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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 !

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    $\begingroup$ I didn't vote to close, yet I think this is not an MO type question. You seem to ask for help understanding a textbook proof. Maybe you should check several sources for this theorem by Dedekind. For instance, it is contained in Lang's Algebra. $\endgroup$ Nov 4, 2013 at 21:32
  • $\begingroup$ Yes, I'm trying to understand the proof (but some steps are just comments). I will see your reference, thanks a lot! $\endgroup$
    – Marco
    Nov 4, 2013 at 22:14
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    $\begingroup$ There is no need to consider the subfield fixed by the decomposition group or the primitive element theorem for the residue field extension. See math.uconn.edu/~kconrad/blurbs/gradnumthy/frobeniuspf.pdf $\endgroup$
    – KConrad
    Nov 4, 2013 at 23:51
  • $\begingroup$ Peter, are you sure this theorem can be found in Lang's algebra? Would it not be rather Lang's algebraic number theory? Another reference that may contain this is Samuel's Théorie des nombres algébriques or its English translation. $\endgroup$
    – Joël
    Nov 5, 2013 at 1:16

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