Showing a prime is inert

Question

This is a part from the book "Generic Polynomials Constructive Aspects of the Inverse Galois Problem" which I don't understand. This is also pretty much the counterexample given in: https://www.jstor.org/stable/1969410?seq=1.

Let $L_2$ be the unramified extension of $\mathbb{Q}_2$ of degree $8$ (that is obtained by adding a $2^8-1$ primitive root of unity). Let $L$ be an extension of $\mathbb{Q}$ also with Galois group $\mathbb{Z}/8\mathbb{Z}$. Let $\mathbb{Q}(\sqrt{D})$ be the quadratic subextension of $L$. Suppose $L_2$ is the compositum of $L$ and $\mathbb{Q}_2$.

I can't figure out why:

$2$ remains inert in $L/\mathbb{Q}$.

$2$ remains inert implies $D\equiv 5\mod 8$.

And finally,

$p|D$ implies the completion $L_p/\mathbb{Q}_p$ is again a $C_8$ extension.

Because of how it's worded, I think $L_2$ being the compositum of $L$ and $\mathbb{Q}_2$ must be why $2$ remains inert. Also maybe, use Kronecker-Weber and try to get some relation between the roots of unity? I would appreciate any hints or references.

Answer 1

Here are some hints:

• Let $\mathfrak p$ be any prime of $L$ above $2$. Then $L_2$ is the completion of $L$ at $\mathfrak p$: it is the smallest complete field that contains both $L$ and $\mathbb Q_2$. In particular, $\mathrm{Gal}(L_2/\mathbb Q_2)$ is isomorphic the decomposition subgroup of $\mathfrak p$ in $\mathrm{Gal}(L/\mathbb Q)$.
• Since $2$ is inert in $L$, it is inert in $\mathbb Q(\sqrt{D})$. Use the Kummer-Dedekind theorem to deduce that $D\equiv 5\pmod 8$.
• If $p\mid D$, then $p$ is ramified in $\mathbb Q(\sqrt D)$. Let $\mathfrak p$ be a prime of $L$ above $p$. Then the inertia subgroup of $\mathfrak p$ is a subgroup $I$ of $\mathrm{Gal}(L/\mathbb Q)$ such that $\mathrm{Gal}(L^I/\mathbb Q)$ is unramified at $p$.