Decomposition group of a prime ideal and root of polynomials

Question

Let $f(x)$ be a monic irreducible polynomial with integer coefficient. Let $K$ be the splitting field of $f$ and $\alpha$ one of its roots. Let $p$ a prime number such that $p$ does not divide $disc(f)$ and suppose that $f$ has a root mod $p$. Then there exists a prime $P$ of $K$ over $p$ such that the decomposition group $G(P)\leq G(K|Q)$ is contained in $G(K|Q(\alpha))$. Why this is true?

Answer 1

$F$ is a separable extension of $Q$. Take $p$ relatively prime with the conductor of $Z[\alpha]$, that is the ideal $\{y \in O_F \ | \ y O_F\subseteq Z[\alpha]\}$. Notice that this ideal measure how far $Z[\alpha]$ is from $O_F$ (that is not monogenic in general). Then by Neukirch, Jürgen, Algebraic Number Theory, pag. 47, we have that, since $f$ has a root $\bmod p$ and then a linear factor, there is a prime $\mathcal{P}$ over $p$ in $F$ such that ${\rm f}(\mathcal{P}|p)=1$. Thus, for any $\mathfrak{p}$ over $\mathcal{P}$ in $K$, we have $f(\mathfrak{p}|p)=f(\mathfrak{p}|\mathcal{P})$.

If $p$ does not divide the discriminant of $K$, then $p$ doesn't ramify, so that $e(\mathfrak{p}|p)=1$. Then $e(\mathfrak{p}|\mathcal{P})=1$.

Now, since $K$ is Galois both over $F$ and $Q$, the order of the decomposition group $$\# D_{K|F}(\mathfrak{p})=e(\mathfrak{p}|\mathcal{P})f(\mathfrak{p}|\mathcal{P})=f(\mathfrak{p}|\mathcal{P})=f(\mathfrak{p}|p)=e(\mathfrak{p}|p)f(\mathfrak{p}|p) =\# D_{K|Q}(\mathfrak{p})$$ and then $$ D_{K|Q}(\mathfrak{p})=D_{K|F}(\mathfrak{p}) \leq G(K|F).$$

The only problem is now to prove that if $p$ does not divide $disc(f)$ then it is relatively prime with the conductor of $Z[\alpha]$ and does not divide the discriminant of $K$.

For this see Lang, Algebraic Number Theory, III.3.