I'm studying the relationship between the Galois group of a polynomial with integer coefficients and the group of his reduction modulo $p$.
More precisely, consider $\mathbb{K}$ a number field such that the extension $\mathbb{K}/\mathbb{Q}$ is Galois. Let $\mathfrak{p}$ a prime ideal of his ring of integers $\mathcal{O}_\mathbb{K}$ such that $\mathfrak{p}|p$. Now define $$D_\mathfrak{p}=\{\sigma \in \mbox{Gal}(\mathbb{K}/\mathbb{Q})\;|\;\sigma(\mathfrak{p})=\mathfrak{p}\}. $$ Then, every ring automorphism of $\mathcal{O}_\mathbb{K}$ defined by $\sigma\in D_\mathfrak{p}$ induces an element in $\mbox{Gal}((\mathcal{O}_\mathbb{K}/\mathfrak{p})/\mathbb{F}_p)$, let's say $\sigma_\mathfrak{p}$ (and in fact, $r_\mathfrak{p}:\sigma\mapsto \sigma_\mathfrak{p}$ is an homomorphism of groups).
The first question is: why this is always surjective?
I have a hint to do this: Consider the subfield of $\mathbb{K}$ fixed by $D_\mathfrak{p}$, $\mathbb{L}=\mathbb{K}^{D_\mathfrak{p}}$ and choose $\theta\in\mathcal{O}_\mathbb{K}-\{0\}$ the image of which in $\mathcal{O}_\mathbb{K}/\mathfrak{p}$ is a generator of that $\mathbb{F}_p$-extension (in the sense of the Primitive Element Theorem), then consider the roots in $\mathbb{K}$ of the minimal polynomial of $\theta$ over $\mathbb{L}$.
The second question is: how to prove that if $P\in \mathbb{Z}[T]$ be such that the reduction $\overline{P}\in \mathbb{F}_p[T]$ is separable and the field $\mathbb{K}$ is the field of decompostion of $P$, then $r_\mathfrak{p}$ is also injective?
Then, in this case, choosing $\mathfrak{p}$ and using $r_\mathfrak{p}^{-1}$ we get a morphism from the Galois group of $\overline{P}$ to the Galois group of $P$.
What I've done:
For the first we notice that $\mbox{Gal}((\mathcal{O}_\mathbb{K}/\mathfrak{p})/\mathbb{F}_p)$ is a cyclic group generated by the Fröbenius map $\tau:x\mapsto x^p$, so we only need to find $\sigma \in D_\mathfrak{p}$ such that $r_\mathfrak{p}(\sigma)=\tau$, i.e., such that $\overline{\sigma(\theta)}=\overline{\theta^p}$.
Now, if we denote by $\mu_\theta^\mathbb{L}\in \mathbb{L}[x]$ the minimal polynomial of $\theta$ over $\mathbb{L}$, then $$\Sigma=\{x\in \mathbb{K}\;|\;\mu_\theta^\mathbb{L}(x)=0\}=\mbox{Gal}(\mathbb{K}/\mathbb{L})\cdot \theta = D_\mathfrak{p}\cdot \theta$$ Then, $\mu_\theta^\mathbb{L}$ must divide $Q=\prod_{\sigma\in D_\mathfrak{p}}(x-\sigma(\theta))$. Are they equal?
For the second one, I think that maybe can be done by argumenting by contradiction and in that way find a root with multiplicity, but I'm not sure.
Thanks a lot for your help !
