I have a potential answer to my question here but I have serious doubt. The question was the following: for a polynomial $P\in Q[X]$ we have a decomposition field $K$ and $G=\text{Gal}(K/\mathbb{Q})$. We can consider $P$ in $\mathbb{Q}_p[X]$ which gives $G_p=\text{Gal}(L/\mathbb{Q}_p)$ if $L$ is the decomposition field of $P$ over $\mathbb{Q}_p$.
My question was: what informations can we get from $G_p$ to $G$?
Potential answer:
First point: let $\mathfrak{p}$ a prime over $p$ then the completion $K_\mathfrak{p}$ is equal to $L$ i.e. is the decomposition field of $P\in\mathbb{Q}_p[X]$. One has $L\subseteq K_\mathfrak{p}$ because $P$ is decomposed in $K$ so it is decomposed in $K_\mathfrak{p}$. One hase $K_\mathfrak{p}\subseteq L$ because $L\cap\overline{\mathbb{Q}}$ decompose $P$ (over $\mathbb{Q}$) because the roots of $P$ are in $\overline{\mathbb{Q}}$ (because $P\in\mathbb{Q}[X]$), hence $K\subseteq L\cap\overline{\mathbb{Q}}$ so $K_\mathfrak{p}\subseteq(L\cap\overline{\mathbb{Q}})_\mathfrak{p}\subseteq L$ (because $L$ is complete).
Second point: if $p$ is inert or ramified in $\mathcal{O}_K$ then there is a unique ideal $\mathfrak{p}$ of $\mathcal{O}_K$ above $p$ so that the decomposition group $D(\mathfrak{p})$ will be equal to $G=\text{Gal}(K/\mathbb{Q})$ (because for all $\sigma\in G$ we'll have $\sigma(\mathfrak{p})=\mathfrak{p}$). But in general $$ \text{Gal}(K_\mathfrak{p}/\mathbb{Q}_p)\hookrightarrow\text{Gal}(K/\mathbb{Q}) $$ with image $D(\mathfrak{p})$. As a result we have should have here $\text{Gal}(K_\mathfrak{p}/\mathbb{Q}_p)\simeq\text{Gal}(K/\mathbb{Q})$.
Problem:
I suspect that there is a problem somewhere above because all irreducible polynomial have a non null discriminant so have ramified $p$ but not all irreducible polynomial have solvable Galois group, which is however always the case for local Galois group so that it should not be possible that $G=G_p$.
My question now: where is the problem above?
PS: if you have an answer for my first question don't hesitate!
