Why do minimal polynomial have coefficients in the integral closured ring?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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I'm reading the following theorem:
enter image description here



My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?










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  • 1




    The as are the result of evaluating the elementary symmetric functions on the alphas.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 4:47











  • I guess I know the answer now...how stupid I am...
    – CYC
    Feb 27 '17 at 4:57






  • 3




    I suggest you write a complete answer answering your question.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 5:25














up vote
2
down vote

favorite












I'm reading the following theorem:
enter image description here



My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?










share|cite|improve this question

















  • 1




    The as are the result of evaluating the elementary symmetric functions on the alphas.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 4:47











  • I guess I know the answer now...how stupid I am...
    – CYC
    Feb 27 '17 at 4:57






  • 3




    I suggest you write a complete answer answering your question.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 5:25












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I'm reading the following theorem:
enter image description here



My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?










share|cite|improve this question













I'm reading the following theorem:
enter image description here



My question is, why are $a_i$s in $A[alpha_1...alpha_n]$?







abstract-algebra field-theory algebraic-number-theory






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share|cite|improve this question











share|cite|improve this question




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asked Feb 27 '17 at 4:39









CYC

940711




940711







  • 1




    The as are the result of evaluating the elementary symmetric functions on the alphas.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 4:47











  • I guess I know the answer now...how stupid I am...
    – CYC
    Feb 27 '17 at 4:57






  • 3




    I suggest you write a complete answer answering your question.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 5:25












  • 1




    The as are the result of evaluating the elementary symmetric functions on the alphas.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 4:47











  • I guess I know the answer now...how stupid I am...
    – CYC
    Feb 27 '17 at 4:57






  • 3




    I suggest you write a complete answer answering your question.
    – Mariano Suárez-Álvarez
    Feb 27 '17 at 5:25







1




1




The as are the result of evaluating the elementary symmetric functions on the alphas.
– Mariano Suárez-Álvarez
Feb 27 '17 at 4:47





The as are the result of evaluating the elementary symmetric functions on the alphas.
– Mariano Suárez-Álvarez
Feb 27 '17 at 4:47













I guess I know the answer now...how stupid I am...
– CYC
Feb 27 '17 at 4:57




I guess I know the answer now...how stupid I am...
– CYC
Feb 27 '17 at 4:57




3




3




I suggest you write a complete answer answering your question.
– Mariano Suárez-Álvarez
Feb 27 '17 at 5:25




I suggest you write a complete answer answering your question.
– Mariano Suárez-Álvarez
Feb 27 '17 at 5:25










1 Answer
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There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.






    share|cite|improve this answer
























      up vote
      0
      down vote



      accepted










      There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.






      share|cite|improve this answer






















        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.






        share|cite|improve this answer












        There's an $sigma$ sending $alpha$ to $alpha_i$ fixing $K$, hence $alpha_i$ are integral over $A$ since $alpha$ is, the rest is easy.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 9 at 11:02









        CYC

        940711




        940711



























             

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