $g(T) \in \mathcal{O}_F[T]$ is the minimal polynomial of $\alpha$ over $F$. If $g$ is Eisensteinian over $F$, then $E \mid F$ is totally ramified (i.e. $[E:F] = e(E:F)$) and $\alpha$ is a uniformizer of the extension $E \mid F$.
But with merely these two facts, it seems hard to calculate the ramification index and the inertia degree of the extensions like $K_1 = \mathbb{Q}_2(\sqrt{3})(\sqrt{7}) \mid \mathbb{Q}_2(\sqrt{3})$. For example, when I want to use Fact 1, the minimal polynomials of $\sqrt{7}$ or $\sqrt{7}+1$ both have multiple root after reduced to the residue field of $\mathbb{Q}_2(\sqrt{3})$, which is merely $\mathbb{F}_2$ since we have shown that $\mathbb{Q}_2(\sqrt{3}) \mid \mathbb{Q}_2$ is totally ramified. When I want to use Fact 2, it is hard to find an Eisensteinian polynomial over $\mathbb{Q}_2(\sqrt{3})$ having $\sqrt{7}$ or something like $\sqrt{7}+1$ as a root, since $e(\mathbb{Q}_2(\sqrt{3}): \mathbb{Q}_2)=2$ and everything in $\mathbb{Q}_2$ has extended valuation $\geq 2$ on $\mathbb{Q}_2(\sqrt{3})$. So I got stuck here.
A similar question is the following:
Question 2: Let $F$ be a complete discrete valued field, $\alpha \in F$, $v_F(\alpha) \geq 1$. Suppose $n$ is a positive integer coprime to $v_F(\alpha)$. Prove that $F(\sqrt[n]{\alpha})$ is totally ramified over $F$ of degree $n$.
My attempts 2: When $F \supset \mathbb{Q}_p$, I can divide $v_F(\alpha)$ on both sides of $T^n – a$ to obtain an Eisenstein polynomial over $F$, and hence the extension is indeed totally ramified. But this "division" seems to be impossible in general $F$ not containing $\mathbb{Q}_p$. What can I do to handle this?
It seems that the two questions are closely related to something that I'm not familiar with when judging unramified extensions and totally ramified extensions.
Thank you for your comments and helps! 🙂
Best Answer
For $\Bbb{Q}_2(\sqrt3,\sqrt7)$ the trick is to say that $-7\equiv 1\bmod 8$ is a square in $\Bbb{Z}_2$ so it is $\Bbb{Q}_2(\sqrt3,i)$ which contains $\zeta_3$.
For $\Bbb{Q}_2(\sqrt3,\sqrt2)$ note that
$$(X-\frac{\sqrt3-1}{\sqrt2}-1)(X-\frac{-\sqrt3-1}{\sqrt2}-1)= X^2+X(\sqrt{2}-2)-\sqrt{2}\in \Bbb{Z}_2[\sqrt{2}][X]$$ is Eisenstein.