# decomposition group and inertia group, the minimal polynomial,surjectivity of the map $D_{M/P}\rightarrow Gal$

Can anyone explain the underlined sentence?

For notation, A:Dedekind domain, K=Frac(A), L/K:Galois extension, B:The integral closure of A in L, M:A maximal ideal of B, P:The intersection of M and A (hence the maximal ideal of A), $$D_{M/P}$$ :the decomposition group.

I reckon the way we take $$\alpha$$ is the key, but cannot make it to the conclusion, 'we find that the only non-zero roots of...'.

I read some of the close questions already answered but none of them was using this type of logic.

$$g(y)$$ is the min. polynomial of $$\overline{\alpha}$$ over $$A/P$$, so it has to divide the polynomial $$\overline{f}(y)=f(y) \,\mathrm{mod}\, P$$, since $$\overline{\alpha}$$ is a root of $$\overline{f}(y)$$ (and $$\overline{f}(y)$$ is nonzero, take $$f(y)$$ monic). From this and the expression $$f(y)=\prod_H(y-\sigma(\alpha))$$ it follows that the roots of $$g(y)$$ are just some of the roots $$\sigma(\alpha)$$ taken modulo $$M$$, and the goal is to identify which ones.
Now $$\alpha$$ was chosen so that $$\alpha \in \sigma(M)$$ whenever $$\sigma \notin D_{M/P}$$, i.e. $$\sigma(M)\neq M$$. Applying $$\sigma^{-1}$$, we have that $$\sigma^{-1}(\alpha) \in M$$ whenever $$\sigma(M)\neq M$$. Changing $$\sigma^{-1}$$ to $$\sigma$$ (note that $$\sigma^{-1} \notin D_{M/P}$$ iff $$\sigma \notin D_{M/P}$$), we have that $$\sigma(\alpha) \in M$$ whenever $$\sigma \notin D_{M/P}$$. And conversely, we have $$\alpha \notin M$$ (because $$\overline{\alpha} \neq 0$$), so given any $$\sigma \in D_{M/P}$$, we have that $$\sigma(\alpha) \notin \sigma(M)=M$$. So altogether: $$\sigma(\alpha) \,\mathrm{mod}\,M$$ is nonzero iff $$\sigma \in D_{M/P}$$. So the roots of $$g(y)$$ can come only from these, i.e. in the form $$\overline{\sigma}(\overline{\alpha})$$ (because $$g(y)$$ cannot have $$0$$ as a root, it's the min. poly. of $$\overline{\alpha}$$). And all of them has to be roots for Galois reasons (all the maps $$\overline{\sigma}$$ are elements of the Galois group of the residue field, and $$\overline{\alpha}$$ is a root of $$g(y)$$).