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.

Thank you in advance.

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Let me expand on the highlighted part:

$g(y)$ 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 $g(y)$).

Hope this helps.

  • $\begingroup$ thank you very much. i think i understand . $\endgroup$ – Clarkson Dec 5 '18 at 8:31

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