What is the relationship between the field fixed by the decomposition group and the embeddings in the p-adic numbers?

Question

I am reviewing stuff about local fields and wanted to clear this relationship before going forward.

Let us first fix some notation. Let $\mathbb{K}$ be a finite galois extension of the rational numbers with Galois group $G$, $p$ a prime number, $D$ its decomposition group, $n$ the degree of the extension, $g$ the index of $D$ in the Galois group and $\mathbb{K}_{D}$ the field fixed by the decomposition group. We also call $f(x)$ the minimal polynomial of a primitive element $\mathbb{K}$.

It is clear to me that the index $g$ gives us the number of possible inequivalent extension of the p-adic absolute value to the field $\mathbb{K}$ and that each of these defines a different embedding of $\mathbb{K}_{D}$ in $\mathbb{Q}_{p}$. But some points are not yet clear to me.

1. Does this mean that the polynomial $f(x)$ splits in $g$ different irreducible factors of the same degree in $\mathbb{Q}_{p}$?
2. The field $\mathbb{K}_{D}$ may not be a galois extension of $\mathbb{Q}$. What does this mean? Does this imply that the irreducibles factor of $f(x)$ gives us inequivalent extensions of $\mathbb{Q}_{p}$? Do the converse hold, i.e., if the irreducibles factors of $f(x)$ gives us different extensions of $\mathbb{Q}_{p}$ then $\mathbb{K}_{D}$ is not Galois over $\mathbb{Q}$?
3. If we take an extension $\mathbb{K}$ with non-solvable Galois group must we have that for any prime numbers $p$, the index $g$ is non-zero and hence all the primes splits?
4. If we take an extension $\mathbb{K}$ with non-solvable and simple Galois group does this mean that, by point 2, $f(x)$ is not irreducible over $\mathbb{Q}_{p}$ for any prime number $p$ and its irreducible factors gives us inequivalent extensions of $\mathbb{Q}_{p}$ for all prime numbers $p$?

All these points make intuitive sense to me, yet I am unable to prove the first two. Also because I am not completely sure they are true. (Except for the third one which should follow from the fact that any extension of $\mathbb{Q}_{p}$ is solvable.)

Answer 1

If $\Bbb{Q}[x]/(f)/\Bbb{Q}$ is Galois then factorize in irreducibles $f = \prod_{j=1}^g f_j\in \Bbb{Q}_p[x]$ each $\Bbb{Q}_p[x]/(f_j)/\Bbb{Q}_p$ is Galois and solvable and isomorphic to the splitting field of $f\in \Bbb{Q}_p[x]$. The subfield of $\Bbb{Q}[x]/(f)$ fixed by $Gal(\Bbb{Q}_p[x]/(f_j)/\Bbb{Q}_p) \subset Gal(\Bbb{Q}[x]/(f)/\Bbb{Q})$ is $\Bbb{Q}_p\cap \Bbb{Q}[x]/(f)$ where the embedding is the one sending $x+(f)$ to $x+(f_j)$.