Tame ramification for non-henselian fields











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I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:
enter image description here
($T$ denotes the maximal unramified subextension of $L$)



And this is the remark:
enter image description hereNamely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?










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    Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
    – Lubin
    Nov 24 at 20:57















up vote
3
down vote

favorite
1












I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:
enter image description here
($T$ denotes the maximal unramified subextension of $L$)



And this is the remark:
enter image description hereNamely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?










share|cite|improve this question




















  • 1




    Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
    – Lubin
    Nov 24 at 20:57













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:
enter image description here
($T$ denotes the maximal unramified subextension of $L$)



And this is the remark:
enter image description hereNamely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?










share|cite|improve this question















I have a question concerning the remark after defintion (7.6) in Chapter 2 of Neukirch’s Algebraic number theory. This is the defintion:
enter image description here
($T$ denotes the maximal unramified subextension of $L$)



And this is the remark:
enter image description hereNamely, he writes that we can use this definition for non-henselian fields as well. But the existence of $T$ relies on henselianity (at least, the proof in book uses it). So what does he really mean and what is the „correct“ definition?







number-theory algebraic-number-theory






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edited Nov 27 at 1:56

























asked Nov 24 at 16:19









Gregg

888




888








  • 1




    Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
    – Lubin
    Nov 24 at 20:57














  • 1




    Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
    – Lubin
    Nov 24 at 20:57








1




1




Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57




Isn’t it the case that when $v$ has a unique extension to $L$ that there is a maximal unramified of $K$?
– Lubin
Nov 24 at 20:57










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1






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1
down vote



accepted










Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.



As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:




  1. If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.

  2. In this case, the integral closure of the valuation ring is the valuation ring of the extension.

  3. Unramified extensions are separable.

  4. A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.






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    Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.



    As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:




    1. If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.

    2. In this case, the integral closure of the valuation ring is the valuation ring of the extension.

    3. Unramified extensions are separable.

    4. A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.



      As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:




      1. If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.

      2. In this case, the integral closure of the valuation ring is the valuation ring of the extension.

      3. Unramified extensions are separable.

      4. A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.



        As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:




        1. If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.

        2. In this case, the integral closure of the valuation ring is the valuation ring of the extension.

        3. Unramified extensions are separable.

        4. A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.






        share|cite|improve this answer














        Just to give a somewhat complete answer, I will expand on the comment of Prof. Lubin.



        As he pointed out, it is indeed true that if the valuation admits a unique extension, then there is the maximal unramified subextension, i.e. the composite of all unramified subextensions. To prove this, one can follow the general strategy of the proof of Proposition (7.2) in Neukirch‘s book and use the following observations:




        1. If the valuation admits a unique extension to an algebraic extension, then it does so for any subextension.

        2. In this case, the integral closure of the valuation ring is the valuation ring of the extension.

        3. Unramified extensions are separable.

        4. A separable extension generated by $alpha$ admits a unique extension of the valuation iff its minimal polynomial is irreducible over the completion.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 at 14:31

























        answered Nov 27 at 1:53









        Gregg

        888




        888






























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