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












4














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.



enter image description here






abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry







share|cite|improve this question





























    4














    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.



    enter image description here






    abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry







    share|cite|improve this question



























      4












      4








      4


      1





      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.



      enter image description here






      abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry







      share|cite|improve this question















      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.



      enter image description here






      abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry




      abstract-algebra number-theory galois-theory algebraic-number-theory arithmetic-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 14:04

























      asked Dec 4 '18 at 2:13









      Kento

      473




      473






















          1 Answer
          1






          active

          oldest

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          5














          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.






          share|cite|improve this answer























          • thank you very much. i think i understand .
            – Kento
            Dec 5 '18 at 8:31













          Your Answer





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          1 Answer
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          active

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          5














          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.






          share|cite|improve this answer























          • thank you very much. i think i understand .
            – Kento
            Dec 5 '18 at 8:31


















          5














          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.






          share|cite|improve this answer























          • thank you very much. i think i understand .
            – Kento
            Dec 5 '18 at 8:31
















          5












          5








          5






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 21:44

























          answered Dec 5 '18 at 5:01









          Pavel Čoupek

          4,44111125




          4,44111125












          • thank you very much. i think i understand .
            – Kento
            Dec 5 '18 at 8:31




















          • thank you very much. i think i understand .
            – Kento
            Dec 5 '18 at 8:31


















          thank you very much. i think i understand .
          – Kento
          Dec 5 '18 at 8:31






          thank you very much. i think i understand .
          – Kento
          Dec 5 '18 at 8:31




















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